Unit 8 Chemical Equilibrium Focusing on Acid-Base Systems

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1 Unit 8 Chemical Equilibrium Focusing on Acid-Base Systems

2 unit 8Chemical Equilibrium Chemical Equilibrium Focusing on Acid Base Systems Equilibrium describes any condition or situation of balance. We recognize equilibrium in a chemical reaction system, oddly enough, by noticing nothing we see no change in any property of the system. The easiest conclusion to draw would be that nothing is happening, but closer study reveals that, at the molecular level, a lot of change is going on. Chemical reaction equilibrium is always a dynamic balance between two opposing changes, which are balanced because they are occurring at equal rates, within a closed system. What we observe directly is the net effect neither an increase nor a decrease in any measurable property. Chemistry involves the study of change in chemical substances. To predict and control chemical change, we must better understand the nature of the system at the molecular level. For instance, to understand why and how bubbles of gas form or dissolve in a liquid, we must take into account the nature of the gas, the nature of the liquid, and the actions of their invisible entities all at the same time. This unit explores the nature of dynamic equilibrium in chemical systems. It explains much more thoroughly and completely many of the chemical change concepts you have already learned. You will examine some very important reactions those involving acids and bases in solution at a higher conceptual level. This knowledge will allow you to describe, explain, and predict many new chemical systems and situations. The continual exploration and improvement of concepts such as these is a critical part of the nature of science. As you progress through the unit, think about these focusing questions: What is happening in a system at equilibrium? How do scientists predict shifts in the equilibrium position of a system? How do Brønsted Lowry acids and bases illustrate equilibrium? 670 Unit 8

3 Unit 8 GENERAL OUTCOMES In this unit, you will explain that there is a dynamic balance of opposing reactions in chemical systems at equilibrium determine quantitative relationships in acid base ionization and other simple systems at equilibrium Chemical Equilibrium Focusing on Acid Base Systems 671

4 Unit 8 Chemical Equilibrium Focusing on Acid Base Systems Prerequisites Concepts collision reaction theory dissociation and ionization amount concentration ion concentration percent reaction stoichiometric calculation net ionic equations acids and bases indicators Skills laboratory safety scientific problem solving You can review prerequisite concepts and skills in the appendices and on the Nelson Web site. A Unit Pre-Test is also available online. GO ARE YOU READY? These questions will help you find out what you already know, and what you need to review, before you continue with this unit. Knowledge 1. A large number of chemical reactions (most notably redox reactions) will only occur in aqueous solution, at least at a rate great enough to be observable in a laboratory. In terms of collision reaction theory (Figure 1), state (a) why a given reaction might only occur in solution (b) why collisions between entities do not necessarily result in reaction (c) the two effects that an increase in temperature has on collisions between entities involved in a reaction (d) the effect that an increase in concentration of one kind of entity has on collisions between all entities involved in a reaction An Ineffective Collision Figure 1 Collision reaction theory considers the numbers, speeds, and orientation of colliding entities to explain the progress of chemical reactions. 2. Assume that each of the following substances is placed in water. Rewrite the formula and the physical state to indicate whether it is very soluble or only slightly soluble in water at SATP. For those substances predicted to produce ions in solution, write symbols for all aqueous ions present after the substance dissociates upon dissolving. (a) MgSO 4 7H 2 O(s) (b) CH 3 COCH 3 (l) (c) CH 2 CH 2 (g) (d) CaCO 3 (s) (e) PbCl 2 (s) (f) FeCl 3 (s) (g) C 3 H 5 (OH) 3 (l) 3. Each of the following substances is mixed with water to form an aqueous solution. Write an equation (using the modified Arrhenius theory) to explain the acidity or basicity of the solution in terms of hydronium or hydroxide ions. Your equations should, when necessary, represent the ionizing of the substance as a reaction with water. (a) HCl(g) (b) CH 3 COOH(l) (c) H 2 SO 4 (l) (d) HNO 3 (l) (e) NaCl(s) (f) NH 3 (g) (g) NaOH(s) 672 Unit 8

5 Unit 8 4. The concentration of chemical substances in solution can vary widely (Figure 2). Concentration affects solution properties such as colour, conductivity, freezing point, and viscosity. Concentration also affects the frequency of particle (entity) collisions; and, thus, will usually affect the observed rate of a reaction. Complete Table 1. Table 1 Concentration of Entities and Quantities of Reagents in Solution dilute solution Mass Solution Concentration Reagent dissolved (g) volume (L) Entity (mol/l) NaOH(s) OH (aq) Al 2 (SO 4 ) 3 (s) 2.00 SO 4 2 (aq) Al 2 (SO 4 ) 3 (s) 2.00 Al 3+ (aq) CaCl 2 (s) 1.00 Cl (aq) concentrated solution 5. Chemical substances may also have widely varying concentrations in the gaseous state. Calculate the amount concentration of (a) 24.0 g of hydrogen in a 2.00 L container (b) 500 kpa of oxygen in a 10.0 L container at 0 C (c) 4.40 mol of carbon dioxide at SATP (d) mol of methane at STP 6. Understanding chemical equilibrium theory often involves using a net ionic equation. For each of the following combinations of reagents, predict the product(s), and write a net ionic reaction equation. Balance the equation with simplest integer coefficients, and include physical states for all substances. (a) Copper(II) chloride and potassium carbonate solutions are mixed. (b) Ethene (ethylene) reacts with hydrogen chloride to form chloroethane. (c) Aluminium foil reacts with hydrochloric acid. (d) Ammonia undergoes simple decomposition. (e) Magnesium metal is placed in an aqueous solution of gold(iii) chloride. (f) Mixing sulfurous acid solution with sodium hypochlorite solution results in a spontaneous redox reaction. 7. The acidity of solutions often has a considerable effect on the type, rate, and extent of the reactions they will undergo. Acids may be classed as strong or weak, depending on the extent of their reaction with water. Complete Table 2, identifying the acids as strong or weak. Figure 2 Concentration affects the rate of chemical reaction as well as physical properties. Table 2 Solution Acidity and Basicity Strength [Acid] % Reaction [H 3 O (aq)] Acid S/W (mol/l) in/with water ph (mol/l) HCl(aq) HBr(aq) HNO 3 (aq) CH 3 COOH(aq) HCN(aq) HNO 2 (aq) Chemical Equilibrium Focusing on Acid Base Systems 673

6 15 chapter Equilibrium In this chapter Exploration: Shakin the Blues Investigation 15.1: The Extent of a Chemical Reaction Web Activity: Equilibrium State Mini Investigation: Modelling Dynamic Equilibrium Lab Exercise 15.A: The Synthesis of an Equilibrium Law Web Activity: Paul Kebarle Lab Exercise 15.B: Determining an Equilibrium Constant Web Activity: Writing Equilibrium Expressions Investigation 15.2: Equilibrium Shifts (Demonstration) Biology Connection: CO 2 Transport Investigation 15.3: Testing Le Châtelier s Principle Case Study: Urea Production in Alberta Lab Exercise 15.C: The Nitrogen Dioxide Dinitrogen Tetroxide Equilibrium Investigation 15.4: Studying a Chemical Equilibrium System Web Activity: Poison Afloat Equilibrium Systems Equilibrium in Chemical Systems The simplest equilibrium systems are static: Nothing is moving or changing to create the balance. A textbook sitting on a level desktop is an example of static equilibrium.it stays motionless because two equal and opposite forces act on it simultaneously. The downward pull of Earth (gravity) on the book is exactly balanced by the upward push by the desktop. A laboratory balance is a common technology that uses this kind of equilibrium. Chemical equilibrium is also a balance between two opposing agents of change, but always in a dynamic system. An expert juggler in performance (Figure 1) is similar to a chemical system at equilibrium. The juggler s act is a dynamic equilibrium, with some balls moving upward and some moving downward at any given moment. There is no net change because the rates of upward movement and downward movement are equal at any given moment. Chemical systems at equilibrium have constant observable properties. Nothing appears to be happening because the internal movement involves entities that are too small to see. A critical task of chemical engineers is to disturb (unbalance) chemical equilibria in industrial reactions. Production of specific desired products is controlled by manipulating the conditions under which reactions occur. Some general concepts apply to all chemical equilibrium systems; these concepts are the focus of this chapter. STARTING Points Answer these questions as best you can with your current knowledge. Then, using the concepts and skills you have learned, you will revise your answers at the end of the chapter. 1. What, precisely, is happening to the chemical entities involved in a reaction while observation shows that products are being formed? 2. Is anything happening to the chemical entities involved in a reaction when observation shows the reaction appears to have stopped, with no more products being formed? 3. Why do some reactions seem to occur partially, and apparently stop while some of all of the reactants are still present, while in other reactions all of the limiting reagent appears to be consumed? 4. Can the chemical amount of product be predicted successfully for reactions that are not quantitative? Career Connection: Food Science Technologist; Chemical Process Engineer 674 Chapter 15

7 Figure 1 How is a juggler similar to a chemical system in equilibrium? Exploration Shakin the Blues When small pieces of zinc are added to a dilute solution of excess hydrochloric acid in an open beaker (Figure 2), a vigorous reaction occurs with lots of gas and heat given off. The zinc continues to react and, when it is completely consumed, the visible signs of reaction come to an end. After seeing many reactions such as this one in your science studies, you may have come to think that all chemical reactions only go one way: from reactants to products. But do they always? Materials: lab apron; eye protection; 400 ml flask and stopper; 250 ml water; 5.0 g potassium hydroxide (KOH(s)); 3.0 g glucose or dextrose; 2% methylene blue; stirring rod Pour 250 ml of water into the flask. Add 6 drops of methylene blue and all of the potassium hydroxide and glucose to the flask. Stir the mixture with the stirring rod until the solids have dissolved. Stopper the flask and set it on the bench. Observe the colour of the solution. Shake the solution vigorously and note any changes (Figure 3). Set the flask on the table and leave it standing until another change is noticed. Repeat the previous two steps many times. Make observations each time. (a) Describe the reaction in the flask in relation to the discussion at the beginning of this activity. (b) What evidence do you have to substantiate your answer to question (a)? (c) Predict whether the colour changes will continue forever. Test your prediction over a reasonable period of time. (d) Evaluate your prediction. Potassium hydroxide is poisonous and corrosive. Keep potassium hydroxide away from skin and eyes. Wear eye protection. Figure 2 Zinc reacts rapidly and quantitatively with hydrochloric acid. Figure 3 How many times does this reaction happen? Equilibrium Systems 675

8 15.1 Explaining Equilibrium Systems Scientists describe chemical systems in terms of empirical properties such as temperature, pressure, volume, and amounts of substances present. Chemical systems are simpler to study when separated from their surroundings by a definite boundary. This separation gives an experimenter control over the system so that no matter can enter or leave. Such a physical arrangement is called a closed system. A reaction in solution in a test tube or a beaker can be considered a closed system, as long as no gas is used or produced in the reaction. Systems involving gases must be closed on all sides by a solid container. Separating a chemical reaction system from its surroundings makes studying its properties, conditions, and changes much simpler. The use of controlled systems is an integral part of scientific study. Figure 1 When the pressure on this equilibrium system changes, the equilibrium is disturbed. Closed Systems at Equilibrium One example of a chemical system at equilibrium is a soft drink in a closed bottle a closed system in equilibrium. Nothing appears to change, until the bottle is opened. Removing the bottle cap and reducing the pressure alters the equilibrium state, as the carbon dioxide is allowed to leave the system (Figure 1). Carbonated drinks that have gone flat because of the decomposition of carbonic acid can be carbonated again by the addition of pressurized carbon dioxide to the solution to reverse the reaction, and then capping the container to restore the original equilibrium. Collision reaction theory is fundamental to the study of chemical systems. As originally introduced in this textbook to provide a basis for stoichiometric calculations, this theory required us to initially assume, for simplicity, that reactions are always spontaneous, rapid, quantitative, and stoichiometric. Common experience, however, shows that this assumption is not always true. Not all reactions are rapid; for example, corrosion of a car body may take years. A study of oxidation reduction reactions soon provides evidence that many reactions are not spontaneous. This chapter will examine and test the assumption of quantitative reaction in detail to significantly increase your understanding of chemical systems. INVESTIGATION 15.1 Introduction The Extent of a Chemical Reaction In Chapter 8, you performed experiments that produced evidence that reactions are quantitative. In a quantitative reaction, the limiting reagent is completely consumed. To identify the limiting reagent, you can test the final reaction mixture for the presence of the original reactants. For example, in a diagnostic test, you might try to precipitate ions from the final reaction mixture that were present in the original reactants. Purpose The purpose of this investigation is to test the validity of the assumption that chemical reactions are quantitative. Report Checklist Purpose Design Analysis Problem Materials Evaluation (1, 2, 3) Hypothesis Procedure Prediction Evidence Problem What are the limiting and excess reagents in the chemical reaction of selected quantities of aqueous sodium sulfate and aqueous calcium chloride? Design Samples of sodium sulfate solution and calcium chloride solution are mixed in different proportions and the final mixture is filtered. Samples of the filtrate are tested for the presence of excess reagents, using diagnostic tests. To perform this investigation, turn to page Chapter 15

9 Section 15.1 Practice 1. The evidence gathered in Investigation 15.1 may be classified as an anomaly an unexpected result that contradicts previous rules or experience. (a) Write the balanced equation for the double replacement reaction of sodium sulfate and calcium chloride solutions. (b) Write a statement describing the anomaly that occurred, using chemical names from the equation. (c) Write the net ionic equation for the reaction. (d) Use chemical names from the net ionic equation to write a statement about the anomaly. (e) Which of the previous statements more accurately describes the chemical system, according to collision reaction theory? 2. When scientists first encounter an apparent anomaly, they carefully evaluate the design, procedure, and technological skills involved in an investigation. One important consideration is the reproducibility of the evidence. Compare your evidence in Investigation 15.1 with the evidence collected by other groups. Is there support for the reproducibility of this evidence? DID YOU KNOW? Anomalies Signals for Change Anomalies, or discrepant events, are important as scientists acquire and develop scientific knowledge. Sometimes these events have been ignored, discredited, or elaborately explained away by scientists who do not wish to question or reconsider accepted laws and theories. Investigating anomalies sometimes leads to the restriction, revision, or replacement of scientific laws and theories. Evidence obtained from many reactions contradicts the assumption that reactions are always quantitative. In Investigation 15.1, there is direct evidence for the presence of both reactants after the reaction appears to have stopped. This apparent anomaly can be explained, in terms of collision reaction theory, by the idea that a reverse reaction can occur: the products, calcium sulfate and sodium chloride, can react to re-form the original reactants. The final state of this chemical system can be explained as a competition between collisions of reactants to form products and collisions of products to re-form reactants. forward Na 2 SO 4 (aq) CaCl 2 (aq) 0 CaSO 4 (s) 2 NaCl(aq) reverse This competition requires that the system be closed so that reactants and products cannot escape from the reaction container. The chemical system in Investigation 15.1 can be considered a closed system, bounded by the volume of the liquid phase. We assume that any closed chemical system with constant macroscopic properties (no observable change occurring) is in a state of equilibrium, usually classified, for convenience, as one of three types. Phase equilibrium involves a single chemical substance existing in more than one phase in a closed system. Water placed in a sealed container evaporates until the water vapour pressure (concentration of water in the gas phase) rises to a maximum value, and then remains constant (Figure 2). Solubility equilibrium involves a single chemical solute interacting with a solvent substance, where excess solute is in contact with the saturated solution (Figure 3). A chemical reaction equilibrium involves several substances: the reactants and products of a chemical reaction. All three types of equilibrium are explained by a theory of dynamic equilibrium a balance between two opposite processes occurring at the same rate. The terms forward and reverse are used to identify which process is being referred to, and are specific to a written equilibrium equation. When any equation is written with arrows to show that the change occurs both ways, the left-to-right change is called the forward reaction, and the right-to-left change is called the reverse reaction. H 2 O(g) 0 H 2 O(l) Figure 2 According to the theory of dynamic equilibrium, as long as the container remains closed at constant temperature, the rate at which molecules in the liquid state evaporate is equal to the rate at which molecules in the gas state condense. H 2 O(l) 0 H 2 O(g) Figure 3 For excess solid copper(ii) sulfate in equilibrium with its saturated aqueous solution, the rates of dissolving and crystallization are equal. CuSO 4 (s) 0 Cu 2 (aq) SO 4 2 (aq) Equilibrium Systems 677

10 WEB Activity Simulation Equilibrium State This simulation illustrates the establishment of a simple dynamic equilibrium by showing how the concentrations of entities change over time, beginning from an initial condition. A textbook can only represent this process as a sequence of static (still) diagrams (such as those in Figure 4) or graphs (such as that in Figure 5). GO mini Investigation Modelling Dynamic Equilibrium This activity models the progress of a chemical reaction to equilibrium, representing concentrations of reactants and products as volumes of water. The establishment of equilibrium is graphed as volume versus number of transfers. This graph then represents a typical concentration time graph of a chemical reaction to equilibrium. Materials: two drinking straws of different diameters; two 25 ml graduated cylinders; graph paper; water; meniscus finder Label one 25 ml graduated cylinder R (for reactants), and fill with water to the 25.0 ml mark. Label the other cylinder P (for products) and leave it empty. Holding a straw in each hand, place one straw in each graduated cylinder so that each straw rests on the bottom of its cylinder. Use the larger straw in the R cylinder. Transfer water simultaneously from each cylinder to the other by placing an index finger over the open end of each straw (to seal it). Then lift the straws out of their original cylinders, move the bottom of each straw over the other cylinder, and lift your index fingers to allow the water in each straw to drain into the cylinder below. Replace the straws in their original cylinders. After each transfer, measure and record the volume of water in each cylinder to 0.1 ml. Also record the number of the transfer. Repeat the transfer step until no significant change in water volumes has occurred for at least three transfers. Graph the volume of water in each cylinder as a dependent (responding) variable against the number of transfers as the independent (manipulated) variable. Repeat the activity, switching the straws used in each cylinder. Plot the values on the same axes as for the first trial. (a) How does the graph change, and how is it the same, when the smaller straw is initially in the R cylinder? (b) How would doing the transfers more slowly affect the final volumes in each cylinder? (c) How would replacing the larger straw with an even larger one affect the final volumes in each cylinder? (d) Express the equilibria of the water transfer trials as percent yields; that is, the percentage of reactant water that is converted to product. (e) Predict the graph s shape if both of the straws chosen had the same diameter. Chemical Reaction Equilibrium Chemical reaction equilibria are more complex than phase or solubility equilibria, due to the variety of possible chemical reactions and the greater number of substances involved. To explain chemical equilibrium systems, we need to combine ideas from atomic theory, kinetic molecular theory, collision reaction theory, and the concepts of reversibility and dynamic equilibrium. Although this synthesis is successful as a description, and also as an explanation, it has only limited application in predicting quantitative properties of an equilibrium system. 678 Chapter 15

11 Section 15.1 The Hydrogen Iodine Reaction System Chemists have studied the reaction of hydrogen gas and iodine gas extensively, because the molecules are simple in structure and the reaction takes place in the gas phase. Once hydrogen and iodine are mixed, the reaction proceeds rapidly at first. The initial dark purple colour of the iodine vapour gradually fades, and then remains constant (Figure 4). I 2 (g) H 2 (g) I 2 (g) H 2 (g) HI(g) I 2 (g) H 2 (g) HI(g) Figure 4 When hydrogen and iodine are added to the flask, the colour of the iodine vapour is the only easily observable (empirical) property. An equilibrium equation describes this evidence theoretically. H 2 (g) I 2 (g) 0 2 HI(g), t 448 C Initially, hydrogen (in excess) and iodine are added to the flask. The colour of the iodine vapour is the only easily observable property. Early in the reaction, hydrogen and iodine form hydrogen iodide faster than hydrogen iodide forms hydrogen and iodine. Overall, the amount of iodine decreases, so the colour of the flask contents appears to lighten. Both hydrogen and hydrogen iodide are colourless. At equilibrium, analysis shows that the flask contains all three substances. The purple colour shows that some iodine remains. The constancy of the colour is evidence that equilibrium exists. Forward and reverse reactions are occurring at equal rates. Table 1 contains data from three experiments with the hydrogen iodine system: one in which hydrogen and iodine are mixed; one in which hydrogen, iodine, and hydrogen iodide are mixed; and one in which only hydrogen iodide is present initially. At a temperature of 448 C, the system quickly reaches an observable equilibrium each time. Chemists use evidence such as that in Table 1 to describe a state of equilibrium in two ways: in terms of percent reaction and in terms of an equilibrium constant. Percent reaction describes the equilibrium for one specific system example only, whereas an equilibrium constant describes all systems of the same reaction at a given temperature. Alternatively, you can draw a graph of the reaction progress, plotting quantity (or concentration) of the reagents versus time (Figure 5). Table 1 The Hydrogen Iodine System at 448 C Initial system concentrations Equilibrium system concentrations System (mmol/l) (mmol/l) H 2 (g) I 2 (g) HI(g) H 2 (g) I 2 (g) HI(g) Quantity Reaction of Hydrogen and Iodine t = 448 C Time (reaction progress) HI H 2 Figure 5 The graph of the quantity of each substance against time shows that the rate of reaction of the reactants decreases as the number of reactant molecules decreases, and the rate at which the product changes back to reactants increases as the number of product molecules increases. These two rates must become equal at some point, after which the quantity of each substance present will not change. I 2 Equilibrium Systems 679

12 DID YOU KNOW? Lavoisier and Closed Systems Antoine Laurent Lavoisier ( ) is recognized as the father of modern chemistry for many reasons. He created the basic nomenclature system we still use for compounds, demonstrated the law of conservation of mass, and explained and clarified the theory of combustion. Most importantly, perhaps, his successes convinced the chemical community of the critical importance of his methods of careful measurement, and of carrying out experiments in closed systems carefully accounting for, and preventing, invisible reactants and products (notably, gases) from escaping. When chemists began to understand the importance of maintaining closed systems in order to draw correct conclusions about reactions, chemistry could finally move forward as a true science. Learning Tip When a reaction is shown to be quantitative (as written in the equation), it means that the reverse reaction happens so little that it can be ignored for all normal purposes. Another way to think of a quantitative reaction is that if the products shown (as written) were mixed together as reactants, there would be no apparent reaction. A reaction that is quantitative in the forward direction is necessarily nonspontaneous in the reverse direction. In this unit, we will (arbitrarily) assume that "quantitative" specifies a reaction that, at equilibrium, is more than 99.9% complete. Another way to think of it is that less than one part per thousand (0.1%) of an original reactant remains unreacted, at equilibrium, in a quantitative reaction. A percent yield is defined as the yield of product measured at equilibrium compared with the maximum possible yield of product. In other words, percent yield can be useful for communicating the position of an equilibrium. The maximum possible yield of product is calculated using the method of stoichiometry, assuming a quantitative forward reaction with no reverse reaction. Percent yield provides an easily understood way to refer to quantities of chemicals present in equilibrium systems. For example, analysis of the evidence in System 1, Table 1, shows that, at 448 C, this particular hydrogen iodine system reaches an equilibrium with a percent yield of 78.0% (Table 2). Table 2 Percent Yield of the Hydrogen Iodine System at 448 C Equilibrium [HI]* Maximum possible [HI]* Percent yield System (mmol/l) (mmol/l) (%) *Square brackets [ ] indicate amount concentration. Equilibrium arrows (0) communicate that an equilibrium exists. To communicate the extent of a reaction, a percent yield may be written above the equilibrium arrows in a chemical equation. The following equation describes the position of a hydrogen iodine equilibrium in System 1, Table 2, at 448 C. 78% H 2 (g) I 2 (g) 0 2 HI(g) t 448 C Scientists now think of all chemical reactions as occurring in both forward and reverse directions. Any reaction falls loosely into one of four categories. Reactions that favour reactants very strongly, that is, reactions that normally have a percent yield of much less than 1%, are simply observed as being nonspontaneous. In these reactions, mixing reactants has no observable result. Reactions producing observable equilibrium conditions may react less or more than 50%, favouring reactants or products respectively. Significant amounts of both reactants and products are always present. Finally, reactions that favour products very strongly, much more than 99%, are observed to be complete (quantitative). The chemical equations for quantitative reactions are generally written with a single arrow to indicate that the effect of the reverse reaction is negligible. Table 3 shows how percent yield may be used to classify equilibrium systems and how the classification may be communicated in reaction equations. Table 3 Classes of Chemical Reaction Equilibria Percent yield Description of equilibrium Position of equilibrium negligible nonspontaneous (no apparent reaction) 50% reactants favoured 50% 0 50% products favoured 50% % quantitative 680 Chapter 15

13 Section 15.1 When considering equilibrium systems, we cannot use the simple assumption of quantitative reaction. When there is no limiting reagent for a reaction, and when we cannot assume complete reaction, stoichiometric calculations require a little more thought. Such calculations may conveniently be set up as an ICE table, meaning that the initial, change, and equilibrium values are arranged in tabular form. SAMPLE problem 15.1 Consider the reaction equation for the formation of hydrogen iodide at 448 C. Assume the reaction is begun with 1.00 mmol/l concentrations of both H 2 (g) and I 2 (g). Construct an ICE table to determine equilibrium concentrations of the reagents. The equilibrium concentration of I 2 (g) (determined by colour intensity) is 0.22 mmol/l. Set up the ICE table as follows: Table 4 The H 2 (g) I 2 (g) 0 2 HI(g) Equilibrium [H 2 (g)] [I 2 (g)] [HI(g)] Concentration (mmol/l) (mmol/l) (mmol/l) Initial Change Equilibrium 0.22 DID YOU KNOW? Diabetes: Blood Sugar Equilibrium For people with diabetes, reaction equilibrium established by sugar in the human body is critically important. Recent advances in the technology allow testing of blood sugar concentration with personal devices such as the One Touch SureStep blood-glucose meter (Figure 6). The meter displays the concentration in mmol/l after analyzing a single drop of blood extracted from a fingertip. Multiple readings, including date and time, are stored electronically and can be displayed at any time. The data can be downloaded to a computer. Begin by calculating the change (decrease) in concentration of iodine. ( ) mmol/l 0.78 mmol/l The changes of the other concentrations may be calculated directly from this value, using stoichiometric ratios from the balanced equation. For hydrogen, the change is also a decrease (negative value) of mmol/l 0.78 mmol/l 1 For hydrogen iodide, the change is an increase (positive value) of mmol/l 1.6 mmol/l 1 Complete the ICE table, using these values to enter the concentrations at equilibrium. Figure 6 This device helps diabetics monitor their blood glucose levels. Table 5 The H 2 (g) I 2 (g) 0 2 HI(g) Equilibrium [H 2 (g)] [I 2 (g)] [HI(g)] Concentration (mmol/l) (mmol/l) (mmol/l) Initial Change Equilibrium In Sample Problem 15.1, notice that every substance in the reaction is a gas. Therefore, all of the stoichiometric calculations can use concentrations directly, rather than chemical amounts, because the volume must be the same for every gaseous substance in a closed container. The volume is a common factor in the calculation step that uses the stoichiometric ratio. This same reasoning means that concentrations can also be used directly for stoichiometric calculation whenever every substance in a reaction is an aqueous entity dissolved in the same volume of solvent. CAREER CONNECTION Food Science Technologist The development and analysis of food products for individuals with diabetes is crucial. Food science technologists carefully measure and conduct tests on carbohydrates so that patients can control their blood sugar equilibrium. Learn more about the many food industries that employ food science specialists. GO Equilibrium Systems 681

14 Learning Tip Chemists often refer to homogeneous reaction systems. This term means that every entity involved in the reaction exists in the same physical state. The reaction from Sample Problem 15.1, where all reactants and products are gases, is a typical case. The other common case involves reactions where all entities involved are in aqueous solution. A system that has more than one phase, such as solid copper reacting in aqueous silver nitrate solution, is a heterogeneous reaction system. Practice 3. For a chemical system at equilibrium: (a) What are the observable characteristics? (b) Why is the equilibrium considered dynamic? (c) What is considered equal about the system? 4. In a gaseous reaction system, 2.00 mol of methane, CH 4 (g), is initially added to mol of chlorine, Cl 2 (g). At equilibrium the system contains 1.40 mol of chloromethane, CH 3 Cl(g), and some hydrogen chloride, HCl(g). (a) Write a balanced reaction equation for this equilibrium and calculate the maximum possible yield of chloromethane product. (b) Calculate the percent yield at this equilibrium and state whether products or reactants are favoured. 5. Combustion reactions, such as the burning of methane, often favour products so strongly that they are written with a single arrow. Assuming the forward reaction has a very low activation energy (Chapter 12) and the reverse reaction has a very high activation energy (to account for the difference in the tendency to occur), sketch a possible potential energy diagram representing the progress of such a reaction. 6. After 4.0 mol of C 2 H 4 (g) and 2.50 mol of Br 2 (g) are placed in a sealed container, the reaction C 2 H 4 (g) Br 2 (g) 0 C 2 H 4 Br 2 (g) establishes an equilibrium. Figure 7 shows the concentration of C 2 H 4 (g) as it changes over time at a fixed high temperature until equilibrium is reached. Concentration (mol/l) Reaction of Ethene and Bromine Time Figure 7 A graph of the reaction of ethene with bromine C 2 H 4 (a) Sketch this graph. Draw lines on your copy to show how the concentration of each of the other two substances changes. (b) Create an ICE table, using reagent amount concentrations. (c) What is the volume of the container? (d) Calculate the percent yield of dibromoethane. 7. Write a balanced net ionic equation for each of the following described reactions, showing appropriate use of equilibrium "arrow" symbols where appropriate, and indicating (with symbols) whether products or reactants are favoured. (a) An excess of solid copper reacts with virtually all of the silver ions in a sample solution. (b) When a solution containing calcium ions is mixed with a solution containing a large excess of sulfate ions, a precipitate forms, but tests indicate that a small quantity of calcium ions remains in solution. (c) When acetic acid is dissolved in water, the acetic acid molecules react with water molecules to form hydronium and acetate ions. Careful ph testing shows that about 980 of every 1000 acetic acid molecules remain in their molecular form, at equilibrium. 682 Chapter 15

15 Section 15.1 LAB EXERCISE 15.A The Synthesis of an Equilibrium Law The following chemical equation represents a chemical equilibrium: Fe 3 (aq) SCN (aq) 0 FeSCN 2 (aq) This equilibrium is convenient to study because the colour of the system characterizes the equilibrium position of the system (Figure 8). Purpose The purpose of this investigation is the synthesis of an equilibrium law. Complete the Analysis of the investigation report. Problem What mathematical formula, using equilibrium concentrations of reactants and products, gives a constant for the iron(iii) thiocyanate reaction system? Report Checklist Purpose Design Analysis Problem Materials Evaluation Hypothesis Procedure Prediction Evidence Design Reactions are performed using various initial concentrations of iron(iii) nitrate and potassium thiocyanate solutions. The equilibrium concentrations of the reactants and the product are determined from the measurement and analysis of the colour intensity using a spectrophotometer. Possible mathematical relationships among the concentrations are tried and analyzed to determine if the mathematical formula gives a constant value. Evidence Table 6 Iron(III) Thiocyanate Equilibrium at SATP [Fe 3 (aq)] [SCN (aq)] [FeSCN 2 (aq)] Trial (mol/l) (mol/l) (mol/l) Fe 3+ (aq) FeSCN 2+ (aq) SCN (aq) Figure 8 The two reactants combine to form a dark red equilibrium mixture. The red colour of the solution is due to the aqueous thiocyanate iron(iii) product, FeSCN 2 (aq). Analysis Test the following mathematical relationships for constancy: 1. [Fe 3 (aq)][scn (aq)][fescn 2 (aq)] 2. [Fe 3 (aq)] [SCN (aq)] [FeSCN 2 (aq)] [FeSCN 2 (aq)] [Fe 3 (aq)][scn (aq)] [Fe 3 (aq)] [FeSCN 2 (aq)] 5. [SCN (aq)] [FeSCN 2 (aq)] WEB Activity Canadian Achievers Paul Kebarle Paul Kebarle (Figure 9) pioneered the measurements of gas-phase ion-molecule equilibria. Kebarle s findings, now significantly expanded by other workers, constitute a central database that is of fundamental importance in many diverse fields of scientific research. 1. What fundamental data did Kebarle and his co-workers obtain from their research? 2. List three fields of research that Kebarle s work has aided. GO Figure 9 Paul Kebarle DID YOU KNOW? Computers Scientists often use computers to analyze numerical evidence in order to establish mathematical relationships among experimental variables. The mathematical formulas derived are useful in understanding chemical processes and in applying these processes to technology. Equilibrium Systems 683

