The solubility of lead (II) chloride in water is 4.50 g/L at 25 degrees C

Question

a. What is the molarity of Pb2+ in a saturated solution
b. What is the molarity of Cl- in the solution in the previous problem
c. What is the equilibrium constant for the reaction:
PbCl2(s) --> Pb2+(aq) + 2Cl-(aq)

Answer 1

mols PbCl2 = grams/molar mass. Since that is g/L, you now have mols/L and that is M.

Look at the equation. (Cl^-) in mols/L must be twice the PbCl2

Plug (Pb^2+) and (Cl^-) into Ksp expression and solve for K. Ksp is the equilibrium constant.