Question

Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.85×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.17×10−4, what is the equilibrium constant Kfinal for the following reaction? PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(aq)

Answer 1

Answer:

The equilibrium constant of the given reaction is 0.01351.

Explanation:

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

The equilibrium constant of the reaction = $K_3=1.85\times 10^{-10}$

$K_3=\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}$...[1]

$AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq)$

The equilibrium constant of the reaction = $K_4=1.17\times 10^{-4}$

$K_4=\frac{[Ag^+][Cl^-]}{[AgCl]}$..[2]

$[Cl^-]=\frac{K_4\times [AgCl]}{[Ag^+]}$

$PbCl_2(aq)+2Ag^+(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq)$

The expression of equilibrium constant of the creation is ;

$K=\frac{[AgCl]^2[Pb^{2}]}{[PbCl_2]][Ag^+]^2}$

Dividing [1] by [2]

$\frac{K_3}{K_4}=\frac{\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}}{\frac{[Ag^+][Cl^-]}{[AgCl]}}$

$\frac{K_3}{K_4}=\frac{[Pb^{2+}][Cl^-][AgCl]}{[PbCl_2][Ag^+]}$

Substituting the value of $[Cl^-]$ from [2] :

$\frac{K_3}{K_4}=\frac{[Pb^{2+}][AgCl]}{[PbCl_2][Ag^+]}\times \frac{K_4\times [AgCl]}{[Ag^+]}$

$\frac{K_3}{K_4}=K_4\times K$

$K=\frac{K_3}{(K_4)^2}=\frac{1.85\times 10^{-10}}{(1.17\times 10^{-4})^2}$

$K=0.01351$

The equilibrium constant of the given reaction is 0.01351.