An enclosed reactor contains a solution of iron sulfide $(Fe{S}_{(s)}).$ The air above the solution in this reactor contains hydrogen sulfide $(H_2(S))$ gas. You know $H_2{S}_{(aq)}($which forms from hydrogen sulfide gas dissolution$)$ is a weak acid and dissociates according to the following equations:

$H_2{S}_{(g)}harr{H}_2{S}_{(aq)},K_H=0\mathrm{.}11\frac{M}{a}tm$

$H_2{S}_{(aq)}harr{H}^{+}+H{S}_{(aq)}^{-},K_{a,1}={10}^{-7\mathrm{.}10}$

$H{S}_{(aq)}^{-}harr{H}^{+}+S_{(aq)}^{2-},K_{a,2}={10}^{-12\mathrm{.}92}$

$Fe{S}_{(s)}harrF{e}_{(aq)}^{2+}+S^{2^{-}(aq)},K_{sp}={10}^{-7\mathrm{.}20}$

a$.$ Write the charge balance expression for this chemical equilibrium reaction system.

b$.$ Solve for the concentrations of: $H_2{S}_{(aq)}$

$H{S}^{-}$

$S^{2-},$ and $F{e}^{2+},$ and $O{H}^{-}$in terms of $P_{H2S},K\text{'}$s and $H^{+}.$

c$.$ You measure the $pH$ of the water in this enclosed reactor to be $8\mathrm{.}5.$ Calculate the partial pressure of $H_{2}S$ gas $($in atm$)$ in the air above the liquid inside this enclosed reactor. Start by re$-$writing your charge balance expression in terms of the species concentrations you solved for in $($b$).$