Question

What is molar solubility of PbI₂ when the equilibrium concentration of CrO₄^2- is 1.50*10^-5 M?

K_sp for PbI₂ = 8.4*10^-9 and K_sp for PbCrO₄ = 2.0*10^-14

Answer 1

Ksp of PbCrO4 = 2.0 x 10^-14

PbCrO4   --------------> Pb+2   +   CrO42-

Ksp = [Pb+2][CrO42-]

2.0 x 10^-14 = [Pb+2] (1.50 x 10^-5)

[Pb+2] = 1.33 x 10^-9 M

PbI2   ------------> Pb+2 +    2 I-

                            1.33 x 10^-9                2S

Ksp = [Pb+2][I-]^2

8.4 x 10^-9 = [1.33x 10^-9] [2S]^2

S = 1.26

Molar solubility of PbI2 = 1.26 M