Question

Given the two reactions

PbCl2⇌Pb2++2Cl−,   K3 = 1.72×10⁻¹⁰, and

AgCl⇌Ag++Cl−,   K4 = 1.12×10⁻⁴,

what is the equilibrium constant Kfinal for the following reaction?

PbCl2+2Ag+⇌2AgCl+Pb2+

Express your answer numerically.

Answer 1

Given that

PbCl2⇌Pb2++2Cl−,   K3 = 1.72×10⁻¹⁰

take reverse of 2nd equation and multiply with 2

2Ag+ + 2Cl− ⇌ 2AgCl K5 = (1/K4)² = ( 1/ 1.12×10⁻⁴)²

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PbCl2⇌ Pb2+ + 2Cl−,   K3 = 1.72×10⁻¹⁰

2Ag+ + 2Cl− ⇌ 2AgCl K5 = (1/K4)² = ( 1/ 1.12×10⁻⁴)²

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Now add both the equations and cancel 2Cl- on both sides

PbCl2 + 2Ag+ ⇌ 2AgCl + Pb2+

K_final = K3.K5

= ( 1.72×10⁻¹⁰ ) ( 1/ 1.12×10⁻⁴)²

= 1.37 X 10^-2

Therefore, K_final = 1.37 X 10^-2