If the concentration of I^-1 is found to be 2.4x10^-3 mol/L in a saturated solution of PbI2, what is the Ksp of PbI2?
The given answer is 6.9x10^-9
........PbI2 ==> Pb^2+ + 2I^-
I......solid.....0........0
C......solid.....x........2x
E......solid.....x........2x
Ksp = (Pb^2+)(I^-)^2
Ksp = (x)(2x)^2
So the problem tells you the value of 2x. Substitute 2x and x into the Ksp expression and solve for Ksp.
To determine the Ksp (solubility product constant) of PbI2, we need to use the given concentration of I^-1 and the stoichiometry of the balanced equation.
The balanced equation for the dissolution of PbI2 is:
PbI2 ⇌ Pb2+ + 2I^-
The stoichiometry of this equation tells us that the concentration of Pb2+ will also be equal to the concentration of I^-1 since the coefficient is 1:2.
Given that the concentration of I^-1 is 2.4x10^-3 mol/L, we can assume that the concentration of Pb2+ is also 2.4x10^-3 mol/L (since it is in a 1:2 stoichiometric ratio).
Now, we can write the expression for the solubility product constant (Ksp) of PbI2:
Ksp = [Pb2+] * [I^-]^2
Substituting the concentrations, we have:
Ksp = (2.4x10^-3) * (2.4x10^-3)^2 = 6.9x10^-9
Therefore, the Ksp of PbI2 is 6.9x10^-9.