Question: How many grams of \(Zn{(CN)_2}(s)(117.44g/mol)\)would be soluble in 100 mL of\({H_2}O\)? Include the balanced reaction and the expression for \({K_{sp}}\)in your answer. The \({K_{sp}}\)value for\(Zn{(CN)_2}(s)\;is\;3.0 \times 1{0^{ - 16}}\).

\({{\rm{K}}_{sp}}{\rm{\;of\;Zn}}{({\rm{CN}})_2}{\rm{\;is\;}}{K_{sp}} = \left( {{\rm{Z}}{{\rm{n}}^{2 + }}} \right) \cdot {\left( {C{N^ - }} \right)^2} = 3.0 \cdot {10^{ - 16}}\)

The molar mass is \({\rm{Zn}}{({\rm{CN}})_2}{\rm{\;is\;}}117.44{\rm{g}}/{\rm{mol}}\)

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left( {Z{n^{2 + }}} \right) \cdot {{\left( {C{N^ - }} \right)}^2}}\\{3.0 \cdot {{10}^{ - 16}} = x \cdot {{(2x)}^2}}\\{4{x^3} = 3.0 \cdot {{10}^{ - 16}}}\\{x = 4.22 \cdot {{10}^{ - 6}}}\\{\left( {Z{n^{2 + }}} \right) = x = 4.22 \cdot {{10}^{ - 6}}{\rm{M}}}\end{array}\)

\(\begin{array}{*{20}{c}}{{n_{Zn{{(CN)}_2}}} = \left( {Z{n^{2 + }}} \right) \cdot 0.100{\rm{L}}}\\{ = 4.22 \cdot {{10}^{ - 6}}{\rm{M}} \cdot 0.100{\rm{L}}}\\{ = 4.22 \cdot {{10}^{ - 7}}{\rm{mol}}}\end{array}\)

Find mass,

\({m_{{Z_n}{{(CN)}_2}}} = 4.22 \cdot {10^{ - 7}}{\rm{mol}} \cdot 117.44{\rm{g}}/{\rm{mol}}\)

=\(4.96 \cdot {10^{ - 5}}{\rm{g}}\)

The solution is there are \(4.96 \cdot {10^{ - 5}}{\rm{g}}\) of \({\rm{Zn}}{({\rm{CN}})_2}({\rm{s}})(117.44{\rm{g}}/{\rm{mol}})\)would be soluble in 100 mL \({{\rm{H}}_2}{\rm{O}}\).