chemistry
posted by Anonymous .
If the molar solubility of lead iodide is 2.4 x 10^3M. Calculate the Ksp.

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.