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?

Qsp = (Pb^2+)(I^-)^2

From Qsp you get all of the answers to the other questions.

The reaction taking place is as follows:

PbI2 (s) ⇌ Pb2+ (aq) + 2I- (aq)

To calculate Q, we can use the expression for the reaction quotient (Q), which is calculated in the same way as the equilibrium constant (Ksp), but using the actual concentrations instead of equilibrium concentrations.

Q = [Pb2+] * [I-]^2

Substituting the given concentrations, we have:

Q = (4.5 x 10^-5) * (6.5 x 10^-4)^2
≈ 1.79 x 10^-13

Comparing Q with the Ksp value of 8.7 x 10^-9, we see that Q is significantly smaller than Ksp. This indicates that the solution is unsaturated.

In an unsaturated solution, the concentration of the dissolved ions is less than their solubility product. Hence, no precipitate will form in this solution.