Any help on this chemistry problem would be appreciated

Question

Lead (II) chromate, PbCrO4, has a Ksp of 2.0 x 10-16 at 25°C.  

(a) Write the balanced chemical equation to show the dissociation of lead (II) chromate in solution AND the equilibrium expression for this reaction.

(b) Calculate the molar solubility of lead (II) chromate at this temperature.

(c) If a potassium chromate solution with a concentration of 0.085 M is added dropwise to the solution of lead (II) chromate, what effect would this have on the equilibrium? What evidence would the chemist see of this shift?

(d) Calculate the concentration of Pb2+ (aq) remaining in solution after the potassium chromate from part (c) is added.

Answer 1

Hello Tenzin,

This is a common problem, and you can expect to see it again, for other substances.

To solve it, first recognize that PbCrO₄ is an ionic compound, with Pb⁺² and CrO₄^2- . So:

PbCrO₄ (s) ↔ Pb⁺² (aq) + CrO₄^2- (aq).

By definition, then, K_sp = [Pb⁺² (aq)][CrO₄^2- (aq)] = 2.0 x 10^-16 at 25°C.

Supposing that some amount of lead chromate dissolves, to give a concentration [x] = [Pb⁺² (aq)] = [CrO₄^2- (aq)]. Then:

K_sp = [x][x]. = [x]² ; and, [x] = 1.4 x 10^-8 , which is the molar solubility of lead chromate.

Adding K₂CrO₄ will shift the equilibrium further (than it already is) toward solid PbCrO₄. You can find the new concentration [x] for the lead by recognizing that now

K_sp = [x][0.085 + x]. = 2.0 x 10^-16 at 25°C.

From our prior math, we already know that [x] is very small, so we can neglect it relative to 0.085M, and get:

K_sp ≈ [x][0.085] ≈ 2.0 x 10^-16 at 25°C.

That gives [x] = 2.4 x 10^-15 which is the concentration of the remaining lead, [Pb⁺²(aq]).

The initial lead concentration is very low, but in principle as the potassium chromate is added, the tiny amount of lead chromate in solution will precipitate, making the solution slightly cloudy.