A 1.00-L solution saturated at 25°C with lead(II) iodide contains 0.54 g of PbI2. Calculate the solubility-product constant for this salt at 25°C.

Answer 1

M PbI2 = moles/L.
mols = grams/molar mass
It is in 1 L. Calculate M and call this x.

PbI2(s) ===> Pb^2+ + 2I^-
...x.........x.........2x

Ksp = (Pb^2+)(I^-)^2
Substitute x for Pb^2+ and 2x for I^- (then square the I^-) and solve for Ksp.

Answer 2

6.4x10^-9