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?
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.
The reaction taking place is the dissociation of PbI2 into its ions in solution. It can be represented by the chemical equation:
PbI2 ⇌ Pb2+ + 2I-
To calculate Q, we need to use the concentrations of the ions involved in the reaction. The concentrations are given as [Pb2+] = 4.5 x 10^-5 and [I-] = 6.5 x 10^-4.
Q = [Pb2+][I-]^2
Substituting the given values:
Q = (4.5 x 10^-5)(6.5 x 10^-4)^2
= 9.845 x 10^-14
Comparing Q to the Ksp, we see that Q (9.845 x 10^-14) is larger than the Ksp (8.7 x 10^-9). This means the solution is supersaturated because Q is greater than the solubility product constant (Ksp).
Since the solution is supersaturated and Q is greater than the Ksp, a precipitate of PbI2 will likely form in the solution.