A saturated solution of lead(II) chloride

A saturated solution of lead(II) chloride, PbCl2 , was prepared by dissolving solid PbCl2 in water. The concentration of Pb2+ ion in the solution was found to be 1.62x10^-2 M . Calculate Ksp for PbCl2.

PbCl2 Pb^2+ 2Cl
I Solid 0 0
C +x +2x
E X 2x


x= 1.62 x 10^02 M

Ksp = [(2x)^2]x
Ksp = 4x^3

plug in the x value for x
Ksp = 4(1.62x10^-2)^3
Ksp= 1.7x10^-5

There are two chlorine ions for each lead ion, so you would multiply the concentration of lead ions by two to get the concentration of chlorine ions.

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

Replace concentrations in terms of x, where x is 1.62 x 10^-2 M.

Ksp = [x][2x]^2
Ksp = (1.62 x 10^-2)(2*1.6 x 10^-2)^2
Ksp = 1.6 x 10^-5

Cheers!