The ksp of pbi2 is 1.4 x 10-8. what is the molar solubility of lead(ii iodide in a solution of 0.400 m sodium iodide?

The solubility product of a substance us calculated by the product of the concentration of the dissociated ions in the solution raise to the stoichiometric coefficient of the ions. Therefore, we need the dissociation reaction. For this, it will have the reaction:

PbI2 = Pb^2+ + 2I-

We solve as follows:

Ksp = [Pb2+][I-]^2 = 1.4 x 10-8
1.4 x 10-8 = x(2x)^2
1.4 x 10-8 = 4x^3
x = 1.5x10^-3 M

The molar solubility would be 1.5x10^-3 M.