CHM 152LL: Le Chatelier s Principle Prepare the prelab in your laboratory notebook BEFORE you attend your lab class. You will need 1 data table for each change made to a system. The headings in the procedure section indicate the number of changes that will be made to each system. Be sure to include the net ionic equation and original color of the equilibrium mixture above the change tables for each system. See page 2 for an example table showing one change. This week you will also take home a small vial container to bring a sample of a beverage you drink often so its ph can be measured next week. Purpose: In this experiment you will observe shifts in equilibrium systems when conditions, such as the concentration and temperature, are changed. You will explain the observed color changes of four reactions in terms of Le Chatelier s principle. Introduction: Equilibrium systems involve reversible reactions in which the forward and reverse reactions are occurring at equal rates. If an equilibrium system remains undisturbed, the concentration of each chemical in the system will remain constant indefinitely. If the system is disturbed, one of the rates (forward or reverse) will temporarily increase, and when the system reaches a new equilibrium state, the concentration of each chemical will be different than the original yet the equilibrium constant remains the same. Le Chatelier predicted how such an equilibrium system would respond to changes imposed on it. In this experiment you will investigate four different equilibrium systems. For the first two systems, colored substances are used to allow you to determine the direction of equilibrium shifts based on color changes. The substances involved in the second two systems are weak acid and weak base equilibrium reactions, so indicators are added to help detect equilibrium shifts. (An indicator is a substance that changes color in a specific ph range so you will also be able to determine whether the mixture is becoming more acidic or basic.) The following format should be used to set up separate data tables (for your pre-lab) for each of the four equilibrium systems. 1) Write the appropriate balanced net ionic reaction above the data tables for each system. Leave space to the right of all 4 net ionic equations to record the original mixture color once reagents are combined. (Note: You only need to write the net ionic reaction one time for each system, then prepare the indicated number of change tables below the reaction.) 2) Label three columns with the following headings: Change Imposed, Observation, and Explanation. a. The change imposed should describe the chemical added or the appropriate temperature change. b. Your observations should describe the color changes observed, e.g., the blue solution turned lighter blue or the red solution turned green. c. For your explanations, answer the following questions: 1) What change was imposed on the system? (i.e. What chemical s concentration in the reaction was increased or decreased, or was heat added or removed?) 2) Does the reaction shift right (forward direction) or left (reverse direction)? 3) As a result of this shift, which chemical has increased or decreased in concentration causing the observed color change? For example, consider the following equilibrium reaction between cobalt(ii) nitrate and potassium chloride. For convenience we are showing the net ionic reaction: Co 2+ (aq) + 4 Cl - (aq) CoCl 2 4 (aq) pink blue When cobalt(ii) nitrate and potassium chloride are added together at room temperature the resulting solution is purple once equilibrium is reached, indicating that both Co 2+ ions and CoCl 4 2- ions (both GCC CHM 152LL: Le Chatelier s Principle page 1 of 8
reactants and products) are present in appreciable amounts. This means this reaction is not heavily reactant or product favored at room temperature. If HCl(aq) were added to the system, the concentration of the Cl - ions would increase (H + ions are spectators), which would cause the equilibrium to shift to the right. As a result of the shift, some of the pink Co 2+ ions would react, while more blue CoCl 2 4 ions would form, thus the solution becomes more blue. An example of the data table is shown below. Co 2+ (aq) + 4 Cl - (aq) CoCl 2 4 (aq) (original mixture: purple color) Pink colorless blue Change Imposed Observation Explanation add HCl(aq) The purple solution turns blue. 1) [Cl - ] 2) Reaction 3) [CoCl 2 4 ], so solution turns blue Drawing Example: When we draw an equilibrium reaction, we must draw all the chemicals present in solution, not just the substances in the net ionic reaction. Thus, the spectator ions are included in the drawing below. Note that this reaction sets up an equilibrium, therefore the resulting solution is not just products, but a mixture of both reactants and products. (To have equilibrium all the chemicals in the reaction must be present!) It is important to note that all the chemicals in the first two test tubes are present in the last test tube because you cannot create or destroy matter! In the example shown below, two units of cobalt(ii) nitrate are combined with a stoichiometric amount of potassium chloride, which forms the resulting equilibrium solution. (Refer to the net ionic reaction above the table to verify this!) Note that one cobalt ion has reacted, while the other one has not. Also, ions may be drawn separately as shown in the two drawings on the left or a coefficient may be written in front of the ion to indicate the number of ions present as show in the drawing on the right. 2 units Co(NO 3 ) 2 8 units KCl equilibrium mixture: (from 4:1 ratio of 1 Co 2+ combined with 4 Cl - to make CoCl 2-4 ; Cl - to Co 2+ in the 1 Co 2+ and 4 Cl - are left unreacted; net ionic reaction!) spectators ions: 8 K + - and 4 NO 3 Procedure Part I. Formation of CuBr 4 2- and Temperature Effects (2 changes = 2 tables) Cu 2+ ions from copper(ii) sulfate can combine with Br ions from potassium bromide to form the yellow CuBr 4 2- complex ion and thus set up equilibrium as shown in the following net ionic reaction: GCC CHM 152LL: Le Chatelier s Principle page 2 of 8
