Question

Given the two reactions H2S?HS?+H+,   K1 = 9.15

Given the two reactions

H2S?HS?+H+,   K1 = 9.15

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Answer #1

part - 1

add both equations ....

H2S + HS- <---------> HS- + H+ + S2- + H+

HS- will cancel out from both sides

H2S <-------> S2- + 2H+

and new equilibrium constant = K1 X K2 = 9.15 X 10^-8 X 1.18 X 10^-19 = 1.0797 X 10^-26

now reverse the equation to get your equation....

S2- + 2H+ <-----------> H2S ......new equilibrium constant = 1 / (1.0797 X 10^-26 ) = 9.26 X 10^25

part - 2

multiply second equation by 2

2AgCl ,<----------> 2Ag+ + 2Cl- .......new equilibrium constnat = K5 = K4^2 = ( 1.13 X 10^-4)^2 = 1.2769 X 10^-8

now subtract above equation from first equation...

PbCl2 - 2AgCl <----------> Pb2+ + 2Cl- - 2Ag+ - 2Cl-

PbCl2 + 2Ag+ <----------> 2AgCl + Pb2+

new equilibrium constant = K3/K5 = 1.81 X 10^-10 / (1.2769 X 10^-8) = 0.01417

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