A solution of PbI2 has [Pb2+] = 4.5 x 10-5 and [I-] = 6.5 x 10-4. PbI2 has Ksp = 8.7 x 10-9. Write down the reaction that is taking place. Calculate Q. Is Q larger or smaller than the Ksp? Is the solution unsaturated, saturated, or supersaturated? Will a precipitate form in this solution?

Answer 1

PbI2 ==> Pb^2+ + 2I^-
Q = ion product = (Pb^2+)(I^-)^2
Q = (4.5E-5)(6.5E-4)^2 = 1.9E-11
That should give you enough information to answer the other questions.