Question

Part B Given the two reactions PbCl2⇌Pb2++2Cl−, K3 = 1.89×10−10, and AgCl⇌Ag++Cl−, K4 = 1.13×10−4, what is the equilibrium constant Kfinal for the following reaction? PbCl2+2Ag+⇌2AgCl+Pb2+ Express your answer numerically.

Answer 1

multiply the cofficient of AgCl <--> Ag+ + Cl- by 2
2AgCl <--> 2Ag+ + 2Cl-
eulibrium comstant will become square let it be K5
K5 = (K4)^2
= (1.13*10^-4)^2
= 1.28*10^-8

now reverse the above reaction
2Ag+ + 2Cl- <--> 2AgCl
when we reverse a reaction equlibrium constant will get inversse
let it be K6
K6 = 1/K5
= 1/(1.28*10^-8)
= 7.81*10^7

now add PbCl2 <--> Pb+2 + 2Cl- to the above reaction
we get
PbCl2 + 2Ag+ <-- > 2AgCl2 + Pb+2
when we add to reaction equlibrium constant of resulting equation is multiple of added reaction
Kfinal = (K6*K3)
= 7.81*10^7*1.89*10^-10
= 1.48*10^-2

Answer : 1.48*10^-2