MnS(s) + 2 H+ --> Mn2+ + H2S(g)

At 25 C the solubility product constant, Ksp, for MnS is 5 x 10-15 and the acid dissociation constants K1 and K2 for H2S are 1 x 10-7 and 1 x 10-13, respectively. What is the equilibrium constant for the reaction represented by the equation above at 25 C?

a) 1x10^-13/5x10^-15
b) 5x10^-15/1x10^-7
c) 1x10^-7/5x10^-20
d)5x10^-15/1x10^-20
e) 1x10^-20/5x10^-15

I know the answer is D but can someone explain why it is this answer?

3 answers

Keq = (Mn^2+)(H2S)/(H^+)^2

Ksp = (Mn^2+)(S^2-) = 5E-15
H2S ==> 2H^+ + S^2-
k1k2 for H2S = 1E-20

Divide Ksp/k1k2 and see that it produces Keq from above.

(Mn^2+)(S^2-)(H2S)/(H^+)^2(S^2-) and (S^2-) csncels to leave Keq.
Ksp/k1k2 =

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