# Solubitity of FeS

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.