16 DID YOU KNOW? Related Interests Two Norwegian chemists, Cato Maximilian Guldberg and Peter Waage, conducted detailed empirical studies of many equilibrium systems in the mid-1800s (Figure 10). By 1864, they had proposed a mathematical description of the equilibrium condition that they called the law of mass action. Analyzing the results of their experiments, Guldberg and Waage noticed that, when they arranged the equilibrium concentrations into a specific form of ratio, the resulting value was the same no matter what combinations of initial concentrations were mixed. The Equilibrium Constant, K c Analysis of the evidence from many experiments such as those in Lab Exercise 15.A (page 683) reveals a mathematical relationship that provides a constant value for a chemical system over a range of amount concentrations. This constant value is called the equilibrium constant, K c, for the reaction system. Evidence and analysis of many equilibrium systems have resulted in the following equilibrium law. For the reaction a A b B 0 c C d D, c d the equilibrium law expression is K c [ C] [ D] a b [ A] [ B] In this mathematical expression, A, B, C, and D represent chemical entity formulas and a, b, c, and d represent their coefficients in the balanced chemical equation. The relationship holds only when amount concentrations are observed to remain constant, in a closed system, at a given temperature. COMMUNICATION example 1 Write the equilibrium law expression for the reaction of nitrogen monoxide gas with oxygen gas to form nitrogen dioxide gas. Solution 2 NO(g) O 2 (g) 0 2NO 2 (g) K c [NO 2 (g)] 2 [NO(g)] 2 [O 2 (g)] Figure 10 Cato Maximilian Guldberg ( ) and Peter Waage ( ) were related by more than their interest in chemistry: They were brothers-in-law! Learning Tip It is common practice (convention) to ignore units and list only the numerical value for an (amount concentration) equilibrium constant. The expression of units is often very complex for K c relationships. But, because each entity concentration is always entered with mol/l units, any entity concentration we calculate from a K c value will always give an answer having mol/l units, so you need only memorize this (simplifying) rule. We use a balanced chemical equation with whole-number coefficients to write the mathematical expression of the equilibrium law. The coefficients of the balanced equation become the exponents of the amount concentrations. If the equation were to be written in reverse, the equilibrium law expression would simply be the reciprocal of the expression above, and the equilibrium constant would be the reciprocal of the one for the reaction as written here. Using the products over reactants convention results in a relationship between the numerical value of K c and the forward extent of the equilibrium that is easier to visualize for the equation as written. The higher the numerical value of the equilibrium constant, the greater the tendency of the system to favour the forward direction; that is, the greater the equilibrium constant, the more the products are favoured at equilibrium. COMMUNICATION example 2 The value of K c for the formation of HI(g) from H 2 (g) and I 2 (g) is 40, at temperature t. Determine the value of K c for the decomposition of HI(g) at the same temperature. Solution 2 Hl(g) 0 H 2 (g) I 2 (g) [H K c 2 (g)][i 2 (g)] = [Hl(g)] 2 Note that the decomposition reaction equation is the reverse of the formation reaction equation, and the value of K c for decomposition is the reciprocal of the K c for formation. 684 Chapter 15

17 Section 15.1 Experiments have shown that the value of the equilibrium constant depends on temperature. The value is also affected by very large changes in the equilibrium concentration of a reactant or a product. A moderate change in the concentration of any one of the reactants or products results in a change in the other concentrations, so that the equilibrium constant remains the same. The equilibrium constant provides only a measure of the equilibrium position of the reaction; it does not provide any information on the rate of the reaction. Because they hold for a significant range of different concentrations, equilibrium constant expressions have been found to be very useful, and K c values for reactions are in common use throughout the scientific community. Equilibrium constants are adjusted to reflect the fact that pure substances in solid or liquid (condensed) states have concentrations that are essentially fixed the chemical amount (number of moles) per unit volume is a constant value. For example, a litre of liquid water at SATP has a mass of 1.00 kg (a chemical amount of 55.5 mol) and, thus, a fixed amount concentration of 55.5 mol/l. The concentration of condensed states is not included in a K c expression we assume that these constant values become part of the expressed equilibrium constant. Substances in a gaseous or dissolved state have variable concentrations, and must always be shown in an equilibrium law expression. COMMUNICATION example 3 Write the equilibrium law expression for the decomposition of solid ammonium chloride to gaseous ammonia and gaseous hydrogen chloride. Solution NH 4 Cl(s) 0 NH 3 (g) HCl(g) K c [NH 3 (g)][hcl(g)] The concentration of solid NH 4 Cl(s) is omitted from the equilibrium law expression. The role of temperature in equilibrium constant expressions is critical, although the temperature is not written in the expression directly. The value of the equilibrium constant, K c, always depends on the temperature. Any stated numerical value for an equilibrium constant, or any calculation using an equilibrium constant expression, must specify the reaction temperature at equilibrium. Since equilibrium depends on the concentrations of reacting substances, these substances must be represented in the expression as they actually exist meaning that ions in solution must be represented as individual entities. Equilibrium constant expressions are always written from the net ionic form of reaction equations, balanced with simplest whole-number (integral) coefficient values unless otherwise specified. DID YOU KNOW? Using Constant Relationships You are already familiar with the usefulness of some other constant mathematical relationships about real phenomena. Finding a relationship that is constant for equilibrium concentrations is just another example. If you examine circles of different sizes carefully, you discover that the distance around any circle divided by the distance across it (at the widest part) always gives the same (constant) answer, no matter how big or small the circle. This constant value, , is so useful that it has been given its own symbol, the Greek letter pi,. This relationship is most usefully expressed as C d because measuring a diameter is much easier than measuring a circumference. Many other relationship expressions produce this same constant, including those usually used to calculate the area of a circle and the time period of a pendulum s swing. + EXTENSION The Meaning of the Equilibrium Constant Try this simulation to deepen your understanding of the equilibrium constant. GO COMMUNICATION example 4 Write the equilibrium law expression for the reaction of zinc in copper(ii) chloride solution. Solution Zn(s) Cu 2 (aq) 0 Cu(s) Zn 2 (aq) K c [Zn2 (aq)] [Cu 2 (aq)] Again, note that the solids, as well as the spectator ions (the chloride ions in this example), are omitted from the equilibrium law expression. Equilibrium Systems 685

18 LAB EXERCISE 15.B Determining an Equilibrium Constant Determining the equilibrium constant for a reaction at specific conditions is often essential for an industrial chemist. This knowledge is necessary before adjusting the reaction conditions in order to optimize production of the desired substances. Complete the Analysis of the investigation report. Purpose The purpose of this investigation is to use the equilibrium law to determine the equilibrium constant at 200 C for the decomposition reaction of the molecular compound phosphorus pentachloride. Report Checklist Purpose Design Analysis Problem Materials Evaluation Hypothesis Procedure Prediction Evidence Problem What is the value of the equilibrium constant for the decomposition of phosphorus pentachloride gas to phosphorus trichloride gas and chlorine gas, at a temperature of 200 C? Evidence equilibrium temperature 200 C equilibrium concentrations: [PCl 3 (g)] [Cl 2 (g)] mol/l [PCl 5 (g)] mol/l SUMMARY Writing Equilibrium Law Expressions Write an equilibrium law expression based on a balanced equation for the reaction system. Use single whole-number coefficients, written in net ionic form, and ignore concentrations of pure solid or liquid phases: If: aa bb 0 cc dd then: K c [C] c [D] d [A] a [B] b An equilibrium constant value always depends on the system temperature is independent of the reagent concentrations is independent of any catalyst present is independent of the time taken to reach equilibrium is normally stated as a numerical value, ignoring any units is greater, the more the system favours the formation of products Predicting Final Equilibrium Concentrations For simple homogeneous systems, it is possible to algebraically predict reagent concentrations at equilibrium using a known value for K c and initial reactant concentration values. For more complex systems, the calculation becomes more difficult such systems are left for more advanced chemistry courses. SAMPLE problem 15.2 In a 500 ml stainless steel reaction vessel at 900 C, carbon monoxide and water vapour react to produce carbon dioxide and hydrogen. Evidence indicates that this reaction establishes an equilibrium with only partial conversion of reactants to products. Initially, 2.00 mol of each reactant is placed in the vessel. K c for this reaction is 4.20 at 900 C. What amount concentration of each substance will be present at equilibrium? Write a balanced equation for the reaction equilibrium. CO(g) H 2 O(g) 0 CO 2 (g) H 2 (g) K c 4.20 at 900 C 686 Chapter 15

19 Section 15.1 Use the balanced equation to write the equilibrium law expression. [CO 2 (g)][h 2 (g)] K c 4.20 [CO(g)][H2 O(g)] The initial amount concentrations of the CO(g) and the H 2 O(g) are the same: c 2.00 mol 4.00 mol/l [CO(g)] [H L 2 O(g)] An ICE table makes it easier to keep track of amount concentration changes that occur during a reaction, and to find the amount concentrations at equilibrium. At equilibrium, let the final amount concentration of the product H 2 (g) be x mol/l (any convenient symbol could be used). Then [CO 2 (g)] must also be x mol/l, since the stoichiometric ratio of the reaction is 1:1:1:1. By this same reasoning, at equilibrium, the initial concentrations of CO(g) and H 2 O(g) must have decreased by x mol/l; so [CO(g)] [H 2 O(g)] (4.00 x) mol/l. When you enter values for Change into the ICE table, you must show increases as, and decreases as. Table 7 The CO(g) H 2 O(g) 0 CO 2 (g) H 2 (g) Equilibrium [CO(g)] [H 2 O(g)] [CO 2 (g)] [H 2 (g)] Concentration (mol/l) (mol/l) (mol/l) (mol/l) Initial Change x x x x Equilibrium (4.00 x) (4.00 x) x x Substitute equilibrium concentrations in the equilibrium law expression. [CO x 2 2 (g)][h 2 (g)] 4.20 (4.00 x) 2 [CO(g)][H2 O(g)] Since the right side of the equation is a perfect square, solving for x is quite straightforward. x (4.00 x) 2 x x x (2.05)(4.00 x) x 3.05x 8.20 x 2.69 Assign positive and negative signs and complete the ICE table. Table 8 The CO(g) H 2 O(g) 0 CO 2 (g) H 2 (g) Equilibrium [CO(g)] [H 2 O(g)] [CO 2 (g)] [H 2 (g)] Concentration (mol/l) (mol/l) (mol/l) (mol/l) Initial Change Equilibrium EXTENSION Equilibrium Constant and Reaction Quotient Is there any way of knowing whether or not a reaction is at equilibrium? If we know the concentrations of reactants and products, and the equilibrium constant at the appropriate temperature, we can use a concept called the "reaction quotient" to predict which way the reaction will proceed, or if it is already at equilibrium. GO At equilibrium, at 900 C, [CO(g)] [H 2 O(g)] 1.31 mol/l and [CO 2 (g)] [H 2 (g)] 2.69 mol/l Note that, if both initial reactant concentrations are not the same, solving the equation for x is more complicated, requiring use of the quadratic formula. Questions in this text are restricted to examples that do not require the quadratic formula for solution. Equilibrium Systems 687

20 WEB Activity Simulation Writing Equilibrium Expressions This simulation allows you to select a reaction type and the initial reactant concentrations, which the program uses to plot the resulting equilibrium graph. You will then be guided through a series of questions. GO Section 15.1 Questions 1. Write a balanced equation with integer coefficients and the expression of the equilibrium law for each of the following reaction systems at fixed temperature. (a) Hydrogen gas reacts with chlorine gas to produce hydrogen chloride gas in the industrial process that eventually produces hydrochloric acid. (b) In the Haber process (Chapter 8), nitrogen reacts with hydrogen to produce ammonia gas. (c) At some time in the future, industry and consumers may make more extensive use of the combustion of hydrogen as an energy source. (d) When aqueous ammonia is added to an aqueous nickel(ii) ion solution, the Ni(NH 3 ) 6 2 (aq) complex ion is formed (Figure 11). Ni 2+ (aq) Ni(NH 3 ) 6 2+ (aq) Figure 11 A Ni 2 (aq) solution is green. Ammonia reacts with the nickel(ii) ion to form the intensely blue hexaamminenickel(ii) ion, Ni(NH 3 ) 6 2 (aq). (e) In the Solvay process for making washing soda (Chapter 7), one reaction involves heating solid calcium carbonate (limestone) to produce solid calcium oxide (quicklime) and carbon dioxide. (f) In Investigation 15.1, aqueous solutions of sodium sulfate and calcium chloride are mixed. (Remember to use a net ionic equation.) (g) In a sealed can of soda, carbonic acid, H 2 CO 3 (aq), decomposes to liquid water and carbon dioxide gas. 2. You can apply the empirical and theoretical concepts of equilibrium to many different chemical reaction systems. Use the generalizations from your study of organic chemistry to predict the position of equilibrium for bromine placed in a reaction container with ethylene at a high temperature. 3. Interpret the graph in Figure 12 to answer the questions about the reaction. Hydrogen and iodine were placed in a reaction vessel, which was then sealed, and heated to 450 C. Concentration (mol/l) Reaction of Hydrogen and Iodine t = 448 C Time Figure 12 The progress of a hydrogen iodine reaction (a) All three substances are gases. If the container has a volume of 2.00 L, what chemical amount of each substance was present initially? (b) What chemical amount of hydrogen iodide had formed at equilibrium? (Create an ICE table.) (c) Describe the rate at which hydrogen is reacting from the moment the reactants are mixed to the time when equilibrium has been established, in terms of collision reaction theory. 4. For each of the following, write the chemical reaction equation with appropriate equilibrium arrow, as shown in Table 3 (page 680). (a) The Haber process is used to manufacture ammonia fertilizer from hydrogen and nitrogen gases. Under lessthan-desirable conditions, only an 11% yield of ammonia is obtained at equilibrium. (b) A mixture of carbon monoxide and hydrogen, known as water gas, is used as a supplementary fuel in many large industries. At high temperatures, the reaction of coke and steam forms an equilibrium mixture in which the products (carbon monoxide and hydrogen gases) are favoured. (Assume that coke is pure carbon.) (c) Because of the cost of silver, many high school science departments recover silver metal from waste solutions containing silver compounds or silver ions. A quantitative reaction of waste silver ion solutions with HI H 2 I Chapter 15

21 Section 15.1 copper metal results in the production of silver metal and copper(ii) ions. (d) One step in the industrial process used to manufacture sulfuric acid is the production of sulfur trioxide from sulfur dioxide and oxygen gases. Under certain conditions, the reaction produces a 65% yield of products. 5. Write the expression of the equilibrium law for the hydrogen iodine hydrogen iodide system at 448 C. Using the evidence for System 1 as reported in Table 1 on page 679, calculate the value of the equilibrium constant. 6. In the Haber process for synthesizing ammonia gas from nitrogen and hydrogen, the value of K c is for the reaction at 500 C. In a sealed container at equilibrium at 500 C, the concentrations of H 2 (g) and of N 2 (g) are measured to be 0.50 mol/l and 1.50 mol/l, respectively. Write the equilibrium law expression and calculate the equilibrium concentration of NH 3 (g). 7. At a certain constant (very high) temperature, 1.00 mol of HBr(g) is introduced into a 2.00 L container. Decomposition of this gas to hydrogen and bromine gases quickly establishes an equilibrium, at which point the amount concentration of HBr(g)is measured to be mol/l. (a) Write a balanced equation for the reaction. (b) Write the equilibrium law expression. (c) Calculate the chemical amount of HBr(g) present at equilibrium. (d) Calculate the chemical amount of HBr(g) that has reacted to form H 2 (g) and Br 2 (g) products when equilibrium is established. (e) Calculate the chemical amounts of H 2 (g) and Br 2 (g) that have been produced, and, thus, are present, when equilibrium is established. (f) Calculate the amount concentration of all substances present at equilibrium. (g) Calculate K c for this reaction at this temperature. 8. To a heated reaction vessel with a volume of 1.00 L, a lab technician adds 6.23 mmol H 2 (g), 4.14 mmol of I 2 (g), and mmol of HI(g). At equilibrium, a spectrophotometer is used to determine that the concentration of iodine vapour is 2.58 mmol/l. Construct an ICE table and find K c for the reaction system H 2 (g) I 2 (g) 0 2 HI(g). 9. Consider the system CO 2 (g) H 2 (g) 0 CO(g) H 2 O(g) Initially, 0.25 mol of water and 0.20 mol of carbon monoxide are placed in the reaction vessel. At equilibrium, spectroscopic evidence shows that 0.10 mol of carbon dioxide is present. Construct an ICE table and find K c for this system. 10. Consider the system 2 HBr(g) 0 H 2 (g) Br 2 (g) Initially, 0.25 mol of hydrogen and 0.25 mol of bromine are placed into a 500 ml electrically heated reaction vessel. K c for the reaction at the temperature used is (a) Find the concentrations of the substances at equilibrium. (b) Calculate the chemical amount of each substance present at equilibrium. 11. Explain briefly how atomic theory, kinetic molecular theory, collision reaction theory, and the concepts of reaction rate and reversible reactions are all necessary to explain chemical reaction equilibrium observations. Extension 12. In a very long-term sense, Earth may be considered a closed system. One equilibrium of concern to scientists is the same one involved in carbonation of soft drinks, on a vastly larger scale. Scientists believe that over time, the carbon dioxide gas in the atmosphere should be in equilibrium with carbon dioxide dissolved in the oceans. They also know that the concentration of CO 2 (g) in the atmosphere has been increased significantly (by about 20%) in the last century, which, they believe, is mostly due to the burning of fossil fuels. Concerns about the consequences of global warming make it imperative that scientists improve their theories about the various cycles, processes, and equilibria involving this greenhouse gas. Research and summarize currently accepted theory about carbon dioxide dissolved in the oceans, and list some other cycles and systems involving reaction or production of CO 2 (g). GO Equilibrium Systems 689

22 15.2 Figure 1 Henri Louis Le Châtelier ( ), French chemist and engineer, worked in chemical industries. To maximize the yield of products, Le Châtelier used systematic trial and error. After measuring properties of equilibrium states in chemical systems, he discovered a pattern and stated it as a generalization. This generalization has been supported extensively by evidence and is now considered a scientific law. By convention, it is known as Le Châtelier s principle. Qualitative Change in Equilibrium Systems Observing the effects of varying system properties on the equilibrium of systems contributes greatly to our understanding of the equilibrium state. From a technological perspective, controlling the extent of equilibrium by manipulating properties is very desirable because control leads to more efficient and economic processes. From a scientific perspective, observing systems at equilibrium leads to improved theories that describe, explain, and predict the nature of equilibrium, thus increasing our understanding. Equilibrium is an area of study where, historically, technology has led science. Reactions were first manipulated in response to some human need, although the reactions responses were not explained until much later by successive theories of increasing validity. This section of the chapter will examine equilibrium manipulation in the same way, with empirical descriptions of equilibrium manipulation given first, followed by theoretical explanations of the observed results. According to Le Châtelier s principle, when a chemical system at equilibrium is disturbed by a change in a property of the system, the system always appears to react in the direction that opposes the change, until a new equilibrium is reached (Figures 1 to 3). The application of Le Châtelier s principle involves a three-stage process: an initial equilibrium state, a shifting non-equilibrium state, and a new equilibrium state. Le Châtelier s principle provides a method of predicting the response of a chemical system to an imposed change. Using this simple and completely empirical approach, chemical engineers could produce more of the desired products, making technological processes more efficient and more economical. For example, Fritz Haber used Le Châtelier s principle to devise a process for the economical production of ammonia from atmospheric nitrogen. (See the Haber process, Chapter 8, page 325.) Figure 2 Fe 3 (aq) SCN (aq) 0 FeSCN 2 (aq) The test tube on the left is at equilibrium, as shown by the constant colour of the FeSCN 2 (aq) ion. The equilibrium is disturbed by the addition of Fe 3 (aq) ions. The system shifts and some of the additional Fe 3+ (aq) reacts to produce more FeSCN 2 (aq), thus establishing a new equilibrium state. When the shift is complete, the concentration of Fe 3+ (aq) is higher than before (only some of the added ions react), the concentration of SCN (aq) is lower, and the concentration of FeSCN 2 (aq) is higher. The higher concentration of FeSCN 2 (aq) is evident from the more intense colour of the solution observed in the test tube on the right. Fe 3 (aq) SCN (aq) 0 FeSCN 2 (aq) 690 Chapter 15

23 Section 15.2 INVESTIGATION 15.2 Introduction Equilibrium Shifts (Demonstration) In this investigation, you will be looking at two equilibrium systems: N 2 O 4 (g) energy 0 2NO 2 (g) colourless reddish brown CO 2 (g) H 2 O(l) 0 H (aq) HCO 3 (aq) The second equilibrium system, produced by the reaction of carbon dioxide gas and water, is commonly found in the human body and in carbonated drinks. A diagnostic test is necessary to detect shifts in this equilibrium. Bromothymol blue, an acid base indicator, can detect an increase or decrease in the hydrogen ion concentration in this system. Bromothymol blue turns blue when the hydrogen ion concentration decreases, and yellow when the hydrogen ion concentration increases. Report Checklist Purpose Design Analysis Problem Materials Evaluation (2, 3) Hypothesis Procedure Prediction Evidence Purpose The purpose of this demonstration is to test Le Châtelier s principle by studying two chemical equilibrium systems: the equilibrium between two oxides of nitrogen, and the equilibrium of carbon dioxide gas and carbonic acid. Problem How does a change in temperature affect the nitrogen dioxide dinitrogen tetroxide equilibrium system? How does a change in pressure affect the carbon dioxide carbonic acid equilibrium system? To perform this investigation, turn to page 700. Le Châtelier s Principle and Concentration Changes Le Châtelier s principle predicts that if the addition of a reactant to a system at equilibrium increases the concentration of that substance, then that system will undergo an equilibrium shift forward (to the right). The effect of the shift is that, temporarily, we observe the reactant concentration decreasing, as some of the added reactant changes to products. This period of change ends with the establishment of a new equilibrium state where, once again, there are no observable changes. The system has changed in such a way as to oppose the change introduced. For example, the production of freon-12, a CFC refrigerant, involves the following equilibrium reaction taking place at a fixed temperature: CCl 4 (l) 2 HF(g) 0 CCl 2 F 2 (g) 2 HCl(g) freon-12 To improve the yield of the primary product, freon-12, more hydrogen fluoride is added to the initial equilibrium system. The additional concentration of reactant disturbs the equilibrium state and the system shifts to the right, consuming some of the added hydrogen fluoride by reaction with carbon tetrachloride. As a result, more freon-12 is produced and a new equilibrium state is reached. In chemical reaction equilibrium shifts, an imposed concentration change is normally only partially counteracted, and the final equilibrium state concentrations of the reactants and products are usually different from the values at the original equilibrium state. See Figure 3 for a graphic interpretation of the freon-12 equilibrium shift. Note that adding more carbon tetrachloride, CCl 4 (l), would have no effect on the equilibrium state in the container. This reactant is (and stays) in liquid form, so its concentration is constant and would not be increased by increasing the amount of CCl 4 (l) present. Adjusting an equilibrium state by adding and/or removing a substance is by far the most common application of Le Châtelier s principle. For industrial chemical reactions, engineers strive to design processes where reactants are added continuously and products are continuously removed, so that an equilibrium is never allowed to establish. If the reaction is always shifting forward, the process is always making product (and, presumably, the industry is always making money). Concentration (mol/l) CCI 4 (I) + 2 HF(g) 0 CCI 2 F 2 (g) + 2 HCI(g) Time [HF] [HCI] [CCI 2 F 2 ] Figure 3 The reaction establishes an equilibrium that is disturbed (at the time indicated by the vertical dotted line) by the addition of HF(g). Some of the added HF reacts, decreasing in concentration, while the concentration of both products increases until a new equilibrium is established and concentrations become constant again. Note that the concentration of HF(g) at the new equilibrium is greater than at the original equilibrium, so the imposed change is only partly counteracted. The initial K c value and final K c value are the same. Equilibrium Systems 691

24 Concentration (mol/l) CCI 4 (I) + 2 HF(g) 0 CCI 2 F 2 (g) + 2 HCI(g) O 2 air O 2 O2 O 2 O 2 O 2 heart Time capillary blood vessels and body cells [HF] [HCI] [CCI 2 F 2 ] Figure 4 The reaction establishes an equilibrium that is disturbed (at the time indicated by the vertical dotted line) by the removal of HCl(g). The equilibrium shifts forward, increasing the concentration of both products while decreasing HF(g) concentration, until a new equilibrium is established. The initial K c value and the final K c value are the same. air-filled sacs in lungs O 2 Figure 5 Oxygenated blood from the lungs is pumped by the heart to body tissues. The deoxygenated blood returns to the heart and is pumped to the lungs. Shifts in equilibrium occur over and over again as oxygen is picked up in the lungs and released throughout the body. The removal of a product (if the removal decreases concentration as well as chemical amount) will also shift an equilibrium forward, producing more product to counteract the change imposed. The freon-12 reaction can be shifted forward by removing either gaseous product, since decreasing the amount of a gas lowers its concentration in any reaction container of fixed size (Figure 4). The following equation represents the final step in the production of nitric acid: 3 NO 2 (g) H 2 O(l) 0 2 HNO 3 (aq) NO(g) In this industrial process, nitrogen monoxide gas is removed from the chemical system by a reaction with oxygen gas. The removal of the nitrogen monoxide causes the system to shift to the right some nitrogen dioxide and water react, replacing some of the removed nitrogen monoxide. As the system shifts, more of the desired product, nitric acid, is produced. Although equilibrium systems are important in industrial chemical production, they are even more vital in biological systems. A particularly important biological equilibrium is that of hemoglobin (a protein in red blood cells), oxygen, and oxygenated hemoglobin. Hb O 2 0 HbO 2 As blood circulates to the lungs, the high concentration of oxygen shifts the equilibrium to the right and the blood becomes oxygenated (Figure 5). As the blood circulates throughout the body, cell reactions consume oxygen. This removal of oxygen shifts the equilibrium to the left and more oxygen is released. Collision Reaction Theory and Concentration Changes Collision reaction theory provides a simple explanation of the equilibrium shift that occurs when a reactant concentration is increased. We assume that the number of reactant entities per unit volume suddenly increases, so that collisions are suddenly much more frequent for the forward reaction. The forward reaction rate, therefore, increases significantly. Since the reverse reaction rate is not changed, the opposing rates are no longer equal, and, for a time, the difference in rates results in an observed increase of products. Of course, as the concentration of products increases, so does the reverse reaction rate. At the same time, the new forward rate decreases as reactant is consumed, until eventually the two rates become equal to each other again. The rates at the new equilibrium are faster than those at the original equilibrium, because the system now contains a larger number of particles (and, therefore, a higher concentration) in dynamic equilibrium. If a substance is removed, causing an equilibrium shift, the explanation is similar except that the initial effect is to suddenly decrease either the forward or the reverse rate by decreasing the concentration. Addition or removal of a reagent present in pure solid or pure liquid state does not change the concentration of that substance. The reaction of condensed phases (solids and liquids) takes place only at an exposed surface and if the surface area exposed is changed, it is always exactly the same change in available area for both forward and reverse reaction collisions. The forward and reverse rates change by exactly the same amount if they change at all, so equilibrium is not disturbed and no shift occurs. 692 Chapter 15

25 Section 15.2 Le Châtelier s Principle and Temperature Changes The heat energy in a chemical equilibrium equation is treated as though it were a reactant or a product. reactants energy 0 products reactants 0 products energy (endothermic in the forward direction) (exothermic in the forward direction) Heating or cooling a system adds or removes heat energy from the system. In either situation, the equilibrium shifts to minimize the change. If the system is cooled, the equilibrium shifts so that more heat energy is produced. If the system is heated, the equilibrium shifts in the direction in which heat energy is absorbed. For example, in the salt sulfuric acid process used to produce hydrochloric acid, the system is heated in order to increase the percent yield of hydrogen chloride gas. 2 NaCl(s) H 2 SO 4 (l) energy 0 2 HCl(g) + Na 2 SO 4 (s) Adding heat energy shifts the system to the right, absorbing some of the added energy. In the production of sulfuric acid, the key reaction step is the equilibrium represented by the following equation. Percent yield of the product is increased at low temperature. 2 SO 2 (g) O 2 (g) 0 2 SO 3 (g) energy Removing heat energy causes the system to shift to the right. This shift yields more sulfur trioxide while partially replacing the heat energy that was removed. Collision Reaction Theory and Energy Changes Collision reaction theory explains the equilibrium shift (that occurs when the heat energy of a system at equilibrium is changed) as the result of an imbalance of reaction rates. Consider the previously mentioned reaction equation a typical exothermic reaction. The reaction energy is shown this time in standard r H notation. 2 SO 2 (g) O 2 (g) 0 2 SO 3 (g) r H 198 kj We explain the result of cooling the system by assuming that both forward and reverse reaction rates are slower at lower temperatures, because the particles move more slowly and collide less frequently. The reverse rate decreases more than the forward rate, however. While the rates remain unequal, the observed result is the production of more product and the release of more heat energy. The shift causes concentration changes that will increase the reverse rate and decrease the forward rate until they become equal again, at a new, lower temperature (Figure 6). Note that industrial exothermic equilibrium reactions are often carried out at high temperatures, even though adding heat energy shifts the equilibrium toward reactants (lowers the percent yield). The Haber process (Case Study, Section 8.3) is a good example. Heat energy is added because the forward and reverse reaction rates are too slow at lower temperatures to allow the reaction to reach equilibrium in a reasonable time. Making large quantities of the marketable product in a short time is much more important to a manufacturer than creating a small increase in the yield of each batch. Whenever possible, chemical engineers try to design a continuous process for an industrial reaction one that shifts the reaction forward by constantly adding reactants, and constantly removing products. This system is no longer a closed system, so the reaction never establishes equilibrium. Such an industrial reaction may run continuously for months or even years. BIOLOGY CONNECTION CO 2 Transport Many biological processes depend on equilibria. For example, the transportation of carbon dioxide depends on the CO 2 0 H 2 CO 3 equilibrium system. In the tissues of the body where carbon dioxide is produced, the equilibrium shifts so that more carbonic acid is formed. In the lungs, the shift is in the reverse direction, as carbon dioxide is released. You will discover other examples of equilibria in a biology textbook. Concentration (mol/l) 2 SO 2 (g) + O 2 (g) 0 2 SO 3 (g) Time GO [SO 2 ] [O 2 ] [SO 3 ] Figure 6 The reaction establishes an equilibrium that is disturbed (at the time indicated by the vertical dotted line) by a decrease in temperature. The equilibrium shifts forward, increasing the concentration of SO 3 product while decreasing the concentration of both reactants, until a new equilibrium is established. Because the temperature is changed, the final K c value for this example is greater than the initial K c value because the shift favours the forward reaction. CAREER CONNECTION Chemical Process Engineer What role do chemical engineers play in designing systems and sequences of reactions for industrial chemical production? Research the education requirements, job prospects, and work assignments of people in this profession. GO Equilibrium Systems 693

26 Le Châtelier s Principle and Gas Volume Changes According to Boyle s law, the amount concentration of a gas in a container is inversely proportional to the volume of the container. Since the amount concentration of a gas is directly proportional to its pressure, we can predict the possible effect of container volume change on the equilibrium position of homogeneous gaseous systems. Decreasing the volume by half doubles the concentration of every gas in the container. To predict whether a change in pressure will affect a system s equilibrium, you must consider the total chemical amount of gas reactants and the total chemical amount of gas products. For example, in the equilibrium reaction of sulfur dioxide and oxygen, three moles of gaseous reactants produce two moles of gaseous products. 2 SO 2 (g) O 2 (g) 0 2 SO 3 (g) Concentration (mol/l) 2 SO 2 (g) + O 2 (g) 0 2 SO 3 (g) Learning Tip Time [SO 2 ] [O 2 ] [SO 3 ] Figure 7 The reaction equilibrium is disturbed by a decrease in container volume (at the time indicated by the vertical dotted line). The equilibrium shifts forward, increasing the concentration of SO 3 while decreasing the concentration of reactants, until a new equilibrium is established. The initial K c value and the final K c value are the same. Since increasing the temperature of any exothermic reaction at equilibrium always shifts the equilibrium left (toward reactants), an increase in temperature must decrease the value of K c for such a reaction. Similarly, increasing the temperature will increase the value of K c for any endothermic reaction. You have already learned that the value of K c is temperature dependent. No other change imposed on a system at equilibrium changes the numerical value of the equilibrium constant. If the volume is decreased, the overall pressure is increased. Increased pressure causes a shift to the right, which decreases the total number of gas molecules (three moles to two moles) and, thus, reduces the pressure. If the volume is increased, the pressure is decreased, and the shift is in the opposite direction. A system with equal numbers of gas molecules on each side of the equation, such as the equilibrium reaction between hydrogen and iodine (page 679), is not affected by a change in volume. Similarly, systems involving only liquids or solids are not affected by changes in pressure. Note that adding a gas that is not involved in the equilibrium (such as an inert gas) to the container will increase the overall pressure in the container, but will not cause a shift in equilibrium. Adding or removing gaseous substances not involved in the reaction does not change the concentrations of the reactant and product gases. Collision Reaction Theory and Gas Volume Changes When a system involving gaseous reactants and products is changed in volume, the resulting equilibrium shift is again explained as an imbalance of reaction rates. 2 SO 2 (g) O 2 (g) 0 2 SO 3 (g) 198 kj Collision reaction theory explains the result of decreasing the volume of this system by assuming that both forward and reverse reaction rates become faster because the concentrations of reactants and products both increase. For this example, however, the forward rate increases more than the reverse rate because there are more particles involved in the forward reaction. Consequently, the increase in the total number of collisions is greater for the forward reaction process. Again, while the rates remain unequal, the observed result is the production of more product. The shift causes concentration changes that gradually increase the reverse rate and decrease the forward rate until they become equal again (Figure 7). Catalysts and Equilibrium Systems Catalysts are used in most industrial chemical systems. A catalyst decreases the time required to reach an equilibrium position, but does not affect the final position of equilibrium. The presence of a catalyst in a chemical reaction system lowers the activation energy for both forward and reverse reactions by an equal amount (Chapter 12), so the equilibrium establishes much more rapidly but at the same position as it would without the catalyst present. Forward and reverse rates increase equally. The final equilibrium concentrations are reached in a shorter time compared with the same, but uncatalyzed, reaction. The value of catalysts in industrial processes is to decrease the time required for equilibrium shifts created by manipulating other variables, allowing a more rapid overall production of the desired product. 694 Chapter 15

