A \({\bf{0}}.{\bf{125}}{\rm{ }}{\bf{M}}\) solution of \({\bf{Mn}}{\left( {{\bf{N}}{{\bf{O}}_{\bf{3}}}} \right)_{\bf{2}}}\) is saturated with\({{\bf{H}}_{\bf{2}}}{\bf{S}}{\rm{ }}\left( {\left[ {{{\bf{H}}_{\bf{2}}}{\bf{S}}} \right]{\rm{ }} = {\rm{ }}{\bf{0}}.{\bf{10}}{\rm{ }}{\bf{M}}} \right)\). At what pH does MnS begin to precipitate?

\(MnS(s) \rightleftharpoons M{n^{2 + }}(aq) + {S^{2 - }}(aq)\quad {K_{sp}} = 4.3 \times 1{0^{ - 22}}\)

\({H_2}S(aq) + 2{H_2}O(l) \rightleftharpoons 2{H_3}{O^ + }(aq) + {S^{2 - }}(aq)\quad K = 1.0 \times 1{0^{ - 26}}\)

A solution of \(0.125 M Mn{\left( {N{O_3}} \right)_2}\)is saturated with \({{\mathop{\rm H}\nolimits} _2}S \left( {\left[ {{H_2}S} \right] = 0.10 M} \right)\)

Let us calculate the minimum pH at which MnS begins to precipitate.

\({\text{MnS}}({\text{s}}) \rightleftharpoons {\text{M}}{{\text{n}}^{2 + }}({\text{aq}}) + {{\text{S}}^{2 - }}({\text{aq}})\)

\({K_{sp}} = 4.3 \times {10^{ - 22}}\)

\({{\text{H}}_2}{\text{S}}({\text{aq}}) + 2{{\text{H}}_2}{\text{O}}({\text{l}}) \rightleftharpoons 2{{\text{H}}_3}{{\text{O}}^ + }({\text{aq}}) + {{\text{S}}^{2 - }}({\text{aq}})\)

\(K = 1.0 \times {10^{ - 26}}\)

First, let us find the concentration of \(\left[ {{S^{2 - }}} \right]\)

\(\begin{align}{{}{}}{{K_{sp}} = \left[ {M{n^{2 + }}} \right] \cdot \left[ {{S^{2 - }}} \right]}\\{\left[ {{S^{2 - }}} \right] = \frac{{{K_{sp}}}}{{\left[ {M{n^{2 + }}} \right]}}}\\{ = \frac{{4.3 \times {{10}^{ - 22}}}}{{0.125}}}\\{ = 3.44 \times {{10}^{ - 21}}{\rm{ M}}}\end{align}\)

Concentration of \(\left[ {{H_3}{O^ + }} \right]\) is

\(\begin{array}{*{20}{c}}{K = \frac{{{{\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]}^2} \cdot \left[ {{{\rm{S}}^{2 - }}} \right]}}{{\left[ {{{\rm{H}}_2}{\rm{S}}} \right]}}}\\{1.0 \cdot {{10}^{ - 26}} = \frac{{{{\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]}^2} \times 3.44 \times {{10}^{ - 21}}}}{{0.10}}}\\{{{\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]}^2} = 2.9 \times {{10}^{ - 7}}}\\{\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] = 5.39 \times {{10}^{ - 4}}{\rm{M}}}\end{array}\)

And the pH value is

\(\begin{aligned}{}pH & = - \log \left[ {{H_3}{O^ + }} \right]\\ &= - \log \left[ {5.39 \cdot {{10}^{ - 4}}} \right]\\ &= {\rm{ }}3.27\end{aligned}\)