Calculate the concentration of each species present in a \(0.050 - M\) solution of \({H_2}\;S. \)

We have two ionization reactions for\({{\bf{H}}_{\bf{2}}}{\bf{S}}\)because it is a diprotic acid:

\(\begin{align}{l}{{\bf{H}}_{\bf{2}}}{\bf{S(aq)}} \to {{\bf{H}}^{\bf{ + }}}{\bf{(aq) + H}}{{\bf{S}}^{\bf{ - }}}{\bf{(aq)}}.\\{\bf{H}}{{\bf{S}}^{\bf{ - }}}{\bf{(aq)}} \to {{\bf{H}}^{\bf{ + }}}{\bf{ + }}{{\bf{S}}^{{\bf{2 - }}}}{\bf{(aq)}}.\end{align}\)

The two ionization constants can be found in the table in Appendix H of the book:

\(\begin{align}{{\bf{K}}_{\bf{1}}}{\bf{ = 8}}{\bf{.9 \times 1}}{{\bf{0}}^{{\bf{ - 8}}}}.\\{{\bf{K}}_{\bf{2}}}{\bf{ = 1}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 19}}}}.\end{align}\)

The first reaction\({{\bf{H}}_{\bf{2}}}{\bf{S(aq)}} \to {{\bf{H}}^{\bf{ + }}}{\bf{(aq) + H}}{{\bf{S}}^{\bf{ - }}}{\bf{(aq)}}.\)can be interpreted in two ways. The first is that the concentrations of\({{\bf{H}}^{\bf{ + }}}\)and\({\bf{H}}{{\bf{S}}^{\bf{ - }}}\)formed are exactly the same. The other is that because the\({{\bf{K}}_{\bf{1}}}\)value is so small, the change in the\({{\bf{H}}_{\bf{2}}}{\bf{S}}\)concentration is minimal.

As a result, we may write:

\(\begin{align}{K_1} &= \frac{{{x^2}}}{{0.050}}\\8.9 \times 1{0^{ - 8}} &= \frac{{{x^2}}}{{0.050}}\\ x &= \left( {{H^ + }} \right)\\ &= \left( {H{S^ - }} \right)\\ & = 6.7 \times 1{0^{ - 5}} M.\end{align}\)

Now, we can compute the concentrations of the second reaction's components. We can assume that the concentrations of \({H^ + }\) and \({S^{2 - }}\) are equal and that the change in the concentration of \(H{S^ - }\) is inconsequential once again.

\(\begin{align}{K_2} &= \frac{{{x^2}}}{{6.7 \times 1{0^{ - 5}}}}\\1.0 \times 1{0^{ - 19}} &= \frac{{{x^2}}}{{6.7 \times 1{0^{ - 5}}}} x &= \left( {{S^{2 - }}} \right)\\ &= \left( {{H^ + }} \right)\\ &= 2.6 \times 1{0^{ - 12}} M.\end{align}\)

The sum of the \({H^ + }\) ions formed in the first and those formed in the second is the overall concentration of the \({H^ + }\)ions.

\(\begin{align}{\left( {{H^ + }} \right)_{total }} &= \left( {6.7 \times 1{0^{ - 5}} + 2.6 \times 1{0^{ - 12}}} \right) M\\ &= 6.7 \times 1{0^{ - 5}} M.\end{align}\)

All of the concentrations are now known. \({H_2}S\) has a concentration of \(0.050 M\), while the \(H{S^ - }\) ion has a concentration of \(6.7 \times 1{0^{ - 5}} M\). The \({H^ + }\) ion has a concentration of \(6.7 \times 1{0^{ - 5}} M\), while the \({S^{2 - }}\) ion has a concentration of \(2.6 \times 1{0^{ - 12}}M\).

\(\begin{align}\left( {{H_2}S} \right) &= 0.050 M.\\\left( {H{S^ - }} \right) &= 6.7 \times 1{0^{ - 5}} M.\\\left( {{S^{2 - }}} \right) &= 2.6 \times 1{0^{ - 12}} M.\\\left( {{H^ + }} \right) &= 6.7 \times 1{0^{ - 5}} M.\end{align}\)