TOPIC-1: CHEMICAL EQUILIBRIUM A MATTER OF BALANCE

Transcription

1 Chemistry 534 CHAPTER 6 Chemical Equilibrium The subject of any investigation is called a system. When the study involves experimentation, the system is said to be real. When the study involves ideas, the system is said to be ideal. The advantage of an ideal system is that it makes possible "study by analysis". That is, an ideal system may be broken down into its constituent parts. Each part becomes a system in itself yet simpler and easier to understand than the whole. On the other hand, in a real system, it is usually difficult or impossible to remove the various parts without causing the system to either malfunction or collapse. TOPIC-1: CHEMICAL EQUILIBRIUM A MATTER OF BALANCE The phenomenon of chemical equilibrium occurs in a variety of different systems in nature, from personal biological systems to global ecological systems. In industry, the control of chemical reactions by the manipulation of their equilibrium conditions is very important in many processes. Equilibrium is a balance, a compromise achieved by two opposing tendencies in nature each attempting to cause change. One compromise is the natural tendency to raise the entropy (disorder) of a system as much as possible. The other compromise is the natural tendency to lower the energy of a system as much as possible. Once equilibrium has been reached, the two tendencies are in a state of balance and the necessity for change no longer exists. In fact, at equilibrium, there is a natural tendency to resist change. As long as a system is away from equilibrium, it will experience a tendency to reach equilibrium. The tendency will be greater the greater the distance from equilibrium. Furthermore, useful work can only be obtained from a system which is on its way to equilibrium. Once at equilibrium, no work can be obtained from a system. Note: All changes in nature are due to the tendency on the part of different systems to reach a state of equilibrium. In this chapter, we learn the principles governing chemical equilibrium. This includes the factors which influence equilibrium as well as the mathematical laws governing its behavior. 1

2 EXPERIMENT 1 Consequences of reversibility A DEMO Objective: To investigate reversible reactions In this demonstration, we will use a mechanical model to simulate the behavior of a reversible reaction. The mechanical system consists of two 1000 ml beakers (labeled beaker-a and beaker-b), a 250 ml beaker, and a 100 ml beaker. At the start, beaker-a is filled with water. The objective is to transfer water back and forth between the two 1000 ml beakers via the two smaller beakers. Note that the 250 ml beaker represents the forward reaction, allowing water to go towards the right, while the 100 ml beaker represents the reverse reaction, allowing water to go towards the left. ã Remember: Since equilibrium is a continuous process, in our mechanical model there must be a continuous exchange of water between beaker-a and beaker-b. Also, the exchange can only take place via the two intermediate beakers. In this particular experiment, due to the size of the containers used, the maximum water which can be transferred towards the right is 250 ml while the maximum water which can be transferred towards the left is 100 ml. We will now verify that, once equilibrium is established, a fixed quantity of water will continuously be exchanged between beaker-a and beaker-b. 2

3 1) How does the rate of reaction vary as the reaction progresses? 2) Which graph illustrates the evolution (or progress) of the forward reaction? 3) What can you say concerning the rate of the reverse reaction during the same period of time. 4) Which graph illustrates the evolution (or progress) of the reverse reaction? 5) What happens when the system reaches equilibrium? a) In terms of the macroscopic properties: b) In terms of the forward and reverse rates of reaction: 3

4 In the last experiment, did you observe that, once equilibrium was reached, the amount of water transferred left and right remained constant? This resulted in beaker-a and beaker-b both retaining a constant (although unequal) amount of water. Note that unlike our mechanical model, in a chemical system the right and left transfers occur simultaneously in a single container. As such, the mass on each side of the equilibrium remains constant. This does not mean that the total mass on the left equals the total mass on the right. What it does mean is that the total mass on the left remains constant and the total mass on the right remains constant. þ IMPORTANT Do not confuse a reaction which goes to completion with a reversible reaction in equilibrium (which does not go to completion). In a "completion" reaction, the system consists of two sections analogous to a "before-and-after" situation. The reactants, which are always written on the left, constitute the " before" situation. The products, which are always written on the right, constitute the " after" situation. During the "before" situation, some or all of the reactants are totally used up. In this type of equation, the total mass on the left (mass of the reactants) equals the total mass on the right (mass of the products). Stated differently, in an ordinary chemical reaction, the total mass before the reaction equals the total mass after the reaction. However, an equilibrium reaction does not go to completion. Such a reaction never ends. Both the substances on the left and the substances on the right are continuously and simultaneously reacting. This is not a "before-and-after" situation, it is a now situation. Thus, it is incorrect to assume that the total mass on the left equals the total mass on the right (as is true for reactions which go to completion). Indeed, in an equilibrium equation, it is quite possible for the mass on the left to be very little while the mass on the right to be very great (or vice-versa). Of course, the total mass of the system (the mass on the left plus the mass on the right) remains constant. To differentiate a reaction which goes to completion from an equilibrium reaction (which never goes to completion), the following special "arrows" are used: 4

5 In effect, at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This is why neither side gains or loses mass. Furthermore, the continuous forward and reverse reactions is what causes the equilibrium to be "dynamic". Here is an illustration of a chemical system in equilibrium: 4 g/s 4 g/s e word "rate" means something is occurring "per unit time". That is, as time progresses, the occurrence continues (accumulation occurs). The symbol for the word "rate" is the slash"/". Thus we may write: 20 $/h 20$ per hour for rate of pay 4 L/s 4 litres per second for rate of flow 60 km/h 60 kilometres per hour for rate of speed 5 mol/s 5 moles per second for rate of reaction 6) Assuming that the following system is in equilibrium, draw the equation symbol for the reaction: X+Y W+Z 7) Assuming that the following equilibrium system starts from the left, draw the equation symbol for the reaction: X+Y W+Z 8) Assuming that the following equilibrium system starts from the right, draw the equation symbol for the reaction: X+Y W+Z 5

6 Activity- A Model for Chemical Equilibrium In this activity you will simulate a system approaching chemical equilibrium. Your task is to determine if the simulation is a good model for a chemical system at equilibrium. Apparatus Two 150 ml beakers, two plastic rulers Two pieces of glass tubing of different diameters, cm long 10 ml graduated cylinder, water with blue food coloring, water with yellow food coloring Method 1. Put some blue colored water in one beaker and some yellow colored water in the other. The only restriction is that the total volume of the two is 100 ml. 2. Work in pairs. Transfer water from B, the blue beaker, using one of the pieces of glass tubing, to Y. the yellow beaker. At the same time your partner should transfer water from Y to B using the other piece of glass tubing. Be sure to keep the glass tube vertical at all times. 3. Use the rulers to record the height of the water in each beaker after every five exchanges. 4. Continue transferring water until you have three successive readings that are the same. 5. When the heights remain constant measure the volume transferred by each glass tube. 6. Repeat steps 1-5 for different volumes of blue and yellow water, but keep the total volume 100 ml. Observations Time (5 exchanges) etc. Height of Water Blue Beaker (mm) Height of Water Yellow Beaker (mm) Discussion 1. What are the following analogous to in a chemical reaction? The height of the water in the two beakers: the transfer of water from one beaker to another: two pieces of glass tubing of different diameters: the coloring in the water; the volume of water transferred: the volume of water transferred when the height remains constant. 2. What evidence suggests that you reached a point of equilibrium in this activity? You should be able to think of two key observations. 3. What property of the system determines the final height of the water in the beakers? What would this be analogous to in a chemical reaction? 4. Plot height of water versus time in units of five exchanges. (a) Describe the graph in words. (b) (c) What is the significance of the horizontal part of the graph? What would be the significance of the slope of a tangent drawn at any point on this graph? 6

7 (d) (e) What happens to the slope of the tangent as the system approaches equilibrium? What does this imply? The implications of (d) contradict what is actually happening: you can still see water being transferred from one beaker to another. Explain how it is possible for the graph to suggest that the reaction has stopped even though you can see it going on. 5. Consider the following gaseous reaction: CO(g) + Cl 2 (g) COCl 2 (g) One mole of Cl 2 was added to 2 mol of CO in a 1 L container. The concentration of the three gases was measured using a spectrophotometer every minute The following data were recorded: Time (min) [CO] mol/l [Cl 2 ] mol/l [COCl 2 ] (a) (b) (c) (d) (e) (f) Plot a graph of the data. Explain why the graph has the shape it has. What is the significance of the horizontal part of the graph? What would be the significance of the slope of a tangent drawn at any point on this graph? What happens to the slope of the tangent as the system approaches equilibrium? What does this imply? Use the analogy to explain the changes in the slope of the tangents. 7

