1. 1 Solubility Equilibria Dissolution M m X x (s)  m M n+ (aq) + x X y- (aq) Precipitation m M n+ (aq) + x X y- (aq)  M m X x (s) For a dissolution process, we give the equilibrium constant expression the name solubility product (constant) K sp. For M m X x (s)  m M n+ (aq) + x X y- (aq) K sp = [M n+ ] m [X y- ] x

2. 2 Problem Write the expressions for the solubility product constant K sp of: a) AgCl b) PbI 2 c) Ca 3 (PO 4 ) 2 d) Cr(OH) 3

3. 3 Measuring K sp and Calculating Molar Solubility from K sp For the dissolution of solid CaF 2 in water, we might find that the concentrations of the ions Ca 2+ and F - at equilibrium are [Ca 2+ ] = 3.3 x 10 -4 mol/L [F - ] = 6.7 x 10 -4 mol/L K sp = [Ca 2+ ] [F - ] 2 = (3.3 x 10 -4 mol/L) (6.7 x 10 -4 mol/L) 2 = 1.5 x 10 -10

4. 4 Measuring K sp and Calculating Molar Solubility from K sp We could also find the K sp of a solid by mixing solutions of known concentrations of the ions, leading to the precipitation of the solid. When the system reaches equilibrium we can measure the ion concentrations to calculate K sp.

5. 5 The advantage of using the dissolution of the solid method is that a clear relationship between the concentrations of the ions exists, based upon their relative stoichiometry in the solid.

6. 6 If we mix two separate solutions that are sources of calcium ions and fluoride ions, no such relationship exists, and we can have infinitely many mixtures that still obey the solubility product expression. Since K sp values are equilibrium constants, they will change with temperature!

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8. 8 Molar solubility and saturation If we know the K sp value for a solid, we can calculate its molar solubility, which is the number of moles of the solid that can dissolve in the solvent before the solution becomes saturated (no more solid will dissolve). Saturated means equilibrium!

9. 9 Problem A saturated solution of Ca 3 (PO 4 ) 2 has [Ca 2+ ] = 2.01 x 10 -8 mol/L and [PO 4 3- ] = 1.6 x 10 -5 mol/L. Calculate K sp for Ca 3 (PO 4 ) 2

10. 10 Problem If a saturated solution of BaSO 4 is prepared by dissolving solid BaSO 4 in water, and [Ba 2+ ] = 1.05 x 10 -5 mol/L, what is the K sp for BaSO 4 ?

11. 11 Problem Which has the greater molar solubility? AgCl with K sp = 1.8 x 10 -10 or Ag 2 CrO 4 with K sp = 1.1 x 10 -12

12. 12 Factors that Affect Solubility The Common-Ion Effect If a solution already has a significant concentration of a common-ion the solution will dissolve LESS of the solid than the same volume of pure water can.

13. 13 The Common-Ion Effect M m X x (s)  m M n+ (aq) + x X y- (aq) K sp = [M n+ ] m [X y- ] x Imagine that instead of dissolving a solid in a solution of common- ion that we add a common ion to a solution created by dissolving the solid in pure water.

14. 14 The Common-Ion Effect M m X x (s)  m M n+ (aq) + x X y- (aq) K sp = [M n+ ] m [X y- ] x From Le Chatalier’s Principle, we know if we add one of the product ions (the common ion we added), the stress on the equilibrium is this added product. To relieve the stress the equilibrium will shift towards the reactants meaning more solid is created.

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16. 16 Problem Calculate the molar solubility of MgF 2 (K sp = 7.4 x 10 -11 ) in pure water and in 0.10 mol/L MgCl 2.

17. 17 Precipitation of Ionic Compounds Can we predict if a solid will precipitate if we mix two solutions of different ions? YES! Consider the mixing of a solution of Ca 2+ ions and a solution of F - ions. A precipitation of solid is the dissolution reaction in reverse, so we can express the reaction as CaF 2 (s)  Ca 2+ (aq) + 2 F - (aq)K sp = [Ca 2+ ][F - ] 2

18. 18 Precipitation of Ionic Compounds When we mix the solutions, the system is most likely not at equilibrium. For solid dissolution / precipitation reactions, we use a procedure similar to the reaction quotient Q c to define the ion product (IP) or Q sp. Q sp = [Ca 2+ ][F - ] 2

19. 19 If Q sp = K sp, the solution is saturated, and the system is at equilibrium. If Q sp > K sp, the solution is supersaturated, so the system is not at equilibrium. The concentration of the ions is greater than it would be at equilibrium, and so the reaction proceeds from ions towards the solid. We expect precipitation to occur!

20. 20 If Q sp < K sp, the solution is unsaturated, so the system is not at equilibrium. The concentration of the ions is less than it would be at equilibrium, and so we expect more solid to be able to dissolve in this solution!

21. 21. Will a precipitate form when 0.150 L of 0.10 mol  L -1 Pb(NO 3 ) 2 and 0.100 L of 0.20 mol  L -1 NaCl are mixed? K sp of PbCl 2 is 1.2 x 10 -5

22. 22 Problem Will a precipitate form on mixing equal volumes of the following solutions? a) 3.0 x 10 -3 mol/L BaCl 2 and 2.0 x 10 -3 mol/L Na 2 CO 3 b) 1.0 x 10 -5 mol/L Ba(NO 3 ) 2 and 4.0 x 10 -5 mol/L Na 2 CO 3 (K sp of BaCO 3 is 2.6 x 10 -9 )