Question

What is the molar solubility of lead(II) chromate in 0.070M Na2S2O3? For PbCrO4, Ksp=2.0e-16; for Pb(S2O3)3^-4, Kf=2.2e6.

Answer 1

The dissociation equation is

PbCrO₄ (s) <=======> Pb²⁺ (aq) + CrO₄^2- (aq); K_sp = [Pb²⁺][CrO₄^2-] = 2.0*10^-16 ……(1)

Pb²⁺ reacts with Na₂S₂O₃ as below

Pb²⁺ (aq) + 3 S₂O₃^2- (aq) ———-> Pb(S₂O₃)₃^4- (aq); K_f = [Pb(S₂O₃)₃]/[Pb²⁺][S₂O₃^2-]³ = 2.2*10⁶ …..(2)

Add (1) and (2) and write

PbCrO₄ (s) + 3 S₂O₃^2- (aq) ———-> Pb(S₂O₃)₃^4- (aq) + CrO₄^2- (aq)

K = [Pb(S₂O₃)₃^4-][CrO₄^2-]/[S₂O₃^2-]³ = [Pb(S₂O₃)₃^4-]/[Pb²⁺][S₂O₃^2-]³*[Pb²⁺][CrO₄^2-] = K_sp*K_f = 2.0*10^-16*2.2*10⁶ = 4.4*10^-10

Set up the ICE cart as below.

PbCrO₄ (s) + 3 S₂O₃^2- (aq) ———-> Pb(S₂O₃)₃^4- (aq) + CrO₄^2- (aq)

Initial 0.070 0 0

Change -3x x x

Equilibrium (0.070 - 3x) x x

Write down the expression for K as below.

K = [Pb(S₂O₃)₃^4-][CrO₄^2-]/[S₂O₃^2-]³ = (x).(x)/(0.070- 3x)³

===> 4.4*10^-10 = x²/(0.070- 3x)³

Use the small x approximation; since K is small, we can assume x << 0.070 and write (0.070 - 3x) = 0.070. Plug in the value and write

4.4*10^-10 = x²/(0.070)³ = x²/(3.43*10^-4)

===> x² = 4.4*10^-10*3.43*10^-4 = 1.5092*10^-13

===> x = 3.8848*10^-7 = 3.885*10^-7

Since x = [Pb²⁺] = [CrO₄^2-] = [PbCrO₄] = 3885*10^-7 M, hence the solubility of PbCrO₄ is 3.885*10^-7 M (ans).