Question

What is the molar solubility of lead(II) chromate in 0.10 M Na2S2O3? For PbCrO4, Ksp = 2.0 x 10–16; for Pb(S2O3)3 4–, Kf = 2.2 x 106.

Answer 1

If you combine the two reactions, you have a new reaction depicting the solubility of PbCrO4 with the S2O3^2- ion present.

(1) PbCrO4(s) <==> Pb^2+(aq) + CrO4^2-(aq)

Ksp = 2.0 x 10^-16

(2) Pb^2+(aq) + 3S2O3^2-(aq) ==> Pb(S2O3)3^4-(aq)

Kf = 2.2 x 10^6
(3) PbCrO4(s) + 3S2O3^2-(aq) <==> Pb(S2O3)3^4-(aq) + CrO4^2-(aq)

Equilibrium constant K3 = Ksp x Kf = 4.4 x 10^-10

Set up an ICE chart. Remember that PbCrO4 is a pure solid and does not appear in the K expression.
PbCrO4(s) + 3S2O3^2-(aq) <==> Pb(S2O3)3^4-(aq) + CrO4^2-(aq)
Initial 0.10 . . . . . . . . . . . . . . . .0 . . . . . . . . . . . . .0
Change -3x . . . . . . . . . . . . . . . . .x . . . . . . . . . . . . .x
Equilibrium 0.10 - 3x . . . . . . . . . . . . . .x . . . . . . . . . . . . .x

K3 = [Pb(S2O3)3^4-][CrO4^2-] / [S2O3^2-]^3 = (x)(x) / (0.10 - 3x)^3 = 4.4 x 10^-10
With such a small K, we can ignore the -x term that follows 0.10 to simplify the math.

x^2 / (0.10)^3 = x^2 / (0.001) = 4.4 x 10^-10
x^2 = 4.4 x 10^-13
x = 6.63 x 10^-7 M = [CrO4^2-] = [PbCrO4]