Cu 2+ (aq) + 4 Br - (aq) CuBr 2-4 (aq) Equation 1 blue yellow You will examine the effect of temperature changes on this equilibrium system. 1. Prepare a hot water bath by filling a 250 ml beaker about half full with tap water and setting it on a hot plate (for use in Step 5 below). Turn the heat setting to 5. While the water bath is heating, you will prepare the solutions below. 2. Place about 15 ml of 0.20 M CuSO 4 (aq) solution into a small beaker. (Note the original colors of the CuSO 4 and KBr solutions.) 3. Add about 5 ml of saturated KBr solution to the same beaker and stir to mix. The equilibrium between the two reactants will set up very quickly. 4. Divide the solution into three roughly equal parts in three test tubes. Keep one tube at room temperature for comparison. Use the other tubes to study the effect of changes in temperature on this equilibrium system. Note: While waiting for steps 5 and 6 below to heat and cool for 15 minutes, you should continue on to the other reactions. 5. Place one of the test tubes in the hot water bath for 15 minutes and record your observations. Make sure to compare to the room temperature sample. 6. Prepare an ice-water/salt bath by adding 1 scoop of rock salt to an ice-water mixture in a 150 ml beaker, and set the third test tube in the ice-water/salt bath to chill. Leave the test tube in the icewater bath for 15 minutes. Compare to the room temperature tube and record your observations. *Discard these solutions in the waste container in the fume hood. Safety: When you have completed each experiment, place rinsed, wet test tubes upside down in the test tube rack to dry. DO NOT attempt to dry the test tubes with a paper towel and glass stir rod since the pressure can break the test tubes. Part II. Formation of FeSCN 2+ (3 changes = 3 tables) Iron(III) ions, Fe 3+, and thiocyanate ions, SCN -, can combine to form the iron(iii) thiocyanate ion, FeSCN 2+, and set up equilibrium as shown in this net ionic equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Equation 2 yellow colorless red You will examine the effects of adding different reagents (FeCl 3, NH 4 SCN, and HgCl 2 ) to this equilibrium system. Note: Hg 2+ ions will react with SCN - ions to form a colorless complex ion, Hg(SCN) 4 2. 1. Place 50 ml of deionized water in a beaker, then add 20 drops of 0.1 M FeCl 3 and 20 drops of 0.1 M NH 4 SCN. Stir the solution until it is thoroughly mixed. Record the color of this equilibrium mixture. The orange-red color is due to the presence of both the red FeSCN 2+ complex ions and the yellow Fe 3+ ions. A change in the concentration of the red-colored ion (FeSCN 2+ ) will result in a change in the depth of color. If the concentration of FeSCN 2+ increases, then the solution will appear a darker orange-red color; whereas, if the concentration of FeSCN 2+ decreases, then the solution will appear lighter, and can be such a pale yellow as to seem almost colorless. 2. Half-fill four test tubes with the equilibrium mixture prepared in step 1. Keep one tube as a color control for comparison. Add 20 drops of the 0.1 M FeCl 3 solution to the second tube, 20 drops of the 0.1 M NH 4 SCN solution to the third tube, and 20 drops of 0.1 M HgCl 2 solution to the fourth tube. Compare the colors of each tube to the control tube and record your observations. *Discard these solutions in the waste container in the fume hood. GCC CHM 152LL: Le Chatelier s Principle page 3 of 8
Part III. Ionization of a Weak Base (1 change = 1 table) The weak base ammonia, NH 3, reacts with water to produce hydroxide ions, OH -, according to the following equilibrium reaction: NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) Equation 3 You will use phenolphthalein as the indicator to study this reaction. Note that phenolphthalein is colorless in acidic solutions and bright pink in basic solutions. 1. Place 15 ml of deionized water in a beaker, then add two drops of 15 M NH 3 and two drops of phenolphthalein indicator (record the original mixture color after addition of indicator). Mix the solution, then divide the solution into two portions by pouring it into two separate test tubes. Keep one tube as a color control for comparison. 2. Add 15 drops of 2 M NH 4 Cl solution to the other tube and gently shake the test tube to mix the contents. Record your observations. *Discard these solutions in the sink. Part IV. Ionization of a Weak Acid (1 change = 1 table) Acetic acid, CH 3 COOH(aq), is a weak acid that undergoes the following ionization reaction with water: CH 3 COOH (aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO - (aq) Equation 4 For this system, you will use the methyl orange indicator, which is red in very acidic solutions having ph < 3.2, yellow in solutions having ph > 4.4, and orange in between. 