1. Solubility Equilibria

2. Write solubility product (K sp ) expressions from balanced chemical equations for salts with low solubility. Solve problems involving K sp. Additional KEY Terms Dissociation Saturated

3. C 6 H 12 O 6(s) C 6 H 12 O 6(aq)

4. There are 3 actions that affect solubility: 1. Nature of the solute and solvent “like dissolves like” Polar / ionic solute dissolve in polar solvent. Non-polar dissolve in non-polar. Even the most insoluble ionic solids are actually soluble in water to a limited extent

5. 2. Temperature Solids in liquids: ↑ temperature - ↑ solubility. Gases in liquids: ↑ in temperature - ↓ solubility. 3. Pressure Does not affect the solubility of (s)/(l). (g): ↑ pressure ↑ solubility.

6. A a B b(s) aA + (aq) + bB¯ (aq) K sp, called the solubility product constant. K sp = [A + ] a [B - ] b Product of ion concentrations in a saturated solution. K c = [A + ] a [B - ] b [A a B b ]

7. Write the dissociation and the product constant equation for the solubility of calcium hydroxide. K sp = [Ca 2+ ][OH - ] 2 Ca(OH) 2 (s ) Pb 3 (PO 4 ) 2(s) 3 Pb 2+ (aq) + 2 PO 4 3- (aq) K sp = [Pb 2+ ] 3 [PO 4 3- ] 2 Write a solubility product expression for Pb 3 (PO 4 ) 2. Ca 2+ (aq) + OH - (aq) 2

8. At equilibrium, the [Ag + ] = 1.3 x 10 -5 M and the [Cl - ] = 1.3 x 10 -5 M, what is the K sp of silver chloride? K sp = [Ag + ][Cl - ] K sp =(1.3 x 10 -5 )(1.3 x 10 -5 ) K sp = 1.7 x 10 -10 AgCl (s ) Ag + (aq) + Cl - (aq) *NOTE: K sp has no units.

9. Solubility And I.C.E. Tables (Yeah!)

10. Calculate K sp of lead (II) chloride if a 1.0 L saturated solution has of lead ions. I---00 C-x +x+2x E0E0 K sp = [Pb +2 ][Cl - ] 2 K sp = [1.62 x 10 -2 ][ 3. 24 x 10 -2 ] 2 K sp = 1.70 x 10 -5 PbCl 2(s) Pb 2+ (aq) + 2 Cl - (aq) 1.62 x 10 -2 M 2(1.62 x 10 -2 ) 1.62 x 10 -2 M

11. The solubility of PbF 2 is. What is the value of the solubility product constant? PbF 2(s) Pb 2+ (aq) + 2 F¯ (aq) 0.466 g 1 L 245.2 g 1 mol = 1.90 x 10 -3 M PbF 2 0.466 g/L 207 + 2 (19) = 245g/mol K sp = [Pb 2+ ][F - ] 2

12. K sp = (1.90 x 10 -3 )(3.80 x 10 -3 ) 2 K sp = 2.74 x 10 -8 PbF 2(s) Pb 2+ (aq) + 2 F¯ (aq) [E][E] 0 1.9 x 10 -3 M [I][I] 00 [C][C]- x+ x+ 2x 3.8 x 10 -3 M Saturated – all solid reactant dissociates.

13. Calculate K sp if 50.0 mL of a saturated solution was found to contain 0.2207 g of lead (II) chloride. I0.0159 M 0 0 C-x +x +2x E0 0.0159 M0.0318 M K sp = [Pb 2+ ][Cl - ] 2 0.2207 g 278.1g 1 mol = 0.0159 M PbCl 2 0.05 L PbCl 2(s) Pb 2+ (aq) + 2 Cl - (aq) K sp = [0.0159][0.0318] 2 = 1.61 x 10 -5

14. K sp of magnesium hydroxide is 8.9 x 10 -12. What are the [equilibrium] of ions in saturated solution? I---00 C-x+x+2x E0 x 2x Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH - (aq) K sp = [Mg 2+ ][OH - ] 2 8.9 x 10 -12 = [x][2x] 2 8.9 x 10 -12 = [x] 4x 2 8.9 x 10 -12 = 4x 3

15. [Mg 2+ ] = x = 1.3 x 10 -4 mol/L [OH - ] = 2x = 2.6 x 10 -4 mol/L 8.9 x 10 -12 = 4x 3 44 2.23 x 10 -12 = x 3 3√3√ 3√3√ 1.3 x 10 -4 = x

16. Precipitation

17. Compare value of Q, with given K sp to determine if an aqueous solution is saturated or unsaturated. Q = K sp Saturated solution, no precipitate. Q >K sp Precipitate forms (“oversaturated”) Q < K sp Solution is unsaturated. Q sp = [A + ] a [B¯] b

18.  Substances which are insoluble are actually slightly soluble.  The solubility product, K sp, describes the product of ion concentrations in saturated solutions.  Solubility can be determined from the solubility product.

19. CAN YOU / HAVE YOU? Write solubility product (K sp ) expressions from balanced chemical equations for salts with low solubility. Solve problems involving K sp. Additional KEY Terms Dissociation Saturated