Question

Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.89×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.25×10−4, what is the equilibrium constant Kfinal for the following reaction? PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(aq) Express your answer numerically.

Answer 1

Answer:

$1.2\times 10^{-2}$

Explanation:

The given reactions are:

PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq)     $K_3 = 1.89\times 10^{-10}$

AgCl(aq)⇌Ag+(aq)+Cl−(aq)           $K_4 = 1.25\times 10^{-4}$

Required reaction is:

PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(aq)

$K_f = \frac{K_3}{K_4^2}\\=\frac{1.89\times 10^{-10}}{1.25\times 10^{-4}}^2\\=1.2\times 10^{-2}$