Calculate the \(K_{s p}\) of the following compounds, given their molar solubilities. (a) \(\mathrm{AgCN}, 7.73 \times 10^{-9} \mathrm{M}\) (b) \(\mathrm{Ni}(\mathrm{OH})_{2}, 5.16 \times 10^{-6} \mathrm{M}\) (c) \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}, 1.67 \times 10^{-8} M\)

For each compound, write the balanced chemical equation for the dissociation in water. Here are the dissociation equations for the given compounds: (a) \(\mathrm{AgCN} \rightleftharpoons \mathrm{Ag}^+ + \mathrm{CN}^-\) (b) \(\mathrm{Ni}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Ni}^{2+} + 2\mathrm{OH}^-\) (c) \(\mathrm{Cu}_3\left(\mathrm{PO}_4\right)_2 \rightleftharpoons 3\mathrm{Cu}^{2+} + 2\mathrm{PO}_4^{3-}\)

For \(\mathrm{AgCN}\), the dissociation equation is: \(\mathrm{AgCN} \rightleftharpoons \mathrm{Ag}^+ + \mathrm{CN}^-\) Let the molar solubility of AgCN be s. Then, the concentration of Ag+ and CN- ions will also be s. The expression for solubility product (\(K_{sp}\)) will be: \(K_{sp}=[\mathrm{Ag}^+][\mathrm{CN}^-]=s^2.\) Given molar solubility, \(s=7.73\times10^{-9}\,\mathrm{M}.\) Now, we can calculate \(K_{sp}\): \(K_{sp}=(7.73\times10^{-9})^2=5.97\times10^{-17}\).

For \(\mathrm{Ni}(\mathrm{OH})_2\), the dissociation equation is: \(\mathrm{Ni}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Ni}^{2+} + 2\mathrm{OH}^-\) Let the molar solubility of Ni(OH)2 be s. Then, the concentration of Ni2+ ions will be s and the concentration of OH- ions will be 2s. The expression for solubility product (\(K_{sp}\)) will be: \(K_{sp}=[\mathrm{Ni}^{2+}][\mathrm{OH}^-]^2=s(2s)^2=4s^3.\) Given molar solubility, \(s=5.16\times10^{-6}\,\mathrm{M}.\) Now, we can calculate \(K_{sp}\): \(K_{sp}=4(5.16\times10^{-6})^3=5.49\times10^{-16}\).

For \(\mathrm{Cu}_3\left(\mathrm{PO}_4\right)_2\), the dissociation equation is: \(\mathrm{Cu}_3\left(\mathrm{PO}_4\right)_2 \rightleftharpoons 3\mathrm{Cu}^{2+} + 2\mathrm{PO}_4^{3-}\) Let the molar solubility of Cu3(PO4)2 be s. Then, the concentration of Cu2+ ions will be 3s and the concentration of PO43- ions will be 2s. The expression for solubility product (\(K_{sp}\)) will be: \(K_{sp}=[\mathrm{Cu}^{2+}]^3[\mathrm{PO}_4^{3-}]^2=(3s)^3(2s)^2=108s^5.\) Given molar solubility, \(s=1.67\times10^{-8}\,\mathrm{M}.\) Now, we can calculate \(K_{sp}\): \(K_{sp}=108(1.67\times10^{-8})^5=3.65\times10^{-34}\). In summary: - \(K_{sp}\) of AgCN: \(5.97 \times 10^{-17}\,\mathrm{M}.\) - \(K_{sp}\) of Ni(OH)2: \(5.49 \times 10^{-16}\,\mathrm{M}.\) - \(K_{sp}\) of Cu3(PO4)2: \(3.65 \times 10^{-34}\,\mathrm{M}.\)