Question

Lead iodide, PbI2, has a Ksp in water of 7.1e-9 M3 at room temperature.

Calculate the solubility of PbI2 in water. Express your answer in grams per liter.

What is the solubility of PbI2 in 0.1 M NaI (aq)? Express your answer in grams per liter.

Answer 1

This is simple Stoichiometry problem

Write down PbI2 -> Pb2+ + 2I-

Therefore

Ksp = [Pb2+][I-]^2

From the stoich you can derive that

Ksp = Cs(2Cs)^2 = 4Cs^3

Hence Cs = (Ksp/4) ^ 1/3

Then multiply the Molar Weight to get grams / Liter unit that you like

Answer 2

The second problem is common ion effect

First see is the added ion concentration more than or less than your Cs.

0.1 seems to be >>>> than Cs so common ion will happen

Ksp = Cs'(0.1)^2 = 0.01Cs'

Hence

Cs' = Ksp/0.01

Then just multiply by Molar weight again to convert to g/L units .

Note use the Molar Weight of PbI2 !!!!