Question

A) Given the two reactions 1) H2S<--> HS- + H+, K1= 9.21x10^-8 and 2) HS- <-->...

A)

Given the two reactions

1) H2S<--> HS- + H+, K1= 9.21x10^-8 and

2) HS- <--> S2- + H+, K2=1.59x10^-19,

what is the equillibrium constant Kfinal for the following reaction?

S2- + 2H+ <--> H2S

B)

Given the two reactions

3) PbCl2 <--> Pb2+ + 2Cl-, K3= 1.87x10^-10, and

4) AgCl <--> Ag+ + Cl-, K4= 1.11x10^4,

what is the equillibrium constant Kfinal for the following reaction?

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

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Homework Answers

Answer #1

A)

Ka = [H2S]/[H+]^2[S2-]

Invert (1) and (2)

this is equally to

1/K1 and 1/K2 (since we invert it, we need to ^-1 each term)

Ka = 1/K2*1/K1 = K2/K1 = 1/(9.21*10^-8)*1/(1.59*10^-19) = 6.828782*10^25

Ka = 6.828782*10^25; makes sense since H2S is a weak acid, it is storngly favoured to H2S instead of it sionzied states

B)

K = [Pb+2]/[Ag+]^2

ignore PbCl2 and AgCl since those are acids

then

we need Ag+ in the left, and two the 2nd power so... 1/K4 then (1/K4)^2

therefore

K = K3*(1/K4)^2 = (1.87*10^-10)/(1.11*10^4) =1.6846847*10^-14

K = 1.6846847*10^-14

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