8 The experiment you have just performed demonstrates that visible changes occur during a chemical reaction. For example, the formation of a gas, the disappearance of the tablets, effervescence occurring at the surface of the liquid, etc. These changes, seen with the naked eye, are called macroscopic (large scale) changes. Conversely, changes that occur which cannot be seen with the naked eye are called microscopic (small scale) changes. For any chemical system to be in equilibrium, the following three conditions are necessary: Œ The system must be closed. There must be no macroscopic changes (no visible or observable changes). Ž There must be some reactants and some products present at all times. Note: The system is said to be in dynamic equilibrium because there is continuous microscopic activity. That is, while there are no observable (macroscopic) changes (such as temperature, pressure, volume, color, etc.), the molecules are constantly in motion. As a result, there are always some reactants forming some products, and, at the same time, some products forming some reactants. This is the nature of "dynamic" equilibrium. 9 ) Tell which of the following are characteristic of an equilibrium system a) A system which is closed. b) There is a gas in the system. c) The reactants have completely transformed into products. d) There is a loss of matter. e) The color of the solution remains constant. f) The mass of undissolved matter equals the mass of dissolved solid. g) The presence of a catalyst. h) There are no observable changes. i) The solute has completely dissolved. 8

9 10) State and explain whether or not each of the following systems are in equilibrium: a) A distillation flask contains a constant quantity of alcohol. b) The flame of a Bunsen burner keeps its shape, height, and color for 30 minutes. c) In a water boiler, the temperature and pressure do not vary. d) The water level of Lac Saint-Jean remains unchanged for a week. e) The column of alcohol inside a thermometer remains steady. 11) Give two examples from every day experience of systems in equilibrium: 9

10 12) To summarize, complete the following phrase by filling in the blanks: A closed system is in equilibrium if its properties are and if both and are always present. Furthermore, the same equilibrium condition may be attained by starting with either the or the. End of experiment The equilibrium system just studied can attain equilibrium either with the reactants or with the products as the starting materials. Such reactions are called "reversible" reactions as they simultaneously proceed in both the forward and reverse directions. To indicate the fact that a reaction is reversible, a special "two-way" arrow is drawn in the equation as shown below: CaCl 2 (aq) + Na 2 SO 4 (aq) CaSO 4 (s) + 2 NaCl(aq) 13) Using your own words, define the following reactions: a) Forward reaction b) Reverse reaction c) Reversible reaction 10

11 TOPIC-2: UNSTABLE EQUILIBRIUM To better understand equilibrium, we must understand the factors which influence its operation. We will investigate four factors and study their effect on chemical equilibrium. The factors are called "stresses" as they place a "stress" on the equilibrium of a system. Œ Variation in the concentration of the reactants Variation in the pressure of the system Ž Variation in the temperature of the system The presence of a catalyst 1Review of concentration l that concentration is the amount of matter per unit volume, similar to density. ever, whereas density refers to the amount of solid matter per unit volume s per unit volume), concentration refers to the number of particles per unit volume of liquid. concentration is the number of particles there are in a specific volume of liquid, the more particles in the liquid, the greater the concentration (and vice-versa). For example, suppose we have a tablespoon of salt. If we pour this salt into a glass of water, we will have a specific concentration of salt water. Now suppose we pour a tablespoon of salt into a huge container of water. Again we have salt water. This time, however, the concentration is much lower (as there are fewer particles per unit volume). ause concentration is mass (moles) per unit volume and the standard volume is the litre, the unit for concentration is moles/litre or mol/l (or M for short). 14) Suppose a group of 10 people visit the following three locations where, at each location, they spread out in the space available: A football field An elevator ƒ A theater a) At which location is the concentration of people the greatest? b) At which location is the concentration of people the least? Note: It is customary to use square brackets "[ ]" to represent the words "the concentration of". Thus, for example, we may write: [people] to mean " the concentration of people" [NaCl] to mean "the concentration of sodium chloride" [HCl] = 1.0 mol/l to mean "the concentration of HCl is 1.0 mol/l" 11

12 Le Chatelier's Principle If a system at equilibrium is subjected to a change, reactions occur to counteract that change and restore the system to a new equilibrium. Factors that can change the equilibrium state: 1. Concentration: If a reactant is added to an equilibrium system, the forward reaction speeds up; if a product is added, the reverse reaction speeds up. Eventually, a new equilibrium is reached. example: [Co(H 2 O) 6 ] Cl - [CoCl 4 ] H 2 O red ions blue ions a) When more chloride ions are added, there is an increase in the number of successful collisions with the red ions and so the rate of the forward reaction increases. The solution turns blue as more products are formed. Very soon, there is an increase in the number of successful collisions between product molecules and the rate of the reverse reaction increases. A new equilibrium is reached when the rate of the reverse reaction "catches up" to the rate of the forward reaction. It is called "new" because both rates are faster than the original rates. b) When more water is added, there is an increase in the number of successful collisions with the blue ions and so the rate of the reverse reaction increases. The solution turns red as more reactants are formed. Very soon, there is an increase in the number of successful collisions between reactant molecules and the rate of the forward reaction increases. A new equilibrium is reached when the rate of the forward reaction 'catches up', with the rate of the reverse. It is called 'new" because both rates are faster than the original rates. NOTE: At any equilibrium state, there are some red ions and some blue ions present. The colour you see depends on which ion is present in the greatest concentration. If there is an equal amount of both ions, the solution is purple. PROBLEMS: i. Consider the following equilibrium system: S0 2 (g) + 1/2 O 2 (g) S0 3 (g) a) If more oxygen is added to the system, in which direction will the equilibrium be shifted? b) How will this shift affect the concentration of each substance in the system? ii. Consider the following equilibrium system: 2 H 2 0(g) 2 H 2 (g) (g) Name 3 concentration changes that would be made that would result in an increase In the amount of water vapour. 2. Temperature: An increase in temperature will result in an increase in the rate of the endothermic reaction example: [Co(H 2 O) 6 ] Cl - + heat [CoCl 4 ] H 2 O red ions blue ions a) An increase in temperature increases the rate of the forward reaction The reactants absorb heat energy and collide more successfully to form products. The solution turns blue. b) A decrease in temperature slows down the rate of the forward reaction Since the reverse reaction is still proceeding at its original rate, the solution turns red. 12

13 Experiment 2: Applying Le Chatelier's Principle In this experiment, reversible reactions will occur; reactants will form products and products will form reactants. A change in the relative amounts of reactants and products can be observed by noting colour changes or the formation of a precipitate. The following equilibrium system will be studied: 2 CrO 4 2- (aq) Cr 2 O 7 2- (aq) The concentration of each ion present in the system depends on the amount of hydrogen ions present. The concentration of hydrogen ions can be increased by adding hydrochloric acid (HCl) and can be decreased by adding sodium hydroxide, a source of OH ions. The OH - ions react with H+ ions to form water. What effect will this stress (ie. adding or removing H+ ions) have on the equilibrium system? Part 1: Step 1: Step 2: Observe the colour of the chromate and dichromate ions. chromate (CrO 4 2- ) dichromate (Cr 2 O 7 2- ) Place 10 drops of each solution into separate test tubes. Add NaOH solution, drop by drop, until a colour change is noted in one of the tubes. Keep these tubes for step 5. CrO 4 2- initial colour final colour Cr 2 O 7 2- Step 3: Place 10 drops of each solution into separate test tubes. Add HCl solution, drop by drop, until a colour change is noted in one of the tubes. Keep these tubes for step 4. initial colour final colour CrO 4 2- Cr 2 O 7 2- Step 4: Step 5: Add NaOH solution, drop by drop, to the test tubes from step 3 until a colour change is noted in one of the tubes. Result: Add HCl solution, drop by drop, to the test tubes from step 2 until a colour change is noted in one of the tubes. Result: 13

