Solubility Products

Question

What is the solubility (in mol/L) of PbCl₂ given that the Ksp is 4.18⋅10^−13.

Answer 1

PbCl₂(s) <==> Pb²⁺⁽aq) + 2Cl^-(aq)

Ksp = [Pb²⁺][Cl^-]²

Let [Pb²⁺] = x

Then, [Cl^-] = 2x

4.18x10^-13 = (x)(2x)² = 4x³

x³ = 1.045x10^-13

x = 4.71x10^-5 M = solubility of PbCl₂