Question

Calculate the solubility of Pbso4 in a water sample, which contain 1.4x10-2M Cacl2?

Given : Ksp for Pbs04= 1.6x10-8M2
Ksp for Cas04= 2.6x10-5M2

Answer 1

I think the idea here is that CaSO4 is so much more soluble than PbSO4 that the CaCl2/CaSO4 equilibrium will determine the sulfate concentration and that in turn will determine the solubility of PbSO4.

.........CaSO4 ==> Ca^2+ + SO4^2-
I........solid.....0........0
C........solid.....x........x
E........solid.....x........x


..........CaCl2 ==> Ca^2+ + 2Cl^-
I........0.014.......0........0
C.......-0.014.....0.014....2*0.014
E.........0........0.014....0.028

Ksp CaSO4 = (Ca^2+)(SO4^2-)
(Ca^2+) = x from CaSO4 + 0.014 from CaCl2. (SO4^2-) is x. Plug those into Ksp expression for CaSO4 and solve for SO4. Note that you may need to solve a quadratic but I can't tell just by looking. Then go to the PbSO4 part of the problem.

.............PbSO4 --> Pb^2+ + SO4^2-
I ...........solid.....0........0
C............solid.....x........x
E............solid.....x........x

Ksp PbSO4 = (Pb^2+)(SO4^2-)
(Pb^2+) = x = solubility
(SO4^2-) = x from PbSO4 + the number for sulfate from CaSO4. Solve for (Pb^2+). Again, you may need to consider a quadratic here.