14 15) Did the addition of hydrogen ions from HCl favour the forward or reverse reaction? 2 CrO 4 2- (aq) Cr 2 O 7 2- (aq) 16) Add H + ions to the equation above and balance the other side by adding water molecules. Explain the effect of adding OH - ions. Part 2: Step 1: Add 10 drops of dichromate solution to each of 7 separate test tubes. Classify the solutions in the table below as acid, base, or neutral. Add drops of solution, one to each of the 7 test tubes containing the dichromate solution until a colour change is noted. Solution Type initial colour final colour CH 3 COOH HNO 3 Ca(OH) 2 H 2 S0 4 KOH C 2 H 5 OH NH 4 OH Step 2: Repeat step 1 using 7 test tubes containing chromate solution. Solution Type initial colour final colour CH 3 COOH HNO 3 Ca(OH) 2 H 2 S0 4 KOH C 2 H 5 OH NH 4 OH 17) Which solutions behaved like HCl? 18) Which solutions behaved like NaOH? 19) How can you account for the behaviour of C 2 H 5 OH? 14

15 Part 3: Step 1: Add 10 drops of chromate solution to a test tube. Add 10 drops of barium nitrate solution, Ba(N0 3 ) 2. Observe the formation of the precipitate. Allow it to settle. Ba 2+ (aq) + CrO (aq) à BaCrO 4 (s) Step 2: Repeat step 1 using dichromate solution. Allow the precipitate to settle. 20) The chromate solution contains both chromate and dichromate ions. The dichromate solution contains both chromate and dichromate ions. The colour that you see depends on which ion is present in the greater concentration. Barium nitrate only forms a precipitate with chromate ions. Analyze your results. Experiment 3: Equilibrium and temperature ÄA DEMO We know that temperature affects the rate of a chemical reaction. Does a change in temperature also affect the state of equilibrium? Let's find out experimentally using the following equilibrium system: 2 N0 2 (g) N 2 O 4 (g) reddish colorless To test for any change in equilibrium, we will set up two systems labeled SYSTEM-1 and SYSTEM-2. Each system will contain the above reaction in a separate glass tube. One system will be kept at room temperature, while the other will be subjected first to a temperature of about 80 0 C and then to a temperature of about 0 0 C. Recall that a change in temperature changes the volume of a gaseous system. Therefore, to keep the volume constant (while we change the temperature), the glass tube is sealed at both ends. Step 1. Submerge a glass tube containing NO 2 gas in a beaker of hot water (about 80 0 C). Observe the gaseous reaction. Step 2. Now replace the beaker of hot water with a beaker of cold water and ice. Again observe the reaction. 21) What are the equilibrium properties of the system at room temperature? 15

16 22) Has an increase in temperature changed the equilibrium state? Explain: 23) Has a decrease in temperature changed the equilibrium state? Explain: 24) Here is the equation of the system we used in terms of energy: 2 N0 2 (g) N (g) kj Complete the following phrase: For this system, a(n) in temperature favors an exothermic or forward reaction. Conversely, a(n) in temperature favors an endothermic or reverse reaction. ÄEnd of experiment-3 Let's summarize the results of experiment-3. The equilibrium reaction was: 2 N0 2 (g) N (g) kj The system remained in equilibrium as long as the temperature remained constant. Each time the temperature was changed (a stress introduced), the system reacted and a new equilibrium position was established. An increase in temperature upset the equilibrium by increasing the kinetic energy of the molecules. To re-establish equilibrium, the system partially counteracted or opposed the stress by consuming some of the added heat energy. It did this by shifting the equilibrium towards the left so as to convert some of the extra kinetic energy (heat) into enthalpy. The shift lowered the kinetic energy of the molecules and gradually re-established equilibrium. A decrease in temperature upset the equilibrium by decreasing the kinetic energy of the molecules. To re-establish equilibrium, the system countered the stress by releasing heat energy. It did this by shifting the equilibrium towards the right so as to release heat. The shift raised the kinetic energy of the molecules and gradually re-established equilibrium. 16

17 â Remember: For any system m equilibrium, an increase in temperature causes the system to shift towards the side opposite the energy in the equation (in the direction which uses up energy). Here are some examples of what happens when we INCREASE the temperature (the reverse occurs when we decrease the temperature): Endothermic reaction A + B + heat C + D System shifts right Exothermic reaction A + B C + D + heat System shifts left EXPERIMENT-4: Equilibrium and pressure ÄA DEMO Objective: To study the effect of pressure on equilibrium To study the effect of pressure on a system in equilibrium we require a gaseous reaction. We'll use nitrogen dioxide gas, N0 2 (g), which produces dinitrogen tetroxide, N (g) with which it forms an equilibrium system. To obtain the nitrogen dioxide gas, concentrated nitric acid, HNO 3, is made to react with powdered copper, Cu(s). The reaction produces nitrogen dioxide gas and water vapor. Cu(s) + 4 HNO 3 (l) 2 NO 2 (g) + Cu(NO 3 ) 2 (s) + 2 H 2 0(g) To remove the water vapor, we pass the mixture of gases through a "drying tube" which contains calcium chloride, CaCl 2. The calcium chloride attracts the water resulting in the following gaseous equilibrium system: 2 NO 2 (g) N (g) reddish colorless For convenience, we will contain the system in a syringe. The syringe forms a closed equilibrium system whose pressure can easily be varied by moving the piston. 17

18 Your teacher or lab technician will prepare the equilibrium system as follows: Step-1: Step-2: Step-3: Step-4: Step-5: Place about 2 g of powdered copper into a 250 ml Erlenmeyer filter flask and seal the flask with a "twistit" rubber stopper. Place about 4 g of calcium chloride into a drying tube and connect one end to the flask. Seal the other end with a rubber cap. Using a 10 ml syringe, slowly inject 4 ml of concentrated nitric acid, HNO 3, into the Erlenmeyer flask thereby starting the reaction. ÄNote: Due to its very reactive nature, handle the nitric acid with caution at all times (avoid inhaling any fumes). When nitrogen dioxide gas is noticeable in the drying tube (reddish color), draw about 70 ml of gas into the syringe. Remove the syringe and immediately seal it (with a rubber stopper or cap). Ä Note: When the syringe is sealed, pull out the piston slightly so that the red color of NO 2 (g) becomes visible. 25) What are the properties of this system in equilibrium? Step-6: Press the piston of the syringe so as to double the pressure. 26) What are the new properties of this system? 27) When the pressure was doubled, a stress acted on the system. The system then responded to the stress (increase in pressure). Did the response by the system favor a forward or reverse reaction? Explain: 18

19 28) Select the correct symbol which represents the temporary reaction as the system adjusts to the increase in pressure: 2 NO 2 (g) N (g) Step-7: Adjust the piston of the syringe so that the volume is increased. 29) What are the new properties of the system? 30) Was the response by the system to favor the forward or reverse reaction? Explain: 31) Select the correct symbol which represents the temporary reaction as the system adjusts to the last change in pressure: 2 N0 2 (g) N (g) End of experiment-4 Let's summarize the results of experiment-4 The system in equilibrium consisted of the following gaseous reaction: 2 N0 2 (g) 1 N (g) reddish colorless C The system remained in equilibrium as long as the pressure remained constant. CEach time the pressure was changed (a stress introduced), the system reacted and a new equilibrium position was established. 19

20 An increase in pressure upset the equilibrium by causing the gas molecules to get closer together. This caused a temporary shift towards the right (forward reaction). The system shifted towards the right because, as the equation shows, two N0 2 (g) molecules transform themselves into one N (g) molecule. This diminishes the total number of molecules in order to compensate for the reduced volume. A decrease in pressure upset the equilibrium by causing the gas molecules to take up more space. This caused a temporary shift towards the left (reverse reaction). The system shifted towards the left because, as the equation shows, one N (g) molecule transforms itself into two N 2 0(g) molecules. This increases the total number of molecules in order to compensate for the larger volume available. Remember: For any system in equilibrium containing gases, if the pressure is increased, the system will shift towards the side having the fewest number of gaseous moles. Here are some examples of what happens when we INCREASE the pressure (the reverse occurs when we decrease the pressure): A(g) + 2B(g) C(g) + D(g) 3 gas moles 2 gas moles System shifts right 2A(g) + B(g) 2C(g) + 3D(g) 3 gas moles 5 gas moles System shifts left 3A(s) + 2 B(g) 2 C(g) + D(g) 2 gas moles 3 gas moles System shifts left A(g) + B(g) C(g) + D(g) 2 gas moles 2 gas moles No effect 3A(s) + 2 B(s) 2C(s) + D(g) 0 gas moles 1 gas mole System shifts left 20

