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
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.
2 answers
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 !!!!
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 !!!!
2 answers