27 Section 15.2 SUMMARY Variables Affecting Chemical Equilibria Variables Imposed Change Response of System concentration increase shifts to consume some of the added reactant or product decrease shifts to replace some of the removed reactant or product temperature increase shifts to absorb some of the added heat energy decrease shifts to replace some of the removed heat energy volume increase shifts toward the side with the larger (gaseous systems (decrease in pressure) total chemical amount of gaseous only) entities decrease shifts toward the side with the smaller (increase in pressure) total chemical amount of gaseous entities Practice 1. What three types of changes shift the position of a chemical equilibrium? 2. For each of the following chemical systems at equilibrium, use Le Châtelier s principle to predict the effect of the change imposed on the chemical system. Indicate the direction in which the equilibrium is expected to shift. For each example, sketch the graph of concentrations versus time, plotted from just before the change to the established new equilibrium. (a) H 2 O(l) energy 0 H 2 O(g) The container is heated. (b) H 2 O(l) 0 H (aq) OH (aq) A few crystals of NaOH(s) are added to the container. (c) CaCO 3 (s) energy 0 CaO(s) CO 2 (g) CO 2 (g) is removed from the container. (d) CH 3 COOH(aq) 0 H (aq) CH 3 COO (aq) A few drops of pure CH 3 COOH(l) are added to the system. 3. Much methanol is produced industrially by the exothermic reaction CO(g) 2H 2 (g) 0 CH 3 OH(l), carried out at high pressure (5 10 MPa) and temperature (250 C) in the presence of several catalyst substances. Methanol is less flammable than gasoline, and so it is a safer fuel. It is the fuel used in open-wheel Champ Car racing, and also in the Indianapolis 500. (a) State, in terms of forward and reverse reaction rates, why using a very high pressure of the reactant gases is economically desirable for the manufacturer. (b) State in which direction a high temperature will shift this reaction equilibrium. (c) Explain why using a high temperature is desirable, in terms of the time required for the reaction to reach equilibrium. (d) Explain, in terms of equilibrium position and equilibrium shift, why this reaction is done in an open system, where reactants are continually added to the pressure vessel and liquid product is continually removed. Equilibrium Systems 695

28 INVESTIGATION 15.3 Introduction Testing Le Châtelier s Principle The equilibria chosen for this investigation involve chemicals that provide coloured solutions. The investigation tests predictions about equilibrium shifts (made using Le Châtelier s principle) by observing colour changes. In order to complete the Prediction section of the report, you must read the Design, Materials, and Procedure carefully. Then make a Prediction about the result of each change made in the Procedure. Purpose The purpose of this investigation is to test Le Châtelier s principle by applying stress to four different chemical equilibria. Report Checklist Purpose Design Analysis Problem Materials Evaluation (2, 3) Hypothesis Procedure Prediction Evidence Problem How does applying changes to conditions of particular chemical equilibria affect the systems? Design Stresses are applied to four chemical equilibrium systems and evidence is gathered to test predictions made using Le Châtelier s principle. Control samples are used in all cases. To perform this investigation, turn to page 701. Case Study Urea Production in Alberta Canada has a vast wealth of natural resources that we can use to produce many chemicals with an incredible variety of uses. Because plants require nitrogen for growth, the primary purpose of some of these chemicals is for agricultural use, such as nitrogen fertilizers. You have already learned that Alberta produces large amounts of ammonia, which can be used directly as a fertilizer. Ammonia is stored as a liquid at high pressure, and injected directly into the soil (see Section 8.3, Figure 7), but it is toxic and corrosive to human tissue, which makes it dangerous to use. Special equipment and care are required. Many food producers prefer to use a highnitrogen, nontoxic, solid compound. Urea (Figure 8) is another simple molecular chemical also a nitrogen fertilizer that is inexpensive, simple to produce, easy to transport, and extraordinarily useful. This chemical is used in the millions of tonnes, for applications as diverse as wastewater plants (for treating effluent) air transportation (for de-icing runways) forestry and agriculture (for fertilizer) livestock feeding (for a protein supplement) woodworking (for making glues and resins) construction supplies (for making insulation) furniture (for making particle board and chipboard) clothing (for making certain dyes) Until the 1900s, the common source of this chemical was stale animal urine. The body forms this compound as its principal means of removing excess nitrogen. In fact, until the early 1800s, it was thought that this, or any other organic chemical extracted from living things, always had to be produced by a living organism. In 1828, however, Friedrich Wöhler, in a famous classic experiment, synthesized urea in his laboratory. His work forever changed the organic concept of chemistry. Early in the last century, an efficient industrial method was Figure 8 Urea is a small and simple molecule, but a critically important nitrogen-containing compound. It has a very high nitrogen percentage by mass, and is very soluble in water both very important points for a compound to be useful as a plant fertilizer. Three different representations of the urea molecule are shown here. H H H N N H urea developed, and has since been used for mass production of this versatile and valuable substance. Urea is produced through the reaction of ammonia, NH 3 (g), with carbon dioxide, CO 2 (g), at high temperature and C O 696 Chapter 15

29 Section 15.2 pressure. Most of the reactants exist in liquid form at the pressures used. A hot concentrated solution, containing about 80% urea by mass, results from this reaction. This hot solution is further concentrated and cooled through evaporation of the water content, to form either granules (angular crystals) or prills (small round pellets) of white solid urea (Figure 9). The overall reaction may be written as CO 2 (l) 2NH 3 (l) 0 NH 2 CONH 2 (aq) H 2 O(l) t C P MPa The science and technology for urea production developed in response to a strong demand for this compound. The process depends on an understanding of reaction equilibrium, and of the effect (on equilibrium) of high temperature and pressure conditions. Also key was the design and construction of reaction containers (vessels) able to withstand high pressures and temperatures. Sometimes, as well as building on existing scientific knowledge, new technology results in scientific advances. A technology, such as equipment to create very high pressures, may allow great leaps forward in science, such as the synthesis of completely new forms of crystalline solids. New observations then lead to new theories and sometimes even new laws. Case Study Questions 1. There are two phase equilibria that shift during the reaction of carbon dioxide and ammonia. Write equilibrium equations to represent these phase changes, and use Le Châtelier s principle to explain how they are continuously being shifted, both by pressure, and also by the effect of the chemical reaction that is occurring in the vessel. 2. Crystallization of urea from aqueous solution can be expressed as a solubility equilibrium, according to the equation NH 2 CONH 2 (aq) 0 NH 2 CONH 2 (s) (a) On which side of this equation would heat energy be written? Explain your reasoning. (b) Based on molecular structure and bonding theory, explain why you would expect urea to be a very highly soluble compound. (c) Urea granules that are bagged and sold for fertilizer use are quite uniform in size (Figure 10). Explain what physical process would likely be used to separate these granules from any larger or smaller granules that form during crystallization. Figure 9 Prills form when sprayed droplets of very hot urea solution are made to fall through air in a huge tower, cooling and evaporating to dryness on the way down. Figure 10 Urea fertilizer is a nitrogen source for crops. Extension 3. Research Friedrich Wöhler s classic experiment of Write the balanced equation for the reaction he performed. 4. Find out how much urea is produced annually in Alberta, where it is produced, and its current price per tonne. Assemble your findings into an attractive presentation. GO GO Equilibrium Systems 697

30 LAB EXERCISE 15.C The Nitrogen Dioxide Dinitrogen Tetroxide Equilibrium Complete the Prediction, Analysis, and Evaluation sections of the report. Purpose The purpose of this problem is to use Le Châtelier s principle to predict the response of an equilibrium to an introduced change in conditions. Problem How does increasing the pressure affect the nitrogen dioxide dinitrogen tetroxide equilibrium? Report Checklist Purpose Design Analysis Problem Materials Evaluation (1, 2, 3) Hypothesis Procedure Prediction Evidence Design A sample of nitrogen dioxide gas is compressed in a syringe and the intensity of the colour is used as evidence to test the Prediction. Evidence The orange-brown nitrogen dioxide gas colour increases in intensity when the plunger on the syringe is depressed, and then decreases in intensity (Figure 11). The final colour is slightly more intense than the original colour (before moving the plunger). Figure 11 2NO 2 (g) 0 N 2 O 4 (g) An increase in pressure on the nitrogen dioxide dinitrogen tetroxide equilibrium in the closed system results initially in a more intense colour followed by a decrease in colour intensity. INVESTIGATION 15.4 Introduction Studying a Chemical Equilibrium System Figure 2, page 690, shows the colours of aqueous solutions of iron(iii), thiocyanate, and iron(iii) thiocyanate ions. Use your knowledge of Le Châtelier s principle to write a Problem statement, and then design and carry out a simple investigation to determine whether the reaction as written is exothermic or endothermic. Purpose The purpose of this investigation is to use Le Châtelier s principle to solve a problem concerning the effect of an energy change on the following equilibrium system. Report Checklist Purpose Design Analysis Problem Materials Evaluation (1, 2, 3) Hypothesis Procedure Prediction Evidence Fe 3 (aq) SCN (aq) 0 FeSCN 2 (aq) almost colourless colourless red To perform this investigation, turn to page 703. WEB Activity Web Quest Poison Afloat Have you ever considered becoming a crime scene investigator? In this Web Quest, you are promoted to Chief Chem Crime Investigator. You are presented with a body and a series of clues The detecting is up to you. You will have a chance to use your knowledge of chemistry to solve this puzzle and gather the evidence to unravel what happened. GO 698 Chapter 15

31 Section 15.2 Section 15.2 Questions 1. The following equation represents part of the industrial production of nitric acid. Predict the direction of the equilibrium shift for each of the following changes. Explain any shift in terms of the changes in forward and reverse reaction rates. 4 NH 3 (g) 5 O 2 (g) 0 4 NO(g) 6 H 2 O(g) energy (a) O 2 (g) is added to the system. (b) The temperature of the system is increased. (c) NO(g) is removed from the system. (d) The pressure of the system is increased by decreasing the volume. 2. The following chemical equilibrium system is part of the Haber process for the production of ammonia. N 2 (g) 3 H 2 (g) 0 2NH 3 (g) energy Suppose you are a chemical process engineer. Use Le Châtelier s principle to predict five specific changes that you might impose on the equilibrium system to increase the yield of ammonia. 3. In a solution of copper(ii) chloride, the following equilibrium exists: CuCl 2 4 (aq) 4 H 2 O(l) 0 Cu(H 2 O) 2 4 (aq) 4 Cl (aq) dark green blue For the following stresses put on the equilibrium, predict the shift in the equilibrium and draw a graph of concentration versus time to communicate the shift. (a) Concentrated hydrochloric acid is added. (b) Saturated aqueous silver nitrate is added, causing a precipitation reaction. 4. Identify the nature of the changes imposed on the following equilibrium system at the four times indicated by coordinates A, B, C, and D (Figure 12). Concentration (mol/l) C 2 H 6 C 2 H 4 H 2 C 2 H 4 (g) + H 2 (g) 0 C 2 H 6 (g) + energy A B C D Time Figure 12 Graph showing four disturbances to an equilibrium system 5. In which of the following cases would an increase in temperature increase the percent yield at equilibrium? (a) H 2 O(l) 0 H 2 O(g) (b) N 2 (g) 3 H 2 (g) 0 2NH 3 (g) (c) KOH(s) 0 K (aq) OH (aq) heat (d) 2 C(s) 2H 2 (g) 0 C 2 H 4 (g) r H 91 kj r H 53 kj 6. Chloromethane (methyl chloride) is manufactured by chlorinating methane. For this reaction system at equilibrium, explain the effect of each of the imposed changes on the position of reaction equilibrium. CH 4 (g) Cl 2 (g) 0 CH 3 Cl(g) HCl(g) r H is negative (a) More methane is injected into the reaction vessel. (b) The container volume is increased. (c) The temperature is lowered. (d) A catalyst is introduced into the system. 7. Ethyne (acetylene) is manufactured by a high-temperature combustion of methane, using a large excess of methane. For this endothermic reaction system at equilibrium, explain the effect of each of the imposed changes on the value of the equilibrium constant. 6 CH 4 (g) O 2 (g) 0 2C 2 H 2 (g) 10 H 2 (g) 2 CO(g) (a) More methane is injected into the reaction vessel. (b) The container volume is decreased. (c) The temperature is lowered. (d) A catalyst is introduced into the system. Extension 8. In a deep lake or in the ocean, the pressure that a human considers normal has doubled by the time a diver reaches a depth of 10 m, and increases by about one atmosphere for every extra 10 m, to a maximum of about 100 MPa at the deepest points in Earth s oceans. This fact is of major concern to scuba divers for several reasons. Pressure inside a scuba diver s lungs must constantly be adjusted to equal outside water pressure, otherwise the lungs could collapse upon diving, and could explode upon rising. In fact, the development of the pressure regulator (Figure 13) was the technology that made scuba diving possible. The gases in a diver s lungs are dissolved to some extent in the blood that passes through. Pressure changes can change several solubility equilibria, with some serious effects. Research the gases dissolved in human blood, and what equilibrium shifts cause the diving conditions called nitrogen narcosis, and the bends. GO Figure 13 The mouthpiece pressure from a diver s tank automatically adjusts for changes in depth. Equilibrium Systems 699

32 Chapter 15 INVESTIGATIONS INVESTIGATION 15.1 The Extent of a Chemical Reaction In Chapter 8, you performed experiments that produced evidence that reactions are quantitative. In a quantitative reaction, the limiting reagent is completely consumed. To identify the limiting reagent, you can test the final reaction mixture for the presence of the original reactants. For example, in a diagnostic test, you might try to precipitate ions from the final reaction mixture that were present in the original reactants. Purpose The purpose of this investigation is to test the validity of the assumption that chemical reactions are quantitative. Problem What are the limiting and excess reagents in the chemical reaction of selected quantities of aqueous sodium sulfate and aqueous calcium chloride? Design Samples of sodium sulfate solution and calcium chloride solution are mixed in different proportions and the final mixture is filtered. Samples of the filtrate are tested for the presence of excess reagents, using the following diagnostic tests: Report Checklist Purpose Design Analysis Problem Materials Evaluation (1, 2, 3) Hypothesis Procedure Prediction Evidence If a few drops of Ba(NO 3 ) 2 (aq) are added to the filtrate and a precipitate forms, then sulfate ions are present. Ba 2 (aq) SO 4 2 (aq) BaSO 4 (s) If a few drops of Na 2 CO 3 (aq) are added to the filtrate and a precipitate forms, then calcium ions are present. Ca 2 (aq) CO 3 2 (aq) CaCO 3 (s) Materials lab apron eye protection 25 ml of 0.50 mol/l CaCl 2 (aq) 25 ml of 0.50 mol/l Na 2 SO 4 (aq) 1.0 mol/l Na 2 CO 3 (aq) in dropper bottle saturated Ba(NO 3 ) 2 (aq) in dropper bottle two 50 ml or 100 ml beakers two small test tubes 10 ml or 25 ml graduated cylinder filtration apparatus filter paper wash bottle stirring rod Soluble barium compounds are toxic. Remember to wash your hands before leaving the laboratory. Barium nitrate in solid form is a strong oxidizing substance. INVESTIGATION 15.2 Equilibrium Shifts (Demonstration) In this investigation, you will be looking at two equilibrium systems: N 2 O 4 (g) energy 0 2 NO 2 (g) colourless reddish brown CO 2 (g) H 2 O(l) 0 H (aq) + HCO 3 (aq) The second equilibrium system, produced by the reaction of carbon dioxide gas and water, is commonly found in the human body and in carbonated drinks. A diagnostic test is necessary to detect some shifts in this equilibrium. Bromothymol blue, an acid base indicator, can detect an increase or decrease in the hydrogen ion concentration in Report Checklist Purpose Design Analysis Problem Materials Evaluation (2, 3) Hypothesis Procedure Prediction Evidence this system. Bromothymol blue turns blue when the hydrogen ion concentration decreases, and yellow when the hydrogen ion concentration increases. Purpose The purpose of this demonstration is to test Le Châtelier s principle by studying two chemical equilibrium systems: the equilibrium between two oxides of nitrogen, and the equilibrium of carbon dioxide gas and carbonic acid. 700 Chapter 15

33 Chapter 15 INVESTIGATION 15.2 continued Problem How does a change in temperature affect the nitrogen dioxide dinitrogen tetroxide equilibrium system? How does a change in pressure affect the carbon dioxide carbonic acid equilibrium system? Materials lab apron eye protection two NO 2 (g)/n 2 O 4 (g) sealed flasks carbon dioxide hydrogen carbonate ion equilibrium mixture (ph 7) bromothymol blue indicator in dropper bottle small syringe with needle removed (5 to 50 ml) solid rubber stopper to seal end of syringe beaker of ice water mixture beaker of hot water Be careful with the flasks containing nitrogen dioxide: this gas is highly toxic. Use in a fume hood in case of breakage. Procedure 1. Place the sealed NO 2 (g)/n 2 O 4 (g) flasks in hot and cold water baths (Figure 1) and record your observations. 2. Place two or three drops of bromothymol blue indicator in the carbon dioxide hydrogen carbonate ion equilibrium mixture. 3. Draw some of the carbon dioxide hydrogen carbonate ion equilibrium mixture into the syringe, and then block the end with a rubber stopper. 4. Slowly move the syringe plunger and record your observations. hot water NO 2 (g) 0N 2 O 4 (g) ice water Figure 1 Each of these flasks contains an equilibrium mixture of dinitrogen tetroxide and nitrogen dioxide. Shifts in equilibrium can be seen when one of the flasks is heated or cooled. INVESTIGATION 15.3 Testing Le Châtelier s Principle The equilibria chosen for this investigation involve chemicals that provide coloured solutions. This investigation tests predictions about equilibrium shifts (made using Le Châtelier s principle) by observing colour changes. For example, if the colour in Reaction 3 becomes more intensely red, the equilibrium has shifted right to produce more FeSCN 2 (aq) ions, increasing their concentration. See Table 1. In order to complete the Prediction section of the report, you must read the Design, Materials, and Procedure carefully. Then, make a Prediction about the result of each change made in the Procedure. Purpose The purpose of this investigation is to test Le Châtelier s principle by applying stress to four different chemical equilibria. Report Checklist Purpose Design Analysis Problem Materials Evaluation (2, 3) Hypothesis Procedure Prediction Evidence Table 1 Solution Colours Ion Colour CoCl 2 4 (alc) blue Co(H 2 O) 2 6 (alc) pink H 2 Tb(aq) red HTb (aq) yellow Tb 2 (aq) blue Fe 3 (aq) pale yellow SCN (aq) colourless FeSCN 2 (aq) red Cu(H 2 O) 2 4 (aq) pale blue Cu(NH 3 ) 2 4 (aq) deep blue Equilibrium Systems 701

34 INVESTIGATION 15.3 continued Problem How does applying stresses to particular chemical equilibria affect the systems? Part I CoCl 2 4 (alc) 6 H 2 O(alc) 0 Co(H 2 O) 2 6 (alc) 4 Cl (alc) energy Part II H 2 Tb(aq) 0 H (aq) HTb (aq) HTb (aq) 0 H (aq) Tb 2 (aq) Part III Fe 3 (aq) SCN (aq) 0 FeSCN 2 (aq) Part IV Cu(H 2 O) 2 4 (aq) 4 NH 3 (aq) 0 Cu(NH 3 ) 2 4 (aq) 4 H 2 O(l) Design Stresses are applied to four chemical equilibrium systems and evidence is gathered to test predictions made using Le Châtelier s principle. Control samples are used in all cases. For example, before adding sodium hydroxide to a new equilibrium solution, split the solution into two samples in order to have a control sample for colour comparison. Part I Cobalt(II) Complexes Water, saturated silver nitrate, and heat are added to, and heat is removed from, samples of the provided equilibrium mixture. Note: This reaction equilibrium is in solution using an alcohol solvent, shown as (alc), so the concentration of water is a variable in this system. Part II Thymol Blue Indicator Hydrochloric acid and sodium hydroxide are added to samples of the provided equilibrium mixture. Part III Iron(III) Thiocyanate Equilibrium Iron(III) nitrate, potassium thiocyanate, and sodium hydroxide are added to samples of the provided equilibrium system. Part IV Copper(II) Complexes Aqueous ammonia and hydrochloric acid are added to samples of the provided equilibrium mixture. Materials lab apron eye protection 100 ml beaker large waste beaker 6 to 12 small test tubes test-tube rack distilled water crushed ice hot water bath cobalt(ii) chloride equilibrium mixture in ethanol dropper bottles containing 0.2 mol/l AgNO 3 (aq) thymol blue indicator 0.1 mol/l HCl(aq) 0.1 mol/l NaOH(aq) iron(iii) thiocyanate equilibrium mixture 0.2 mol/l Fe(NO 3 ) 3 (aq) 0.2 mol/l KSCN(aq) 6.0 mol/l NaOH(aq) 0.1 mol/l CuSO 4 (aq) 1.0 mol/l NH 3 (aq) 1.0 mol/l HCl(aq) The chemicals used may be corrosive or poisonous, and may cause other toxic effects. Exercise great care when using the chemicals and avoid skin and eye contact. Immediately rinse the skin if there is any contact. If any chemicals get in the eyes, flush eyes for a minimum of 15 min and inform the teacher. Ethanol is flammable. Make sure there are no open flames in the laboratory when using the ethanol solution of cobalt(ii) chloride. Procedure Part I Cobalt(II) Complexes 1. Obtain 25 ml of the equilibrium mixture with the cobalt(ii) chloride complex ions. 2. Place a small amount of the mixture into each of five small test tubes. Use the fifth test tube as a control for comparison purposes. 3. Add drops of water to one test tube until a change is evident. Record the evidence. 4. Add drops of 0.2 mol/l silver nitrate to another test tube and record the evidence. 5. Heat another equilibrium mixture in a hot water bath and record the evidence. 6. Cool an equilibrium mixture in an ice bath and record the evidence. Part II Thymol Blue Indicator 7. Add about 5 ml of distilled water to each of two small test tubes. 8. Add 1 to 3 drops of thymol blue indicator to the water in each test tube to obtain a noticeable colour. Use one test tube of solution as a control. 9. Add drops of 0.1 mol/l HCl(aq) to the experimental test tube to test for the predicted colour changes. 10. Add drops of 0.1 mol/l NaOH(aq) to the same tube to test for the predicted colour changes. 702 Chapter 15

35 Chapter 15 INVESTIGATION 15.3 continued Part III Iron(III) Thiocyanate Equilibrium 11. Obtain about 20 ml of the iron(iii) thiocyanate equilibrium mixture. 12. Place about 5 ml of the equilibrium mixture in each of three test tubes. Use one test tube as a control. 13. Add drops of Fe(NO 3 ) 3 (aq) to one test tube until a change is evident. 14. Add drops of 6.0 mol/l NaOH(aq) to this new equilibrium mixture until a change occurs. (Iron(III) hydroxide has very low solubility.) 15. Add drops of KSCN(aq) to another equilibrium mixture until a change is evident. Part IV Copper(II) Complexes 16. Obtain 2 ml of 0.1 mol/l CuSO 4 (aq) in a small test tube. 17. Add three drops of 1.0 mol/l NH 3 (aq) to establish the equilibrium mixture. 18. Add more 1.0 mol/l NH 3 (aq) to the above equilibrium mixture and record the results. 19. Add 1.0 mol/l HCl(aq) to the equilibrium mixture from step 18 and record the results. Dispose of the chemicals as directed by your teacher. Identify each as toxic (to be collected) or nontoxic (disposable in the sink). Ensure that all equipment and surfaces are clean and wash your hands thoroughly before leaving the laboratory. INVESTIGATION 15.4 Studying a Chemical Equilibrium System Figure 1 shows the colours of aqueous solutions of iron(iii), thiocyanate, and iron(iii) thiocyanate ions. Use your knowledge of Le Châtelier s principle to write a Problem statement, and then design and carry out a simple investigation to determine whether the reaction as written is exothermic or endothermic. Report Checklist Purpose Design Analysis Problem Materials Evaluation (1, 2, 3) Hypothesis Procedure Prediction Evidence Purpose The purpose of this investigation is to use Le Châtelier s principle to solve a problem concerning the effect of an energy change on the following equilibrium system. Fe 3 (aq) SCN (aq) 0 FeSCN 2 (aq) almost colourless colourless red Iron(III) compounds are irritants. Thiocyanate ion solutions are toxic. Avoid skin and eye contact. If there is any skin or eye contact, immediately rinse with plenty of water. Flush the eyes for at least 15 min and inform the teacher. Figure 1 The iron thiocyanate reaction Equilibrium Systems 703

36 Chapter 15 SUMMARY Outcomes Knowledge define equilibrium and state the criteria that apply to a chemical system in equilibrium (15.1) identify, write, and interpret chemical equations for systems at equilibrium (15.1, 15.2) predict, qualitatively, using Le Châtelier s principle, shifts in equilibrium caused by changes in temperature, pressure, volume, concentration, or the addition of a catalyst, and describe how these changes affect the equilibrium constant (15.2) define K c and write equilibrium law expressions for given chemical equations, using lowest whole-number coefficients (15.1) calculate equilibrium constants and concentrations for homogeneous systems when concentrations at equilibrium are known, when initial concentrations and one equilibrium concentration are known, and when the equilibrium constant and one equilibrium concentration are known (15.1) STS state that the goal of science is knowledge about the natural world (15.1, 15.2) list the characteristics of empirical and theoretical knowledge (15.2) state that a goal of technology is to solve practical problems (15.2) Skills initiating and planning: predict variables that can cause a shift in equilibrium (15.2); design an experiment to show equilibrium shifts (15.2); describe procedures for safe handling, storage, and disposal of materials used in the laboratory (15.1, 15.2) performing and recording: perform an experiment to test, qualitatively, predictions of equilibrium shifts (15.2) analyzing and interpreting: write the equilibrium law expression for a given equation (15.1); analyze, qualitatively, the changes in concentrations of reactants and products after an equilibrium shift (15.2); interpret data from a graph to determine when equilibrium is established, and determine the cause of a stress on the system (15.2) communication and teamwork: work collaboratively in addressing problems and communicate effectively (15.1, 15.2) + EXTENSION Key Terms 15.1 closed system equilibrium phase equilibrium solubility equilibrium chemical reaction equilibrium dynamic equilibrium 15.2 Le Châtelier s principle Key Equations forward reaction reverse reaction ICE table equilibrium constant, K c equilibrium law equilibrium shift For the reaction a A b B 0 c C d D, the equilibrium law is K c [C]c [D] d [A] a [B] b MAKE a summary 1. Make a concept map, beginning with the word Equilibrium in the centre of a page. Link all of the Key Terms from this chapter, together with points of your own, to explain and illustrate how connections among these terms include the equilibrium law expression, ICE table format, Le Châtelier s principle, and points from the section Summaries. 2. Refer back to your answers to the Starting Points questions at the beginning of this chapter. How has your thinking changed? Go To The following components are available on the Nelson Web site. Follow the links for Nelson Chemistry Alberta an interactive Self Quiz for Chapter 15 additional Diploma Exam-style Review questions Illustrated Glossary additional IB-related material There is more information on the Web site wherever you see the Go icon in this chapter. GO Plague and the Little Ice Age A Dutch researcher discusses his theory that the atmospheric carbon dioxide equilibrium shifted when 40% of Europe s human population was killed by a plague in the 14th century. Could this have caused the little ice age? GO 704 Chapter 15

37 Chapter 15 REVIEW Chapter 15 Many of these questions are in the style of the Diploma Exam. You will find guidance for writing Diploma Exams in Appendix H. Exam study tips and test-taking suggestions are on the Nelson Web site. Science Directing Words used in Diploma Exams are in bold type. GO DO NOT WRITE IN THIS TEXTBOOK. Part 1 1. A closed system for a chemical equilibrium means that A. the reaction container must be solid and sealed, with a fixed volume B. no substance involved in the equilibrium must be able to enter or leave C. pressure and temperature in the reaction vessel must be constant D. no chemical of any kind may be added to the reaction vessel 2. The possible equilibrium that is not dynamic is the one between A. the liquid and gas phases of octane B. chromate and dichromate ions in aqueous solution C. the downward force of gravity exerted by Earth on an object, and the upward force of a balance exerted on the object D. the oxygen dissolved in water in a lake and the oxygen dissolved in nitrogen in the atmosphere 3. In a reaction, 4.22 mol of product has formed but there are no more visible signs of change. Calculations show that as much as 6.00 mol of product could have formed from the chemical amounts of reactants used. The percent yield is %. NR Use this information to answer questions 4 to 6. Sulfuric acid is the most common commercial acid, with millions of tonnes produced each year (Figure 1). The second step in the contact process for industrial production of sulfuric acid involves the oxidation of sulfur dioxide gas catalyzed by contact with V 2 O 5 (s) powder. The reaction equation is 2SO 2 (g) O 2 (g) 0 2SO 3 (g) H 198 kj Figure 1 Most sulfuric acid is produced in plants such as this one by the contact process, which includes two exothermic combustion reactions. Sulfur reacts with oxygen, forming sulfur dioxide; then, sulfur dioxide, in contact with a catalyst, reacts with oxygen, forming sulfur trioxide. Sulfur trioxide and water form sulfuric acid. 4. Which is the correct form of the equilibrium law for this reaction, as the equation is written? A. K c B. K c [SO 2 (g)] 2 [O 2 (g)][so 3 (g)] 2 C. K c D. K c [SO 2 (g)] 2 [O 2 (g)] [SO3 (g)] [SO 2 (g)] 2 [O 2 (g)] [SO3 (g)] 2 [SO 3 (g)] 2 [SO2 (g)] 2 [O 2 (g)] 5. The imposed condition that does not shift equilibrium toward the product is A. adding the vanadium pentoxide catalyst B. decreasing the temperature C. decreasing the container volume D. adding oxygen to the system container 6. If the equilibrium constant, K c, for the above oxidation reaction at a given temperature is , then the constant for the reverse reaction (written as the decomposition of sulfur trioxide) is A B C D Equilibrium Systems 705

38 7. If excess copper reacts in a solution of silver nitrate, it is correct to state that A. the reaction is not considered quantitative B. the equilibrium constant at SATP will have a numerical value between 1 and 100 C. water is not written in the equilibrium law expression because it is a spectator species D. silver nitrate is not written in the equilibrium law expression because it is a species with a constant concentration Use this information to answer questions 8 to 10. Consider the following reaction to produce hydrogen, done as a first step in the industrial process to make ammonia. Methane (from natural gas) reacts with steam over a nickel powder catalyst. The reaction equation is CH 4 (g) 2H 2 O(g) heat energy 0 CO 2 (g) 4 H 2 (g) 8. This reaction is done in a laboratory autoclave (a stainless steel pressure vessel) and allowed to reach equilibrium. More methane is then injected into the autoclave. We can predict that, when a new equilibrium is reached (at the same temperature), the concentration of every reagent in the equation will have increased, except that of A. methane B. water C. carbon dioxide D. hydrogen 9. When this gaseous reaction system at equilibrium is disturbed by heating the autoclave, we theorize that A. both forward and reverse reaction rates increase, but the forward rate increases more B. the forward reaction rate does not change, but the reverse reaction rate increases C. the reverse reaction rate does not change, but the forward reaction rate increases D. both reaction rates increase equally, so the equilibrium position is unchanged 10. A test reaction is done starting with only methane and NR excess water in the autoclave. If the initial concentration of methane is mol/l, and the methane concentration at equilibrium is mol/l, then the equilibrium concentration of hydrogen is mol/l. Part Define chemical equilibrium empirically. 12. What main idea explains chemical equilibrium? 13. What phrase is used to describe a reaction equilibrium in which the proportion of reactants to products is quite high? 14. Describe and explain a situation in which a carbonated soft drink is in (a) a non-equilibrium state (b) an equilibrium state 15. Predict whether adding a catalyst affects a state of equilibrium. What does the catalyst do? 16. For each of the following descriptions, write a chemical equation for the system at equilibrium. Communicate the position of the equilibrium with equilibrium arrows. Then write a mathematical expression of the equilibrium law for each chemical system. (a) A combination of low pressure and high temperature provides a percent yield of less than 10% for the formation of ammonia in the Haber process. (b) At high temperatures, the formation of water vapour from hydrogen and oxygen is quantitative. (c) The reaction of carbon monoxide with water vapour to produce carbon dioxide and hydrogen has a percent yield of 67% at 500 C. 17. Scientists and technologists are particularly interested in the use of hydrogen as a fuel. Interpret this reaction equation by predicting the relative proportions of reactants and products in this system at equilibrium. 2H 2 (g) O 2 (g) 0 2H 2 O(g) K c at SATP Use this information to answer questions 18 to 24. The solubility of pure oxygen in contact with liquid water at SATP is very low: only about 42 ppm, or 42 mg/l. The solubility equilibrium when water is in contact with air (21% oxygen) at SATP, is even lower: about 8.7 mg/l. The equilibrium equation is O 2 (g) 0 O 2 (aq) 18. Write the equilibrium law expression for a saturated solution of oxygen in water. 19. If the gas in the closed system is pure oxygen, the solubility is higher than it is if the gas is air. Express the solubility of pure oxygen in water at SATP as an amount concentration. 20. Express the concentration of pure oxygen gas at SATP as an amount concentration. (Recall that the molar volume of gases at SATP is 24.8 L/mol.) 21. Use the answers to the previous two questions to determine the value of K c for an equilibrium of pure oxygen gas in contact with its saturated solution at SATP. 22. Predict whether the value of K c for this equilibrium will be different for an equilibrium of air in contact with water at 25 C. Predict which system condition changes will, and which will not, change the value of an equilibrium constant. 706 Chapter 15