21 Experiment 5: Equilibrium and a Catalyst Objective: To investigate the effects of a catalyst on equilibrium Recall that a catalyst increases the rate of a reaction. Does a catalyst also affect the equilibrium state of a system? Let's find out. (NH 4 ) 2 S 2 O 8 (aq) + 2 KI(aq) I 2 (aq) + K 2 S0 4 (aq) + (NH 4 ) 2 S0 4 (aq) Our system consists of equal volumes of ammonium peroxydisulfate, (NH 4 ) 2 S (aq) and potassium iodide, KI(aq). After a short time, the system reaches equilibrium. To find out whether or not a catalyst affects equilibrium, we divide the mixture into two equal parts. One part will be left untouched, while copper ions, Cu 2+, will be added as a catalyst to the other part. Step-1: Pour 20 ml of solution-a and 20 ml of solution-b into a 100-mL beaker. Step-2: Wait approximately 5 minutes to allow the system to reach equilibrium. Ä Note: The solution will turn yellowish. Step-3: To the remaining 20 ml of solution-a, add the remaining 20 ml of solution-b. and add one drop of copper nitrate, Cu(NO 3 ) 2, the catalyst. (Note the time it takes for the reaction to reach equilibrium) Step-4: Observe the reaction and compare it with the uncatalyzed reaction. 32) Describe the properties of the reaction before the addition of a catalyst: 33) Describe the properties of the reaction after the addition of a catalyst: 34) Concerning the catalyst: a) Did the catalyst affect the equilibrium position of the system? Explain: b) Did the catalyst affect the rate at which equilibrium is reached? Explain: ÿ End of experiment-5 21

22 Let's summarize the results of experiment-5 which used the following reaction: (NH 4 ) 2 S 2 O 8 (aq) + 2 KI(aq) I 2 (aq) + K 2 S0 4 (aq) + (NH 4 ) 2 S0 4 (aq) The addition of a catalyst did not affect the equilibrium of the reaction. In effect a catalyst increased both the forward and the reverse reactions equally. As a result, the system arrived at equilibrium faster. A catalyst, therefore, brings the system to equilibrium faster but does not shift the equilibrium. The same equilibrium concentration of reactants and products are present with or without the catalyst. A catalyst brings the reaction to equilibrium faster by lowering equally the activation energy for the forward and the reverse reactions. 35) Complete the following by filling-in the blanks: If an external stress acts on a system in equilibrium, the system has three possibilities of counteracting the stress. One is to favor the reaction by shifting the equilibrium towards the side of the equilibrium equation, the other is to favor the reaction by shifting the equilibrium towards the side of the equilibrium equation or no change (pressure) if there are an equal number of moles (for gases only). Analyzing the results of our experiments on equilibrium, we can make a generalization which helps predict the shift in equilibrium caused by an external stress. This generalization was first discovered and stated by Henri Louis Le Chatelier and, because it is applicable to such a large number of systems, it is called Le Chatelier's Principle. LE CHATELIER'S PRINCIPLE: When a system in equilibrium is subjected to a change, processes occur that tend to counteract the imposed change and the system reaches a new state of equilibrium. Henri Louis Le Chatelier was born in Paris in He was a chemist, a metallurgist and a teacher at Coll ge de France in Paris. He contributed significantly towards the development of plaster and cement. In 1884 Henri Louis Le Chatelier formulated the now famous law bearing his name. 22

23 USING LE CHATELIER'S PRINCIPLE TO PREDICT CHANGES IN EQUILIBRIUM 36) Which reaction, forward or reverse, is favored by the addition of H + (aq) ions to each of the following systems: a) H + (aq) + F - (aq) HF(aq) b) CH 3 COOH(aq) CH 3 COO - (aq) + H + (aq) c) H 2 C0 3 (aq) CO 3 2- (aq) + 2 H + (aq) 37) Here are more systems in equilibrium. For each, one of the species (reactants) has an arrow above it indicating that its concentration has been decreased. Describe the direction of the equilibrium shift and the effect which this change has on the concentration of the other species (reactants). Þ a) NH 4 + (aq) + OH - (aq) H 2 0(l) + NH 3 (g) Þ b) CO 3 2- (aq) + NH 3 (aq) NH 4 + (aq) + OH - (aq) Þ c) HCl(aq) + NH 3 (aq) NH 4 + (aq) + Cl - (aq) Þ d) N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 38) FeSCN 2+ (aq) ions are responsible for the reddish color of the following solution: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) yellow colourless red Explain what happens to the color of the solution if we increase the concentration of the FeSCN 2+ (aq) ions: 23

24 39) The following system in equilibrium is subjected to an increase in pressure: 2 H 2 (g) + O 2 (g) 2 H 2 0(g) a) Which reaction is favored, forward or reverse? Explain: b) What effect does this change have on the quantity of hydrogen molecules in the system? Explain: 40) Suppose we decrease the volume of each of the systems below. Predict in which direction, forward or reverse, the equilibrium shifts. a) N 2 (g) 3 H 2 (g) 2 NH 3 (g) b) SO 2 (g) + H 2 0(l) H 2 S0 3 (aq) c) 4 HCl(g) (g) 2 H 2 O(g) + C1 2 (g) d) CH 3 COOH(aq) H + (aq) + CH 3 COO - (aq) e) H 2 (g) + I 2 (g) 2 HI(g) 41) The fermentation of champagne produces carbon dioxide gas, C0 2 (g), in solution. When the bottle is sealed, a closed system is formed. The fermentation reaction, however, continues producing carbon dioxide gas in the champagne. 24 C0 2 (g) in the champagne C0 2 (g) in the air inside bottle When the bottle is opened, explain why bubbles form. 42) The following aqueous solution is in equilibrium: 2 NaI(aq) + (NH 4 ) 2 S 2 O 8 (aq) (NH 4 ) 2 SO 4 (aq) + I 2 (aq ) + Na 2 SO 4 (aq) colorless colorless colorless yellowish colorless If we add a catalyst to the system, predict the change in color of the solution.

25 43) State two ways of shifting the equilibrium in the following reaction to favor the formation of water vapour. H 2 0(g) H 2 (g) + 1/2 0 2 (g))h = kj Determining Equilibrium Concentrations Often, given some initial concentrations, we want to know the final concentrations of the species at equilibrium. One technique involves the use of a table with the values of the known and unknown concentrations. The number of columns in the table depends upon the number of species in the equation. Here is a typical concentration table for the reaction: A B + C Equation: A B + C Initial 0 0 Reaction Equilibrium ÄImportant: Make sure that the equation is balanced. In the table, the three rows below the equation represent the concentration of the species at three different stages of the reaction: Initial is the initial concentration of the species before equilibrium starts. Reaction is the concentration ratio of the species during the forward reaction (equilibrium being established, reactants used/products produced) Equilibrium is the concentration of the species when equilibrium is established. In using this technique, note that the Initial line represents the situation before the reaction begins. As such, the concentration of the products in the Initial line is always zero (as shown above). Furthermore, the concentrations of the species in the Reaction line are always in proportion with the coefficients of the equation. The reason for this is because they are used and produced according to the coefficients of the equation. What this means is that, if we know any one of the concentrations in the Reaction line, using ratio and proportion we can easily calculate the concentration of the other species for this line (simply by looking at the coefficients of the 25

