Question

Calculate the molar solubility of lead(II) chloride in a 0.10 M calcium chloride solution.

I keep getting 0.0017, but that is incorrect.

Answer 1

For the reaction -

           PbCl₂ (s) <-------> Pb⁺² (aq) + 2 Cl^- (aq)

I (M)       -                        0               0.10

C           -                         +s              +2s

Eq          -                         s             (0.10+2s)

Since - Ksp = [Pb⁺²] [Cl^-]² = (s)(0.10+2s)²

        1.7*10^-5 = (s)(0.10)²

Therefore, s = [Pb²⁺] = 0.0017 M