39 Chapter In terms of the equilibrium law, explain why more oxygen dissolves in water when the gas above it is pure oxygen than when the gas above it is air. 24. The quality of surface water in lakes and streams (Figure 2) is of critical importance to society. Dissolved oxygen content of the surface fresh water in Canada averages 10 ppm. Explain what stress is placed upon fish and other organisms in streams, lakes, and wetlands if climate change increases the average temperature of the water. 27. H 2 (g) Br 2 (g) 0 2 HBr(g) K c 12.0 at t C DE (a) 8.00 mol of hydrogen and 8.00 mol of bromine are added to a 2.00 L reaction container. Construct an ICE table and use it to predict the concentrations at equilibrium. (b) 12.0 mol of hydrogen and 12.0 mol of bromine are added to a 2.00 L reaction container. Construct an ICE table and use it to predict the concentrations at equilibrium. 28. CO(g) H 2 O(g) 0 CO 2 (g) H 2 (g) K c 4.00 at 900 C In a container, carbon monoxide and water vapour react to produce carbon dioxide and hydrogen. The equilibrium concentrations are [H 2 O(g)] 2.00 mol/l, [CO 2 (g)] 4.00 mol/l, and [H 2 (g)] 2.00 mol/l. Determine the equilibrium concentration of carbon monoxide. 29. Write a statement of Le Châtelier s principle. 30. What variables are commonly manipulated to shift a chemical equilibrium system? 31. Describe how a change in volume of a closed system containing a gaseous reaction at equilibrium affects the pressure of the system. Figure 2 The Bow River begins as a cold, glacier-fed mountain stream with a higher-than-average oxygen content, and is world famous for its trout fly fishery. It also supplies the city of Calgary with water for a million people daily, as well as providing water for agriculture in southern Alberta. 25. In many processes in industry, engineers try to maximize the yield of a product. Outline how concentration can be manipulated in order to increase the yield of a product. 26. In a container at high temperature, ethyne (acetylene) and DE hydrogen react to produce ethene (ethylene). No ethene is initially present. Later, at equilibrium, the concentration of ethene is mol/l. C 2 H 2 (g) H 2 (g) 0 C 2 H 4 (g) The initial concentrations of both acetylene and hydrogen are 1.00 mol/l. (a) Use an ICE table to determine the equilibrium constant. (b) Sketch a reaction progress graph to show the change in concentration values over time, from the beginning of the reaction to equilibrium. 32. In a sealed container, nitrogen dioxide is in equilibrium with DE dinitrogen tetroxide. 2NO 2 (g) 0 N 2 O 4 (g) K c 1.15, t 55 C (a) Write the mathematical expression for the equilibrium law applied to this chemical system. (b) If the equilibrium concentration of nitrogen dioxide is mol/l, predict the concentration of dinitrogen tetroxide. (c) Predict the shift in equilibrium that will occur when the concentration of nitrogen dioxide is increased. 33. Predict the shift in the following equilibrium system resulting from each of the following changes: 4 HCl(g) O 2 (g) 0 2H 2 O(g) 2Cl 2 (g) 113 kj (a) (b) (c) (d) an increase in the temperature of the system a decrease in the system s total pressure due to an increase in the volume of the container an increase in the concentration of oxygen the addition of a catalyst 34. Chemical engineers use Le Châtelier s principle to predict shifts in chemical systems at equilibrium resulting from changes in the reaction conditions. Predict the changes necessary to maximize the yield of product in each of the following industrial chemical systems: (a) the production of ethene (ethylene) C 2 H 6 (g) energy 0 C 2 H 4 (g) H 2 (g) (b) the production of methanol CO(g) 2H 2 (g) 0 CH 3 OH(g) energy Equilibrium Systems 707

40 35. Apply Le Châtelier s principle to predict whether, and in which direction, the following established equilibrium would be shifted by the change imposed: 2 CO(g) O 2 (g) 0 2CO 2 (g) heat energy (a) temperature is increased (b) vessel volume is increased (c) oxygen is added (d) platinum catalyst is added (e) carbon dioxide is removed 36. For each example, predict whether, and in which direction, an established equilibrium would be shifted by the change imposed. Explain any shift in terms of changes in forward and reverse reaction rates. (a) Cu 2 (aq) 4 NH 3 (g) 0 Cu(NH 3 ) 2 4 (aq) CuSO 4 (s) is added (b) CaCO 3 (s) energy 0 CaO(s) CO 2 (g) temperature is decreased (c) Na 2 CO 3 (s) energy 0 Na 2 O(s) CO 2 (g) sodium carbonate is added (d) H 2 CO 3 (aq) energy 0 CO 2 (g) H 2 O(l) vessel volume is decreased (e) KCl(s) 0 K (aq) Cl (aq) AgNO 3 (s) is added (f) CO 2 (g) NO(g) 0 CO(g) NO 2 (g) vessel volume is increased (g) Fe 3 (aq) SCN (aq) 0 FeSCN 2 (aq) Fe(NO 3 ) 3 (s) is added 37. Predict in which of the following equilibria a decrease in temperature favours the forward reaction. (a) Br 2 (l) 0 Br 2 (g) (b) N 2 (g) 3 H 2 (g) 0 2NH 3 (g) r H is negative (c) LiCl(s) 0 Li (aq) Cl (aq) heat (d) 6 C(s) 3 H 2 (g) 0 C 6 H 6 (l) r H 49 kj (e) CaCO 3 (s) energy 0 CaO(s) CO 2 (g) Use this information to answer questions 38 and 39. Alberta s petroleum industry has a chronic problem all fossil fuels found in the province contain some sulfur. If not removed, this sulfur will react upon burning to release SO 2 (g). Sulfur dioxide is very irritating to lung tissue and is highly corrosive; thus, it is a major contributor to air pollution, acid rain, and respiratory disease. Furthermore, sulfur impurities may damage the fuel injection and anti-pollution systems of modern internal combustion engines if not removed from gasoline and diesel fuels. For these reasons, recent Canadian legislation requires that sulfur content in diesel fuels sold for on-road use must be less than 15 ppm (15 mg/kg) by July, In a refinery or bitumen upgrader, the sulfur is first removed from fossil fuel feedstock by cracking and/or hydrogenation, resulting in reaction of the sulfur to produce (extremely toxic) H 2 S(g). Standard industry technology to remove hydrogen sulfide gas from petrochemical gas stream mixtures involves the use of an amine scrubber unit, in a two-step process that depends on two kinds of equilibrium. For simplicity, assume that an amine scrubber reaction vessel contains a 25% aqueous solution of diethanolamine (C 2 H 4 OH) 2 NH(aq), which approximates the actual solution used in the various oil sands plants in Alberta. 38. The scrubbing operation involves a gas mixture of lowmolar-mass hydrocarbons, carbon dioxide, and hydrogen DE sulfide. This mixture of gases is injected into the scrubber unit at high pressure, with the scrubber solution at about 40 C. For hydrogen sulfide, first the toxic gas dissolves H 2 S(g) 0 H 2 S(aq) (negative r H) which is followed immediately by the chemical reaction (C 2 H 4 OH) 2 NH(aq) + H 2 S(aq) 0 (C 2 H 4 OH) 2 NH + 2 (aq) + HS (aq) (negative r H) (a) Draw a structural formula for diethanolamine. (b) Explain, using Le Châtelier s principle, how the relatively low temperature and high pressure act to help the scrubber solution absorb toxic hydrogen sulfide. 39. When the unabsorbed hydrocarbon gases are removed from the scrubber unit, the sulfur atoms remain behind, trapped in solution as hydrogen sulfide ions. In the next process step, the scrubber solution is regenerated : the absorbed toxic compound is now emitted from solution and removed from the reaction vessel. Use Le Châtelier s principle to describe how conditions should be altered in the scrubber vessel, to shift equilibrium positions to make this process as efficient as possible. 708 Chapter 15

41 Chapter When carbon dioxide gas dissolves in water, the process is more correctly thought of as being an exothermic chemical reaction with water, to form aqueous carbonic acid. Soft drink beverages are carbonated in this way, at high pressure. Write and balance an equilibrium equation for this reaction. 41. Carbonic acid, H 2 CO 3 (aq), then reacts exothermically with basic aqueous diethanolamine in essentially the same way that hydrosulfuric acid, H 2 S(aq), does. Write and balance the equilibrium equation for this reaction, and use it to explain whether the process conditions for scrubbing H 2 S(g) should work to remove CO 2 (g) as well. Extension 42. Work cooperatively to research, assemble, and present a DE more complete and accurate summary of the operation of a typical Alberta industrial amine scrubber system. Your presentation should include a description of the role of chemical equilibrium in the system information on applications of this process throughout Alberta the use of the best features of any available word processing or slideshow software Research and describe the role of temperature in the operation of a halogen lamp. For example, how is it possible for a halogen lamp to operate with the filament at 2700 C when the tungsten normally would not last very long at this high temperature? Why is such a high temperature desirable? 44. When the Olympic Games were held in Mexico in 1968, DE many athletes arrived early to train in the higher altitude (2.3 km) and lower atmospheric pressure of Mexico City. Exertion at high altitudes, for people who are not acclimatized, may make them dizzy or lightheaded from lack of oxygen. Explain this observation. Your explanation should include the theory of dynamic equilibrium Le Châtelier s principle a description of how people who normally live at high altitudes are physiologically adapted to their reducedpressure environment GO GO GO 43. A halogen light bulb contains a tungsten (wolfram) filament, W(s), in a mixed atmosphere of a noble gas and a halogen; for example, Ar(g) and I 2 (g) (Figure 3). The operation of a halogen lamp depends, in part, on the equilibrium system W(s) I 2 (g) 0 WI 2 (g) Figure 3 Halogen light bulbs are more efficient than ordinary incandescent bulbs, but they burn so hot they may require special fixtures, and must be treated with extra care. Equilibrium Systems 709

42 16 chapter Equilibrium In this chapter Exploration: Salty Acid or Acidic Salt? Lab Exercise 16.A: The Chromate Dichromate Equilibrium Web Activity: Edgar Steacie Investigation 16.1: Creating an Acid Base Strength Table Lab Exercise 16.B: Predicting Acid Base Equilibria Web Activity: Pool Chemistry Lab Exercise 16.C: Aqueous Bicarbonate Ion Acid Base Reactions Investigation 16.2: Testing Brønsted Lowry Reaction Predictions Lab Exercise 16.D: Creating an Acid Base Table Case Study: Changing Ideas on Acids and Bases The Evolution of a Scientific Theory Web Activity: Titration of Polyprotic Acids and Bases Biology Connection: Homeostasis Web Activity: Preparation of Buffer Solutions Web Activity: Maud Menten Investigation 16.3: Testing a Buffer Effect Equilibrium in Acid Base Systems The nature of science involves constant questioning and testing of theories and so it is with theories about acids and bases. A great many chemical reaction systems involve acids in some way, including the one that begins the decomposition (digestion) of the food you eat, and, thus provides you with energy. Many other types of aqueous reaction systems have rates that are easily controlled by adjusting the level of acidity. Because such systems are commonly found both in nature and in industry, scientists seek to work with the most complete and successful acid base concepts which in turn means seeking new ways to test previously accepted theories. Theories that do not describe, explain, and predict the chemistry of acids and bases well enough will initially be restricted to only those situations where they work. As a result of further testing, scientists will either revise the theory, or replace it altogether. This chapter presents new hypotheses, evidence, and analyses to help you develop a more comprehensive understanding of both the aqueous reaction environment, and the activity of acids and bases within that environment. Your knowledge of chemical equilibrium (Chapter 15) allows you to explore these questions from a new perspective, and, in turn, will allow you to form a more complete and less restrictive theory of acids and bases. In fact, few topics in chemistry illustrate this scientific principle of ongoing theory testing and development so well. These underlying principles that theories must be supported by evidence, and that understanding is increased by always questioning and testing existing knowledge are the basis of the uniquely productive way of knowing about the natural world that we call science. To these principles, and to the enormous accumulation of knowledge they have made possible, we owe most aspects of our present technological civilization. STARTING Points Answer these questions as best you can with your current knowledge. Then, using the concepts and skills you have learned, you will revise your answers at the end of the chapter. 1. How can some substances neutralize both acids and bases? 2. Can acid base reactions and their products be predicted? Explain. 3. Can ph curves for titrations of weak acids and weak bases predict equivalence points as strong acid strong base ph curves do? Explain. 4. How is the buffering of some medications related to stomach fluid acidity? Career Connection: Environmental Engineer; Chemistry Researcher; Microbiologist 710 Chapter 16

43 Figure 1 Testing your concepts is a continual part of doing science. Exploration Salty Acid or Acidic Salt? In Chapter 5, you learned about saturated solutions as examples of dynamic equilibrium, and prepared a saturated sodium chloride solution. The strong acid, HCl(aq), shares half of its chemical formula with NaCl(aq). How might the two solutions interact? Materials: two 10 ml graduated cylinders; saturated sodium chloride solution, NaCl(aq); concentrated hydrochloric acid, HCl(aq) (a) Write and balance a chemical reaction equation for any reaction you can predict when sodium chloride solution and concentrated hydrochloric acid are mixed. (b) Write an equation to express the nature of a saturated sodium chloride aqueous solution as a dynamic equilibrium. (c) Write the equilibrium law expression for a saturated aqueous sodium chloride solution. Identify the substance with a constant concentration. Hydrochloric acid is corrosive. Wear appropriate eye protection, lab gloves, and a lab apron. Measure approximately 8 ml of saturated sodium chloride solution into the larger graduated cylinder. Measure approximately 2 ml of concentrated hydrochloric acid into the smaller graduated cylinder. Pour the concentrated hydrochloric acid into the cylinder containing the saturated sodium chloride solution. Record your observations. Dispose of all substances down the sink, using lots of water. (d) Explain what has apparently happened to the saturated NaCl(aq) equilibrium. (e) Explain which ion concentration in the large cylinder must have changed, whether it was increased or decreased, and what principle you use to know this answer. (f) Which initial solution was more concentrated? Use Le Châtelier s principle to explain how you know this answer. (g) Was the initial HCl(aq) a saturated solution? Was it at equilibrium? Was it at equilibrium before being removed from its closed storage container? Explain how you can use the basic principles of equilibrium to know these answers. Equilibrium in Acid Base Systems 711

44 CrO 4 2 _ (aq) 16.1 Cr 2 O 7 2 _ (aq) Figure 1 The chromate dichromate aqueous ion system, in equilibria at high and low ph Water Ionization and Acid Base Strength In many of the preceding units, you have studied examples of chemical reactions and systems that in some way depend on the nature of acids and bases and/or the ph of solution. For example, aqueous permanganate ions are powerful oxidizing agents, but can act to oxidize other reagents only in the presence of hydrogen (hydronium) ions. Almost any soluble R COOH organic compound will make an aqueous solution with a ph below 7, because the hydrogen atom of such a group is relatively easily removed. The electrolysis of potassium iodide solution produces a solution that is strongly basic. Understanding these connections involves considering equilibrium effects, as shown by the chromate dichromate aqueous ion system (Figure 1). You have already learned several concepts about the strengths and properties of acids and bases (see Chapter 6 Summary, page 262). If we now combine equilibrium concepts with these acid base concepts, we can develop a much more comprehensive understanding of acids and bases. This understanding, in turn, will allow you to better explain and predict how acids and bases behave. For instance, the reaction examined in Lab Exercise 16.A is an easily explained example of how the acidity of a solution can directly affect other ion equilibria. LAB EXERCISE 16.A The Chromate Dichromate Equilibrium In an aqueous solution, chromate ions are in equilibrium with dichromate ions (Figure 1). 2 CrO 2 4 (aq) 2H (aq) 0 Cr 2 O 2 7 (aq) H 2 O(l) Complete the Prediction and Design (including diagnostic tests) of the investigation report. Purpose The scientific purpose of this investigation is to test a Design for varying the acidity of an equilibrium. Report Checklist Purpose Design Analysis Problem Materials Evaluation Hypothesis Procedure Prediction Evidence Problem How does changing the hydrogen ion concentration affect the chromate dichromate equilibrium? Hypothesis The position of this equilibrium depends on the acidity of the solution. For simplicity, a great many chemical reaction equations use H (aq) to represent an aqueous hydrogen ion. This representation often works very well, as in the redox equations used in Unit 7, or the titration analysis equations used for stoichiometric calculations in Chapter 8. For other purposes, however, it is necessary to represent this ion more accurately in order to understand and explain the theoretical nature of the reaction. As you learned in Chapter 6, it is more consistent with evidence to think of this entity as a hydronium ion, and to represent it as H 3 O (aq). Acid base equilibrium theory necessarily involves collision reaction theory for a variety of entities. This chapter will, therefore, use the hydronium ion convention almost exclusively. Before aqueous acidic or basic solution equilibrium can be investigated further, the equilibrium nature of the ions of the solvent (water) must first be examined, understood, and taken into consideration. 712 Chapter 16

45 Section 16.1 The Water Ionization Constant, K w Even highly purified water has a very slight conductivity that is only observable if measurements are made with very sensitive instruments (Figure 2). According to Arrhenius theory, conductivity is due to the presence of ions (Figure 3). Therefore, the conductivity observed in pure water must be the result of ions produced by the ionization of some water molecules into hydronium ions and hydroxide ions. Because the conductivity is so slight, the equilibrium at SATP must greatly favour the water molecules. K c H H O H O H [H 3 O (aq)][oh (aq)] [H 2 O(l)][H 2 O(l)] H O H or K c [H 3 O (aq)] [OH (aq)] Evidence indicates that, at 25 C, at any given moment, fewer than two of every billion molecules in pure liquid water exist in ionized form! When we write an equilibrium constant expression for this ionization, the value of K c is an extremely small number. Recall that, because pure liquid water (or the water in any dilute aqueous solution) has an essentially constant concentration, it does not appear in the expression; it is simply incorporated into the equilibrium constant value. The water ionization equilibrium relationship is so important in chemistry that this particular K c constant is given its own special symbol and name. This new constant is called the ion product or ionization constant for water, K w. K w [H 3 O (aq)][oh (aq)] at SATP The equilibrium equation for the ionization of water shows that hydronium ions and hydroxide ions form in a 1:1 ratio. Therefore, the concentration of hydronium ions and hydroxide ions in pure water must be equal. This equality must also be true for any neutral aqueous solution. Using the mathematical expression for K w, and the value of K w at SATP, the concentrations of H 3 O (aq) and OH (aq) can be calculated by taking the square root of the K w value. Recall (from Chapter 15) that in all entity concentration calculations from any K c value, the entity concentration is simply assumed to have units of mol/l. [H 3 O (aq)] [OH (aq)] mol/l H The ionization of water is especially important in the empirical and theoretical study of acidic and basic solutions. Recall from Chapter 6 that, according to the modified Arrhenius theory, an acid is a substance that reacts with water to produce hydronium ions. The additional hydronium ions provided by the acid increase the hydronium ion concentration. Since the hydronium ion concentration is greater than 10 7 mol/l, the solution is acidic.a basic solution is one in which the hydroxide ion concentration is greater O H H + O H H 10 6 % H 2 O(l) H 2 O(l) 0 H 3 O (aq) OH (aq) Figure 3 Collisions of water molecules very occasionally produce this cation anion pair. H O Figure 2 A sensitive multimeter is required to detect the electrical conductivity of the highly purified water that is typically used in a chemistry laboratory. (See Appendix C.3 for guidance on using a multimeter.) DID YOU KNOW? Successful Collisions A collision that successfully forms hydronium and hydroxide ions is very rare. This is not because collisions are rare each water molecule collides with others tens of trillions of times every second! But, the chance that any given collision will both have sufficient energy, and also be at exactly the right orientation, is very, very small indeed. Ordinarily, an equilibrium so strongly favouring the reverse reaction would just be ignored thought of as not happening at all but, in this case, the ions are uniquely important because of the effects they have on all reactions in aqueous solution. Conversely, you already know that the reverse reaction (hydronium ions with hydroxide ions) is quantitative. Almost every such ion collision will be effective because the required energy is very low, and the attraction of opposite ion charges acts to orient the entities correctly. Modelling water molecules this way as space-filling models with superimposed atomic symbols and lines to represent bonds gives an overall representation of the process that is logically consistent with experimental evidence. Equilibrium in Acid Base Systems 713

46 Learning Tip Keep in mind that the value for K w is subject to the same restrictions as any other equilibrium constant: one being that it will change if the temperature changes. For example, in pure water at 20 C, K w has a numerical value of , whereas at 30 C its value is K w will also change enough to be invalid in any aqueous solution with a very high solute concentration because then the assumption that the concentration of the water solvent is at or near a constant value (55.5 mol/l) no longer holds true. CAREER CONNECTION Environmental Engineer Monitoring and protecting water quality is an essential part of the work of environmental engineers. They design systems and treatment processes that control ph levels, temperature, and amounts of dissolved oxygen for industries that use water. Would you like to become an environmental engineer? Look into salaries, employment opportunities, and educational programs at at least two different educational institutions. Learning Tip GO These communication examples show the application of the usual units convention for simplifying calculations from equilibrium constants. Units for the constant are ignored, and other concentration units are always entered in mol/l. Then, since the units for the calculated value are always mol/l, they are just written in with the answer. than 10 7 mol/l. Basic solutions are produced in two ways: either by the complete dissociation (upon dissolving in water) of an ionic hydroxide, or by partial reaction of some weak base entity (ion or molecule) with water to produce hydroxide ions. The most important point about K w is that it applies to pure water, and also to any solution that is mostly water. This means that this ionization equilibrium will be involved in any other reaction going on in aqueous solution, if that reaction involves hydronium ions or hydroxide ions in any way. As an example, consider the equilibrium reaction just studied in Lab Exercise 16.A. Since that reaction involves hydronium (hydrogen) ions, the chromate dichromate equilibrium can be controlled (shifted) easily, simply by adjusting either the hydronium ion or the hydroxide ion concentration; which really just means deliberately shifting the water ionization equilibrium. A great many chemical reactions are dependent in this way on the water ionization equilibrium. Many of them (those containing H 3 O (aq) or OH (aq) ions) are evident in the Relative Strengths of Oxidizing and Reducing Agents table (Appendix I). Note that, since the mathematical relationship is simple, we can easily use K w to calculate either the hydronium ion amount concentration or the hydroxide ion amount concentration in an aqueous solution, if the other concentration is known. Since [H 3 O (aq)][oh (aq)] K w then [H 3 O (aq)] and [OH (aq)] K w [OH (aq)] K w [H3 O (aq)] In ordinary dilute aqueous acidic or basic solutions, the presence of substances other than water decreases the certainty of the K w value at 25 C to two significant digits. All questions and examples in this text assume temperatures of 25 C, and aqueous solutions that are not highly concentrated, with a K w value of , unless specifically stated otherwise. COMMUNICATION example 1 A 0.15 mol/l solution of hydrochloric acid at 25 C is found to have a hydronium ion concentration of 0.15 mol/l. Calculate the amount concentration of the hydroxide ions. Solution HCl(aq) H 2 O(l) 0 H 3 O (aq) Cl (aq) [OH (aq) K w [H3 O (aq)] mol/l mol/l (entity concentration units assumed to be mol/l) Using the K w relationship, the hydronium ion concentration is mol/l. 714 Chapter 16

47 Section 16.1 COMMUNICATION example 2 Calculate the amount concentration of the hydronium ion in a 0.25 mol/l solution of barium hydroxide. Solution Ba(OH) 2 (s) Ba 2 (aq) 2OH (aq) [OH (aq)] 2 [Ba(OH) 2 (aq)] [H 3 O (aq)] mol/l 0.50 mol/l K w [OH (aq)] mol/l mol/l Using the K w relationship, the hydronium ion concentration is mol/l. COMMUNICATION example 3 Determine the hydronium ion and hydroxide ion amount concentrations in 500 ml of an aqueous solution for home soap-making containing 2.6 g of dissolved sodium hydroxide. Solution 1 mol n NaOH 2.6 g mol g mol [NaOH(aq)] 0.13 mol/l L NaOH(s) Na (aq) OH (aq) [OH (aq)] [NaOH(aq)] 0.13 mol/l [H 3 O (aq)] K w [OH (aq)] mol/l mol/l Using the K w relationship, the hydronium ion concentration is mol/l, and the hydroxide ion concentration is 0.13 mol/l. DID YOU KNOW? Warnings on Packaging Many households, in previous centuries, kept supplies of lye (sodium hydroxide) on hand for making soap. Because it looked like sugar, it was occasionally swallowed by curious children, causing terrible injuries to their throats. A prominent American physician, Dr. Chevalier Jackson ( ) realized that warnings on the packaging would encourage parents to keep this dangerous substance out of the reach of their children. This is one of the earliest instances of warning labelling on packaging. Partially because of Dr. Jackson's efforts, the United States Congress passed the Federal Caustic Labelling Act in In Canada, we have the Consumer Chemicals and Containers Regulations. Equilibrium in Acid Base Systems 715

48 COMMUNICATION example 4 Calculate the amount concentration of hydronium ions in a mol/l aqueous solution of ammonia that is used in a spray bottle for window cleaning solution (Figure 4). A reference states that there is 2.1% reaction of dissolved ammonia with water (ionization) at this concentration, at SATP. Solution NH 3 (aq) H 2 O(l) 0 NH 4 (aq) OH (aq) [OH (aq)] mol/l Figure 4 Many solutions sold for cleaning windows contain ammonia. [H 3 O (aq)] mol/l K w [OH (aq)] mol/l mol/l Using the K w relationship, in mol/l aqueous ammonia, the hydronium ion concentration is mol/l. Practice 1. The hydronium ion concentration in an industrial effluent is 4.40 mmol/l. Determine the concentration of hydroxide ions in the effluent. 2. The hydroxide ion concentration in a household cleaning solution is mmol/l. Calculate the hydronium ion concentration in the cleaning solution. 3. Calculate the hydroxide ion amount concentration in a solution prepared by dissolving 0.37 g of hydrogen chloride in 250 ml of water. 4. Calculate the hydronium ion amount concentration in a saturated solution of calcium hydroxide (limewater) that has a solubility of 6.9 mmol/l. 5. What is the hydronium ion amount concentration in a solution made by dissolving 20.0 g of potassium hydroxide in water to form 500 ml of solution? 6. Calculate the percent ionization of water at SATP. Recall that L of water has a mass of 1000 g. Communicating Concentrations: ph and poh Recall (Chapter 6) that the enormous range of aqueous solution hydronium ion concentrations is more easily expressed using the logarithmic ph scale (Figure 5). Mathematically, ph log [H 3 O (aq)] and, inversely, [H 3 O (aq)] = 10 ph For basic solutions, it is sometimes more useful to use a scale based on the amount concentration of hydroxide ions. Recall that the definition of poh follows the same format and the same certainty rule as ph. poh log[oh (aq)] and, inversely, [OH (aq)] 10 poh 716 Chapter 16

49 Section 16.1 battery acid vinegar soft drink normal rain blood sea water antacid solution household lye ammonia ph poh Figure 5 The ph scale increasing acidity ph < 7 increasing basicity ph > 7 neutral (pure water) The mathematics of logarithms allows us to express a simple relationship between ph and poh. According to the rules of logarithms, log(ab) log(a) log(b) Using the equilibrium law for the ionization of water, [H 3 O (aq)][oh (aq)] K w log[h 3 O (aq)] log[oh (aq)] log(k w ) ( ph) ( poh) ph poh (at SATP) This relationship allows for a quick conversion between ph and poh values. SUMMARY Water Ionization Conversions and Values (at SATP) K w [H 3 O (aq)][oh (aq)] ph log[h 3 O (aq)] poh log[oh (aq)] [H 3 O (aq)] 10 ph [OH (aq)] 10 poh ph poh Equilibrium in Acid Base Systems 717

50 Learning Tip Recall this convenient rule of thumb for keeping track of significant digits in (logarithmic) calculations of ph or poh. The number of digits following the decimal point in the ph or poh (logarithm) value should be equal to the number of significant digits shown in the amount concentration of the ion. For example, a hydronium ion concentration of mol/l is expressed as a ph of 2.57, and a poh of 4.3 is expressed as a hydroxide ion concentration of mol/l. Practice 7. Food scientists and dieticians measure the ph of foods when they devise recipes and special diets. (a) Copy and complete Table 1. Table 1 Acidity of Foods [H 3 O (aq)] [OH (aq)] Food (mol/l) (mol/l) ph poh oranges asparagus 5.6 olives blackberries 10.6 (b) Based on ph only, predict which of the foods would taste most sour. 8. To clean a clogged drain, 26 g of sodium hydroxide is added to water to make 150 ml of solution. What are the ph and poh values for the solution? 9. What mass of potassium hydroxide is contained in 500 ml of solution that has a ph of 11.5? Comment on the degree of certainty of your answer. Learning Tip For any aqueous solution of an acid, percent reaction of that acid with water is also commonly called its percent ionization in chemistry references, because the net effect is the same as if the acid molecules simply ionize, as was once (simplistically) assumed in Arrhenius original theory. You should consider reaction with water and ionization in water to be equivalent terms for acid base solution theory, and may expect to see either term used routinely in text and in questions. Acid Strength as an Equilibrium Position In Chapter 6, you learned that acidic solutions of different substances at the same concentration do not possess acid properties to the same degree. The ph of a 1.00 mol/l solution of an acid can vary anywhere from a value of nearly 7 to a value of nearly 0, depending on the specific acid in the solution. Other properties can also vary. For example, acetic acid does not conduct an electric current nearly as well as hydrochloric acid of equal concentration (Figure 6). When we observe chemical reactions of these acids, it is apparent that acetic acid, although it reacts in the same manner and amount as hydrochloric acid, does not react as quickly. The concepts of strong and weak acids were developed to describe and explain these differences in properties of acids. An acid is described as weak if its characteristic properties are less than those of a common strong acid, such as hydrochloric acid. Weak acids are weaker electrolytes and react at a slower rate than strong acids do; the ph of solutions of weak acids is closer to 7 than the ph of strong acids of equal concentration. In Chapter 6, strong acids were explained as ionizing quantitatively by reacting with water to form hydronium ions, whereas weak acids were explained as ionizing only partially (usually 50%). The empirical distinction between strong and weak acids can be explained much more completely now by combining the modified Arrhenius theory with equilibrium theory. A strong acid is explained as an acid that reacts quantitatively with water to form hydronium ions. For example, the reaction of dissolved hydrogen chloride (hydrochloric acid) with water is virtually complete. Even though the equation could be written with double equilibrium arrows and the extent shown with a 99.9% note, it is simpler, and much more common, to just use a single arrow to show that the reaction is quantitative. HCl(aq) H 2 O(l) H 3 O (aq) Cl (aq) A weak acid is an acid that reacts partially with water to form hydronium ions. Measurements of ph indicate that most weak acids react less than 50%. For example, acetic acid reacts only 1.3% in solution at 25 C and 0.10 mol/l concentration. 718 Chapter 16

