Question

Calculate the molar solubility of PbCl₂ in 0.332M CaCl_2. K_sp=1.72 x10^-5

Answer 1

Ksp of PbCl2 = 1.72 x 10^-5

concentration of CaCl2 = 0.332 M

concentration of Cl- = 2 x 0.332 = 0.664 M

PbCl2   ------------> Pb+2   +   2 Cl-

                                 S            0.664

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

1.72 x 10^-5 = S x (0.664)^2

S = 3.90 x 10^-5 M

Molar solubility of PbCl2 = 3.90 x 10^-5 M