1. Equilibrium and Solubility 1.Solubility rules for common ions 2.Using the solubility product, K sp, to calculate solubility - molar solubility, gram solubility 2.The common ion effect 3.Precipitation and the solubility product - the criterion for solubility 4.Effects of pH on solubility - solubility of hydroxides - influence of hydrolysis on solubility

2. Solubility Rules Soluble: all Group I (Li +, Na +, K +, Rb +, Cs + ) and NH 4 + salts all nitrates, most acetates, and most sulfates most halides, except as noted below Slightly Soluble: fluorides: SrF 2, BaF 2, PbF 2, other halides: PbCl 2, PbBr 2, HgBr 2 sulfates: CaSO 4, Ag 2 SO 4, Hg 2 SO 4 hydroxides: Ca(OH) 2, Sr(OH) 2 Insoluble (except as noted above): sulfides, carbonates, sulfites, phosphates, hydroxides MgF 2, CaF 2, AgCl, AgBr, AgI, HgI 2 SrSO 4, BaSO 4, PbSO 4

3. Solubility and K sp The equilibrium constant for the solubility reaction has a special name, the solubility product, K sp. BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq)K sp = 1.1 x10 -10 Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2- (aq)K sp = 1.2 x10 -12 Example 1:What is the solubility of Ag 2 CrO 4 (s) in water? K sp = 1.2 x10 -12 = [Ag + ] 2 [CrO 4 2- ] 2x x = [2x] 2 [x] (x = moles per Liter of CrO 4 2- ) x = (0.3 x 10 -12 ) 1/3 = 6.7 x 10 -5 M molar solubility = 6.7 x 10 -5 M gram solubility = 2.2 x 10 -2 g-L -1

4. The Common Ion Effect Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2- (aq)K sp = 1.2 x10 -12 Example 2:What is the solubility of Ag 2 CrO 4 (s) is 0.100 M AgNO 3 ? K sp = 1.2 x10 -12 = [Ag + ] 2 [CrO 4 2- ] 0.100+2x x = (0.100 + 2x) 2 (x) (x = moles per Liter of CrO 4 2- ) x = 1.2 x 10 -10 M molar solubility = 1.2 x 10 -10 M gram solubility = 4.0 x 10 -8 g-L -1 ≈ (0.100) 2 (x) An example of the common ion effect: the solubility of Ag 2 CrO 4 (s) is lowered in the presence of added Ag + (aq)

5. Precipitation and K sp Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2- (aq)K sp = 1.2 x10 -12 Example 3: Will Ag 2 CrO 4 (s) precipitate if 300 mL of 1.2 x10 -5 M AgNO 3 and 200 mL of 3.5 x10 -4 M Na 2 CrO 4 are mixed? Precipitation does not occur, since the ion product is less than K sp. Ans:yes, if Q 0 = [Ag + ] 0 2 [CrO 4 2- ] 0 exceeds K sp in the mixture. (0.300 x1.2 x10 -5 )/(.500) (0.200 x3.5 x10 -4 )/(.500) Q 0 = (0.72 x10 -5 ) 2 (1.4 x10 -4 ) = 7.3 x10 -16

6. Effects of pH on Solubility of Hydroxides Example 4: What is the solubility of Mg(OH) 2 at pH 10? The solubility of Mg(OH) 2 increases as pH decreases. Solubility varies as (pH) -2 in solutions more basic than pH 8. Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH - (aq)K sp = 2 x10 -13 1 x10 -4 M x K sp = 2 x10 -13 = [Mg 2+ ] [OH - ] 2 = x (1 x10 -4 ) 2 molar solubility = 2 x 10 -5 M at pH 10 = 2 x 10 -9 M at pH 12 = 20 M at pH 7

7. Effects of pH on Solubilities of Salts (1) Example 5: What is the solubility of CaCO 3 at pH 8.32? at pH 10.32? When pH is more acidic than pK a2, the solubility of CaCO 3 increases because most of the carbonate present is in the form of bicarbonate. When pH<10, solubility varies linearly with (1/pH). CaCO 3 (s) Ca 2+ (aq) + CO 3 -2 (aq)K sp = 8.7 x10 -9 depends on pH x K sp = 2 x10 -13 = [Ca 2+ ] [CO 3 2- ] pH = pK a2 + log 10 [CO 3 2- ] [HCO 3 - ] x depends on pH 10.32

8. Effects of pH on Solubilities of Salts (2) Example 5: What is the solubility of CaCO 3 at pH 8.32? at pH 10.32? CaCO 3 (s) Ca 2+ (aq) + CO 3 2- (aq)K sp = 8.7 x10 -9 0.01 x x (1) at pH 8.32, [CO 3 2- ] = 0.01* [HCO 3 - ] CaCO 3 (s) Ca 2+ (aq) + CO 3 2- (aq)K sp = 8.7 x10 -9 0.5 x x (2) at pH 10.32, [CO 3 2- ] = [HCO 3 - ] solubility = 9 x10 -4 M solubility = 9 x10 -5 M

9. Effects of pH on Solubilities of Salts (3) Example 6: Describe how the solubility of CaF 2 depends on pH. When pH is more acidic than pK a, the solubility of CaF 2 increases because most of the fluorine present is in the form of HF rather than F -. CaF 2 (s) Ca 2+ (aq) + 2 F - (aq)K sp = 3.9 x10 -11 depends on pH x K sp = 3.9 x10 -11 = [Ca 2+ ] [F - ] 2 pH = pK a + log 10 [F - ] [HF] x depends on pH 3.18