1. Equilibrium questions

2. Question 1 For the system 2 SO 2 (g) + O 2 (g)  2 SO 3 (g),  H is negative for the production of SO 3. Assume that one has an equilibrium mixture of these substances. Predict the effect of each of the following changes on the value of the equilibrium constant and on the number of moles of SO 3 present in the mixture at equilibrium. Briefly account for each of your predictions. (Assume that in each case all other factors remain constant.) (a)Decreasing the volume of the system. (b)Adding oxygen to the equilibrium mixture. (c)Raising the temperature of the system.

3. Answer 1 (a)As volume decreases, pressure increases and the reaction shifts in the direction of fewer molecules (less volume; more SO 3 ) to relieve the stress. Value of K eq does not change. (b)Additional O 2 disturbs the equilibrium and SO 3 is formed to relieve the stress. Value of K eq does not change. (c)Increase in temperature shifts the reaction to the left to “use up” some of the added heat. Less SO 3 remains. Value of K eq decreases due to the relative greater increase in the rate of the endothermic reaction (reaction to the left).

4. Question 2 At 25ºC the solubility product constant, K sp, for strontium sulfate, SrSO 4, is 7.6×10 -7. The solubility product constant for strontium fluoride, SrF 2, is 7.9´10 -10. (a)What is the molar solubility of SrSO 4 in pure water at 25ºC? (b)What is the molar solubility of SrF 2 in pure water at 25ºC?

5. Answer 2 (a)SrSO 4 (s)  Sr 2+ (aq) + SO 4 2- (aq) At equilibrium: [Sr 2+ ] = X M = [SO 4 2- ] X 2 = K sp = 7.6x10 -7 X = 8.7´10 -4 mol/L, solubility of SrSO 4 (b)SrF 2 (s)  Sr 2+ (aq) + 2 F - (aq) At equilibrium: [Sr 2+ ] = X M = [F - ] = 2X M K SP = [Sr 2+ ][F - ] 2 = (X)(2X) 2 = 7.9´10 -10 X = 5.8x10 -4 mol/L, solubility of SrF 2

6. Question 3 Answer the following questions that relate to solubility of salts of lead and barium. (a)A saturated solution is prepared by adding excess PbI 2 (s) to distilled water to form 1.0 L of solution at 25˚C. The concentration of Pb 2+ (aq) in the saturated solution is found to be 1.3  10 –3 M. The chemical equation for the dissolution of PbI 2 (s) in water is shown below. PbI 2 (s)  Pb 2+ (aq) + 2 I – (aq) (i)Write the equilibrium-constant expression for the equation. (ii)Calculate the molar concentration of I – (aq) in the solution. (iii)Calculate the value of the equilibrium constant, K sp. (b)A saturated solution is prepared by adding PbI 2 (s) to distilled water to form 2.0 L of solution at 25˚C. What are the molar concentrations of Pb 2+ (aq) and I – (aq) in the solution? Justify your answer. (c)Solid NaI is added to a saturated solution of PbI 2 at 25˚C. Assuming that the volume of the solution does not change, does the molar concentration of Pb 2+ (aq) in the solution increase, decrease, or remain the same? Justify your answer.

7. Answer 3 (a)(i) K sp = [Pb 2+ ][I – ] 2 (ii) [I – ] = 2 [Pb 2+ ] = 2(1.3  10 –3 ) = 2.6  10 –3 (iii) K sp = (1.3  10 –3 )(2.6  10 –3 ) 2 = 8.8  10 –9 (b)[Pb 2+ ] = 1.3  10 –3 M & [I – ]= 2.6  10 –3 ; a saturated solution has the same concentration no matter the volume (c)[Pb 2+ ] decreases; an increase in the concentration of the iodide ion causes an increase in the speed of the reverse reaction, lead ions bond with the added iodide ions to create more solid lead iodide; LeChatelier’s Principle shift to the left

8. Question 4 Answer the following questions relating to the solubility of the chlorides of silver and lead. (a)At 10  C, 8.9  10 -5 g of AgCl(s) will dissolve in 100. mL of water. (i)Write the equation for the dissociation of AgCl(s) in water. (ii)Calculate the solubility, in mol L –1, of AgCl(s) in water at 10  C. (iii)Calculate the value of the solubility-product constant, K sp for AgCl(s) at 10  C.

9. Answer 4 (a)(i) AgCl  Ag + + Cl – (ii)  = 6.2  10 -6 M (iii) K sp = [Ag + ][Cl – ] = (6.2  10 -6 ) 2 = 3.9  10 -11