26 balanced equation). To find the concentration at equilibrium, note that the reactants (the species on the left) are subtracted (Initial - Reaction), while the products (the species on the right) are added (Initial + Reaction). 46) In a 1.0 L flask, a chemist places 4.0 mol of substance A and 4.0 mol of substance B When the system reaches equilibrium, there are 2.0 mol of substance X. Fill in the concentration table below given the reaction: A(g) + 2B(g) 2X(g) + 4Y(g) ÄNote: Concentration must be expressed as moles/litre. In this case, the concentrations are not given directly. What is given is the number of moles and a volume of one-litre. Thus, the given concentrations are: 4 mol/l, 4 mol/l and 2.0 mol/l respectively. Equation: A(g) + 2 B(g) 2 X(g) + 4Y(g) Initial Reaction Equilibrium 47) In a 10 litre container, a chemist places 6.0 mol of N 2 (g) and 10 mol of H 2 (g). At equilibrium, 4.0 mol of N 2 (g) remains. Fill in the concentration table below given the reaction: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) ÄNote: Again, the concentrations are not given directly. We are given the number of moles of the reactants and a total volume. From this, we can calculate the concentrations in mol/l as follows: For the 6.0 mol of N 2 (g), we have 6 mol/l0. litres = 0.60 mol/l For the 10 mol of H 2 (g), we have 10 mol/l0. litres = 1.0 mol/l For the 4.0 mol N 2 (g), we have 4 mol/l0. litres = 0.40 mol/l Equation N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Initial Reaction Equilibrium ã Remember: Concentration must be expressed as moles per litre, mol/l. 26

27 48) Five moles of PCl 5 (g) are sealed in a one litre container. At equilibrium, an analysis shows that the mixture contains two moles of C1 2 (g). Complete the equilibrium concentration table below: PCl 5 (g) PCl 3 (g) + C1 2 (g) Equation: PCl 5 (g) PCl 3 (g) + C1 2 (g) Initial Reaction Equilibrium 49) At room temperature, mol of S0 2 (g) and 0.01 mol of 0 2 (g) were placed in a 100 ml container. At equilibrium, mol of S0 3 (g) were formed. Fill-in the equilibrium concentration table for this reaction: 2 S0 2 (g) (g) 2 S0 3 (g) ÄNote: The volume here is 100 ml (or L). Since the unit for concentration is moles per LITRE, we need to express the given concentrations as mol/l. Thus; for the mol of S0 2 (g), we have mol/0.100 litres = mol/l for the 0.01 mol of 0 2 (g), we have 0.01 mol/0.100 litres = mol/l for the mol of S0 3 (g), we have mol/0.100 litres = mol/l Equation: 2 S0 2 (g) (g) 2 S0 3 (g) Initial Reaction Equilibrium 27

28 TOPIC-3: THE EQUILIBRIUM CONSTANT Using Le Chatelier's Principle, we can predict qualitatively the effect of a change (stress) when one of the three factors disrupts equilibrium: Œ A change in concentration A change in temperature Ž A change in pressure It would be useful to predict quantitatively the effects of a stress. For example, while qualitatively we can predict the shift in equilibrium caused by a change in concentration, it would be useful to predict the quantitative effect. Mathematical predictions are derived from mathematical formulas. Thus, we need to develop a formula (in accordance with Le Chatelier's Principle) to mathematically predict the effects on equilibrium caused by a specific external stress. EXPERIMENT-6: Discovering a mathematical formula Objective: To discover a mathematical relationship between the concentration of the species (reactants) and their effects on a system in equilibrium. The system we'll study is the dissociation of acetic acid (vinegar): CH 3 COOH(aq) H + (aq) + CH 3 COO - (aq) We have two different concentrations of acetic acid (CH 3 COOH or HCH 3 COO). For convenience, they have been labeled SYSTEM-1 and SYSTEM-2.. SYSTEM-1: CH 3 COOH 1.0 x 10-1 mol/l SYSTEM-2: CH 3 COOH 1.0 x 10-3 mol/l Record your data in the table below. Step-1: Step-2: Step-3: Record the temperature of the systems (room temperature) Using ph paper (or a ph meter if available), determine the ph of SYSTEM-1. Determine the ph of SYSTEM-2. 28

29 50) Record your experimental data here: At Start At Equilibrium System Concentration (mol/l) Temperature ( o C) ph [H + ] by calculation x x ) Now calculate and fill-in the concentration of hydrogen ions, the [H + ] column in the table above, for SYSTEM-1 and SYSTEM-2. (Convert and record your answers in scientific notation.) 52) Complete the concentration table below for SYSTEM-1 by: a) Filling-in the Initial line with your experimental data (from previous table). b) Filling-in the [H + ] in the Equilibrium line (with data from previous table). c) Using ratio and proportion as well as calculation to complete the table. Equation: CH 3 COOH H + + CH 3 COO - Initial Reaction Equilibrium 29

30 53) Complete the concentration table below for SYSTEM-2 (same as above): Equation: CH 3 COOH H + + CH 3 COO - Initial Reaction Equilibrium 54) Summarize the Equilibrium concentrations of SYSTEM-1 and SYSTEM-2 here: Equilibrium Concentrations (mol/l) System CH 3 COOH H + CH 3 COO Now that we have the concentrations at equilibrium for both systems, let's try to discover a mathematical relationship between concentration and equilibrium. 55) Using the concentration values from the "summary" table above, calculate the value of the following mathematical combinations (for both systems): SYSTEM-1 SYSTEM-2 Œ [CH 3 COOH] x [H + ] x [CH 3 COO - ] [H + ] x [CH 3 COO - ] [CH 3 COOH] Ž [CH 3 COOH] + [H + ] + [CH 3 COO - ] 56) Study the results of your calculations for SYSTEM-1 and SYSTEM-2. How do they compare with the results of other students? 30

31 57) What general conclusion can you make from this experiment? End of experiment-6 äthe LAW OF MASS-ACTION Cato Maximilian Guldberg and his brother-in-law, Peter Waage, studied equilibrium reactions at Christiana University in Oslo, Norway. They arrived at a conclusion similar to yours. They concluded that there is a definite mathematical relationship governing the behavior of equilibrium systems and concentration changes. They called their conclusion the "Law of Mass- Action". Although mathematically simple to use, in words, the law sounds complicated. It states: The multiplication of the molar concentration of the products, raised to the power of their coefficients, divided by the multiplication of the molar concentration of the reactants, raised to the power of their coefficients, results in a constant value provided the temperature does not change. The constant is known as the equilibrium constant and is designated by the letter "K" and is sometimes written as Keq or Kc. For example: In the following reaction, which occurs at a temperature of C, the calculation of K is: 2NH 3(g) 3H 2 (g) + 1N 2 (g) K = [H 2 ] 3 x [N 2 ]' [NH 3 ] 2 K = 6.02 x 10-2 mol 2 /L 2 Coupled with Le Chatelier's Principle, the Law of Mass-Action is very useful. 31

32 According to Le Chatelier's Principle, when we modify the conditions of a system in equilibrium, the system tries to undo the modification as much as possible. Moreover, the Law of Mass-Action tells us that, in reacting to the modification, the system will adjust the concentrations of the species in such a way that the value of the equilibrium constant (K) does not change. In the reaction above, if we change the concentration of any species, the equilibrium constant remains at 6.02 x 10-2 (provided the temperature is not changed). * IMPORTANT: When using the equilibrium formula, ignore the concentration of solids and of water in the liquid state (as their concentration remains constant). Consider the dissociation of ammonia: 2 NH 3 (g) 3 H 2 (g) + 1 N 2 (g) The Law of Mass-Action tells us that we may change the concentration of any species without changing the value of the equilibrium constant. Thus, for example, we may add hydrogen gas, H 2 (g), to this system. As we do, the system responds by reducing the concentration of hydrogen gas. It does this by shifting the equilibrium towards the left (thereby consuming more hydrogen gas). Moreover, the adjustment takes place in such a way that the value of the equilibrium constant remains "constant". 58) State the expression of the equilibrium constant for each system below: a) CO(g) + 2 H 2 (g) CH 3 OH(g) K = b) H 2 (g) + C1 2 (g) 2 HCl(g) K = c) 2 N0 2 (g) 2 NO(g) (g) K = d) 0 2 (g) 0 2 (aq) kj K = e) 4 HCl(g) (g) 2 H 2 0(g) + 2 C1 2 (g) K = Since there are different types of equilibrium systems, it is convenient to group them into families. To differentiate the equilibrium constants of one family from another, it is customary to use a subscript. 32