51 Section 16.1 Recall, from Chapter 15, that any equilibrium position depends on concentration(s) as well as on temperature; so, this 1.3% ionization value for acetic acid is only valid for a 0.10 mol/l solution at 25 C. Laboratory ph experiments show that the higher the concentration of a weak acid solution, the lower its percent ionization becomes. 1.3% CH 3 COOH(aq) H 2 O(l) 0 H 3 O (aq) CH 3 COO (aq) The hydronium ion concentration of any acid solution can be calculated by multiplying the percent reaction by the initial amount concentration of the acid solute. For example, in HCl(aq) solution, virtually 100% of the HCl molecules react with water molecules at equilibrium. HCl(aq) H 2 O(l) H 3 O (aq) Cl (aq) [H 3 O (aq)] mol/l mol/l There are six acids ordinarily classed as strong : hydrochloric, nitric, sulfuric, hydrobromic, hydroiodic, and perchloric acid solutions. Only the first three are common. For any aqueous strong acid, we can simply assume that the concentration of hydronium ions in solution is equal to the initial concentration of the acid dissolved. For weak acids (the majority of examples), we always need to calculate the hydronium ion concentration in solution, because only a small proportion of the initial acid concentration will be converted to ions at equilibrium. Figure 6 In solutions of equal concentration, a weak acid such as acetic acid conducts electricity to a lesser extent than does a strong acid such as hydrochloric acid. Equilibrium in Acid Base Systems 719

52 COMMUNICATION example 5 In a 0.10 mol/l solution of acetic acid, only 1.3% of the CH 3 COOH molecules have reacted at equilibrium to form hydronium ions. Calculate the hydronium ion amount concentration. Solution 1.3% CH 3 COOH(aq) H 2 O(l) 0 H 3 O (aq) CH 3 COO (aq) [H 3 O 1.3 (aq)] 0.10 mol/l mol/l The hydronium ion concentration in 0.10 mol/l acetic acid is mol/l. As explained in Chapter 6, we can easily compare the strengths of different acids by comparing the measured ph values for aqueous solutions of equal concentration. The lower the ph, the higher the hydronium ion concentration, the greater the percent reaction, and thus the stronger the acid. Furthermore, we can find the percent reaction for ionization of any weak acid solution from the measured ph of a solution of known initial concentration. Learning Tip Percent ionization: [H p 3 O (aq)] 100 [HA(aq)] Rearrange the equation to solve for hydronium ion concentration: [H 3 O p (aq)] [HA(aq)] 100 where p percent ionization and [HA(aq)] initial concentration of weak acid COMMUNICATION example 6 The ph of a 0.10 mol/l methanoic acid solution is Calculate the percent reaction for ionization of methanoic acid. Solution [H 3 O (aq)] 10 ph [H 3 O (aq)] mol/l mol/l p 100 [HCOOH(aq)] [H p 3 O (aq)] 100 [HCOOH(aq)] mol/l mol/l 4.2% The percent ionization of 0.10 mol/l aqueous methanoic acid is 4.2%. 720 Chapter 16

53 Section 16.1 WEB Activity Canadian Achievers Edgar Steacie Edgar Steacie (Figure 7) was an internationally acclaimed research scientist and a senior administrator of the National Research Council. 1. What was Steacie s main area of research? 2. Why was Steacie known as a statesman of science for Canada? 3. What is a Steacie Fellowship? GO Figure 7 Edgar Steacie ( ) Section 16.1 Questions 1. How does the hydronium ion concentration compare with the hydroxide ion concentration if a solution is (a) neutral? (b) acidic? (c) basic? 2. What two diagnostic tests can distinguish a weak acid from a strong acid? 3. According to Arrhenius original theory, what do all bases have in common? 4. Hydrocyanic acid is a very weak acid. (a) Write an equilibrium reaction equation for the ionization of 0.10 mol/l HCN(aq). The percent ionization at SATP is %. (b) Calculate the hydronium ion concentration and the ph of a 0.10 mol/l solution of HCN(aq). 5. At 25 C, the hydronium ion concentration in vinegar is 1.3 mmol/l. Calculate the hydroxide ion concentration. 6. At 25 C, the hydroxide ion concentration in normal human blood is mol/l. Calculate the hydronium ion concentration and the ph of blood. 7. Acid rain has a ph less than that of normal rain. The presence of dissolved carbon dioxide, which forms carbonic acid, gives normal rain a ph of 5.6. What is the hydronium ion concentration in normal rain? 8. If the ph of a solution changes by 3 ph units as a result of adding a weak acid, by how much does the hydronium ion concentration change? 9. If 8.50 g of sodium hydroxide is dissolved to make 500 ml of cleaning solution, determine the poh of the solution. 10. What mass of hydrogen chloride gas is required to produce 250 ml of a hydrochloric acid solution with a ph of 1.57? 11. Determine the ph of a 0.10 mol/l hypochlorous acid solution, which has 0.054% ionization at 25 C. 12. Calculate the ph and poh of a hydrochloric acid solution prepared by dissolving 30.5 kg of hydrogen chloride gas to make 806 L of solution. What assumption is made when doing this calculation? 13. Acetic (ethanoic) acid is the most common weak acid used in industry. Determine the ph and poh of an acetic acid solution prepared by dissolving 60.0 kg of pure, liquid acetic acid to make 1.25 kl of solution. The percent reaction with water at this concentration is 0.48%. 14. Determine the mass of sodium hydroxide that must be dissolved to make 2.00 L of a solution with a ph of Write an experimental design for the identification of four colourless solutions: a strong acid solution, a weak acid solution, a neutral molecular solution, and a neutral ionic solution. Write sentences, create a flow chart, or design a table to describe the required diagnostic tests. 16. Sketch a flow chart or concept map that summarizes the conversion of [H 3 O (aq)] to and from [OH (aq)], ph, and percent reaction (ionization) of acid solute. Make your flow chart large enough that you can write the procedure between the quantity symbols in the diagram. Equilibrium in Acid Base Systems 721

54 (a) (b) 16.2 The Brønsted Lowry Acid Base Concept By now, our much-revised acid and base theory includes concepts of hydronium ions, reaction with water, reaction equilibrium, and the ionization equilibrium of water. Our modified theory is, thus, much more comprehensive, and much better at describing, explaining, and predicting acid base reactions, than the original theory proposed by Arrhenius. We find, however, that there are still problems, and our theory must still be considered too restrictive. There is no provision in it for reactions that do not occur in aqueous solution. In addition, we find that there are some substances that seem to have both acid and base properties. As a common example, sodium hydrogen carbonate (sodium bicarbonate, baking soda) forms a basic solution (raises the ph) in water, but the same compound will partly neutralize (lower the ph of) a sodium hydroxide (lye) solution. When we observe that NaHCO 3 (s) forms a basic aqueous solution, we conclude that this happens because some hydrogen carbonate ions react with water molecules to produce hydroxide ions. We know from many other observations that sodium ions have no acidic or basic properties. The following reaction equation seems to explain our observation easily: HCO 3 (aq) H 2 O(l) 0 H 2 CO 3 (aq) OH (aq) But, if adding sodium hydrogen carbonate makes a (strongly basic) sodium hydroxide solution less basic, the concepts we are using lead us to conclude that something must be reacting to decrease the concentration of the hydroxide ions. It must be the hydrogen carbonate ions because there are no other entities in this system except sodium ions and water molecules. We can easily write an equation to explain this observation, as well: HCO 3 (aq) OH (aq) 0 CO 3 2 (aq) H 2 O(l) This equation seems to explain how the hydroxide ions can be partially consumed. But, if it is correct, it begs the question, Is sodium bicarbonate a base, or is it an acid? Clearly, our theory still needs some work! We need a broader, more comprehensive concept to successfully explain these kinds of observations. Figure 1 Johannes Brønsted ( ) (a) and Thomas Lowry ( ) (b) independently created new theoretical definitions for acids and bases, based upon proton transfer during a reaction. The Proton Transfer Concept New theories in science usually result from looking at the evidence in a way that has not occurred to other observers. In 1923, two European scientists independently developed a new approach to acids and bases (Figure 1). These scientists focused on the role of an acid and a base in a reaction rather than on the acidic or basic properties of their aqueous solutions. An acid, such as hydrogen chloride, functions in a way opposite to a base, such as ammonia. According to the Brønsted Lowry concept, hydrogen chloride, upon dissolving, donates a proton to a water molecule: HCl(aq) H 2 O(l) H 3 O (aq) Cl (aq) acid H and ammonia, upon dissolving, accepts a proton from a water molecule. H NH 3 (aq) H 2 O(l) 0 OH (aq) NH 4 (aq) base 722 Chapter 16

55 Section 16.2 Water does not have to be one of the reactants. For example, hydronium ions present in a hydrochloric acid solution can react directly with dissolved ammonia molecules. H 3 O (aq) NH 3 (aq) H 2 O(l) NH 4 (aq) acid H base We can describe this reaction as NH 3 molecules removing protons from H 3 O ions. Hydronium ions act as the acid, and ammonia molecules act as the base. Water is present as the solvent but not as a primary reactant. In fact, water does not even have to be present, as evidenced by the reaction of hydrogen chloride and ammonia gases (Figure 2). HCl(g) NH 3 (g) NH 4 Cl(s) acid According to the Brønsted Lowry concept, a Brønsted Lowry acid is a proton donor and a Brønsted Lowry base is a proton acceptor. A Brønsted Lowry neutralization is a competition for protons that results in a proton transfer from the strongest acid present to the strongest base present. A Brønsted Lowry reaction equation is an equation written to show an acid base reaction involving the transfer of a proton from one entity (an acid) to another (a base). The Brønsted Lowry concept does away with defining a substance as being an acid or base. Only an entity that is involved in a proton transfer in a reaction can be defined as an acid or base and only for that particular reaction. This last point is extremely important: protons may be gained in a reaction with one entity, but lost in a reaction with another entity. For example, in the reaction of HCl with water shown on the previous page, water acts as the base; whereas, in its reaction with NH 3, water acts as the acid. A substance that appears to act as a Brønsted Lowry acid in some reactions and as a Brønsted Lowry base in other reactions is called amphoteric, or sometimes (incorrectly) amphiprotic. The empirical term, amphoteric, properly refers to a chemical substance with the ability to react as either an acid or a base. The theoretical term, amphiprotic, describes an entity (ion or molecule) having the ability to either accept or donate a proton. The hydrogen carbonate ion in baking soda (Figure 3), like every other hydrogen polyatomic ion, is amphiprotic, as shown by the following reactions: HCO 3 (aq) H 2 O(l) 0 OH (aq) H 2 CO 3 (aq) base acid HCO 3 (aq) H 2 O(l) 0 H 3 O (aq) CO 3 2 (aq) acid H base base K c mol/l K c mol/l When bicarbonate ions are in aqueous solution, some react with the water molecules by acting as an acid, and some react by acting as a base. K c values given for these two equilibriums show that one of them predominates so much so that the other reaction is simply inconsequential. The number of ions reacting as a base is over 2000 times more than the number reacting as an acid. As you might expect, the resulting solution is basic. We can get a clearer idea of the amphoteric nature of baking soda (caused by the amphiprotic nature of the hydrogen carbonate ion) by noting the evidence expressed in the next two equations. Note that bicarbonate ions partly neutralize a strong acid, but they can also partly neutralize a strong base. HCO 3 (aq) H 3 O (aq) 0 H 2 CO 3 (aq) H 2 O(l) (raises the ph of a strong base acid acid solution) HCO 3 (aq) OH (aq) 0 CO 2 3 (aq) H 2 O(l) (lowers the ph of a acid base strong base solution) Figure 2 Invisible fumes of ammonia gas and hydrogen chloride gas mix and react above these beakers, producing tiny visible particles of solid ammonium chloride that are suspended in the air. Figure 3 Baking soda is a common household chemical, but it requires an uncommonly sophisticated theory to describe, explain, or predict all of its properties. Equilibrium in Acid Base Systems 723

56 DID YOU KNOW? Superacids In aqueous solution, all strong acids are equal in strength, since they all react instantly and completely with water to form the strongest possible acid entity that can exist in water the hydronium ion. Scientists have long suspected, however, that the well-known strong acids, HCl and H 2 SO 4, may not be equal in strength when no water is present and that much stronger acids could exist. Dr. Ronald Gillespie of McMaster University has done extensive research on non-aqueous strong acids. His definition of a superacid one that is stronger than pure sulfuric acid is now generally accepted by scientists. Perchloric acid, the only common superacid, easily loses protons to H 2 SO 4 (l) molecules. Fluorosulfonic acid, HSO 3 F(l), is more than one thousand times stronger than H 2 SO 4 (l). It is the strongest Brønsted Lowry acid known. Scientists consider the Brønsted Lowry concept to be a theoretical definition. It falls short of being a comprehensive theory because it does not explain why a proton is donated or accepted, and cannot predict theoretically which reaction will occur for a given entity in any given new situation. The advantage of the Brønsted Lowry definitions is that they enable us to define acids and bases in terms of chemical reactions rather than simply as substances that form acidic and basic aqueous solutions. A definition of acids and bases in terms of chemical reactions allows us to describe, explain, and predict a great many more reactions, whether they take place in aqueous solution, in solution in some other solvent, or between two undissolved entities in pure chemical states. Practice 1. Theories in science develop over a period of time. Illustrate this development by writing a theoretical definition of an acid, using the following concepts. Begin your answer with, According to [authority], acids are substances that. (a) Arrhenius original theory (b) the modified Arrhenius theory (c) the Brønsted Lowry concept 2. How does the definition of a base according to the modified Arrhenius theory compare with the Brønsted Lowry definition? 3. Classify each reactant in the following equations as a Brønsted Lowry acid or base. (a) HF(aq) SO 3 2 (aq) 0 F (aq) HSO 3 (aq) (b) CO 3 2 (aq) CH 3 COOH(aq) 0 CH 3 COO (aq) HCO 3 (aq) (c) H 3 PO 4 (aq) OCl (aq) 0 H 2 PO 4 (aq) HOCl(aq) (d) HCO 3 (aq) HSO 4 (aq) 0 SO 4 2 (aq) H 2 CO 3 (aq) 4. Evidence indicates that the hydrogen sulfite ion is amphiprotic. A sodium hydrogen sulfite solution can partly neutralize either a sodium hydroxide spill or a hydrochloric acid spill. (a) Write a net ionic equation for the reaction of aqueous hydrogen sulfite ions with the hydroxide ions in solution. Label the reactants as acids or bases. (b) Write a net ionic equation for the reaction of hydrogen sulfite ions with the hydronium ions from a hydrochloric acid solution. Label the reactants as acids or bases. 5. What restrictions to acid base reactions do the Brønsted Lowry definitions remove? 6. Why is the Brønsted Lowry concept labelled a theoretical definition rather than a theory? Figure 4 White vinegar is a 5% acetic (ethanoic) acid solution. The amount concentration is 0.83 mol/l. Such a solution is only about 0.43% ionized at 25 C, making acetic acid a typical weak acid, by definition. Conjugate Acids and Bases According to the Brønsted Lowry concept, acid base reactions involve the transfer of a proton. These reactions are universally reversible and always result in an acid base equilibrium. In a proton transfer reaction at equilibrium, both forward and reverse reactions involve Brønsted-Lowry acids and bases. For example, in an acetic acid solution (Figure 4), the forward reaction is explained as a proton transfer from acetic acid to water molecules, and the reverse reaction is a proton transfer from hydronium to acetate ions. base acid CH 3 COOH(aq) H 2 O(l) 0 CH 3 COO (aq) H 3 O (aq) acid H base H 724 Chapter 16

57 Section 16.2 This equilibrium is typical of all acid base reactions. There will always be two acids (in the example, CH 3 COOH and H 3 O ) and two bases (in the example, H 2 O and CH 3 COO ) in any acid base reaction equilibrium. Furthermore, the base on the right (CH 3 COO ) is formed by removal of a proton from the acid on the left (CH 3 COOH). The acid on the right (H 3 O ) is formed by the addition of a proton to the base on the left (H 2 O). A pair of substances with formulas that differ only by a proton is called a conjugate acid base pair. An acetic acid molecule and an acetate ion are a conjugate acid base pair. Acetic acid is the conjugate acid of the acetate ion, and the acetate ion is the conjugate base of acetic acid. The equation shows that there must always be two conjugate acid base pairs in any proton transfer reaction. The hydronium ion and water are the second conjugate acid base pair in this equilibrium. Conjugate acid base pairs appear opposite each other in a table of acids and bases, such as that in Appendix I. conjugate pair 1.3% CH 3 COOH(aq) H 2 O(l) 0 CH 3 COO (aq) H 3 O (aq) (0.10 mol/l at 25 C) conjugate pair H + CH 3 COO O H H Figure 5 This pictorial analogy is used to explain why acetic acid is a weak acid in aqueous solution. We believe that the proton (H ) of the carboxyl group is attracted more strongly by the rest of the acetic acid molecule than it is by the water molecule. We infer this conclusion because, at equilibrium, ph evidence indicates that very few of the acetic acid molecules have lost protons in their collisions with water molecules. At equilibrium, only 1.3% of the CH 3 COOH molecules have reacted with water in a 0.10 mol/l solution at SATP. It appears that the ability of the CH 3 COO part of the acetic acid molecule to keep its proton (H ) is much greater than the ability of H 2 O to attract the proton away (Figure 5). This means that CH 3 COO is a stronger base (it has a greater attraction for protons) than H 2 O. When HCl molecules react with water (Figure 6), the Cl of each HCl molecule has a much weaker attraction for its proton (H ) than any colliding water molecule has. The water molecules win this competition for protons overwhelmingly. At equilibrium, essentially all of the HCl molecules have lost protons to water molecules. In this case, the transfer of protons is quantitative, and, because its molecules have an extremely weak attraction for their protons, HCl(aq) is called (somewhat confusingly) a strong acid. Remember: The stronger the base, the more it attracts another proton. The stronger an acid, the less it attracts its own proton. Cl H + H O H H H O + H + Cl Figure 6 When gaseous hydrogen chloride dissolves in water, the HCl molecules are thought to collide and react quantitatively with water molecules to form hydronium and chloride ions. Equilibrium in Acid Base Systems 725

58 Learning Tip Acids were studied much earlier than bases were, and sometimes the old terminology used to describe acid base situations can be misleading because it always focuses on the nature of the acid. Strong acids are very reactive, but we believe it is because they are very weak proton attractors. It is common to speak of acids donating protons, and of bases accepting them, but this terminology gives an unrealistic view of what we believe is really happening. It is like saying that a bank robber accepts money donated by a bank teller. The tug-of-war analogy in Figure 5 is a more realistic way to think of proton transfer reactions as a competition between two bases, both attracting the same proton. The terms strong acid and weak acid can be explained in terms of the Brønsted Lowry concept, and also by comparing the reactions of different acids with the same base for example, water. Using HA as the general symbol for any acid and A as its conjugate base, the empirically derived Relative Strengths of Aqueous Acids and Bases table (Appendix I) lists the position of equilibrium of aqueous solutions of many different acids. They are ordered according to how much they ionize in (react with) the water solvent. HA(aq) H 2 O(1) 0 H 3 O (aq) A (aq) The extent of the proton transfer between HA and H 2 O determines the relative strength of HA(aq). In Brønsted Lowry terms, when a strong acid reacts with water, an almost complete transfer of protons results for the forward reaction and almost no transfer of protons occurs for the reverse reaction; a nearly 100% reaction with water. The equilibrium constant value is found to be very large. Theoretically, a strong acid holds its proton very weakly, and easily loses the proton to any base, even very weak bases such as water. This result leads to the interpretation that the conjugate base, A, of a strong acid must have a very weak (negligible) attraction for protons. A useful generalization regarding the relative strengths of an acid base conjugate pair is: the stronger an acid, the weaker its conjugate base; and conversely, the weaker an acid, the stronger its conjugate base. (See the Relative Strengths of Aqueous Acids and Bases table in Appendix I.) Chemists have no simple explanation, in terms of forces or bonds, for the differing abilities of acids to donate protons or of bases to accept them. The inability to predict acid and base strengths for an entity not already included in an empirically determined table of acids and bases is a major deficiency of all acid base theories. SUMMARY Brønsted Lowry Definitions An acid is a proton donor and a base is a proton acceptor, in a specific reaction. An acid base reaction involves a single proton transfer from one entity (the acid) to another (the base). An amphiprotic entity (amphoteric substance) is one that acts as a Brønsted Lowry acid in some reactions and as a Brønsted Lowry base in other reactions. A conjugate acid base pair consists of two entities with formulas that differ only by a proton. A strong acid has a very weak attraction for protons. A strong base has a very strong attraction for protons. The stronger an acid, the more weakly it holds its proton. The stronger a base, the more it attracts another proton. The stronger an acid, the weaker is its conjugate base. The stronger a base, the weaker is its conjugate acid. Practice 7. Use the Brønsted Lowry definitions to identify the two conjugate acid base pairs in each of the following acid base reactions. (a) HCO 3 (aq) S 2 (aq) 0 HS (aq) CO 3 2 (aq) (b) H 2 CO 3 (aq) OH (aq) 0 HCO 3 (aq) H 2 O(l) (c) HSO 4 (aq) HPO 4 2 (aq) 0 H 2 PO 4 (aq) SO 4 2 (aq) (d) H 2 O(l) H 2 O(l) 0 H 3 O (aq) OH (aq) 8. Some ions can form more than one conjugate acid base pair. List the two conjugate acid base pairs involving a hydrogen carbonate ion in the reactions in question Chapter 16

59 Section 16.2 INVESTIGATION 16.1 Introduction Creating an Acid Base Strength Table An acid base table organizes common acids (and their conjugate bases) in order of decreasing acid strength. Acid strength can be tested several ways, including by a carefully designed use of indicators. Predict the order of strengths using the Relative Strengths of Aqueous Acids and Bases table (Appendix I). Use the indicators provided to create a valid and efficient Design, in which you clearly identify the relevant variables. Evaluate the Design (only), and suggest improvements if any problems are identified. Report Checklist Purpose Design Analysis Problem Materials Evaluation (1) Hypothesis Procedure Prediction Evidence Purpose The purpose of this investigation is to test an experimental design for using indicators to create a table of relative strengths of acids and bases. Problem Can the indicators available be used to rank the acids and bases provided in order of strength? To perform this investigation, turn to page 768. Predicting Acid Base Reaction Equilibria The Brønsted Lowry concept unfortunately does not include any theoretical explanation about why any given entity attracts a proton more or less strongly. To predict the outcome of any acid base combination, we must rely on empirical evidence, gained by measuring and recording the relative strengths of acid and base entities. Predictions must be restricted to only those acid base combinations for which we already have data. To help us, we can now look for a simple generalization that might allow us to predict the approximate position of equilibrium in an acid base proton transfer. LAB EXERCISE 16.B Predicting Acid Base Equilibria Is it possible to predict how far a reaction will proceed? Use the table of Relative Strengths of Aqueous Acids and Bases in Appendix I and the evidence of position of equilibrium to complete the Analysis of the investigation report. Purpose The purpose of this investigation is to develop a generalization for predicting the position of acid base equilibria. Problem How do the positions of the reactant acid and base in the acid base table relate to the position of equilibrium? Report Checklist Purpose Design Analysis Problem Materials Evaluation Hypothesis Procedure Prediction Evidence 50% 1. CH 3 COOH(aq) H 2 O(l) 0 H 3 O (aq) CH 3 COO (aq) 99% 2. HCl(aq) H 2 O(l) 0 H 3 O (aq) Cl (aq) 50% 3. CH 3 COO (aq) H 2 O(l) 0 CH 3 COOH(aq) OH (aq) 50% 4. H 3 PO 4 (aq) NH 3 (aq) 0 H 2 PO 4 (aq) NH 4 (aq) 50% 5. HCO 3 (aq) SO 3 2 (aq) 0 HSO 3 (aq) CO 3 2 (aq) 6. H 3 O (aq) OH (aq) H 2 O(l) H 2 O(l) WEB Activity Web Quest Pool Chemistry A swimming pool can be an enjoyable place to spend a hot summer afternoon. Most of us are familiar with the dangers of pools. This Web Quest introduces an unfamiliar risk: pool gas. What is pool gas and how does it form? What are the dangers of pool gas, and how can they be reduced? GO Equilibrium in Acid Base Systems 727

60 CAREER CONNECTION Chemistry Researcher Farideh Jalilehvand (Figure 7) is an associate professor of chemistry at the University of Calgary. Her research involves a lot of X-ray absorption spectroscopy. Find out how acidity due to sulfur accumulation in water-logged wood is connected to ancient shipwrecks by visiting her information site, and check out her curriculum vitae (CV). Science is international in scope! Figure 7 Farideh Jalilehvand GO Predicting Acid Base Reactions When making complex predictions, scientists often combine a variety of empirical and theoretical concepts. We need to use a combination of concepts to predict both the products and the extent of acid base reactions. According to the collision reaction theory, a proton transfer may result from a collision between an acid and a base. In a system that is a mixture of several different acid and/or base entities, there are countless random collisions of all the entities present. So, in theory, in any such system, there are many different possible acid base reactions, and all of these reactions occur (to some extent) all the time. Evidence indicates that one reaction predominates: it occurs to an extent so much greater than the other reactions that it is the only observable reaction. We will be able to explain which reaction will predominate if we use a combination of the collision reaction theory and the Brønsted Lowry concept. According to collision reaction theory, in an acid base system, collisions of all entities present are constantly occurring. According to the Brønsted Lowry concept, a proton will only transfer if an acid entity collides with a base entity that is a better proton attractor than itself. A proton could theoretically transfer several times (if there were several different acids and bases in the system), transferring each time to a stronger proton attractor. Once gained by an entity of the strongest base present, a proton will remain there, because there is no other base present in the system that can attract that proton strongly enough to remove it. By the same logic, once any entity of the strongest acid present has lost its proton, its (remaining) conjugate base cannot gain one back from any other entity present, because it is a weaker proton attractor than anything else in the system. Overall, theory suggests that the predominant acid base reaction should be the one that involves proton transfer from the strongest acid to the strongest base present in the system. Experimental evidence indicates that this explanation is correct. Other possible proton transfer reactions occur to such a small extent that they have a negligible effect on the reactants and products, so they are normally ignored. We theorize that the only significant reaction in any acid base system involves proton transfer from the strongest acid present to the strongest base present. For an aqueous solution system, we first must list all the entities present as they exist in aqueous solution. Table 1 summarizes how to represent the entities that are present. Table 1 Predominant Entities Present in Aqueous Solution Substance dissolved Kinds of entities Predominant entities (example) in solution present (example) Ionic compounds cations, anions Ca 2 (aq) and HCO 3 (aq) Ca(HCO 3 ) 2 (aq) Ionic oxides cations, hydroxide ions Na (aq), OH (aq) or Na 2 O(aq), CaO(aq) Ca 2 (aq), OH (aq) Strong acids H 3 O (aq), conjugate base H 3 O (aq), Cl (aq) or HCl(aq) or HNO 3 (aq) H 3 O (aq), NO 3 (aq) Weak acids molecules or ions HF(aq) or HSO 3 (aq) HF(aq) or HSO 3 (aq) Weak bases molecules or ions NH 3 (aq) or HCO 3 (aq) NH 3 (aq) or HCO 3 (aq) The process for determining the nature and extent of the predominant proton transfer reaction in an aqueous acid base system can be thought of as five distinct steps, for convenience, as shown in the following Sample Problem. 728 Chapter 16

61 Section 16.2 SAMPLE problem 16.1 What will be the predominant reaction if spilled drain cleaner (sodium hydroxide) solution is neutralized with vinegar (Figure 8)? Are reactants or products favoured in this reaction? Step 1: List all entities present as they exist in aqueous solution. Refer to Table 1 if necessary. The entity list for this situation is Na (aq) OH (aq) CH 3 COOH(aq) H 2 O(l) Step 2: Use the entity lists of the Relative Strengths of Aqueous Acids and Bases table to identify and label each entity present as a Brønsted Lowry acid or base. Amphiprotic entities are labelled for both possibilities. Conjugate bases of strong acids are not included or labelled as bases because they cannot act as bases in aqueous solution. Metal ions are treated as spectators. A A Na (aq) OH (aq) CH 3 COOH(aq) H 2 O(l) B B Step 3: Use the order of the entities in the Relative Strengths of Aqueous Acids and Bases table to identify and label the strongest Brønsted Lowry acid (the highest one on the table) and the strongest Brønsted Lowry base (the lowest one on the table) that are present in the solution. SA A Na (aq) OH (aq) CH 3 COOH(aq) H 2 O(l) SB B Step 4: Write a balanced equation to show a proton transfer from the strongest acid to the strongest base, assuming that their respective conjugates are the reaction products. H Figure 8 The drain cleaner shown is a concentrated solution of a very strong base. A spill would be highly corrosive and quite hazardous. Excess weak acid, such as the vinegar (5% acetic acid), is the best choice to neutralize a spilled strong base. The final solution will not actually be neutral, but it will be only mildly acidic, and much safer to handle. CH 3 COOH(aq) OH (aq) 0 CH 3 COO (aq) H 2 O(l) Step 5: Predict the position of equilibrium using the generalization developed in Lab Exercise 16.B and illustrated in the margin Learning Tip. For this reaction, the strongest acid is positioned higher on the table than the strongest base, so products are favoured. Under certain assumed conditions (see the following discussion), the equilibrium percent reaction may be labelled as greater than 50%. 50% CH 3 COOH(aq) OH (aq) 0 CH 3 COO (aq) H 2 O(l) The nature of water complicates the prediction of outcomes of acid base reactions in aqueous solution. We need to consider the specific restrictions that apply when predicting proton transfer reactions in aqueous solution, because they are much more common than acid base reactions in any other environment. Hydronium ion is the strongest acid entity that can exist in aqueous solution. If a stronger acid than hydronium ion is dissolved in water, it reacts instantly and completely with water molecules to form hydronium ions. For this reason, the six strong acids are all written as H 3 O (aq) when in aqueous solution. Hydroxide ion is the strongest base entity that can exist in aqueous solution. If a stronger base than hydroxide ion is dissolved in water, it reacts instantly and completely to form hydroxide ions. The only common example is the dissolving of soluble ionic oxide compounds such as Na 2 O(s). In such cases, the oxide ion is written as OH (aq) when in aqueous solution. Learning Tip The relative positions of the strongest acid and the strongest base on an acid base table can be used to approximately determine the position of an acid base equilibrium. Products Favoured SA SB Reactants Favoured SA SB 50% 0 50% 0 Equilibrium in Acid Base Systems 729

62 Learning Tip Since hydroxide ion is the strongest possible base in aqueous solution, reaction of any acid with this ion in aqueous solution is automatically one that favours products. Similarly, the reaction of any base with the strongest possible acid, hydronium ion, must favour products. Of course, the reaction of hydroxide ion with hydronium ion is always quantitative. No entity in aqueous solution can react as a base if it is a weaker base than water. For this reason, the conjugate bases of the six strong acids are not considered as bases, in aqueous solution. Even though you find that products are favoured (as in Sample Problem 16.1), for the predominant reaction, you cannot accurately predict the actual equilibrium position of all cases of any specific acid base reaction by this method. It simply means you know that the forward reaction happens more readily than the reverse reaction, all things being equal. But, recall (from Chapter 15) that the initial concentration of entities plays a very large part in determining the percent reaction at equilibrium. To be able to predict something meaningful about the position of an acid base reaction equilibrium, we must restrict the reaction conditions. Unless you are specifically informed otherwise in a question, assume in this text that, for the predominant reaction in an acid base system, the equation represents a single proton transfer between two entities, neither of which is water, and has a stoichiometric ratio of 1:1:1:1 the strongest acid and the strongest base present in the reaction system are both present in significant chemical amounts, in approximately equal concentrations Under these circumstances (and only then), you may assume that a Brønsted Lowry acid base reaction where products are favoured has a percent reaction greater than 50%, or that a reaction where reactants are favoured may be labelled as 50%. It is also correct (given these restrictions) to state that the K c value for any such reaction is 1 if products are favoured, and 1 if reactants are favoured. It is not possible to be more precise about the equilibrium position, however, at our current level of theory. Note that the above restrictions exclude any reaction equation written to describe the ionization (reaction with water) of any single acid or base entity dissolved to make an aqueous solution. In these cases, water is acting as one of the reactants, and the amount concentration of the water is very much greater than that of the other entity. Finally, recall that the reaction of hydronium ions with hydroxide ions is always quantitative. This equation should always be written with a single arrow. There are many other acid base reactions that are quantitative, particularly those where either hydronium or hydroxide ions are involved, but it is not possible to easily predict whether any acid base reaction will be quantitative from the table of Relative Strengths of Aqueous Acids and Bases. In this text, other quantitative reactions will always be identified for you, as in the following Communication Example. COMMUNICATION example Ammonium nitrate fertilizer is produced by the quantitative reaction of aqueous ammonia with nitric acid. Write a balanced acid base equilibrium equation. Solution A SA NH 3 (aq), H 2 O(l), H 3 O (aq) SB B NH 3 (aq) H 3 O (aq) NH 4 (aq) H 2 O(l) 730 Chapter 16

