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

Answer 1

Molar mass PbI₂ = 461 g / mol

0.54 g / L x 1 mol / 461 g = 1.17x10^-3 mol / L = 1.17x10^-3 M

PbI₂(s) ==> Pb²⁺(aq) + 2I^-(aq)

Ksp = [Pb²⁺][I^-]²

[Pb²⁺] = 1.17x10^-3 M

[I^-] = 2 x 1.17x10^-3 M = 2.34x10^-3 M

Ksp = (1.17x10^-3 )(2.34x10^-3)²

Ksp = 6.42x10^-9