Determine the molar solubility of lead(II) iodide, PbI2, in 0.050 mol/L NaI. Ksp of PbI2 = 9.8 x 10^(-9)

Answer 1

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

The other ionization is NaI and that is ionized 100% (completely that is).
........NaI ==> Na^+ + I^-
I.....0.05M......0.....0
C....-0.05.....0.05...0.05
E......0.......0.05...0.05

Ksp = (Pb^2+)(I^-)^2
Substitute the E line of each of the above into the Ksp expression.
Ksp you have.
(Pb^2+) = x
(I^-) = 2x for solubility of PbI2 and 0.05 from NaI; therefore, (I^-) = (2x + 0.05). Don't forget to square that. x is the solubility.