63 Section 16.2 SUMMARY A Five-Step Method for Predicting the Predominant Acid Base Reaction 1. List all entities (ions, atoms, or molecules including H 2 O(l)) initially present as they exist in aqueous solution. (Refer to Table 1, page 728.) 2. Identify and label all possible aqueous acids and bases, using the Brønsted Lowry definitions. 3. Identify the strongest acid and the strongest base present, using the table of Relative Strengths of Aqueous Acids and Bases (Appendix I). 4. Write an equation showing a transfer of one proton from the strongest acid to the strongest base, and predict the conjugate base and the conjugate acid to be the products. 5. Predict the approximate position of the equilibrium, using the generalization developed in Lab Exercise 16.B on page 727 and the table of Relative Strengths of Aqueous Acids and Bases (Appendix I). Practice Use the five-step method to predict the predominant reactions in the following chemical systems: 9. Hydrofluoric acid and an aqueous solution of sodium sulfate are mixed to test the five-step method of predicting acid base reactions. 10. Strong acids, such as perchloric acid, have been shown to react quantitatively with strong bases, such as sodium hydroxide. 11. Methanoic acid is added to an aqueous solution of sodium hydrogen sulfide. 12. A student mixes solutions of ammonium chloride and sodium nitrite in a chemistry laboratory. 13. Empirical work has shown that nitric acid reacts quantitatively with a sodium acetate solution. 14. A consumer attempts to neutralize an aqueous sodium hydrogen sulfate cleaner with a solution of lye. (See Appendix J if you do not remember what lye is.) 15. Can ammonium nitrate fertilizer, added to water, be used to neutralize a muriatic acid (hydrochloric acid) spill? 16. Predict the acid base reaction of bleach with vinegar (Figure 9). 17. Commercial laundry bleach (Figure 9) is made by reacting chlorine gas with a sodium hydroxide solution, according to the equation Cl 2 (g) 2OH (aq) 0 OCl (aq) Cl (aq) H 2 O(l) As sold, the ph of laundry bleach solutions is always well above 8. You know that the element chlorine has very low solubility in pure water. Explain, using Le Châtelier s principle, why bleach bottle labels warn so strongly against mixing bleach with other cleaning agents, such as acidic toilet bowl cleaners. Note: Household bleach also produces toxic chloramines if mixed with basic ammonia cleaning solutions. The best rule is, NEVER mix bleach directly with any other cleaning powder or solution. It is arguably the most dangerous common household chemical. Figure 9 Bottles of household bleach display a warning against mixing the bleach (aqueous sodium hypochlorite) with acids. Does your prediction of the reaction between vinegar and hypochlorite ions provide any clues about the reason for the warning? Equilibrium in Acid Base Systems 731

64 LAB EXERCISE 16.C Aqueous Bicarbonate Ion Acid Base Reactions An acid base table organizes common acids (and their conjugate bases) in a way that enables us to predict predominant acid base reactions. Evaluate the reliability of the five-step method (only) using information from this investigation, by comparing your analysis of the evidence with your predictions. Report Checklist Purpose Design Analysis Problem Materials Evaluation (2, 3) Hypothesis Procedure Prediction Evidence Purpose The purpose of this investigation is to test the five-step method for predicting reactions in acid base systems. Problem What are the products and position of the equilibrium for sodium hydrogen carbonate (Figure 10) with stomach acid, vinegar, household ammonia, and lye, respectively? Evidence Table 2 The Addition of Baking Soda to Various Solutions Reactant Bubbles Odour ph HCl(aq) yes none increases CH 3 COOH(aq) yes disappears increases NH 3 (aq) no remains decreases NaOH(aq) no none decreases Figure 10 The versatility of baking soda is demonstrated by its use in extinguishing fires, in baking biscuits, and in neutralizing excess stomach acid. It is also used as a medium for local anesthetics baking soda reduces stinging sensations by neutralizing the acidity of the anesthetic, with the result that the speed and efficiency of the anesthetic are improved. The broad range of uses for baking soda results, in part, from its amphiprotic character. INVESTIGATION 16.2 Introduction Testing Brønsted Lowry Reaction Predictions When predicting products for this investigation, list all entities present as they normally exist in an aqueous environment. For those reactants that are added in solid state, assume that they will dissolve. Use the resulting entities for prediction. Evaluate the predictions, the Brønsted Lowry concept, and the five-step method for acid base reaction prediction. Purpose The purpose of this investigation is to test the Brønsted Lowry concept and the five-step method for reaction prediction from a table of relative acid base strength. Report Checklist Purpose Design Analysis Problem Materials Evaluation (2, 3) Hypothesis Procedure Prediction Evidence Problem What reactions occur when various pairs of substances are mixed? Design A prediction is made for each of eleven pairs of substances. The prediction is then tested using one or more diagnostic tests, complete with controls. Additional diagnostic tests increase the certainty of the evaluation. To perform this investigation, turn to page Chapter 16

65 Section 16.2 LAB EXERCISE 16.D Creating an Acid Base Table Complete the Analysis of the investigation report, including a short table of the four acids and bases involved. Use entity position generalizations in the acid base table for reactions that favour products or reactants. The evidence for reaction 1 is interpreted as: Acids CH 3 COOH(aq) Bases Report Checklist Purpose Design Analysis Problem Materials Evaluation Hypothesis Procedure Prediction Evidence Problem What is the order of acid strength for the first four members of the carboxylic acid family? Evidence 50% CH 3 COOH(aq) C 3 H 7 COO (aq) 0 CH3 COO (aq) C 3 H 7 COOH(aq) > 50% C 3 H 7 COO (aq) Purpose The purpose of this investigation is to test an experimental design for using equilibrium position to create a table of relative strengths of acids and bases. 50% HCOOH(aq) CH 3 COO (aq) 0 HCOO (aq) CH 3 COOH(aq) 50% C 2 H 5 COOH(aq) C 3 H 7 COO (aq) 0 C 2 H 5 COO (aq) C 3 H 7 COOH(aq) 50% C 2 H 5 COOH(aq) HCOO (aq) 0 C 2 H 5 COO (aq) HCOOH(aq) Case Study Changing Ideas on Acids and Bases The Evolution of a Scientific Theory Historically, chemists have known the empirical properties of substances long before any theory was developed to explain and predict those properties. For example, chemists were familiar with the distinguishing properties of several acids and bases, and used them routinely, by the middle of the 17th century. Early attempts at an acid base theory tended to focus on acids and to ignore bases. Over time, several theories passed through cycles of formulation, testing, acceptance, further testing, and eventual rejection. Following is a brief historical summary of acid base theories and the evidence that led to their revision. Antoine Lavoisier ( ) (Figure 11) believed that the properties of acids could be traced back to a single substance. Lavoisier studied the combustion of phosphorus and sulfur and determined that these elements combine with something in the atmosphere to produce compounds that form acidic solutions when dissolved in water. When Joseph Priestley identified the component of the atmosphere that actively supports combustion, Lavoisier (1777) named the gas oxygen, meaning acid maker. Lavoisier assumed that oxygen was the substance responsible for the generic properties of acids. There were soon problems with this theory when it was found that several acids, such as muriatic acid (HCl), do not contain oxygen. Furthermore, many substances that form basic solutions (such as lime, CaO) were found to contain oxygen. This evidence led to the rejection of the oxygen theory, but it is historically important because it is the first systematic attempt to chemically characterize acids and bases. The generalization that nonmetallic oxides form acidic solutions is still useful in chemistry. Sir Humphry Davy ( ) (Figure 12) conducted the experiments that demonstrated the absence of oxygen in muriatic acid (HCl), which led to the rejection of Lavoisier s theory. Davy (1810) advanced his own theory that the Figure 11 Antoine Lavoisier Figure 12 Humphry Davy Equilibrium in Acid Base Systems 733

66 presence of hydrogen gives compounds acidic properties. This theory, however, did not explain why many compounds containing hydrogen have neutral properties (for example, CH 4 ) or basic properties (for example, NH 3 ). Justus von Liebig ( ) revised Davy s idea to define acids as substances in which the hydrogen could be replaced by a metal. This revision meant that acids could be thought of as ionic compounds in which hydrogen had replaced the metal ion. Liebig had no corresponding theoretical definition for bases, which were still identified empirically as substances that neutralized acids. Svante Arrhenius ( ) (Figure 13) developed a theory in 1887 that provided the first useful theoretical definitions of acids and bases. He defined acids as substances that ionize in aqueous solution to form hydrogen ions, H (aq). Similarly, he defined bases as substances that dissociate to form hydroxide ions, OH (aq), in solution. This theory explained the process of Figure 13 neutralization as the Svante Arrhenius combination of hydrogen ions and hydroxide ions to form water H (aq) OH (aq) H 2 O(l) Arrhenius explained the strengths of acids in terms of degrees of ionization. While these ideas were a major development in chemists understanding of the properties of acids and bases, there were some problems with Arrhenius theory. In Arrhenius theory, an acid is expected to be an acid in any solvent, which was found not to be the case. For example, when dissolved in water, HCl supposedly breaks up into hydrogen ions and chloride ions, but when it is dissolved in benzene, the HCl remains as intact molecules. The nature of the solvent, therefore, had to play a critical role in acid base properties of substances. The need for hydroxide to always be the base led Arrhenius to propose formulas such as NH 4 OH(aq) for the formula for ammonia in water, which led to the misconception that NH 4 OH(aq) was the base, not NH 3 (aq). According to Arrhenius theory, all salts (ionic solids) should produce neutral solutions, but many do not. For example, solutions of ammonium chloride are acidic and solutions of sodium acetate are basic. Bonding theory suggested that it was very unlikely that a single proton could exist in aqueous solution without being bonded to at least one water molecule. Johannes Brønsted ( ) (Figure 14) and Thomas Lowry ( ) (Figure 15) independently published (1923) essentially the same concept about how acids and bases behave. These scientists focused on the Figure 14 Johannes Brønsted role of an acid and a base in a reaction rather than on the properties of their aqueous solutions. According to the Brønsted Lowry concept, an acid is a proton (H ) donor and a base is a proton acceptor. The solvent has a central role in the Brønsted Lowry concept. Water can be considered an acid or a base since it can lose a proton to form a hydroxide ion (OH ) or accept a proton to form a hydronium ion (H 3 O ). In their view, neutralization is a competition for protons that results in a proton transfer from the strongest acid present to the strongest base present. For example, in the reaction HSO 4 (aq) NH 3 (aq) SO 2 4 (aq) NH 4 (aq) the hydrogen sulfate ion acts as the acid (because it donates a proton) and ammonia acts as the base (because it accepts a proton). A weakness of the Brønsted Lowry concept is its limitation to solutions (gaseous or liquid). Gilbert Lewis ( ) (Figure 16) developed an acid base theory in 1923 that includes all previous theories and definitions of acids and bases. He viewed acids and bases in terms of the covalent bond, a theory he had developed in Lewis defined acids as electron pair acceptors and bases as electron pair donors. It is important to note that in Lewis acid base theory, no hydrogen ion and no solvent need be involved. The Lewis Figure 15 Thomas Lowry Figure 16 Gilbert Lewis definition is broader than all previous definitions and explains more inorganic and organic reactions (Figure 17). For example, in the reaction BF 3 (g) NH 3 (g) H 3 NBF 3 (g) boron trifluoride acts as a Lewis acid because it accepts (forms a bond with) a pair of electrons from the ammonia, which acts as the Lewis base. 734 Chapter 16

67 Section 16.2 Lewis Brønsted- Lowry Arrhenius hydrogen oxygen Figure 17 This Venn diagram shows how comprehensive some different acid base theories are considered to be. In science, it is unwise to assume that any scientific concept is complete. Whenever scientists assume that they understand a concept, two things usually happen. First, conceptual knowledge tends to remain static for a while because little conflicting evidence exists, or because conflicting evidence (being somewhat discomfiting) is ignored. Second, when enough conflicting evidence accumulates, a change in thinking occurs within the scientific community in which the current theory is drastically revised or entirely replaced. Revolutionary concepts most often tend to be formed in a moment of insight, usually following a long period of incredibly hard work done by a great many people. Case Study Questions 1. Identify a word or phrase that describes the central idea in each of the theories described above. 2. For each theory, describe the weakness that led to its being replaced. 3. What was the first theoretical definition of a base, and who developed it? 4. Write Lewis formulas for each substance in the reaction of boron trifluoride with ammonia, to illustrate the electron pair transfer concept. Section 16.2 Questions 1. Aqueous solutions of nitric acid and nitrous acid of the same concentration are prepared. (a) Predict how their ph values compare. (b) Explain your answer using the Brønsted Lowry concept. 2. Briefly state the five steps involved in predicting the predominant reaction and the approximate position of equilibrium in an acid base reaction system. 3. According to the Brønsted Lowry concept, what determines the position of equilibrium in an acid base reaction? 4. What generalization from the table of Relative Strengths of Aqueous Acids and Bases (Appendix I) can be used to predict the position of an acid base equilibrium? 5. State two examples of conjugate acid base pairs, each involving the hydrogen sulfite ion. 6. Predict, with reasoning, including Brønsted Lowry equations, whether each of the following chemical systems will be acidic, basic, or neutral. (a) aqueous hydrogen bromide (b) aqueous potassium nitrite (c) aqueous ammonia (d) aqueous sodium hydrogen sulfate (e) carbonated beverages (f) limewater (g) vinegar 7. Write an experimental design to test each of the predictions in question Write two experimental designs to rank a group of bases in order of strength. Equilibrium in Acid Base Systems 735

68 9. Ammonia molecules and hydronium ions have remarkably similar shapes and structures (Figure 18). Both have three hydrogen atoms bonded in a pyramidal structure to a small central atom, which has one lone pair of electrons. Oxygen and nitrogen atoms both have very high electronegativities, and they are adjacent on the periodic table. Explain why an ammonia molecule is a good proton attractor, whereas a hydronium ion is an extremely poor proton attractor. Figure 18 Evidence indicates that the ammonia molecule, NH 3, modelled here, has the same pyramidal shape as the hydronium ion, H 3 O. 10. Many household cleaning solutions claim to dissolve away rust stains and hard water lime deposits (Figure 19). These cleaners are simply acidic solutions (as is vinegar) that react with slightly soluble substances such as Fe 2 O 3 (s) and CaCO 3 (s). The word react, however, sounds dangerous to consumers, and is hardly ever used in advertising. These cleaning solutions must be used carefully, and you should always read all of the label instructions, or you may find the cleaner also reacting with the item you are trying to clean, or even with you! These cleaners have a variety of formulations, but most contain hydroxyacetic acid (glycolic acid), CH 2 OHCOOH(aq). A 0.10 mol/l glycolic acid solution is 3.9% ionized at 25 C, compared to 0.10 mol/l acetic acid, which is 1.3% ionized at the same temperature. (a) Write a Brønsted Lowry equation for the reaction of aqueous glycolic acid with carbonate ions (from hard water scum ), and label both conjugate acid base pairs. (b) Explain whether the equilibrium would favour products more, or less, if acetic acid (vinegar) were used to do the same cleaning job. Is it likely that any difference would be significant? State your reasoning. (c) Explain whether glycolic acid would react with a given chemical amount of rust more quickly or more slowly than an equal volume of equally concentrated acetic acid solution. (d) Explain whether glycolic acid would react with a greater, or lesser, chemical amount of rust than an equal volume of equally concentrated acetic acid solution. 11. Since ancient times, people have known that strong heating of natural limestone (calcium carbonate) will produce quicklime (calcium oxide). This very useful substance is both reactive and corrosive with many organic substances because it has a high affinity for water. Adding water to quicklime causes the formation of slaked lime (calcium hydroxide) the common name was derived from the idea that the lime was slaking its thirst. This reaction also produces a very considerable quantity of heat! (a) Write a balanced chemical equation for the decomposition of calcium carbonate to carbon dioxide and calcium hydroxide. (b) Write a balanced chemical equation for the addition of water to calcium oxide to produce calcium hydroxide. (c) Write a Brønsted Lowry equation for the instantaneous and overwhelmingly quantitative (and highly exothermic) reaction of aqueous oxide ions with water. Figure 19 Commercial rust removing solutions must be used with care, but can be a very effective illustration of the old TV slogan, better living, through chemistry. 736 Chapter 16

69 Acid Base Strength and the Equilibrium Law Modifying Arrhenius original concept of acids and bases allowed us to explain the behaviour of many substances in solution that could not be explained by his original theory. Then, problems with varying degrees of properties led to considering reactions of acids and bases as examples of equilibrium, which allowed a much more comprehensive understanding of relative acid or base strengths. Apparent conflicts in the definition of acids and bases then led us to define them by what they do in a reaction (in terms of proton transfer), rather than by what they are (in terms of some typical structure). This process of continually testing (identifying problems) and then expanding and improving concepts is typical of science with the goal always being the formation of a more complete and comprehensive theory. Science assumes that no theory ever formed by the human mind can be perfect, and that any theory, no matter how well supported by evidence, must always be questioned and tested. We also realize that every step moves us closer to a more complete understanding. Testing, and then restricting, revising, and (perhaps) replacing concepts, and then testing again is the constant cycle of the scientific method. Next, we explore the possible value of using the equilibrium law to further increase our understanding of acid base behaviours. The Acid Ionization Constant, K a One important way that chemists communicate the strength of any weak acid is by using the equilibrium constant expression for a Brønsted Lowry reaction equation showing the acid ionizing in (reacting with) water. This expression is another very important special case of equilibrium. The constant is given its own unique name and symbol: the acid ionization constant, K a. See the K a values listed in the Relative Strengths of Aqueous Acids and Bases data table (Appendix I). Note that in this table, the first six acids have K a values given only as very large. These acids are collectively called the strong acids because they all react quantitatively ( 99.9%) with water to form hydronium ions. Because no acid stronger than H 3 O (aq) can exist in aqueous solution, all of these acids are considered equivalent. For each of them, the actual acid species present is H 3 O (aq). All of the other acids listed on the table are considered to be weak acids, and they vary greatly in extent of reaction with water at equilibrium. To empirically rank any weak acid as stronger or weaker than any other, we normally compare the ionization of solutions of the same concentration, to see which is more strongly acidic (has a lower ph). This is how Appendix I was created. By considering the equilibrium established in solutions of weak acid to be a Brønsted Lowry proton transfer, we gain a more complete understanding about what is actually going on among the entities reacting. By using K a equilibrium constants, we can derive and predict quantities that are very useful when working with acid base reactions. Consider a solution of hydrofluoric acid, HF(aq). The equation for the reaction in water (ionization) of this weak acid is HF(aq) H 2 O(l) 0 H 3 O (aq) F (aq) The equilibrium law expression is [H K a 3 O (aq)][f (aq)] [HF(aq)] Note that the concentration of liquid water is omitted from the K a equilibrium law expression. We make an assumption that this value will remain essentially constant for EXTENSION Acid Base Reactions This quick simulation lets you explore the percent reaction when various acids and bases react together. The acid ionization constants are provided for two strong and three weak acids. Learning Tip GO Use of ionization terminology (strong and weak) can create a problem, caused by the common use of the word strong. For an acid, strength refers only to the ease with which a base can remove its proton, and has nothing to do with the quantity present in the solution. Concentration refers to the chemical quantity per unit volume of solution, and has nothing to do with strength. Both the strength and the concentration of an acid affect how rapidly, and to what extent, that acid will react. Equilibrium in Acid Base Systems 737

70 Learning Tip The great advantage of K a values over percent ionization values is that, once determined, K a values are valid over a wide range of acid solution concentrations. A percent ionization value is useful only for one specific concentration of one specific weak acid. aqueous solutions that are not highly concentrated. This assumption is not exactly true, because the water is not a separate liquid phase in an aqueous solution. However, as long as the amount of water solvent is much greater than the amount of acid solute, making this assumption will cause negligible change in (and will not add uncertainty to) the answers for any calculations. All K a values are calculated by making this assumption. For this reason (and for other measurement reasons too involved to explain here), K a values are necessarily somewhat inaccurate, and become less accurate as the acid concentration increases. It is important to know that K a values given in tables are normally given to only two significant digits because they are usually only known to a certainty of about ± 5%. Two calculations involving the K a constant are common for weak acid solutions: calculating a K a value from measured (empirical) amount concentration data using a K a value to predict a concentration of hydronium ions for an aqueous solution where the initial weak acid amount concentration is known Calculating K a from Amount Concentrations SAMPLE problem 16.2 The ph of a 1.00 mol/l solution of acetic acid is carefully measured to be 2.38 at SATP. What is the value of K a for acetic acid? Use the balanced equation to write the equilibrium law expression. CH 3 COOH(aq) H 2 O(l) 0 H 3 O (aq) CH 3 COO (aq) Learning Tip Because the value for the initial acid concentration is precise only to two decimal places, the calculation of the equilibrium concentration of the acid (1.00 mol/l mol/l) rounds to 1.00 mol/l. (See the Precision Rule for Calculations, Appendix F.3.) In other words, the decrease in the initial acid concentration is negligible, in this particular case. A negligible change in concentration is quite often the case for K a calculations for weak acids, but not always. Recall that amount concentrations must be used to calculate any K c value and, by convention, units for the equilibrium constant value are simply ignored. [H K a 3 O (aq)][ch 3 COO (aq)] [CH 3 COOH(aq)] First, find the equilibrium concentration of aqueous hydronium ion from the ph. [H 3 O (aq)] 10 ph mol/l Next, use the hydronium ion concentration and the balanced chemical equation to calculate the concentration of the acetate ion. Since one acetate ion forms for each hydronium ion, the equilibrium concentration of acetate ion must be identical to that of the hydronium ion. The stoichiometric ratio is 1:1. [CH 3 COO (aq)] [H 3 O (aq)] mol/l An ICE table is useful to find the equilibrium concentration of acetic acid molecules. Table 1 ICE Table for CH 3 COOH(aq) H 2 O(l) 0 H 3 O (aq) CH 3 COO (aq) [CH 3 COOH(aq)] [H 3 O (aq)] [CH 3 COO (aq)] Concentration (mol/l) (mol/l) (mol/l) Initial Change Equilibrium ( mol/l)( mol/l) K a (1.00 mol/l) Regardless of numerical size, K a values are usually expressed in scientific notation. The calculated K a for acetic acid is Chapter 16

71 Section 16.3 COMMUNICATION example 1 A student measures the ph of a 0.25 mol/l solution of carbonic acid to be Calculate the K a for carbonic acid from this evidence. Solution H 2 CO 3 (aq) H 2 O(l) 0 H 3 O (aq) HCO 3 (aq) K a [H 3 O (aq)][hco 3 (aq)] [H 2 CO 3 (aq)] At equilibrium, [H 3 O (aq)] 10 ph mol/l [HCO 3 (aq)] [H 3 O (aq)] mol/l [H 2 CO 3 (aq)] ( ) mol/l 0.25 mol/l Table 2 ICE Table for H 2 CO 3 (aq) H 2 O(l) 0 H 3 O (aq) HCO 3 (aq) Amount [H 2 CO 3 (aq)] [H 3 O (aq)] [HCO 3 (aq)] concentration (mol/l) (mol/l) (mol/l) Initial Change Equilibrium ( mol/l)( mol/l) K a (0.25 mol/l) According to the equilibrium law, the K a for carbonic acid is Learning Tip Because a Brønsted-Lowry equation is written to show a single proton transfer, the stoichiometric ratio will always be 1:1:1:1. Thus, the units for K a values will always be mol/l. Because we can make this assumption, units are often not included with K a values in reference tables or in final answer statements. COMMUNICATION example 2 The ph of a mol/l solution of sulfurous acid is measured to be Calculate the K a for sulfurous acid from this evidence. Solution H 2 SO 3 (aq) H 2 O(l) 0 H 3 O (aq) HSO 3 (aq) K a [H 3 O (aq)][hso 3 (aq)] [H 2 SO 3 (aq)] At equilibrium, [H 3 O (aq)] 10 ph mol/l [HSO 3 (aq)] [H 3 O (aq)] mol/l [H 2 SO 3 (aq)] ( ) mol/l mol/l Equilibrium in Acid Base Systems 739

72 Table 3 ICE Table for H 2 SO 3 (aq) H 2 O(l) 0 H 3 O (aq) HSO 3 (aq) Amount [H 2 SO 3 (aq)] [H 3 O (aq)] [HSO 3 (aq)] concentration (mol/l) (mol/l) (mol/l) Initial Change Equilibrium (0.068 mol/l)(0.068 mol/l) K a (0.332 mol/l) According to the equilibrium law, the K a for sulfurous acid is Calculating [H 3 O (aq)] from K a The second type of calculation involving a K a value allows us to predict the acidity of any weak acid solution. We can calculate the concentration of hydronium ions from the initial acid concentration and the K a value as follows. SAMPLE problem 16.3 Predict the [H 3 O (aq)] and ph for a mol/l aqueous solution of methanoic acid. Look up the value of K a for methanoic (formic) acid in the Relative Strengths of Aqueous Acids and Bases data table (Appendix I). The value is given as mol/l. Use the balanced equation to write the equilibrium law expression. HCOOH(aq) H 2 O(l) 0 H 3 O (aq) HCOO (aq) K a Use x to represent the numerical value for the equilibrium amount concentration of aqueous hydronium ions. [H 3 O (aq)] x mol/l An ICE table is useful to keep track of the concentration values. Using the 1:1:1:1 stoichiometric ratio, calculate concentration increases and decreases for the reaction to equilibrium, and note them in the ICE table. Substituting into the equilibrium law expression, ignoring units, [H 3 O (aq)][hcoo (aq)] [HCOOH(aq)] Table 4 ICE Table for HCOOH(aq) H 2 O(l) 0 H 3 O (aq) HCOO (aq) [HCOOH(aq)] [H 3 O (aq)] [HCOO (aq)] Concentration (mol/l) (mol/l) (mol/l) Initial Change x x x Equilibrium (0.200 x)* x x [H 3 O (aq)][hcoo (aq)] [HCOOH(aq)] (x) (x) (0.200 x) 740 Chapter 16

73 Section 16.3 * Note: Technically, solving for x (the amount concentration of hydronium ion) from this b b ac 2 4 expression requires the use of the quadratic formula, x, 2a which is quite tedious, unless you have a calculator that is preprogrammed to solve such formulas. Conveniently, for many (but not all) weak acid solutions, we can use an approximation that greatly simplifies solving for x. If the percent reaction of the weak acid is quite small, we can assume that the initial concentration of the acid decreases by so little that the numerical value of the acid amount concentration will remain effectively unchanged at equilibrium. The problem, of course, is deciding when a percent reaction is small enough to allow this assumption to be used. In this text, we will apply the mathematically simplest (and most convenient) rule of thumb that is commonly used to make this decision. If the initial amount concentration of the acid is numerically at least 1000 times its K a value, then you may assume that the initial and equilibrium acid amount concentrations are numerically equal. This limiting assumption is consistent with the 5% certainty of most K a values, meaning it ignores those amount concentration changes that would change the K a value by less than the uncertainty that already exists. Note that the equation to be solved contains a squared value. x (0.200 x) Multiplying both sides by (0.200 x), collecting terms, and equating to zero gives x x (expressed in the form of a quadratic equation). Notice what happens when we test our present example for the simplifying assumption. [HCOOH(aq)] initial 1111, which is greater than 1000 Ka The test allows us to use the assumption (initial acid concentration equilibrium acid concentration, or [HA(aq)] initial [HA(aq)] equilibrium ). In this specific case, so little acid has reacted with water at equilibrium that the initial acid concentration is decreased by a negligible amount less than the uncertainty of the initial value. Numerically, for this example, we may assume that (0.200 x) 0.200, and we can then simplify the equation to x , which makes solving for x quick and easy DID YOU KNOW? Rule of Thumb The very old English phrase rule of thumb, meaning a rule for any quick way to approximate a quantity without measuring, probably comes directly from use of human thumbs. Since the width of an adult s thumb is approximately one inch, and the length of a thumb is approximately four inches, you can make a fairly good estimate of the length of something (in Imperial units) by simply placing your thumbs along the object, one after the other. What is the length of this textbook in (your) thumb widths? What do equestrians mean when they say a horse stands 16 hands high? A practical rule of thumb for Americans driving their cars into Mexico or Canada is to multiply the posted speed by 6 and drop the last digit, in order to quickly convert the km/h speeds on road signs into a fairly close miles-perhour equivalent. Isolate x and take the square root of both sides of the equation: x so, and [H 3 O (aq)] mol/l or mol/l ph log [H 3 O (aq)] log (0.0060) 2.22 Equilibrium in Acid Base Systems 741

74 DID YOU KNOW? pk a For purposes of easy comparison, the K a values of different acids are sometimes expressed as pk a values. A pk a value is the negative logarithm of the K a. A reported pk a value of 12 would represent an acid with a K a of , a very weak acid. In comparing any two weak acids, the one with the lower pk a value is the stronger of the two acids. As shown in the next Communication Example, when the negligible ionization (percent reaction) assumption holds true, predictions of [H 3 O (aq)] using K a values become simple and straightforward. Note that you can perform the 1000:1 ratio test for the simplifying assumption before starting the problem, which is very convenient. COMMUNICATION example 3 Predict the [H 3 O (aq)] and ph for a mol/l aqueous solution of hydrocyanic acid. Solution Test the simplifying assumption: [HCN(aq)] initial (greater than 1000) Ka The assumption holds, and may be used. K a At equilibrium: Let x [H 3 O (aq)] [CN (aq)] [H 3 O (aq)][cn (aq)] [HCN(aq)] Then [HCN(aq)] (0.500 x) (using the assumption) Table 5 ICE Table for HCN(aq) H 2 O(l) 0 H 3 O (aq) CN (aq) Amount [HCN(aq)] [H 3 O (aq)] [CN (aq)] concentration (mol/l) (mol/l) (mol/l) Initial Change x x x Equilibrium (0.500 x) x x x x so, [H 3 O (aq)] mol/l and ph log [H 3 O (aq)] log ( ) 4.75 According to the equilibrium law, the hydronium ion amount concentration of mol/l hydrocyanic acid is mol/l, and the ph of the solution is Note: No Practice, Section, or Review question in this text will require you to use the quadratic formula to solve acid ionization constant problems. However, you may be asked to state whether a given problem would require use of this formula for solution (in other words, whether the ionization assumption test fails). You should develop the habit of always initially testing the assumption and making a qualifying statement before completing the solution to any K a question where this assumption holds. 742 Chapter 16

75 Section 16.3 Practice 1. (a) Write a theoretical definition for the strength of an acid. (b) State the empirical properties that provide evidence for differing acid strengths. (c) Explain the difference in meaning between strength and concentration, as these terms are used to refer to aqueous acids. (d) Does the stronger of two acids, when dissolved, necessarily make a more acidic aqueous solution? Explain. 2. Refer to Appendix I for required information to make the following predictions. For each case, first decide and state whether a solution would require use of the quadratic formula. For all cases where use of the quadratic formula is not required, communicate the full solution. (a) Predict [H 3 O (aq)] for 0.20 mol/l hydrobromic acid. (b) Predict [H 3 O (aq)] for 0.20 mol/l hydrofluoric acid. (c) Predict [H 3 O (aq)] for 0.20 mol/l ethanoic acid. (d) Predict the ph of 2.3 mmol/l nitric acid. (e) Predict the ph of 2.3 mmol/l nitrous acid. (f) Predict the ph of 2.3 mmol/l hydrosulfuric acid. 3. For all of the cases (a f) in question 2 where a prediction could be calculated, rank the acid solutions in order of decreasing acidity. 4. For all of the cases (a f) in question 2, rank the acids in order of decreasing acid strength. Which two acids have essentially the same strength? 5. The hydronium ion concentration in mol/l propanoic acid is determined (from a ph measurement) to be mol/l. (a) Calculate the percent reaction (ionization) of this particular weak acid solution. (b) Calculate K a for aqueous propanoic acid. (c) Is this K a value constant for propanoic acid? Explain. 6. A 0.10 mol/l solution of lactic acid, a weak acid found in milk, has a measured ph of The chemical name for this very common organic compound is 2-hydroxypropanoic acid. (a) Find the percent reaction of this lactic acid solution. (b) Calculate the K a value for aqueous lactic acid. (c) Compare the K a values for 2-hydroxypropanoic acid and for propanoic acid. What effect does adding an OH group to the central carbon atom of this molecule have on the ability of the COOH group to attract a proton? 7. Unlike all other aqueous hydrogen halides, hydrofluoric acid is not a strong acid. It does, however, have a special chemical property: it reacts with glass. This property is used to etch frosted patterns on glassware (Figure 1). (a) Write the K a expression for hydrofluoric acid. (b) Calculate the hydronium and fluoride ion amount concentrations, the ph, and the percent reaction in a commercial 2.0 mol/l HF(aq) solution at 25 C. 8. Phosphoric acid is used in rust-remover solutions. The aqueous acid is available for purchase by high schools in concentrations of about 15 mol/l. (a) Predict the hydronium ion amount concentration, the ph, and the percent reaction of a 10 mol/l solution of phosphoric acid. (b) Suggest a reason (having to do with K a values and solution concentrations) why you might well suspect that these values could be inaccurate. 9. Ascorbic acid is the chemical name for Vitamin C (Figure 2). A student prepares a mol/l aqueous solution of ascorbic acid, tests its ph, and reads a value of 2.40 from the ph meter. (a) From the student s evidence, calculate the K a for ascorbic acid. (b) Compare your result to the value listed in the table of Relative Strengths of Aqueous Acids and Bases (Appendix I). The K a value for ascorbic acid increases with increasing temperature. State whether the student s aqueous solution was likely warmer or colder than standard temperature (25 C) when the ph was tested. Figure 1 Hydrofluoric acid is a weak acid but it etches glass. The etching is not due to the presence of hydronium ions, because, as you know, even the strongest acids are routinely stored in glass containers. Figure 2 Vitamin C (ascorbic acid) is present in many fresh foods, particularly citrus fruits. It is essential to the body to promote healing of injuries and to fight infection. In the winter of 1535, the men of Jacques Cartier s expedition were suffering from scurvy (vitamin C deficiency) in their fort at Stadacona (now Montreal). Cartier lost 25 men before learning of a simple cure from the Iroquois. The Aboriginal people prepared a tea made from white cedar needles rich in ascorbic acid that cured sufferers within days. Equilibrium in Acid Base Systems 743