1. Place 15 ml of deionized water in a beaker, then add two drops of 2 M CH 3 COOH and two drops of methyl orange indicator (record the original mixture color after addition of indicator). Mix the solution, then divide the solution into two portions by pouring it into two separate test tubes. Keep one tube as a color control for comparison. 2. Add 5 drops of 2 M NaCH 3 COO solution to the other tube, and gently shake the test tube to mix the contents. Record your observations. *Discard these solutions in the sink. Le Chatelier s Principle Lab Report Sections: 1) Title, date, name, partners 2) Purpose statement in complete sentences (not copied). 3) Data/results: the balanced net ionic reactions and original mixture colors for the four equilibrium systems, and the change tables providing the observations and explanations for each change. 4) Questions on page 5. Use complete sentences for your answers. 5) Drawings on pages 6 8. 6) Conclusion a paragraph summarizing the findings for the four equilibrium systems. Make sure to include specific examples from each reaction in your discussion! 7) Prelab page with instructor s initials Note: you may NEATLY hand-write the first 3 sections and the conclusion in your lab notebook. GCC CHM 152LL: Le Chatelier s Principle page 4 of 8
Questions 1. Copper(II) sulfate solution is blue and potassium bromide solution is colorless. Why does the solution in Part I turn green when they are added together? What substances are in the resulting solution that make it green? 2. Is the reaction in Part I endothermic or exothermic in the forward direction? (It can t be both!) Explain based on your observations for the hot and cold baths. 3. The chemical HgCl 2 is not part of the reaction in Part II, so explain how its addition caused the equilibrium to shift. 4. Is the solution in Part III becoming more acidic or basic upon addition of NH 4 Cl? Explain. (Hint: what is happening to the hydroxide ion concentration, [OH - ]?) 5. Is the solution in Part IV becoming more acidic or basic upon addition of NaCH 3 COO? Explain. (Hint: what is happening to the hydronium ion concentration, [H 3 O + ]?) 6. Consider the following equilibrium reaction: 2 SO 2 (g) + 2 H 2 O(g) 2 H 2 S(g) + 3 O 2 (g) H = + 1037 kj For a-d, predict the effect of the following changes on the position of equilibrium (left, right, no change), when each of the following changes is made: a. Adding H 2 S(g) b. Decreasing the pressure c. Removing SO 2 (g) d. Increasing the temperature GCC CHM 152LL: Le Chatelier s Principle page 5 of 8
Molecular-level Drawings of the Equilibrium Systems I. Formation of CuBr 4 2- (Hint: See page 2 for a similar example!) Cu 2+ (aq) + 4 Br - (aq) CuBr 4 2- (aq) Equation 1 blue yellow The reaction above is a net ionic reaction which means the spectator ions are not shown. But it is important to remember that the spectator ions are still present in the solution so we will include them in our drawings of reactions in solution. However, we will not draw the solvent water molecules, even though they are also present in all aqueous solutions. To create this equilibrium system we began with the first reactant: the copper(ii) sulfate solution. Draw a representation of two units of copper(ii) sulfate in the first test tube. Draw the other reactant, potassium bromide, in the second test tube in the correct stoichiometric amount required to be added to the two units of copper(ii) sulfate. Refer to to Equation 1 above for the correct mol-mol ratio for the reactants! Then draw the resulting equilibrium mixture in the last test tube and don t forget to include the spectator ions. Remember reactions that set up equilibrium don t go to completion, there are always some reactants and some products in the mixture, so your drawing should represent this. 2 units units copper(ii) sulfate potassium bromide equilibrium mixture GCC CHM 152LL: Le Chatelier s Principle page 6 of 8
II. Formation of FeSCN 2+ Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Equation 2 yellow colorless red The above reaction is a net ionic reaction. To create this equilibrium system we began with the first reactant: the iron (III) chloride solution. Draw a representation of two units of iron(iii) chloride in the first test tube. Then we added the other reactant: the ammonium thiocyanate solution. Draw the correct stoichiometric amount of ammonium thiocyanate in the second test tube. Refer to equation 2 above for the correct mol-mol ratio for the reactants! Then draw the resulting equilibrium mixture in the last test tube and don t forget the spectator ions. Make sure the equilibrium drawing shows all of the substances present in the equilibrium mixture and include proper charges for all ions! 2 units units iron(iii) chloride ammonium thiocyanate equilibrium mixture GCC CHM 152LL: Le Chatelier s Principle page 7 of 8
III. Ionization of a Weak Base NH 3 (aq) + H 2 O(l) NH + 4 (aq) + OH - (aq) Equation 3 Draw a representation of the resulting equilibrium mixture when ammonia is added to water. The equilibrium constant, K, is a small number for this reaction so represent that in your drawing. Do not draw the water molecules. Draw at least 10 molecules and/or ions in the beaker below. IV. Ionization of a Weak Acid CH 3 COOH (aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO - (aq) Equation 4 Draw a representation of the resulting equilibrium mixture when acetic acid is added to water. The equilibrium constant, K, is a small number for this reaction so represent that in your drawing. Do not draw the water molecules. Draw at least 10 molecules and/or ions in the beaker below. GCC CHM 152LL: Le Chatelier s Principle page 8 of 8