33 Here are some examples of equilibrium families and the designation of their equilibrium constants: CHEMICAL EQUATION Constant COMMENT CH 3 COOH(aq) CH 3 COO - (aq) + H + (aq) Ka Dissociation of substances HF(aq) F - (aq) + H +( aq) which liberate H + (aq) ions HNO 2 (aq ) NO - 2 (aq ) + H + (aq) (acids). BaSO 4 (s) Ba 2+ (aq) + SO 2-4 (aq) K sp Equilibrium between a Pbl 2 (s) Pb 2+ (aq) + 2 I - (aq) solid and its ions in an AgCl(s) Ag + (aq) + Cl - (aq) aqueous solution. H 2 O(l) H + (aq) + OH - (aq) K w Dissociation of water N 2 (g) + 3 H 2 (g) 2 NH 3 (g) K Other types of equilibrium H 2 (g) + I 2 (g) 2 HI(g) systems. PCl 5 (g) PCl 3 (g) +C1 2 (g) 0 2 (g) O 2 (aq) 59) State the expression of the equilibrium constant for each system below: a) AgC1(s) Ag + (aq) + Cl - (aq) K sp = b) H 2 0(1) H + (aq) + OH - (aq) K w = c) 2 N0 2 (g) 2 NO(g) (g) K = d) 0 2 (g) 0 2 (aq) kj K = 60) The equilibrium system below occurs at a temperature of C. Given the equilibrium concentrations, find the equilibrium constant (K). PC1 5 (g) PCl 3 (g) + C1 2 (g) [PCl 5 (g )] = mol/l [PCl 3 (g)] = mol/l [C1 2 (g)] = mol/l 33

34 61) Acetic acid dissolves in water according to the following equation: CH 3 COOH(aq) H + (aq) + CH 3 COO - (aq) Calculate the value of the equilibrium constant, at 25 o C, given the concentrations of the species at equilibrium to be: [CH 3 COOH(aq)] = 0.20 mol/l [CH 3 COO - (aq)] = mol/l [H + (aq)] = mol/l 62) The temperature of a saturated solution of strontium chromate, SrCrO 4 (s), is 25 o C. Calculate the value of the equilibrium constant if the concentration of the species at equilibrium are: [Sr 2+ (aq)] = mol/l [CrO 4 2- (aq)] = mol/l The equation of the reaction is: SrCrO 4 (s) Sr 2+ (aq) + CrO 4 2- (aq) 63) A saturated solution of lead iodide, PbI 2, is represented by this equation: PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) Calculate the equilibrium constant for this reaction, given that at 25 0 C the concentrations of the products are: [Pb 2+ (aq)] = 0.00l3 mol/l [I - (aq)] = mol/l 34

35 64) An equilibrium system contains, at C, 0.25 moles of H 2 (g), 0.25 moles of I 2 (g) and 1.75 moles of HI(g). Calculate the equilibrium constant at this temperature knowing that the volume of the system is 0.50 litres. H 2 (g) + I 2 (g) 2 HI(g) + heat 65) A 10 litre sealed flask contains 0.40 moles of nitrogen gas (N 2 ), 3.6 moles of hydrogen gas (H 2 ), and moles of ammonia gas (NH 3 ) at a temperature of C. Determine the equilibrium constant for this system. Reminder: When calculating the equilibrium constant, concentration is expressed as mol/l. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 66) At 25 0 C, the concentrations of sulfate ions (S ) and barium ions (Ba 2+ ) are 1.02 x 10-5 mol/l. Calculate the equilibrium constant for this system. BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) 67) Contrary to what one might think, pure water does not contain only H 2 0(1) molecules. Actually, at 25 0 C, for every litre of water, there are 1.0 x 10-7 moles of water molecules which dissociate according to this equation: H 2 0(l) H + (aq) + OH - (aq) Calculate the value of the equilibrium constant for this system. 35

36 t TOPIC 4: THE CONSTANT RULES The reactions representing the dissociation of acids in a solution form a particular family of equilibrium systems whose constant is known as the "acidity constant", K a. We can represent these reactions with the following general equation: HB(aq) H + (aq) + B - (aq) where B represents an anion (negative ion). Let's study the acidity constant (K a ) of two different acids. EXPERIMENT 7: Is K a A Characteristic Property? Objective: To determine and compare the acidity constants (K a ) of two different acids. 68) Outline an experimental procedure to determine the acidity constants of 1.00 x 10-3 M acetic acid (CH 3 COOH) and 1.00 x 10-3 M Lactic acid (CH 3 CHOHCO 2 H). 69) Record your experimental results: 70) Determine the value of the equilibrium constants: 71) Classify the acids in increasing order of their acidity constant, K a. ΠACID K a CONSTANT 36

37 72) Which acid has dissociated the most, thereby providing the greatest number of H + (aq) ions? Explain. 73) Define the expression, degree of dissociation of an acid. ÄEnd of experiment 7 74) Describe the relationship between: a) The acidity constant (K a ) and the concentration of H + (aq) ions. b) The acidity constant (K a ) and the degree of dissociation. 75) Which of the following acids has the greatest tendency to dissociate in an aqueous solution? Explain: HCHO 2 K a = 2.l x l0-4 HNO 2 K a = 4.5 x l0-4 ƒ HCN K a = 7.0 x l ) A strong acid is a substance which, in an aqueous solution, highly dissociates to produce H+(aq) ions. Which of the following acids is the strongest? Explain: HCIO 2 K a = l.2 x l0-2 HClCH 2 CO 2 K a = l.4 x l0-3 ƒ HC1O K a = l.0 x ) Which acid produces a solution with the highest [H + ] level? The highest ph level? Explain: HF K a = 6.7 x l0-4 HCN K a = 7.0 x l

38 Review (Acids and Bases). ACIDS-BASES An acid is a compound which when dissolved in water produces a solution that: Conducts electricity Reacts with metals (such as Zn and Mg) to give off hydrogen gas (H 2 ) ƒ Changes blue litmus paper to red Tastes sour (example; vinegar) A base is a compound which when dissolved in water produces a solution that: Conducts electricity Reacts with acids to neutralize its properties ƒ Changes red litmus paper to blue ( F Hint: remember Base Blue) Tastes bitter and feels slippery (example: javel water) WATER In pure water, the concentration of hydrogen ions equals the concentration of hydroxide ions. That is, [H + ] = [OH - ] and the water is said to be neutral. ÄNote: It has been found experimentally that K w = [H + ] x [OH - ] = 1.0 x l0-14 Therefore; [H + ] = 1.0 x 10-7 mol/l [OH - ] = 1.0 x l0-7 mol/l. NEUTRALIZATION REACTION A reaction between an acid and a base neutralizes each other forming a salt and water. ACID + BASE à SALT + WATER Example: HCl + NaOH à NaCl + H 2 0 ÄNote: All of the hydrogen ions of the acid (H + ) combine with all of the hydroxide ions (OH - ) of the base. Net ionic equation: H + (aq) + OH - (aq) à H 2 O(l) 38

39 Review (continued) ph SCALE (by definition ph = -log[h + ] A scale used to conveniently specify how acidic or basic a solution is: TITRATION Acid Neutral Vinegar (2.5) Blood (7.4) Base This is an experimental technique to find the concentration of an unknown solution by reaction with a known solution. The process involves neutralizing the solution (adding an acid to a basic solution or vice-versa). By measuring the amount of acid (or base) added whose concentration we know, we can find the concentration of the unknown solution. Remember: ACID BASE (H + concentration)(volume) = (OH - concentration)(volume) ÄNote: Be careful about the particular acid or base used. Write the dissociation equations of both the acid and the base in order to determine the correct H + and OH - concentrations. If the ph is less than 7, the solution is acidic. If the ph is greater than 7, the solution is basic. A ph of 7 is a neutral solution. Some common acids: HCl (hydrochloric acid, also known as muriatic acid)) H 2 S0 4 (sulfuric acid) HN0 3 (nitric acid) HC 2 H or CH 3 COOH (acetic acid or vinegar) BUFFERS Buffers are mixtures of chemicals which make a solution resist a change in its ph. Solutions which have a resistance to changes in their ph value because of the presence of buffers are called buffer solutions. 39

40 ACID-BASE INDICATORS Modern chemistry labs have ph meters which conveniently measure the ph of a solution. Traditionally, however, ph papers are used. When ph paper is dipped into a solution, the wetted portion of the paper changes color. By color comparison, the ph value can be determined. Certain liquids are also used as indicators. ÄNote: When the ph is specified as a whole number, the [H+] equals one times ten to the negative power of the ph value. For example, if the ph = 4, then [H+] = 1 x l0-4 mol/l. However, if the ph is a fractional number, for example, 6.4, then you can use a scientific calculator to obtain the [H+] like this: Once you have calculated the [H+] using one of the methods above, convert your answer to scientific notation. For example, if the ph is 4.2, [H+] = 6.3 x l0-5 M DISPLAY Press: 1 X 1 0 Y X 4.2 +/- = = 6.3 x

