Question

The molar solubility of calcium carbonate in a 0.260 M calcium nitrate solution is

The maximum amount of lead hydroxide that will dissolve in a 0.220 M lead acetate solution is

Answer 1

1)

Ksp of CaCO3 = 3.8 x 10^-9

CaCO3 (s)   -------------> Ca2+ (aq) +   CO32- (aq)

                                        0.260                 s

Ksp = [Ca2+][CO32-]

3.8 x 10^-9 = 0.260 x s

s = 1.46 x 10^-8

molar solubility = 1.46 x 10^-8 M

2)

Ksp of Pb(OH)2 = 2.8 x 10^-16

Pb(OH)2    ----------------> Pb2+ (aq) + 2 OH- (aq)

                                       0.220                2s

Ksp = [Pb2+][OH-]^2

2.8 x 10^-16 = 0.220 x 4s^2

s = 1.78 x 10^-8

maximum amount = 1.78 x 10^-8 M