76 DID YOU KNOW? Sailors and Scurvy Vitamin C deficiency (from a dietary lack of fresh fruits and vegetables) causes scurvy, a disease where the body cannot fight infections, and cuts, sores, and bruises do not heal. Until the last half of the 18th century, this disease was a terrible problem for sailors, and a major cause of death on long voyages. For example, on his famous voyage of 1498, Vasco da Gama lost 100 of 160 crewmen to scurvy. The diet of European sailors was primarily salted meat and hard biscuits items selected for their resistance to spoilage. In 1747, James Lind, a Scottish physician, wrote that feeding citrus fruit to scurvy victims effected an amazingly rapid cure. Captain Cook tried this remedy on his crew during his famous voyages of exploration in the 1770s, and reported that he was astounded to have lost only one man on a three-year voyage! By 1795, the British Royal Navy (after decades of deliberation) formally adopted the practice of providing lemon juice (incorrectly referred to as lime juice) for all hands. Scurvy in the fleet was wiped out, and British sailors have been called limeys ever since. Note that this story illustrates another case of technology leading science: The cure for scurvy was known long before the cause of the disease was explained. Interestingly, sailors on the long sea voyages undertaken by the Chinese during the Ming Dynasty ( ) had no problems with scurvy because their traditional diet included fresh germinated soya beans, the shoots of which are naturally rich in Vitamin C. Base Strength and the Ionization Constant, K b The Brønsted Lowry concept specifies that the strongest base possible in aqueous solution must be hydroxide ions. So, hydroxide ion is considered to be the strong base somewhat parallel to the strong acids. A difference is that many ionic substances contain hydroxide ions initially, before being dissolved; whereas hydronium ions are always formed by the reaction of some entity with water, after a substance dissolves. Recall that for solutions of all ionic hydroxides, such as NaOH(aq) or Ca(OH) 2 (aq), we assume that the compound dissociates completely upon dissolving. Therefore, finding the hydroxide ion concentration does not involve any reaction with water or any reaction equilibrium. Rather, in these cases, we find the hydroxide ion concentration directly from a dissociation equation, as shown in the next Communication Example. COMMUNICATION example 4 Find the hydroxide ion concentration of a mol/l solution of barium hydroxide. Solution Ba(OH) 2 (aq) Ba 2 (aq) 2OH (aq) (complete dissociation) [OH (aq)] [Ba(OH) 2 (aq)] mol/l mol/l According to the stoichiometric ratio, the hydroxide ion concentration is 0.13 mol/l. Other entities that act as bases are collectively called weak bases, because empirical evidence indicates that they attract protons less than hydroxide ions do. Turn to the table of Relative Strengths of Aqueous Acids and Bases (Appendix I), and observe that, to the right of every acid entity formula listed, there is a formula for an entity that is identical to the acid s formula, except that one proton (represented as H ) is missing. As you have learned, such an entity is called the conjugate base of the acid. While the list of these entities (reading upward) can be taken to be a list of bases and their (decreasing) relative strengths, note that there is no corresponding value given to show the extent of reaction of these entities with water. This convention is common in chemistry information about bases in solution must be derived from a table of acid values. Bases vary in strength much as acids do, and the system used to identify, classify, and rank the strength of bases is quite similar to the one you have just learned for acids. Note that the table in Appendix I shows that the weaker an acid is, the stronger its conjugate base (and vice versa). This observation is common sense. If it is very easy to remove the proton from an acid, then the entity remaining (the conjugate base) must not attract protons very well. As you know, weak bases react only partially (usually much less than 50%) with water in aqueous solution to produce hydroxide ions. We can communicate the strength of any weak base using the equilibrium constant expression for its reaction with water in dilute aqueous solution. This equilibrium constant for weak bases is another special case, where the K c constant is called the base ionization constant, K b. Consider a solution of sodium citrate, Na 3 C 6 H 5 O 7 (aq). We assume that in aqueous solution, any ionic compound dissociates completely into its component ions. In this case, 744 Chapter 16

77 Section 16.3 the ions are aqueous sodium and citrate ions: Na (aq) and C 6 H 5 O 7 3 (aq). Sodium ions are spectators, with no apparent acidic or basic properties. The citrate ions, however, are found on the Relative Strengths of Aqueous Acids and Bases data table (Appendix I) in the conjugate bases column. So the citrate ion is a weak Brønsted Lowry base and reacts with water to form a basic solution at equilibrium. The reaction equilibrium equation for aqueous sodium citrate is C 6 H 5 O 7 3 (aq) H 2 O(l) 0 HC 6 H 5 O 7 2 (aq) OH (aq) The equilibrium law expression is K b [HC 6 H 5 O 7 2 (aq)][oh (aq)] [C 6 H 5 O 7 3 (aq)] Just as with K a values, there are two kinds of weak base solution calculations that involve the K b constant: calculating a K b value from empirical (measured) amount concentration data using a K b value to predict an amount concentration of hydroxide ions for an aqueous solution where the initial weak base concentration is known Calculating K b from Amount Concentrations Calculating K b for a weak base uses essentially the same method as calculating K a for a weak acid. Often, an extra (simple) calculation step must be included because measured ph values may need to be converted to find the equilibrium hydroxide ion concentration, as shown in the following Communication Example. COMMUNICATION example 5 A student measures the ph of a mol/l solution of aqueous ammonia and finds it to be Calculate the K b for ammonia from this evidence. Show the establishment of the reaction equilibrium using an ICE table. Learning Tip The concentration of liquid water is omitted from K b equilibrium law expressions for the same reasons that it is omitted from K a expressions. As with K a values, K b values are also only certain to about ±5%. For weak base K b values, we also assume that aqueous solutions are not highly concentrated, just as we do for weak acids. Just as with K a values, K b value units are ignored, by convention. Solution NH 3 (aq) H 2 O(l) 0 NH 4 (aq) OH (aq) K b = [NH 4 (aq)][oh (aq)] [NH 3 (aq)] At equilibrium: [H 3 O (aq)] 10 ph mol/l K w [H 3 O (aq)][oh (aq)] [OH (aq)] K w [H3 O (aq)] mol/l mol/l Equilibrium in Acid Base Systems 745

78 [NH 4 (aq)] [OH (aq)] mol/l [NH 3 (aq)] ( ) mol/l mol/l Table 6 ICE Table for NH 3 (aq) H 2 O(l) 0 NH 4 (aq) OH (aq) [NH 3 (aq)] [NH 4 (aq)] [OH (aq)] Concentration (mol/l) (mol/l) (mol/l) Initial Change Equilibrium [ mol/l][ mol/l] K b = [0.248 mol/l] From this evidence, and according to the equilibrium law, the K b for aqueous ammonia is Figure 3 Dyes with molecular structures based on aniline make possible a variety of bright, stable colours for fabrics and leathers. Learning Tip The use of chemical (IUPAC) and common names can be confusing if you have not memorized a few examples. In this textbook, the normal practice is to give the chemical name, followed by a common name in parentheses. A few substances are so widely used, however, that the common name is ubiquitous (used everywhere, even by chemists). Baking soda (sodium bicarbonate, sodium hydrogen carbonate) and vinegar (acetic acid, ethanoic acid) are two examples of such chemicals; the common name is often given, rather than the IUPAC name. You are expected to be familiar with such examples, and others listed in Appendix J. Practice 10. List some empirical properties that would be useful when distinguishing strong bases from weak bases. 11. For each of the following weak bases, write the chemical equilibrium equation and the equilibrium law expression for K b. (a) CN (aq) (b) SO 2 4 (aq) 12. The hydroxide ion concentration in a mol/l solution of sodium propanoate, NaC 2 H 5 COO(aq), is found to be mol/l. Calculate the base ionization constant for the propanoate ion. 13. Aniline, C 6 H 5 NH 2, is used to make a wide variety of drugs and dyes (Figure 3). It has the structure of an ammonia molecule, with a phenyl group substituted for one hydrogen, and, like ammonia, acts as a weak base. If the ph of a 0.10 mol/l aniline solution was found to be 8.81, what is its K b? Calculating [OH (aq)] from K b The second type of calculation involving a K b value is the prediction of the basicity of any weak base solution. We can calculate the concentration of hydroxide ions from the initial weak base concentration and the K b value. There is an automatic problem, however, because tables of relative acid base strengths do not normally list K b values. We use the automatic relationship between conjugate acid base pairs to deal with this problem. The K a K b Relationship for Conjugate Acid Base Pairs To develop the next useful concept, we use the common weak acid, acetic acid, CH 3 COOH(aq), and its conjugate base, the acetate ion, CH 3 COO (aq). Of course, it would be just as correct to state that we will use the weak base, acetate ion, and its conjugate acid, acetic acid. 746 Chapter 16

79 Section 16.3 The equilibrium reaction and law expression for aqueous acetic acid are CH 3 COOH(aq) H 2 O(l) 0 H 3 O (aq) CH 3 COO (aq) [H K a 3 O (aq)][ch 3 COO (aq)] [CH 3 COOH(aq)] For the weak base, aqueous acetate ion, CH 3 COO (aq) H 2 O(l) 0 CH 3 COOH(aq) OH (aq) [CH K b 3 COOH(aq)][OH (aq)] [CH 3 COO (aq)] Notice what happens when we multiply these equilibrium constant expressions: K a K b [H 3 O (aq)][ch 3 COO (aq)] [CH 3 COOH(aq)] [H 3 O (aq)][oh (aq)] K w [CH 3 COOH(aq)][OH (aq)] [CH 3 COO (aq)] (Does this expression look familiar?) So, for any conjugate acid base pair, K w K a K b and K b K. Ka This last equation is particularly useful because now, to find a K b value for any base listed on the Relative Strengths of Aqueous Acids and Bases table (Appendix I), you need only identify its conjugate acid, and then divide K w, , by the K a value given for that conjugate acid. SAMPLE problem 16.4 Solid sodium benzoate forms a basic solution. Determine K b for the weak base present. NaC 6 H 5 COO(s) Na (aq) C 6 H 5 COO (aq) (complete dissociation) First identify, from the Relative Strengths of Aqueous Acids and Bases table, and the entities present in solution, which entity is reacting as a base. The benzoate ion must be the weak base entity. Its equilibrium reaction with water is C 6 H 5 COO (aq) H 2 O(l) 0 C 6 H 5 COOH(aq) OH (aq) From the equilibrium equation, the conjugate acid of the benzoate ion is identified as benzoic acid. From the acid base table, K a for C 6 H 5 COOH(aq) Using the K w relationship for conjugate acid base pairs, find K b for C 6 H 5 COO (aq). w DID YOU KNOW? Why Acidity? It should be obvious by now that ph values are used much more than poh values, and K a constants are listed in tables rather than K b constants. You might wonder why we choose to emphasize the hydronium ion properties of aqueous solutions after all, hydroxide ions play exactly the opposite (and an equally important) role. But historically, chemists have always been much more concerned with acidic properties than with basic properties. Acids dissolve (react with) many metals to form common ionic compounds and hydrogen. Hydrogen gas was a fascinating product to early chemists, both because it is lighter than air and because it is violently explosive. The stronger acids tend to have painfully sharp odours and flavours, and react destructively with human tissue all properties that make acids quite noticeable. Even when it became clearly understood that acids and bases were opposite aspects of the same concept, it seemed easier to choose acid properties as the initial basis for studying acid base relationships. We talk about the acidity of something routinely; it is a common term, found in any pocket dictionary. What is the equivalent term for how basic something is? There is an accepted word for it, but you ll need a really good (very big) dictionary to find it. Or a good Chemistry text, of course... K b K w Ka According to the K w relationship, the benzoate ion has a K b value of Equilibrium in Acid Base Systems 747

80 COMMUNICATION example 6 Calculate K b for the weak base aqueous ammonia, commonly used in commercial window cleaning solutions. Write the equation for the equilibrium. Solution NH 3 (aq) H 2 O(l) 0 NH 4 (aq) OH (aq) For NH 3 (aq), NH 4 (aq) is the conjugate acid: K a K b K w Ka According to the table of Relative Strengths of Aqueous Acids and Bases, and the K a K b relationship, the K b value for aqueous ammonia is Now we can easily predict how basic a solution of known concentration will be from its K b value. Predicting basicity always involves calculating the hydroxide ion concentration, and may also involve calculating a value for poh, ph, and/or percent reaction. The first step may be determining the K b value by using a table of K a values. The form of the calculation then follows the same format as the equivalent type of question for weak acids, with a similar assumption made, and the same restrictions. Assume that the initial weak base aqueous concentration decreases so little that it is numerically unchanged at equilibrium, but only if the initial amount concentration of the weak base is at least 1000 times greater than its K b value. For questions in this text, you are not required to do calculations using the quadratic formula. The Sample Problem that follows shows the format for calculations of this type, and also an example of each conversion that may be required. Recall, from Section 16.2, that, in aqueous solution, the amphiprotic ion HCO 3 (aq) reacts with water as a base to a very much greater extent than it does as an acid. You may, therefore, assume that the solution is basic because the ion reacts as a base, and ignore its negligible reaction extent as an acid. SAMPLE problem 16.5 Find the hydroxide ion amount concentration, poh, ph, and percent reaction (ionization) of a 1.20 mol/l solution of baking soda. The compound is sodium hydrogen carbonate, NaHCO 3 (s). The weak base is the HCO 3 (aq) ion, produced by dissolving NaHCO 3 (s) in water. For HCO 3 (aq), H 2 CO 3 (aq) is the conjugate acid: K a Find the K b value using the K a K b K w relationship. K b K w Ka Chapter 16

81 Section 16.3 Write the equation for the reaction and the equilibrium expression for K b for hydrogen carbonate ion. HCO 3 (aq) H 2 O(l) 0 H 2 CO 3 (aq) OH (aq) [H K b 2 CO 3 (aq)][oh (aq)] [HCO 3 (aq)] At equilibrium, let x be the numerical value of [OH (aq)] and also [H 2 CO 3 (aq)]. Table 7 ICE Table for HCO 3 (aq) H 2 O(l) 0 H 2 CO 3 (aq) OH (aq) Amount [HCO 3 (aq)] [H 2 CO 3 (aq)] [OH (aq)] concentration (mol/l) (mol/l) (mol/l) Initial Change x x x Equilibrium (1.20 x) x x Then [HCO 3 (aq)] (1.20 x) Test the assumption that (1.20 x) is numerically equal to [HCO 3 (aq)] 1.20 initial (much greater than 1000) Kb The assumption holds, so substitute into the K b expression and solve for x x x [OH (aq)] mol/l Use [OH (aq)] to find the value of poh. poh log [OH (aq)] log ( ) 3.80 Use the poh value to find ph. ph poh Percent reaction is the percent ratio of the equilibrium concentration of produced hydroxide ions to the initial concentration of hydrogen carbonate ions. DID YOU KNOW? Visualizing Reaction Extents From any percent reaction value, you can get a better idea of the reaction extent at equilibrium by converting mentally to a wholenumber ratio. If bicarbonate ions react with water 0.013% at equilibrium, the ratio is 0.013:100, or 13: This means that at any given instant, about 13 of every initial bicarbonate ions have changed to carbonic acid molecules, but of them have not. The few ions that have changed are responsible for all the basicity of the solution! If you really want to test your powers of imagination, try to visualize that for every bicarbonate ions in this aqueous solution, there are approximately water molecules. Now think about all of these entities constantly moving at speeds averaging about 1500 km/h, and then imagine every one of them colliding with another molecule or ion at a rate of approximately times every second [OH (aq)] percent reaction equilibrium 100% [HCO3 (aq)] initial mol/l 100% 1.20 mol/l 0.013% A 1.20 mol/l sodium hydrogen carbonate solution has a hydroxide ion concentration of mol/l, a poh of 3.80, a ph of 10.20, and a percent reaction of 0.013%. The Effect of Amphoteric Entities Finally, we deal with the problem of amphoteric entities. If an entity can react as either a Brønsted Lowry acid or base, how do you know which will be the predominant reaction in its aqueous solution? You can determine the answer with one simple calculation. Equilibrium in Acid Base Systems 749

82 Read the K a value for the entity from the table of Relative Strengths of Aqueous Acids and Bases. Calculate the K b value for the entity from K w and the K a value (from the table) of its conjugate acid, as shown in Sample Problem The higher of the K values tells you which reaction predominates, and, therefore, whether the entity reacts as an acid or as a base in aqueous solution. COMMUNICATION example 7 Which reaction predominates when NaHSO 3 (s) is dissolved in water to produce HSO 3 (aq) solution? Figure 4 Codeine (methylmorphine, C 18 H 21 NO 3 ) is available by prescription as an ingredient in some analgesic (pain relief) medicines. This narcotic alkaloid was initially extracted from opium. Other alkaloids include nicotine, quinine, cocaine, strychnine, and caffeine. All are common plantderived organic compounds based on nitrogen-containing ring structures. Solution K a The conjugate acid is H 2 SO 3 (aq), with K a So, for HSO 3 (aq), K K b w Ka The K a value for HSO 3 (aq) far exceeds its calculated K b value. According to the K w relation and the equilibrium law, an aqueous solution of this substance will be acidic because a hydrogen sulfite ion will react predominately as a Brønsted Lowry acid. Section 16.3 Questions 1. Codeine, an ingredient in these migraine pills (Figure 4), has a K b of Calculate the ph of a mol/l codeine solution (use Cod as a chemical shorthand symbol for this complex weak base entity.). 2. What is the ph of a 0.18 mol/l cyanide ion solution? 3. Acetylsalicylic acid (ASA) is a painkiller used in many headache tablets. This drug forms an acidic solution that attacks the digestive system lining. The Merck Index lists its K a at 25 C to be Explain the difficulty in calculating the ph of a saturated mol/l solution of acetylsalicylic acid, C 6 H 4 COOCH 3 COOH(aq). 4. Boric acid is used for weatherproofing wood and fireproofing fabrics. Very dilute aqueous boric acid is used in eyewash solution as a preservative. Predict the ph of a 0.50 mol/l solution of boric acid. 5. Salicylic acid (2-hydroxybenzoic acid, C 6 H 5 OHCOOH(s)) is an active ingredient of medications, such as Clearasil, that are used to treat acne. Since the K a for this acid was not listed in any convenient references, a student tried to determine the value experimentally. She kept adding water slowly to a 1.00 g sample, with constant stirring, until the crystals all dissolved. The volume of solution was 460 ml. She found that the ph of this solution of salicylic acid was 2.40 at 25 C. Calculate the ionization constant for this acid. 6. Sodium ascorbate is the sodium salt of ascorbic acid and is used as an antioxidant in food products. The ph of a 0.15 mol/l solution of the ascorbate ion, HC 6 H 6 O 6 (aq), is Calculate the K b of the ascorbate ion. 7. Sodium hypochlorite is a strong oxidizing agent that is a fire hazard when in contact with organic materials. Solutions of sodium hypochlorite are used as bleach and disinfectant. Determine the hydroxide ion amount concentration of a sodium hypochlorite solution sold as household bleach. The bleach bottle label reads 5.25% (by mass) when packed. 8. Write the equilibrium law expression for acetic acid reacting (ionizing) in water. Use Le Châtelier s principle to explain, in terms of equilibrium shift, why dilute acetic acid solutions have a higher percent reaction than more concentrated solutions. Consider that diluting a solution containing reacting aqueous entities is very similar to increasing the volume of a container of reacting gaseous entities. 9. Predict and write the predominant Brønsted Lowry reaction in aqueous solutions of the following substances: (a) K 2 HPO 4 (s) (b) NaH 2 PO 4 (s) (c) Na 2 HC 6 H 5 O 7 (s) 10. Calculate the ph of mol/l solutions of each of the reagents in question Chapter 16

83 Interpreting ph Curves For many acid base reactions, the appearance of the products resembles that of the reactants, so we cannot directly observe the progress of a reaction. Also, acids cannot easily be distinguished from bases except by measuring ph. As you first learned in Chapter 8, a graph showing the continuous change of ph during an acid base titration, continued until the titrant is in great excess, is called a ph curve (titration curve) for the reaction. The ph values and changes provide important information about the nature of acids and bases, the properties of conjugate acid base pairs and indicators, and the stoichiometric relationships in acid base reactions. Buffering Regions, Endpoints, and Indicators The ph curves for acid base reactions have characteristic shapes. You have learned that a common reason for plotting a ph curve is to determine the value of the solution s ph when an acid base reaction reaches its equivalence point. We can then use this information to select an appropriate indicator one that will produce an easily seen endpoint when a solution reaches this ph value. Then we can use the same acid base reaction, done as a titration to that indicator s endpoint, for titration analysis. In all titration analyses, the critical measured quantity is the titrant volume required to reach the titration endpoint. For accurate analysis, this value needs to be as close as possible to the theoretically exact value needed to reach the reaction equivalence point. Analysis titrations, like all empirical work, necessarily involve some uncertainty. We call any variation between endpoint values and equivalence point values the titration error of an experiment. Good technique and careful application of knowledge can ensure that this variation is as small as possible preferably negligible. Remember: Endpoint refers to that point in a titration analysis where the addition of titrant is stopped. The endpoint is defined (empirically) by the observed colour change of an indicator. Equivalence point refers to that point in any chemical reaction where chemically equivalent amounts of the reactants have been combined. The equivalence point is defined (theoretically) by the stoichiometric ratio from the reaction equation. Endpoints are easily detectable because ph changes a great deal, and very abruptly, as the reaction solution changes (at the equivalence point) from a tiny excess of acid to a tiny excess of base (or vice versa). But this effect begs the question: Why, for most of a titration, does the ph hardly change at all? To better understand the information that a ph curve provides, we will use a familiar reaction about which we already know a great deal. The ph curve for the strong acid strong base reaction (Figure 2) has three regions of interest. In the course of this excess titration, the ph first changes very slowly, then very rapidly, and finally very slowly again, as titrant is steadily added. Notice that the addition of acid titrant to the base sample initially has very little effect on the ph of the solution in the flask. In fact, the high ph has not changed much even after enough acid has been added to react with 90% of the original amount of the base sample. This nearly level region on a ph curve identifies a buffering region. Buffering is the property of some solutions (often called buffer solutions) of resisting (counteracting) any significant change in ph when an acid or a base is added. In this particular case, buffering occurs because any strong acid added immediately reacts with excess hydroxide ions. Consider that before any acid is added, the sample solution is primarily water molecules and hydroxide ions Learning Tip When you began to study acids and bases, the term neutralize was understood to refer to an acid base reaction that had the same final ph as neutral pure water: a ph of 7. Later, you learned that the ph at the equivalence point for an acid base reaction is almost never 7, except for the strong acid strong base reaction of hydronium ions with hydroxide ions. As commonly used, the word neutralize is taken to have a more general meaning: that one substance can completely or partially lessen the acidic or basic properties of another substance. When we say sodium carbonate can be used to neutralize spilled nitric acid (Figure 1), we mean that adding the base brings the ph value up much closer to neutral. Figure 1 Soda ash (sodium carbonate) is blown onto nitric acid that has spilled from some railway tank cars, to neutralize the highly reactive and dangerous acid. Equilibrium in Acid Base Systems 751

84 ml of 0.48 mol/l NaOH(aq) Titrated with 0.50 mol/l HCl(aq) buffering region ph 8 6 the curve inflection point represents the reaction equivalence point Learning Tip Recall from Unit 4 that the inflection point of any plotted curve is the point where the curvature of the line changes direction. For instance, on the strong acid strong base ph curve in Figure 2, the line curves downward until about 24 ml of acid has been added. At some observable point, the line curvature changes to upward. For this specific titration, the change in ph is so large and so rapid that the ph plot becomes a nearly vertical line. In such a case the equivalence point can also be identified as the midpoint of the steep drop in the plotted line. Also recall that, for this particular reaction (only), we already know (from memory) that the equivalence point ph must be 7.0 (at SATP), without actually needing to plot a titration curve: The reaction of hydroxide and hydronium ions is highly quantitative, and the only product is water Volume of HCl (ml) Figure 2 This ph curve for the continuous and excess addition of 0.50 mol/l HCl(aq) to a 25 ml sample of 0.48 mol/l NaOH(aq) helps chemists to understand the nature of acid base reactions. As long as a large amount of hydroxide ions is present, any added acid is immediately converted to water, producing a solution that still consists primarily of water molecules and hydroxide ions. The hydroxide ion concentration decreases a little, but that affects the solution ph value only very slightly. This ph levelling effect finally fails near the equivalence point when the hydroxide ions in the solution become almost completely consumed. But, until then, the solution maintains a (nearly) constant ph. This buffering property turns out to be critically important for a great number of reactions in applied chemistry and biology. Note that once excess acid has been added, the solution consists primarily of water molecules and hydronium ions. Again, the ph remains stable because adding more hydronium ions now does not change the nature of the solution; it only increases the hydronium ion concentration slightly. Another buffering region is established this time at a low ph. You will learn more about the causes of buffering regions and the nature of buffer solutions later in this section. Following the first buffering region there is a very rapid change in ph for a very small additional volume of the titrant. The inflection point on the ph curve represents the equivalence point of this reaction: At this point, we know from theory that the ph must be 7. We also know from experience (Chapter 8) that this reaction can be used for titration analysis, if an indicator is selected that changes colour in solution at a ph very close to 7. As shown in Figure 3, bromothymol blue indicator has a colour change ph range with a middle value very close to (just slightly below) ph 7, which makes it an appropriate indicator for this particular reaction. The values below show that the (theoretical) titrant volume required for complete reaction, as predicted by calculation, should be 24 ml. Figure 3 shows that if the colour change of bromothymol blue to a green intermediate colour is the endpoint, then the titrant volume actually measured at the endpoint (24 ml) is negligibly different from the theoretical predicted value. So, this titration analysis gives very accurate results. OH (aq) H 3 O (aq) 2 H 2 O(l) 25 ml 0.48 mol/l 24 ml 0.50 mol/l 12 mmol 12 mmol buffering region 752 Chapter 16

85 Section 16.4 ph ml of 0.48 mol/l NaOH(aq) Titrated with 0.50 mol/l HCl(aq) alizarin yellow bromothymol blue orange IV Volume of HCl (ml) Figure 3 Alizarin yellow is not a suitable indicator because it will change colour long before the equivalence point of this strong acid strong base reaction, which theoretically has a ph of exactly 7. Orange IV is also unsuitable; its colour change would occur too late. The ph at the middle of the colour change range for bromothymol blue is 6.8, which very closely matches the equivalence point ph; so, a titration analysis endpoint for this reaction, as indicated by bromothymol blue, should give accurate results. Acid Base Indicator Equilibrium Acid base indicators may be better understood and explained by using Brønsted Lowry definitions. Any acid base indicator is really two entities for which we use the same name: a Brønsted Lowry conjugate acid base pair. At least one (most often both) of the entities is visibly coloured, so you can tell simply by looking when it forms or is consumed in a reaction. Phenolphthalein is a common example (Figure 4), where the conjugate base form is bright red and the conjugate acid form is colourless. Typically, common indicators such as methyl red (Figure 5) have quite complex molecular formulas, so we use a (very simplified) shorthand to identify them, for convenience. For example, we symbolize the (invisible) acid form of phenolphthalein as HPh, and the (bright red) base form as Ph. When an indicator equilibrium is shifted to the point where equal concentrations of both forms exist, an intermediate colour may be observed. For example, equal concentrations of the blue and yellow forms of bromothymol blue (at a ph of 6.8) appear green to the human eye. The explanation of the behaviour of acid base indicators depends, in part, on both the Brønsted Lowry concept and the equilibrium concept. An indicator is a conjugate weak acid weak base pair formed when an indicator dye dissolves in water. Using HIn(aq) to Figure 4 Sodium hydroxide solution has been added to hydrochloric acid containing phenolphthalein indicator, which is colourless in acids and red in bases. The red colour, indicating the temporary presence of some unreacted sodium hydroxide, is rapidly disappearing as the flask contents swirl and mix. Figure 5 A few common acid base indicators are shown here. Each indicator has its own ph range over which it changes colour from the acid form (HIn) at the lower ph value to the base form (In ) at the higher ph value. Material Safety Data Sheets are available for these chemicals. Equilibrium in Acid Base Systems 753

86 represent the acid form and In (aq) to represent the base form of any indicator, we can write the following equilibrium equation. (Litmus colours are given below the equation as an example.) conjugate pair Learning Tip Typically, acid base indicators such as methyl red (Figure 6) are large molecules with quite complex formulas. Smaller molecules and ions are less likely to interact with light waves in the visible spectrum. The shorthand symbolism we use to represent indicators is partly for convenience, but also partly to reinforce the concept that indicators are conjugate acid base pairs. Consider the actual formulas for the methyl red entities, which we represent as HMr for the acid form, and Mr for the base form. This simplified symbolism makes the proton transfer, which causes the colour change, more obvious. HIn(aq) H 2 O(l) 0 In (aq) H 3 O (aq) acid base base acid red (litmus colour) blue According to Le Châtelier s principle, an increase in the hydronium ion concentration will shift the equilibrium to the left. Then more indicator will change to the colour of the acid form (HIn(aq)). This change happens, for example, when litmus is added to an acidic solution. Similarly, in basic solutions the hydroxide ions remove hydronium ions by reacting with them to form water, with the result that the equilibrium shifts to the right. Then the base colour of the indicator (In ) predominates. Since different indicators have different acid strengths, the acidity or ph of the solution at which an indicator changes colour varies (Figures 5 and 6). These ph values have been measured and are reported in the table of acid base indicators on the inside back cover of this book. HMr Mr H 3 C H 3 C H 3 C H 3 C N N N N COOH + H 2 O ph < 4.8 N N COO + H 3 O + ph > 6.0 Figure 6 The visible colour of methyl red indicator depends on the equilibrium proportions of its two coloured forms at the ph of the solution in which it is placed. Methyl red exists predominantly in its red (acid) form at ph values less than 4.8, and in its yellow (base) form at ph values greater than 6.0. Between ph values of 4.8 and 6.0, intermediate orange colours occur, as both forms of the indicator are present in detectable quantities. Practice 1. The shape of a ph curve is interpreted to describe the change of properties throughout the course of an acid base reaction. (a) In terms of curve shape, describe the characteristics of a buffering region. (b) In terms of ph change and titrant volume, explain what a buffering region represents. (c) In terms of curve shape, describe the characteristics near an equivalence point. (d) In terms of ph change and titrant volume, explain what an equivalence point represents. Use this information to answer questions 2 to 4. According to the Brønsted Lowry concept, acid base indicators are simply coloured conjugate acid base pairs. As for all other weak acids, the conjugate acid forms of these indicators must differ in strength for different entities. 2. For each indicator following, write a Brønsted Lowry equilibrium equation for the reaction of the acid form of the indicator with water, to form a hydronium ion and the 754 Chapter 16