41 t TOPIC 5: ACID-BASE TITRATION Recall that a neutralization reaction is one in which a base and an acid neutralize each other forming a salt and water. What exactly happens during the process of neutralization? Let's find out with an experiment. We will analyze the behavior of an acid solution by the gradual addition of a basic solution. Recall that during a neutralization reaction, the H + (aq) ions of the acid react with the OH - (aq) ions of the base to form water. H + (aq) + OH - (aq) H 2 0(l) EXPERIMENT-8: The acid- base competition Objective: To study acid-base reactions. This experiment consists of two parts: PART-A: Neutralization of a strong acid PART-B: Titration techniques PART-A: Neutralization of a strong acid We will perform the neutralization of hydrochloric acid, HCl(aq), by gradually adding (from a burette) sodium hydroxide, NaOH. Note that both the acid and the base have a concentration of 0.10 mol/l. Step-1: Using a funnel, fill a 50 ml burette with 0.10 M NaOH ÄNote: For convenience, add slightly more NaOH than required then carefully drain the excess into a "waste" beaker to obtain the desired level. Step-2: As accurately as possible using a pipette, pour 10.0 ml of HCl into a 50 ml Erlenmeyer flask. ÄHint: If a pipette is not available, measure the HCl using another burette or a 10 ml graduated cylinder Step 3: Add three drops of phenolphthalein to the acid solution and stir gently. (The phenolphthalein will be used as a ph indicator) 41

42 Step-4: From the burette, you will add 1.0 ml portions of NaOH (one at a time) into the HCl flask. (You will be adding 20 separate portions in all stirring gently between each portion.) Step-5: After each addition of NaOH, dip a stirring rod into the flask. Wet a ph indicator paper and check the ph of the solution. Record the ph value and your observation in the table below. (If available, use a ph meter) Ä Note: Rinse the ph meter with distilled water after each use. Step-6: Stop the neutralization when 20.0 ml of NaOH have been added. Vol. of HCl (ml) 1. Total Vol. of NaOH added (ml) ph Observations ) In general, how does the ph of an acid vary with the addition of a base? 79) In your experiment, how did the ph vary as you added the NaOH solution? 42

43 80) Using your experimental data, trace the graph of the ph (y-axis) as a function of the volume (x-axis) of the base added. ÄNote: For an acid to be neutralized by a base (of the same concentration), it is necessary that the quantity of the acid equals the quantity of the base. We can identify this "equivalence" from the graph as it represents that section having the steepest slope and is known as the equivalence point. At this point, the solution loses its acidic and basic properties as the number of H + (aq) ions equals the number of OH - (aq) ions. 81) From the graph, determine the ph at the equivalence point. 82) What is the ph value for the turning point in your experiment? 43

44 PART-B: Titration techniques A titration is an accurate method of determining the concentration of an unknown solution using a solution of known concentration. In a titration, we carefully determine the equivalence point. Once we have determined the equivalence point, we can calculate the concentration of the unknown solution. To perform the titration, follow the steps below. When finished, record the volume and concentration of the base added. Ü Be prepared to be patient! Step 1: Step 2: Step 3: Step 4: Place 10.0 ml of an HCl acid solution whose concentration is unknown into an Erlenmeyer flask. Add three drops of phenolphthalein to the acid solution and stir gently. (The phenolphthalein will be used as a ph indicator) Adjust the level of NaOH in the burette to the nearest whole number. Using the burette, add the NaOH base to the solution in the Erlenmeyer flask by portions of about 1 ml. ÄNote: When you notice the solution has a tendency of remaining pink in color, you are approaching the equivalence point. When this happens, rather than use 1 ml portions, add the NaOH drop by drop (so as not to over do it). IMPORTANT: As you stir the solution, the pink color should disappear. If the color is very pink, you have passed the end point and must re-do the titration. Step 5: Continue adding "drops" of NaOH (stirring gently each time). The "end point" is reached when the entire solution remains slightly pink. ÄHint: Be sure to record the volume of NaOH added. This way, should you add too much and need to start over, you will know how much is "too much". 83) Record your data here: Volume of NaOH: Concentration of NaOH 84) Determine the number of moles of NaOH: 44

45 85) At the equivalence point: The number of moles of base = the number of moles of acid = 86) Concerning the unknown acid, from your titration: a) How many moles of acid were you given? b) What was the concentration of this acid? (See below) ÄEnd of experiment-8 Review of concentration calculations Note that we can mathematically find the concentration of an acid or a base. In doing so, four variables are involved: the concentration and volume of the acid (C A and V A ), and the concentration and volume of the base (C B and V B ). Knowing that Concentration = moles/volume, moles = Concentration x Volume And, since moles of acid (H + ) = moles of base (OH - ), we have: cone. of acid x volume of acid = conc. of base x volume of base or: C A x V A = C B x V B [H + ] x V A = [OH - ] x V B [H + ] = [OH - ] x V B V A = (0.5)(20) = 0.2 mol/l (50) Now using the equation for the dissociation of H 2 S0 4, we get: 1H 2 S0 4 2H + + SO 4 2- x 0.2 2x = (1)(0.2) x = [H 2 S0 4 ] = 0.1 mol/l 45

46 t TOPIC 6: ENRICHMENT An example calculating the acidity constant (K a ) for hydrogen sulfide acid will familiarize you with the technique used to solve such problems. Sample Problem Suppose we have a solution of hydrogen sulfide acid, H 2 S, whose concentration is 1.0 x 10-1 mol/l. We determine experimentally that, at equilibrium, [H + ] ions is 1.0 x l0-4 mol/l. What is the value of the acidity constant of H 2 S? H 2 S(aq) H + (aq) + HS - (aq) Our technique in finding the equilibrium constant involves creating an equilibrium concentration table illustrating the three stages of equilibrium. After filling-in the table, we use the Law of Mass Action to find the value of the acidity constant (K a ). Recall that in the concentration table, the balanced equation is written on the first line, the Initial line represents the concentration of the system before the reaction starts, the Reaction line represents the concentration of the reactants used and the products produced (equilibrium being established), and the Equilibrium line represents the concentrations of the substances at equilibrium. Like this: Equation: H 2 S (aq) H + (aq) + HS - (aq) Initial Reaction Equilibrium From the given concentrations, we fill-in the table accordingly. In this case, we are given that at the start of the reaction, the concentration of H 2 S is 1.0 x 10-1 mol/l and that at equilibrium, the concentration of H + is 1.0 x 10-4 mol/l. Thus, we have: Equation: H 2 S (aq) H + (aq) + HS - (aq) Initial 1.0 x Reaction Equilibrium 1.0 x

47 We can now complete the H + column this way: Initial = 0 Reaction = +? Equilibrium = 1.0 x 10-4 Our table now looks like this: Equation: H 2 S (aq) H + (aq) + HS - (aq) Initial 1.0 x Reaction x Equilibrium 1.0 x 10-4 Remember that the concentrations on the Reaction line are in proportion with the coefficients of the equation. This means that if we know any concentration on this line, we can easily find the rest of the concentration (on the Reaction line). In this case, the concentrations are in the ratio of 1 to 1 to 1 Thus, we can fill-in these values in the table: Equation: H 2 S (aq) H + (aq) + HS - (aq) Initial 1.0 x Reaction x x x 10-4 Equilibrium 1.0 x 10-4 We can now fill in the Equilibrium line by subtraction in the H 2 S column and by addition in the HS - column so that our completed table looks like this: Equation: H 2 S (aq) H + (aq) + HS - (aq) Initial 1.0 x Reaction x x x 10-4 Equilibrium 9.9 x x x 10-4 Using the concentrations in the Equilibrium line, we solve for K a : K a = [H + (aq)] x [HS - (aq)] = (l.0 x l0-4 ) x (l.0 x l0 4 ) = 1.0 x10-7 [H 2 S(aq)] (9.99 x 10-2 ) 47

