What role do “foreign” ions have on the solubility of a slightly soluble salt like lead iodide?

Answer 1

If you can find an ion that will form a soluble complex with the Pb2+ (aq) ions, then it will appear to increase the solubility of the lead iodide. Try OH-.

Answer 2

If they react with the ions formed when the salt dissolves then the solubility will increase.

If they don't react then the dominant effect will be that of increasing the ionic strength of the solution. The equilibrium will be best described by the Kαp and not the Ksp so if s is the molar solubility you will have

.. .. .. .. .. PbI2 <=> Pb+2 +2I-
Dissolve .. s
At Equil .. .. .. .. .. .. .. s .. .. 2s

Kαp = α(Pb+2)*α(I-)^2 = γ(+)*[Pb+2] (γ(-)*[I-])^2 =
=(γ(+)γ(-)^2)*[Pb+2][I-]^2 = γ(+)γ(-)^2*s*(2s)^2 =
=γ(+)γ(-)^2 *4s^3

If you have no foreign ions present then you have a certain value for the activity coefficients e.g. γ1(+) and γ1(-).
When you have foreign ions present the ionic strength increases, thus the activity coefficients decrease and become e.g. γ2(+) and γ2(-).

Kαp is constant thus
Kαp = γ1(+)γ1(-)^2 *4s1^3 = γ2(+)γ2(-)^2 *4s2^3 =>

s2/s1 = cuberoot (γ1(+)γ1(-)^2 / (γ2(+)γ2(-)^2) )
but γ1(+)> γ2(+)>0 and γ1(-)> γ2(-) >0 thus the cuberoot is >1 and s2>s1

Thus the solubility should increase.