Question

The Ksp for PbCl2= 1.7x10^-5

Calculate the molar solubility of PbCl2 in a solution that is 0.1 M of the soluble salt MgCl2

Answer 1

Let 'S'(mol/L) be the solubility of PbCl₂

PbCl₂ -----> Pb²⁺ + 2Cl^-

MgCl₂ -----> Mg²⁺ + 2Cl^-

[ Pb²⁺] = S

[Cl^-] = [Cl^-]_PbCl2 + [Cl^-]_MgCl2

      = 2S + (2x0.1)

      = 2S + 0.2

[ Pb²⁺] = S

Solubility product constant , K_sp = [ Pb²⁺] [Cl^-]² = 1.7x10^-5

                                               S x ( 2S+0.2)² = 1.7x10^-5

                                                   S = 4.25x10^-4 mol/L

Therefore the molar solubility of PbCl2 is 4.25x10^-4 mol/L