48 87) In a 4.0 L sealed container, we place 4.0 mol of dinitrogen tetroxide, N (g). The gas partially dissociates and establishes equilibrium at a temperature of 27 0 C. If, at equilibrium, there is 1.6 mol of nitrogen dioxide N0 2 (g), calculate the equilibrium constant for this system. a) Determination of concentrations: ÄNote: Be sure to specify the given concentrations as moles/litre. This means the volume of the container must be one litre. Otherwise, if the given volume is greater than one litre, divide the number of moles by the number of litres, if the given volume is specified in ml, divide the number of moles by Equation: N (g) 2 N0 2 (g) Initial: Reaction: Equilibrium: b) Determination of the equilibrium constant: 88) At 25 0 C, NOBr(g) partially decomposes. Knowing that initially there were 3.0 mol of NOBr(g) and that at equilibrium there remains 2.0 mol, determine the value of the equilibrium constant. The volume of the system is 1.0 L. Equation: 2 NOBr(g) 2 NO(g) + Br 2 (g) Initial: Reaction: Equilibrium: 48

49 89 At 25 0 C, a saturated solution is obtained by adding 2.0 mol of calcium sulfate (CaSO 4 ) to 1.0 L of water. Upon analysis, it is determined that the concentration of the calcium ions (Ca 2+ ) is 7.81 x l0-3 mol/l. Determine the equilibrium constant (also known as the solubility constant, K sp ). CaSO 4 (s) Ca 2+ (aq) + SO 4 2- (aq) 90) In a given solution at 25 0 C, the concentration of H + (aq) ions is 2.5 x 10-4 mol/l. What is the concentration of the OH - (aq) ions if the equilibrium constant is 1.0 x 10-14? H 2 0(1) H + (aq) + OH - (aq) 49

50 91) Acetic acid (CH 3 COOH) dissociates partially in water at a temperature of 25 0 C. At equilibrium, the concentration of the acid is mol/l. What is the concentration of the H + (aq) ions in this solution if the equilibrium constant is 1.8 x 10-5? CH 3 COOH(aq) H + (aq) + CH 3 COO - (aq) 92) In a 1.0 L balloon, we place 5.0 mol of nitrogen dioxide gas (NO 2 ). At equilibrium, we find 1.5 mol of oxygen gas (O 2 ). What is the value of the equilibrium constant for this system? 2 N0 2 (g) 2 NO(g) + O 2 (g) 50

51 Supplementary Problems Part 1: Neutralisation Problems 1. Exactly 12.0 ml of M NaOH are required to neutralize 20.0 ml of HCl solution. What is the concentration of HCl? 2. What is the concentration of acetic acid in vinegar when 32.5 ml of 0.56 M NaOH are required to neutralize 15.0 ml of vinegar? 3. A sample of Saniflush containing 81.0 g of NaHSO 4 is dissolved in 4.0 L of water and 50.0 ml of it is titrated with 2.0 M KOH.. What is the volume of KOH required for a complete reaction? 4. An effluent from a research station was suspected of containing phosphoric acid H 3 P0 4 To determine the acid concentration, it was titrated with 1.0 M NaOH. If 17.3 ml of NaOH were required to remove the first H + in a 10.0 ml sample of the acid, what is the molarity of the acid? 5. A 25.0 ml sample of muriatic acid (slightly impure HCl sold as a cleanser) reacted exactly with 45.0 ml of NaOH. If the NaOH had a ph of 12.0, find the ph of the acid. 6. Oxalic acid, H 2 C 2 0 4, is a solid acid that is slightly soluble in water. When 6.25 g of acid are dissolved in water containing the indicator phenol red, the solution turns yellow. If 32.2 ml of KOH are required to change the indicator to red at the equivalence point, what is the concentration of the KOH? Part 2: Le Chatelier's Principle Problems 7. Acetic acid is a weak carboxylic acid which ionizes in water as follows: CH 3 COOH CH 3 COO - + H + If one drop of HCl is added to this system: a) in which direction will the equilibrium shift? b) what effect will this have on the concentration of each substance in the system? 8. If one drop of NaOH is added to the system in question 7, a) in which direction will the equilibrium shift? b) what effect will this have on the concentration of each substance in the system? 9. In large, public swimming pools, Cl 2 (g) can be added to the water to produce hypochlorous acid, HOCl, which prevents the growth of algae and bacteria. The equilibrium system is: Cl 2 (g) + H 2 O(l) Cl - (aq) + H + (aq) + HOCl (aq) a) If the ph increases, what happens to the [H + ] in the system? b) If the ph increases, in which direction will the equilibrium shift? 51

52 10. Tooth decay results in the dissolving of tooth enamel, Ca 5 (PO 4 ) 3 OH, in the mouth. The following equilibrium is set up: Ca 5 (PO 4 ) 3 OH(s) 5 Ca 2+ (aq) + 3 PO OH - (aq) When sugar ferments on teeth, H + ions are added to the system. a) In which direction does this change cause the equilibrium to shift? b) What effect does this shift have on the tooth enamel? 11. Chicken eggs are formed by the precipitation of calcium carbonate, CaCO 3. Chickens, like dogs, do not perspire and, therefore, in hot weather they must resort to panting". This releases water and carbon dioxide from their bodies. Using the following sequence of steps: 1. H 2 0( l ) + C0 2 (g) CO 2 (aq) H 2 CO 3 (aq) H + (aq) + HCO 3 - (aq) 2. HCO 3 - (aq) H + (aq) + CO 3 2- (aq) 3. Ca 2+ (aq) + CO 3 2- (aq) CaCO 3 (s) a) explain why chickens lay eggs with thinner shells in summer. (hint: When they pant, which way does the equilibrium shift?) b) explain why giving the chickens soda water (carbonated water) to drink solves the whole problem. Part 3: Equilibrium Constant Problems 12. a) Write the equilibrium equation for the ionization of one proton (H + ) from H 2 SeO 3 (aq). b) Write the equilibrium constant expression. 13. The ph of a 0.1 M solution of cyanic acid (HCNO) is 4.0. a) Write the equation for the ionization of cyanic acid b) Write the K a expression and calculate the value of K a. 14. The K a for hypoiodous acid (HOI) is 2.5 x What would you expect the [H + ] of a M solution of this acid to be? 15. A M solution of HOBr has a ph of Find K a. 16. A 0.25 M solution of benzoic acid (C 6 H 5 COOH) has a [H + ] of 4.0 x 10-3 M. Find K a. 17. K a = 1.3 x 10-2 K a = 6.7 x 10-4 K a = 6.6 x a) Which acid is the strongest? b) Which acid has the highest ph? 52

53 18. The initial concentration of an aspirin solution (HC 9 H ) is x 10-3 M. At equilibrium, the ph is Find K a. PART 4: ph Problems 19. Solution X has a ph of 2 and solution Y has a ph of 4. Which of the following statements is true? a) The hydrogen ion concentration of X is half that of Y. b) The hydrogen ion concentration of X is twice that of Y. c) The hydrogen ion concentration of X is 100 times that of Y. d) The hydrogen ion concentration of X is 1/100 that of Y. 20. If the ph value of an electrolyte changes from 4 to 3, what has happened to the hydrogen ion concentration and the hydroxide ion concentration? 21. a) What is the ph of a 0.01 M nitric acid solution? Nitric acid is a strong acid (ie. 100% dissociated). b) What is the ph of a 0.01 M solution of hydrofluoric acid that is 8% ionized? 22. Calculate the ph of a M solution of potassium hydroxide, a strong base. 23. A solution of hydrobromic acid has a ph of Find the concentration of hydrogen ions. 24. a) Complete the following table: ph [H + ] 4.26?? 6.25 x 10-3 M 2.95?? 1.80 x 10-1 M b) List the acids in order from strongest to weakest. 25. Vinegar is about 100 times more acidic than distilled water. What is the approximate ph of vinegar? 26. Ammonia is about 100 times more basic than distilled water. What is the approximate ph of ammonia? 27. If the ph of a solution decreases by a factor of 1, the concentration of hydrogen ions by a factor of. 28. A strong acid is one which has a ph value and a K a value. 29. A weak acid is one which has a ph value and a K a value. 30. In pure water, the [H+] = [OH - ] = 1.0 x 10-7 M. The ph value = -log[h+] =? 53