A solution of \({\bf{0}}.{\bf{075}}{\rm{ }}{\bf{M}}{\rm{ }}{\bf{CoB}}{{\bf{r}}_{\bf{2}}}\) is saturated with\({{\bf{H}}_{\bf{2}}}{\bf{S}}{\rm{ }}\left( {\left[ {{{\bf{H}}_{\bf{2}}}{\bf{S}}} \right]{\rm{ }} = {\rm{ }}{\bf{0}}.{\bf{10}}{\rm{ }}{\bf{M}}} \right)\). What is the minimum pH at which CoS begins to precipitate?

\(\begin{array}{*{20}{c}}{CoS(s) \rightleftharpoons C{o^{2 + }}(aq) + {S^{2 - }}(aq)\quad \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{K_{sp}} = 4.5 \times 1{0^{ - 27}}} \\ {{H_2}S(aq) + 2{H_2}O(l) \rightleftharpoons 2{H_3}{O^ + }(aq) + {S^{2 - }}(aq)\quad \;\;\;\;\;\;\;\;\;K = 1.0 \times 1{0^{ - 26}}} \end{array}\;\;\)

A solution of \(0.075 M CoB{r_2}\) 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 \({\mathop{\rm CoS}\nolimits} \) begins to precipitate.

\({\text{CoS}}({\text{s}}) \rightleftharpoons {\text{C}}{{\text{o}}^{2 + }}({\text{aq}}) + {{\text{S}}^{2 - }}({\text{aq}})\)

\({K_{sp}} = 4.5 \times {10^{ - 27}}\)

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

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

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

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left[ {{\rm{C}}{{\rm{o}}^{2 + }}} \right] \cdot \left[ {{S^{2 - }}} \right]}\\{\left[ {{S^{2 - }}} \right] = \frac{{{K_{sp}}}}{{\left[ {C{o^{2 + }}} \right]}}}\\{ = \frac{{4.5 \times {{10}^{ - 27}}}}{{0.075}}}\\{ = 6 \times {{10}^{ - 26}}{\rm{M}}}\end{array}\)

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

\(\begin{array}{*{20}{c}}{K = \frac{{{{\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]}^2} \cdot \left[ {{{\rm{S}}^{2 - }}} \right]}}{{\left[ {{{\rm{H}}_2}{\rm{S}}} \right]}}}\\{1.0 \times {{10}^{ - 26}} = \frac{{{{\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]}^2} \times 6 \times {{10}^{ - 26}}}}{{0.10}}}\\{{{\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]}^2} = 0.0167}\\{\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right] = 0.129{\rm{M}}}\end{array}\)

And the pH value is

\(\begin{array}{c}pH = - \log \left[ {{H_3}{O^ + }} \right]\\ = - \log [0.129]\\ = 0.89\end{array}\)