87 Section 16.4 base form of the indicator. Use the same indicator symbolism as used in the table of Acid Base Indicators on the inside back cover of this book. Identify both conjugate acid base pairs for each equation, and label all entities acid or base. (a) methyl orange (b) indigo carmine (c) thymol blue (the entity with two H s in the symbolized formula) (d) thymol blue (the entity with one H in the symbolized formula) (e) bromothymol blue 3. Which indicator entity, in your answers to question 2, can behave as both an acid and a base? What chemical term is used to describe this property of an entity? 4. (a) Refer to the table of Acid Base Indicators to determine which of the five conjugate acid forms for the indicators in question 2 is the strongest acid. (Note the colour change ph ranges.) Then rank the other four indicators beneath it in order of decreasing strength, as in a relative strengths of acids table. (b) Suppose that five identical samples of an aqueous hydroxide ion solution each have four drops of a different indicator (from question 2) added. If each sample is now titrated with the same HCl(aq) titrant, in which indicator/sample combination will the least amount of titrant cause a colour change? Polyprotic Entities and Sequential Reactions Expanding on what you learned in Chapter 8, and adding the Brønsted Lowry concept, we now examine some acids and bases that can lose (or gain) more than one proton. Polyprotic acids can lose more than one proton, and polyprotic bases can gain more than one proton, in Brønsted Lowry transfers. If more than one proton transfer actually occurs in the course of a titration, chemists believe the process occurs as a series of single-proton transfer reactions. Typical ph curves for reactions of diprotic or triprotic acids and bases differ from those of monoprotic acids and bases and can provide useful information about the reactions going on. We observe (Figure 7) that the ph curve for the addition of HCl(aq) to Na 2 CO 3 (aq) shows two equivalence points two significant changes in ph. Chemists interpret such curves as indicating how many quantitative reactions have occurred, sequentially, for that particular acid base titration. Here, for example, two successive quantitative reactions have occurred. The two equivalence points evident in Figure 7 can be explained by two different proton transfer equations ml of 0.50 mol/l Na 2 CO 3 (aq) Titrated with 0.50 mol/l HCl(aq) first reaction equivalence point ph second reaction equivalence point methyl orange Volume of HCl (ml) 75 Figure 7 A ph curve for the addition of 0.50 mol/l HCl(aq) to a 25.0 ml sample of 0.50 mol/l Na 2 CO 3 (aq) can be used to select an appropriate indicator for this reaction done as a titration analysis. The colour change of methyl orange (from ph 4.4 to ph 3.2) means it will show an endpoint that corresponds closely to the second reaction equivalence point. Equilibrium in Acid Base Systems 755

88 First, protons transfer from hydronium ions to carbonate ions, the strongest base present in the initial sample. The first significant drop in ph is evidence that the first reaction is quantitative. We write the Brønsted Lowry equation in the usual way, starting with the acid and base entities present in the sample solution, and also listing the H 3 O (aq) that is being added. nitric acid, HNO 3 (aq) sulfuric acid, H 2 SO 4 (aq) phosphoric acid, H 3 PO 4 (aq) Figure 8 For oxyacids, the protons that transfer in Brønsted Lowry reactions are covalently bonded to oxygen atoms (identified by a red colour in the models above). Of these three examples, the monoprotic and diprotic acids are strong acids, and the triprotic acid is a weak acid. Acid formulas were originally derived from stoichiometric evidence. Arguably, it would be more informative to write the molecular formulas differently, as NO 2 OH, SO 2 (OH) 2, and PO(OH) 3, to clarify the bonding concept. In both science and society, however, once conventions are established, they are difficult to change. SA H 3 O (aq), Cl (aq), Na (aq), CO 2 3 (aq), H 2 O(l) SB B H 3 O (aq) CO 3 2 (aq) H 2 O(l) HCO 3 (aq) Because the first reaction is quantitative, essentially all the carbonate ions will have been consumed at the equivalence point, and an equal quantity of (new) hydrogen carbonate ions will have been formed. Then, in a second reaction, protons transfer from additional (continuosly added) hydronium ions to the hydrogen carbonate ions that were formed in the first reaction. SA A A H 3 O (aq), Cl (aq), Na (aq), H 2 O(l), HCO 3 (aq) B SB H 3 O (aq) HCO 3 (aq) H 2 O(l) H 2 CO 3 (aq) Because the second reaction is also quantitative, all of the hydrogen carbonate ions have reacted at the equivalence point, and only newly formed carbonic acid molecules remain in the solution. Continuing to add more hydronium ion after this point will cause no significant further change (no third reaction), because the strongest base in the system now is water. The carbonate ion is a diprotic base because it can accept a total of two protons. Other polyprotic bases include sulfide ions and phosphate ions. The conjugate entities formed when these bases gain successive protons are: S 2 (aq) HS (aq) H 2 S(aq) PO 4 3 (aq) HPO 4 2 (aq) H 2 PO 4 (aq) H 3 PO 4 (aq) Other polyprotic acids include oxalic acid and phosphoric acid (Figure 8). HOOCCOOH(aq) HOOCCOO (aq) OOCCOO 2 (aq) H 3 PO 4 (aq) H 2 PO 4 (aq) HPO 4 2 (aq) PO 4 3 (aq) Evidence from ph measurements clearly shows that, for every proton transferred by a polyprotic entity, the strength of the new acid or base entity formed greatly decreases. The new entity may be anywhere from 100 to times weaker! Chemists believe that electric charge and electrostatic attraction explain this effect. For example, it should logically be much easier to pull a positively charged proton away from a neutral oxalic acid molecule, than it is to pull one away from a negatively charged hydrogen oxalate ion. Figure 9 shows the ph curve for phosphoric acid titrated with sodium hydroxide. Phosphoric acid is triprotic, so three reactions should be possible. However, note that only two equivalence points are evident. We describe the process as before. At the first equivalence point (ph 4), equal chemical amounts of H 3 PO 4 (aq) and OH (aq) have been combined in solution. As the last of the H 3 PO 4 (aq) reacts, the ph rises abruptly, because when the reaction moves past the equivalence point, only newly formed H 2 PO 4 (aq) is now present, and it is a much weaker acid than H 3 PO 4 (aq). A 756 Chapter 16

89 Section 16.4 ph ml of 0.50 mol/l H 3 PO 4 (aq) Titrated with 0.48 mol/l NaOH(aq) Volume of NaOH (ml) Figure 9 A ph curve for the addition of NaOH(aq) to a sample of H 3 PO 4 (aq) displays only two rapid changes in ph. This result is interpreted as indicating that there are only two quantitative proton transfer reactions of phosphoric acid with hydroxide ions. The third proton transfer reaction never goes to completion, but instead establishes an equilibrium. To write the reaction for the transfer of the first proton, start by listing the entities. SA Na (aq), OH (aq), H 3 PO 4 (aq), H 2 O(l) SB B OH (aq) H 3 PO 4 (aq) H 2 O(l) H 2 PO 4 (aq) A Since all the H 3 PO 4 (aq) has reacted when the reaction has passed the first equivalence point, the second plateau must represent the reaction of OH (aq) with H 2 PO 4 (aq). The second equivalence point (ph 9) corresponds to the completion of the reaction of H 2 PO 4 (aq) with additional OH (aq) solution. As the last of the H 2 PO 4 (aq) reacts, the ph begins to rise abruptly again because, when the reaction has passed this (second) equivalence point, only newly formed HPO 4 2 (aq) is left, and it is an even weaker acid. A SA Na (aq), OH (aq), H 2 O(l), H 2 PO 4 (aq) SB B B OH (aq) H 2 PO 4 (aq) H 2 O(l) HPO 4 2 (aq) No ph endpoint is apparent for the possible reaction of HPO 4 2 (aq) with additional OH (aq). A clue to this lack of a third endpoint can be obtained from the table of acids and bases. The hydrogen phosphate ion is an extremely weak acid and apparently does not quantitatively lose protons to OH (aq). For the third reaction, an equilibrium is established that gradually shifts right as more base is added. There is no ph increase at a third equivalence point because the reaction never goes to completion. A SA Na (aq), OH (aq), H 2 O(l), HPO 4 2 (aq) SB B B OH (aq) H 2 PO 4 (aq) 0 H 2 O(l) HPO 4 2 (aq) Learning Tip When any weak polyprotic acid or base is initially dissolved in water, we always assume that the ionization equilibrium only involves the first proton transfer to or from water. This is essentially true because the tendency of any entity that is formed by a first ionization to lose or gain a (second) proton is very much less than that of the original polyprotic entity dissolved; so much so that it can safely be considered negligible. >50% HPO 4 2 (aq) OH (aq) 0 PO 4 3 (aq) H 2 O(l) As a general rule, only quantitative reactions produce detectable equivalence points in an acid base titration. Equilibrium in Acid Base Systems 757

90 DID YOU KNOW? Sulfuric Acid Sulfuric acid (Figure 10) is probably the world s most important industrial chemical. It is used for so many things that the industrial development and standard of living of a country may be thought of as being proportional to its sulfuric acid production. Canada produces approximately four million tonnes every year. U.S. production (the world s largest) is about 10 times as much. The main direct consumer use of sulfuric acid is the 4.5 mol/l solution inside every standard car and truck battery. Sulfuric acid plants (Figure 11) can be quite compact, and are assembled from off-the-shelf technology. They involve only a few simple reactions, starting with sulfur (in Alberta, mostly extracted from fossil fuels) and oxygen. Figure 10 A model of sulfuric acid, showing relative atom size, bonding, and molecular shape Figure 11 More sulfuric acid is manufactured in North America than any other chemical. Sulfuric acid, H 2 SO 4 (aq), is a unique polyprotic acid because it is the only common one for which the first proton loss is already quantitative in aqueous solution; that is, it is the only strong acid that is polyprotic. Sulfuric acid acts like any of the other five common strong acids, except that its 100% ionization produces hydrogen sulfate ions, HSO 4 (aq), along with hydronium ions, H 3 O (aq). Recall that of all the anions of this type (like HCO 3 (aq), or HPO 4 2 (aq)), the hydrogen sulfate ion is the only one that is a weaker base than water, so it cannot react as a base in aqueous solution. When reacting as an acid, however, hydrogen sulfate ion is one of the stronger weak acids, and usually reacts quantitatively with bases (except for very weak bases, of course). So sulfuric acid will usually, but not always, react with a base (assuming the base is in excess) in two complete proton transfer reactions, one after the other. When you first began a study of reactions of acids and bases, you would have written a complete reaction of sulfuric acid with sodium hydroxide as H 2 SO 4 (aq) 2 NaOH(aq) Na 2 SO 4 (aq) 2 H 2 O(l) This reaction equation, if you think about it now, implies that you assumed that both of this acid s hydrogens (protons) were replaced (transferred). You had no information at that time, however, about how this transfer occurs. For simple stoichiometric calculations based on this reaction, this limited understanding (and simplified equation) worked perfectly well. Explaining and predicting other reactions of sulfuric acid, however, is a different story. To do so, understanding how the neutralization process occurs (as two distinct and sequential reactions) is necessary, because both reactions may not be quantitative when a different (weaker) base is involved, and stoichiometric calculation can only be done for a reaction that is quantitative. There are almost no overall reactions of common polyprotic acids or bases that have more than two definite endpoints. For any such reactions that are actually done as analysis titrations, an indicator can be selected so that the titration can be stopped at either chosen equivalence point. For the titration of sodium carbonate solution with hydrochloric acid, this would mean titrating through two sequential quantitative reactions to the second equivalence point, as shown in Figure 7. For this titration, the second equivalence point involves a greater ph change, and thus is easier to detect (more accurate). As we found earlier, methyl orange indictor is suitable for detecting this second equivalence point. For purposes of stoichiometric calculation, we can combine sequential quantitative Brønsted Lowry reaction equations into a single overall reaction equation. This example is for the titration of sodium carbonate solution with hydrochloric acid. 1. H 3 O (aq) CO 3 2 (aq) HCO 3 (aq) H 2 O(l) followed by 2. H 3 O (aq) HCO 3 (aq) H 2 CO 3 (aq) H 2 O(l) totals to 2 H 3 O (aq) CO 3 2 (aq) H 2 CO 3 2 H 2 O(l) (overall reaction) For purposes of stoichiometric calculation, this equation is equivalent to writing 2 HCl(aq) Na 2 CO 3 (aq) H 2 CO 3 (aq) 2 NaCl(aq) because, either way, the reactant acid base stoichiometric ratio is found to be 2:1. Chemists normally use the simplest representation that will be useful, so Brønsted Lowry equations are more often used for correctly predicting products and equilibria, whereas standard chemical formula equations are more often used for doing stoichiometry as, for example, in titration analyses. 758 Chapter 16

91 Section 16.4 SUMMARY ph Curve Reaction Information Empirical ph curves provide a wealth of information: initial ph of sample solutions ph when excess titrant is added number of quantitative reactions non-quantitative (equilibrium) reactions equivalence point(s) for indicator selection buffering regions Practice 5. How is buffering action displayed on a ph curve? 6. How are quantitative reactions displayed on a ph curve? 7. How is a ph curve used to choose an indicator for a titration? 8. An acetic acid sample is titrated with sodium hydroxide (Figure 12.) 14 CH 3 COOH(aq) Titrated with NaOH(aq) ph Volume of NaOH (ml) Figure 12 The ph curve for the addition of 0.48 mol/l NaOH(aq) to 25.0 ml of 0.49 mol/l CH 3 COOH(aq) illustrates ph changes during the reaction of a weak acid with a strong base. (a) Based on Figure 12, estimate the ph at the equivalence point. (b) Choose an appropriate indicator for this titration. (c) Write a Brønsted Lowry equation for this reaction. (d) At the very beginning of the titration, before the curve levels off, it rises. Explain this rise in terms of entities present in the mixture before and after beginning the titration. 9. A sodium phosphate solution is titrated with hydrochloric acid (Figure 13 on next page). (a) Why are only two equivalence points evident? (b) Write three Brønsted Lowry equations for the sequential reactions shown on the ph curve in Figure 13. Communicate the position of equilibrium for each of the three reactions. 10. Oxalic acid reacts quantitatively in a two-step overall reaction with a sodium hydroxide solution. Assuming that an excess of sodium hydroxide is added, sketch a ph curve (without any numbers) for all possible reactions. Equilibrium in Acid Base Systems 759

92 ml of 0.51 mol/l Na 3 PO 4 (aq) Titrated with 0.50 mol/l HCl(aq) ph Volume of HCl (ml) Figure 13 The ph curve for the addition of HCl(aq) to Na 3 PO 4 (aq) can be interpreted using the Brønsted Lowry acid base concept. Learning Tip Recall that, for the examples you have studied, bromothymol blue has been an appropriate indicator for the SA SB reaction, because it detects an endpoint with a ph of 7 very well. Methyl orange detects endpoint ph values around 4 well, and is, therefore, useful for many SA WB reactions. Finally, phenolphthalein detects endpoints with ph values around 10 well, and is, therefore, useful for many WA SB titrations. ph Curve Shape versus Acid and Base Strength You have learned that the reaction of the strongest possible acid in aqueous solution, H 3 O (aq), and the strongest possible base in aqueous solution, OH (aq), is overwhelmingly quantitative, and always results in an equivalence point ph of 7.00 at 25 C. Not all titrations involve the hydronium and the hydroxide ions, however. ph curves show that some weak acids (such as acetic acid) react quantitatively with OH (aq) (Figure 12). The reactions of some weak bases (such as PO 4 3 (aq) and HPO 4 2 (aq)) with H 3 O (aq), can also produce a quantitative reaction (Figure 13). The strongest of the weak acids also react quantitatively with the strongest of the weak bases. For example, the reaction of HSO 4 (aq) with CO 3 2 (aq) is a quantitative reaction. The farther apart the reacting acid and base are on the Relative Strengths of Aqueous Acids and Bases table (i.e., the larger the difference in K a ), the more likely it is that the reaction will be quantitative. After plotting many ph curves, chemists have developed the general ph curve reference diagram shown in Figure 14. Weak and Strong Acids and Bases SB Figure 14 Laboratory evidence is generalized here to show approximations of the ph curves for quantitative reactions of weak and strong acids and bases. Note the initial change in ph for curves involving weak entities, due to the immediate change of the kinds of entities present in solution when the titration begins. ph >7 WA SB 7 <7 WA SA SA SB SA WB Volume of base WB SA strong acid WA weak acid SB strong base WB weak base 760 Chapter 16

93 Section 16.4 The (stoichiometric) equivalence point ph for a strong acid strong base (SA SB) reaction is seven (7). The equivalence point ph of any quantitative strong acid weak base (SA WB) reaction is less than seven ( 7), whereas that of any quantitative weak acid strong base reaction (WA SB) is greater than seven ( 7). The composite curves in Figure 14 are not acceptable (do not predict well) for any reactions between a weak acid and a weak base. Weak acid weak base reactions often do not have a detectable equivalence point because they are usually not quantitative. For all intents and purposes, analysis reactions and stoichiometric calculations are not done for weak acid weak base reactions, so a ph curve for such a reaction is simply irrelevant. The ph of a solution at the equivalence point of an acid base reaction may be explained by considering the nature of those entities that are present in significant quantities at that point. The entities present at an equivalence point can be predicted either by following the five-step Brønsted Lowry method, or simply by converting a chemical substance formula equation into a net ionic equation. Now let s examine a case of each of the common acid base reaction types, and explain the equivalence point ph in each case. Each of the examples used is a quantitative reaction. Strong Acid Strong Base Reaction For example, nitric acid reacts with potassium hydroxide: SA SB ph 7. or SA H 3 O (aq), NO 3 (aq), K (aq), OH (aq), H 2 O(l) SB B H 3 O (aq) OH (aq) 2 H 2 O(l) HNO 3 (aq) KOH(aq) H 2 O(l) KNO 3 (aq) H (aq) NO 3 (aq) K (aq) OH (aq) H 2 O(l) K (aq) NO 3 (aq) H (aq) OH (aq) H 2 O(l) At the equivalence point, water is the only acid or base entity present, which produces a neutral solution (ph 7). Strong Acid Weak Base Reaction For example, hydrochloric acid reacts with aqueous ammonia: SA WB ph 7. A SA H 3 O (aq), Cl (aq), NH 3 (aq), H 2 O(l) SB B or A H 3 O (aq) NH 3 (aq) H 2 O(l) NH 4 (aq) HCl(aq) NH 3 (aq) NH 4 Cl(aq) H (aq) Cl (aq) NH 3 (aq) NH 4 (aq) Cl (aq) H (aq) NH 3 (aq) NH 4 (aq) At the equivalence point, the only acid or base entity present (other than water) is the ammonium ion, which is a weak acid, resulting in an acidic solution (ph 7). Learning Tip Recall that entities common to both sides of the equation are cancelled out and the coefficients of the remaining entities are reduced to the simplest ratio, in a net ionic equation. Equilibrium in Acid Base Systems 761

94 Weak Acid Strong Base Reaction For example, acetic acid reacts with barium hydroxide: WA SB ph 7. SA CH 3 COOH(aq), Ba 2 (aq), OH (aq), H 2 O(l) SB B CH 3 COOH(aq) OH (aq) H 2 O(l) CH 3 COO (aq) or 2 CH 3 COOH(aq) Ba(OH) 2 (aq) 2 H 2 O(l) Ba(CH 3 COO) 2 (aq) A 2 CH 3 COOH(aq) Ba 2 (aq) 2 OH (aq) 2 H 2 O(l) Ba 2 (aq) 2 CH 3 COO (aq) CH 3 COOH(aq) OH (aq) H 2 O(l) CH 3 COO (aq) At the equivalence point, the only acid or base entity present (other than water) is the acetate ion, which is a weak base, resulting in a basic solution (ph >7). Learning Tip In an equation for a titration, a single arrow,, is used. The reaction represented by the equation must be quantitative in order for any stoichiometric calculations based on the equation to be valid. In such cases, chemical formula equations are often the most convenient form. SUMMARY Titration Generalizations Strong acid strong base reactions are quantitative (100%) and have an equivalence point ph 7. Strong acid weak base quantitative reaction equivalence points have a ph 7. Weak acid strong base quantitative reaction equivalence points have a ph 7. Polyprotic entity samples produce sequential reactions in titrations, each of which may or may not be quantitative. WEB Activity Simulation Titration of Polyprotic Acids and Bases These two simulations enable you to explore the ph curves that result from various titrations. Of course, polyprotic acids and bases each have more than one K value, so the shapes of their titration curves will reflect this. Predict the shapes of the curves before working through the exercises. GO Practice 11. For the first quantitative reaction in each of the following acid base titrations, predict (where possible) whether the equivalence point ph will be greater than, less than, or equal to 7. (a) hydroiodic acid aqueous sodium hydrogen phosphate (b) boric acid aqueous sodium hydroxide (c) aqueous sodium hydrogen sulfate aqueous potassium hydroxide (d) hydrochloric acid solid magnesium hydroxide (e) hydrosulfuric acid aqueous sodium hydrogen carbonate (f) sulfuric acid aqueous ammonia 762 Chapter 16

95 Section Assume only three indicator solutions are available in your lab: methyl orange, bromothymol blue, and phenolphthalein. Choose the most suitable indicator for detecting the equivalence point of each of the following acid base titrations (for the first quantitative reaction only). (a) HBr(aq) Ca(OH) 2 (s) (b) HNO 3 (aq) Na 2 CO 3 (aq) (c) KOH(aq) HNO 2 (aq) 13. The following formulas each represent a solution at an acid base reaction equivalence point. From the entities present in each solution, predict the observed ph as being greater than, less than, or equal to 7. (a) NH 4 Cl(aq) (b) Na 2 S(aq) (c) KNO 3 (aq) (d) NaHSO 4 (aq) 14. Use the five-step Brønsted Lowry method to predict the overall reaction net ionic equation when the following chemicals are mixed. (a) solutions of perchloric acid and sodium carbonate (A ph curve shows two protons are transferred quantitatively in successive reactions.) (b) solutions of nitrous acid and potassium hydroxide (c) solutions of phosphoric acid and sodium hydroxide (A ph curve shows two protons are transferred quantitatively in successive reactions.) 15. Write a standard chemical formula equation, and also the net ionic equation, for the predominant reaction that occurs when each of the following pairs of reagents mix. (a) Stomach acid is neutralized by solid magnesium hydroxide. (b) Aqueous ammonia is added to sulfurous acid. (c) A solution of sulfuric acid is neutralized by sodium hydroxide. (A ph curve indicates two quantitative reactions.) ph Curve Buffering Regions and Buffer Solutions A titration ph curve for any quantitative acid base reaction shows at least one region where buffering action occurs between the beginning of the titration and the equivalence point. If the titration is continued much past the equivalence point, it enters another region of buffering. Such a region represents a solution in which partial reaction has occurred, and, therefore, contains significant amounts of both entities of a conjugate acid base pair. The term buffer refers specifically to the combination of any weak acid with its conjugate base, in the same solution. Buffer solutions have a specific ph due to the nature and concentration of the entities present, and change ph very little when other acids or bases are added provided that the quantity added is much less than the quantities of the conjugate pair entities present in the buffer. For example, consider the titration of acetic acid with sodium hydroxide (Figure 15). The solution ph is approximately 4.8 when 10 ml of sodium hydroxide solution has been added. This volume represents one-half of the (calculated) equivalence point volume of titrant. At this point in the reaction, one-half of the acetic acid originally present will have reacted (changed to acetate ions), according to the following equation: OH (aq) CH 3 COOH(aq) H 2 O(l) CH 3 COO (aq) The mixture, therefore, now contains approximately equal amounts of the remaining unreacted weak acid, CH 3 COOH, and of the conjugate base, CH 3 COO, produced in the reaction. Chemists would describe this mixture as an acetic acid acetate ion buffer. Learning Tip You have observed (Figure 2, page 752) that the strong acid strong base reaction ph curve shows that the conjugate pair H 3 O (aq) H 2 O(l) will maintain a stable (very low) ph; and that the H 2 O(l) OH (aq) pair will stabilize ph at a very high value. Even though these solutions certainly do produce a buffering effect, they are not technically considered to be buffers, because in each case one of the two conjugate pair entities (water) is also the solution solvent, and is present in great excess. A buffer is normally thought of as an aqueous solution mixture of a weak acid and its conjugate base, with both entities of the conjugate pair present in significant quantity, but with both quantities much less than the quantity of the solution solvent (water) present. Equilibrium in Acid Base Systems 763

96 Figure 15 The highlighted plateau shows an effective buffering region during this titration of aqueous acetic acid with sodium hydroxide. At the point halfway to the equivalence point, the solution is a 1:1 ratio mix of an acetic acid acetate ion buffer. (a) ph (b) ph Learning Tip Buffers are normally prepared as mixtures of a weak acid with a solution of some salt of that weak acid. Thus, dissolving solid sodium acetate in a solution of acetic acid would produce a buffer solution with properties the same as those of the highlighted buffering region of Figure 15. Acetic Acid Buffer buffer capacity Volume of NaOH(aq) Acetic Acid Buffer buffer capacity Volume of HCI(aq) Figure 16 When an initial chemical amount of a buffer is eventually depleted, the ph then changes very quickly. ph Titration Curve for Titrating mol/l CH 3 COOH(aq) with mol/l NaOH(aq) Buffering can be readily explained using Brønsted Lowry equations. Suppose a small amount of NaOH(aq) is added to the acetic acid acetate ion buffer. Using the five-step method for predicting the predominant acid base reaction (page 731), the following equation is obtained: SA equivalence point Volume of NaOH(aq) (ml) Na (aq), OH (aq), CH 3 COOH(aq), CH 3 COO (aq), H 2 O(l) SB B B OH (aq) CH 3 COOH(aq) H 2 O(l) CH 3 COO (aq) Figure 15 shows that this reaction is quantitative. A small amount of OH would convert a small amount of acetic acid to acetate ions. The overall effect will be just a small decrease in the ratio of acetic acid to acetate ions in the buffer and a slight increase in the ph. The very small change in concentration of each entity of the acid base conjugate pair present, and the complete consumption of the added hydroxide ions in the process, explains why the ph change is small. This buffer would work equally well if a small amount of a strong acid, such as HCl(aq), were to be added. Evidence indicates that the reaction that occurs in that case is also quantitative. SA A A H 3 O (aq), Cl (aq), CH 3 COOH(aq), CH 3 COO (aq), H 2 O(l) SB B H 3 O (aq) CH 3 COO (aq) H 2 O(l) CH 3 COOH(aq) Added hydronium ion is consumed and the mixture then has a slightly higher ratio of acetic acid to acetate ions and a slightly lower ph. Figure 16 illustrates the concept of buffer capacity the limit of the ability of a buffer to maintain a ph level. When the entity of the conjugate acid base pair that reacts with an added reagent is completely consumed, the buffering fails and the ph changes dramatically. The ability of buffers to maintain a relatively constant ph is important in many biological processes where certain chemical reactions can only occur at a specific ph value. Many aspects of cell functions and metabolism in living organisms are very sensitive to ph changes. For example, each enzyme carries out its function optimally over a small ph range. One essential buffer operates to maintain a stabilized ph in the internal fluid of all living cells. This critically important buffer is a mixture of dihydrogen phosphate ions and hydrogen phosphate ions. In mammals, cellular fluid has a ph in the range A 764 Chapter 16

97 Section 16.4 of 6.9 to 7.4. The H 2 PO 4 (aq) HPO 4 2 (aq) buffer system has a ph of 7.2 when the concentrations of these two conjugate entities are equal. With small variations in concentration, this buffer is effective in maintaining optimum ph levels for the innumerable reactions going on within any given cell. The major buffer system in the blood and other body fluids (except the cytoplasm within cells) is the conjugate acid base pair H 2 CO 3 (aq) HCO 3 (aq). Blood plasma has a remarkable buffering ability, as shown by the empirical results in Table 1. Since production of the entities of this buffer in the body is a continuous process, the buffer is not a limiting reagent, and you do not ordinarily have to worry about using up your buffering capacity. Table 1 Buffering Action of Neutral Saline Solution and of Blood Plasma Final ph after adding Solution (1.0 L) Initial ph of mixture 1 ml of 10 mol/l HCI neutral saline blood plasma Human blood plasma normally has a ph of about 7.4. Any change of more than 0.4 ph units, induced by poisoning or disease, can be lethal. If the blood were not buffered, the acid absorbed from a glass of orange juice would probably be fatal. WEB Activity Simulation Preparation of Buffer Solutions This simulation shows you how to select an appropriate acid base conjugate pair to make a buffer of the desired ph. GO Buffers are also important in many consumer, commercial, and industrial applications. Fermentation and the manufacture of antibiotics require buffering to optimize yields and to avoid undesirable side reactions. The production of various cheeses, yogurt, and sour cream are very dependent on controlling ph levels, since an optimum ph is needed to control the growth of micro-organisms and to allow enzymes to catalyze fermentation processes. Sodium nitrite and vinegar are widely used to preserve food; part of their function is to prevent the fermentation that takes place only at certain ph values. The CRC Handbook of Chemistry and Physics provides recipes for preparing buffer solutions. For example, a buffer with a ph of can be prepared by mixing 50 ml of mol/l NaHCO 3 (aq) with 10.7 ml of 0.10 mol/l NaOH(aq). The reaction to establish the buffer is described in the same way as all other acid base reactions. SA Na (aq), HCO 3 (aq), OH (aq), H 2 O(l) B SB B HCO 3 (aq) OH (aq) H 2 O(l) CO 3 2 (aq) A The buffer preparation recipe reaction converts about 3 5 of the initial hydrogen carbonate ions into carbonate ions. An aqueous mixture of a 3:2 ratio of these two acid base conjugates at the concentrations specified has been found empirically to have an equilibrium ph of precisely at 25 C. Buffer mixtures like this one with highly precise ph values are routinely used to calibrate ph meters for accuracy. BIOLOGY CONNECTION Homeostasis Many reactions in the human body produce or consume carbonic acid or hydrogen carbonate ions. When one of these entities is depleted in the blood, Le Châtelier s principle comes into effect, and more of that entity is produced by an equilibrium shift to keep everything in balance. The carbonic acid in solution in blood is really in equilibrium with dissolved carbon dioxide: CO 2 (aq) H 2 O(l) 0 H 2 CO 2 (aq) The enzyme carbonic anhydrase acts as a catalyst to allow this equilibrium to establish rapidly, or reestablish quickly, once shifted. It is possible to make your blood temporarily too basic by deliberately breathing heavily for a long time, thereby causing your body to lose too much carbon dioxide. Doctors call this condition respiratory alkalosis. There are countless equilibrium reactions interconnected in this way in and around living cells. Biologists refer to this interdependent network of reaction equilibria as an example of homeostasis the condition of automatic continual readjustment of cells, systems, and whole organisms to very specific conditions. Biology courses will tell you a lot more about homeostasis and the factors that affect it. GO + EXTENSION The Hydrogen Carbonate Buffer System The ph of your blood is maintained at a constant level, courtesy of a series of equilibrium reactions, including a buffer system. This extension outlines the reactions, and gives an example of how climbing at high altitude can upset this delicate balance. GO Equilibrium in Acid Base Systems 765

98 WEB Activity Canadian Achievers Maud Menten Maud Menten (Figure 17) was a world-renowned pioneer in explaining the chemical action of enzymes. What key concept was her chief contribution to understanding enzyme activity? GO Figure 17 Maud Menten ( ) DID YOU KNOW? Commercial Buffers Buffers are fairly common, commercially. You can buy buffer mixtures to adjust the ph of aquarium water to suit the type of fish you have. Buffer mixes are sold to add to hot tubs to control the ph of the water. Many food recipes mix ingredients that form natural buffers within the dough, batter, or sauce. Practice 16. Give both an empirical and a theoretical definition of a buffer. 17. List two buffers that help maintain a normal ph level in your body. 18. Use the five-step method to predict the quantitative reaction of a carbonic acid hydrogen carbonate ion buffer (a) when a small amount of HCl(aq) is added (b) when a small amount of NaOH(aq) is added 19. What happens if a large (excess) amount of a strong acid or base is added to a buffer? 20. Use Le Châtelier s principle to predict what will happen to a benzoic acid benzoate ion buffer when a small amount of each of the following substances is added: (a) HCl(aq) (b) NaOH(aq) 21. Which of the following solution pairs, when mixed in equal quantities, will not form an effective buffer? (a) HNO 3 (aq) and NaNO 3 (aq) (c) NH 3 (aq) and NH 4 Cl(aq) (b) C 6 H 5 COOH(aq) and NaC 6 H 5 COO(aq) (d) HCl(aq) and NaOH(aq) INVESTIGATION 16.3 Introduction Testing a Buffer Effect References provide recipes for preparing standard buffer solutions of any desired ph from 1.0 to The one used in this investigation has a ph of precisely 7.0, and might be used to calibrate a ph meter, for example. If our theory of buffers is correct, this solution should resist significant change in ph upon gradual addition of outside acid or base entities. Purpose The purpose of this investigation is to test our concept of buffers. Write the Design, Materials, and table of evidence to match the Procedure that is provided. The Materials list should include the size of the equipment used. The buffer is prepared by a reaction communicated by the following chemical equation: Report Checklist Purpose Design Analysis Problem Materials Evaluation (2, 3) Hypothesis Procedure Prediction Evidence H 2 PO 4 (aq) OH (aq) HPO 2 4 (aq) H 2 O(l) excess limiting (base part (acid part reagent of buffer) of buffer) Problem How does the ph change when a strong acid and a strong base are slowly added separately to an H 2 PO 4 (aq) HPO 4 2 (aq) buffer? To perform this investigation, turn to page 769. CAREER CONNECTION Microbiologist Microbiologists study, test for, and isolate microorganisms such as bacteria, fungi, and viruses. As part of a medical or research team, their work is essential for public health and safety. GO SUMMARY Brønsted Lowry Is a Unifying Concept The five-step Brønsted Lowry method to explain and predict acid base reactions is a preferred, acceptable method because it works for all quantitative and non-quantitative reactions studied so far: neutralization reactions buffer reactions excess reactions indicator reactions polyprotic reactions titration reactions 766 Chapter 16

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