If the molar solubility of lead iodide is 2.4 x 10^-3M. Calculate the Ksp.

Answer 1

PbI2 ==> Pb^+2 + 2I^-

Ksp = (Pb^+2)(I^-)^2
Therefore, (Pb^+2) = 2.4 x 10^-3M
and (I^-) = 4.8 x 10^-3 M.
Plug into Ksp expression and solve for Ksp.