1. Lecture 62/2/05

2. Solubility vs. Solubility constant (K sp ) BaSO 4 (s) ↔ Ba 2+ (aq) + SO 4 2- (aq) K sp = [Ba 2+ ][SO 4 2- ] Ba 2+ (aq) + SO 4 2- (aq) ↔ BaSO 4 (s) _1_ ____1____ K = K sp = [Ba 2+ ][SO 4 2- ] Ag 2 CO 3 (s) ↔ 2Ag + (aq) + CO 3 2- (aq) K sp = [Ag + ] 2 [CO 3 ]

4. What is the K sp for AgI if the solubility equals 9.2 x 10 -9 M at 25˚ C? AgI (s) ↔ Ag + (aq) + I - (aq) K sp = [Ag + ][I - ] K sp = (9.2 x 10 -9 )(9.2 x 10 -9 ) = 8.5 x 10 -17

5. The K sp of AgBr at 100 ˚ C is 5 x 10 -10. Calculate the solubility of AgBr at that temperature in moles per liter. AgBr (s) ↔ Ag + (aq) + Br - (aq) K sp = [Ag + ][Br - ] 5 x 10 -10 = (X)(X) 5 x 10 -10 = X 2 2.2 x 10 -5 M = X AgBr (s) ↔ Ag + (aq) + Br - (aq) I 00 C + X E XX

6. A saturated solution of silver oxalate (Ag 2 C 2 O 4 ) contains 6.9 x 10 -5 M of C 2 O 4 2- at 25˚ C. Calculate the K sp of silver oxalate at that temperature. Ag 2 C 2 O 4 (s) ↔ 2Ag + (aq) + C 2 O 4 2- (aq) K sp = [Ag + ] 2 [C 2 O 4 2- ] Use stoichiometry, if [C 2 O 4 2- ] = 6.9 x 10 -5 M, then [Ag + ] = 1.4 x 10 -4 M K sp = [Ag + ] 2 [C 2 O 4 2- ] K sp = (1.4 x 10 -4 ) 2 (6.9 x 10 -5 ) K sp = 1.4 x 10 -12

7. The K sp of MgF 2 = 5.2 x 10 -11. Calculate the solubility in molarity and grams per liter. MgF 2 (s) ↔ Mg 2+ (aq) + 2F - (aq) K sp = [Mg 2+ ][F - ] 2 5.2 x 10 -11 = (X)(2X) 2 5.2 x 10 -11 = 4X 3 2.4 x 10 -4 M = X Solubility of MgF 2 = 2.4 x 10 -4 M Solubility of MgF 2 = 2.4 x 10 -4 M (62.3 g/mol) = 0.15 g /L MgF 2 (s) ↔ Mg 2+ (aq) + 2F - (aq) I 00 C + X+ 2X E X2X

8. Comparison of solubility based on K sp AgI (K sp = 8.5 x 10 -17 ) < AgBr (K sp = 5.4 x 10 -13 ) < AgCl (K sp = 1.8 x 10 -10 ) PbI 2 (K sp = 9.8 x 10 -9 ) < PbBr 2 (K sp = 6.6 x 10 -6 ) < PbCl 2 (K sp = 1.7 x 10 -5 ) Can only compare salts with same ion ratio AgCl (K sp = 1.8 x 10 -10 ) vs. Ag 2 CO 3 (K sp = 8.5 x 10 -12 ) Solubility of AgCl = 1.3 x 10 -5 mol/L Solubility of Ag 2 CO 3 = 1.3 x 10 -4 mol/L