Solubility Products

Question

The molar solubility of PbCl₂ is 4.71⋅10^−5 M. Calculate the Ksp for PbCl₂. 

Answer 1

PbCl₂(s) <==> Pb²⁺(aq) + 2Cl^-(aq)

Ksp = [Pb²⁺][Cl^-]²

[Pb²⁺] = 4.71x10^-5 M and [Cl^-] = 2 x [Pb²⁺] = 9.42x10^-5 M

Ksp = (4.71x10^-5)(9.42x10^-5)²

Ksp = 4.18x10^-13