Part A Given the two reactions H 2 SH S H K 1 919108 and

Part A: Given the two reactions

H 2 SH S + H + K 1 = 9.19×108

and

H S S 2 + H +    K 2 = 1.28×1019

what is the equilibrium constant K final for the following reaction? S 2 +2 H + H 2 S

Enter your answer numerically.

Part B: Given the two reactions

PbCl2Pb2++2Cl,   K3 = 1.87×10¹⁰

and

AgClAg++Cl,   K4 = 1.26×10⁴,

what is the equilibrium constant Kfinal for the following reaction?

PbCl2+2Ag+2AgCl+Pb2+

Express your answer numerically.

Solution

Part A-

H2S <--> HS^- + H⁺. K1 = 9.19*10^-8

Inversing the reaction-

HS^- + H⁺ <--> H2S . K1 = 1/9.19*10^-8 = 1.09*10⁷....(1)

HS^- <---> S^2- + H⁺². K2 = 1.28*10^-19

Inversing the reaction-

S^2- + H⁺ <---> HS^- . K2 = 1/1.28*10^-19 = 7.81*10¹⁸...(2)

Adding 1 and 2 -

S^2- + 2H⁺ <--->H2S .

K = K1*K2 = 1.09*10⁷*7.81*10¹⁸ = 8.51*10²⁵

Part B -

PbCl2 <---> Pb²⁺ + 2Cl^-. K3 = 1.87*10^-10.....(3)

AgCl<---> Ag⁺ + Cl^- .K4 = 1.26*10^-4

Multiplying by 2-

2AgCl <--> 2Ag⁺ + 2Cl^- .K4 = (1.26*10^-4)² = 1.59*10^-8

Inversing the reaction-

2Ag⁺ + 2Cl^- <--> 2AgCl. K4 = 1/1.59*10^-8 = 6.28*10⁷..(4)

Adding 3 and 4-.

PbCl2 + 2Ag⁺ <---> 2AgCl + Pb²⁺

K = K3*K4 = 1.87*10^-10*6.28*10⁷ = 11.7*10^-3