I have some questions related to a problem of multiple equilibria and solubility. I have to calculate the solubility of FeS in pure water considering that the second stage of hydrolisis of $\ce{S^2-}$, producing $\ce{H2S}$, can not be neglected.

For this problem the book that I was consulting gives me the values of the dissociation constants for $\ce{H2S}$ ($K_\mathrm{a1} = \pu{1.0e-7}$ and $K_\mathrm{a2} = \pu{1.2e-13}$) and the solubility product of $\ce{FeS}$ ($K_\mathrm{sp} = \pu{8e-9}$). I proposed the following reactions: \begin{align} \ce{FeS &<=> Fe^2+ + S^2-}\tag{R1}\\ \ce{H2S + H2O &<=> HS- + H3O+}\tag{R2}\\ \ce{HS- + H2O &<=> S^2- + H3O+}\tag{R3}\\ \ce{2H2O &<=> OH- + H3O+}\tag{R4} \end{align} From these ones I got the mass and charge balance equations: \begin{align} [\ce{Fe^2+}] &= [\ce{HS-}] + [\ce{S^2-}] + [\ce{H2S}]\tag{1}\\ 2[\ce{Fe^2+}] + [\ce{H3O+}] &= 2[\ce{S^2-}] + [\ce{HS-}] + [\ce{OH-}]\tag{2} \end{align} I also get other equations from the equilibrium of the acid in the two stages of its dissociation: \begin{align} K_\mathrm{a1} &= \frac{[\ce{HS-}][\ce{H3O+}]}{[\ce{H2S}]}\tag{3}\\ K_\mathrm{a1} &= \frac{[\ce{S^2-}][\ce{H3O+}]}{[\ce{HS-}]}\tag{4} \end{align} The solubility product of salt and the ionic product of water: \begin{align} K_\mathrm{sp} &= [\ce{Fe^2+}][\ce{S^2-}]\tag{5}\\ K_\mathrm{w} &= [\ce{H3O+}][\ce{OH-}]\tag{6} \end{align} I have six equation in total. So my doubt is: how to simplify these equations? Well, what relation could I get from the mass or charge balance equation in order to simplify the process of resolution?

In my book the answer is $\pu{4e-7}$, but I don't get the necessary relation to resolve all of this problem.

I would really appreciate some advice about it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.