A solution with a volume of 1·00 dm3 is saturated with lead iodide, Pbl2. The concentration of iodide ions is 2·7 mol dm–3.Determine the solubility product of PbI2. 

This is a common ion problem.

.....PbI2 ==> Pb^2+ + 2I^-
I....solid....0.......0
C....solid....x.......2x
E....solid....x.......2

Ksp = (Pb^2+)(I^-)^2
Ksp = ?
(I^-_ is 2.7 M.
(Pb^2+) = 1/2 (I^-)
Substitute and solve for Ksp.

To determine the solubility product (Ksp) of PbI2, we need to use the information given in the question.

The solubility product equation for PbI2 can be written as:
PbI2 ↔ Pb2+ + 2I-

From the question, we are given the concentration of iodide ions (I-) as 2.7 mol dm–3.

We can assume that at equilibrium, x mol dm–3 of PbI2 will dissolve, and thus, the concentrations of Pb2+ and I- will both be x mol dm–3.

Now, using the solubility product equation, we can write the expression for the solubility product (Ksp):

Ksp = [Pb2+][I-]^2

Substituting the known values into the equation, we get:

Ksp = (x)(2.7)^2 = 7.29x

Therefore, the solubility product of PbI2 is 7.29x.