calculate the solubility of PbI2 in 0.0123M CaI2. Ksp=6.81x10^-9

Answer 1

..........PbI2 ==> Pb^2+ + 2I^-
I.........solid....0........0
C.........solid....x........2x
E.........solid....x........2x

......CaI2 ==> Ca^2+ + 2I^-
I...0.0123.....0........0
C..-0.0123....0.0123..2*0.0123
E.......0.....0.0123...0.0246

Ksp - (Pb^2+)(I^-)^2
(Pb^2+) = x
(I^-) = 2x from PbI2 and 0.0246 from CaI so ((I^-) = x+0.0246
Substitute and solve for x = (Pb^2+) = (PbI2) in solution.