Question

A 1.0 M H¿S solution has a pH = 3.75 at equilibrium. What is the value of K

Answer 1

you have this \(H_2S + H_2O\rightleftharpoons H_3O^++HS^-\)

with the equilibrium constant expression: \(K=\dfrac{[H_3O^+][HS^-]}{[H_2S][H_2O]}\)

Im not sure if they want you to ignore the water (because it's staying constant, it's not normally used since it doesn't change significantly), but lets say they do. The you have:

\(K=\dfrac{[H_3O^+][HS^-]}{[H_2S]}\)

they give you \([H_3O^+]\) the question because \(pH=-log[H_3O^+]\).

So find their concentrations at equilibrium and plug them into the expression.