Question: What are the concentrations of Ag⁺, CN^–, and Ag (CN)₂⁻ in a saturated solution of AgCN?

The solubility of AgCN

\({K_{sp}} = \left( {A{g^ + }} \right) \cdot \left( {C{N^ - }} \right) = 1.2 \cdot {10^{ - 16}}\)

The reaction of formation of\({\left( {{\rm{Ag}}{{({\rm{CN}})}_2}} \right)^ - }\)

\({K_f} = \frac{{\left( {{\rm{Ag}}({\rm{CN}})_2^ - } \right)}}{{\left( {{\rm{A}}{{\rm{g}}^ + }} \right).{{\left( {C{{\rm{N}}^ - }} \right)}^2}}} = 1 \cdot {10^{21}}\)

Let us calculate the concentrations of Ag⁺, CN^–, and Ag(CN)₂⁻ in a saturated solution of AgCN.

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left( {A{g^ + }} \right) \cdot \left( {C{N^ - }} \right)}\\{\left( {C{N^ - }} \right) = \frac{{{K_{sp}}}}{{\left( {A{g^ + }} \right)}}}\\{{K_f} = \frac{{\left( {Ag(CN)_2^ - } \right)}}{{\left( {A{g^ + }} \right) \cdot {{\left( {C{N^ - }} \right)}^2}}}}\\{{K_f} = \frac{{\left( {Ag(CN)_2^ - } \right)}}{{\left( {A{g^ + }} \right) \cdot \frac{{K_{sp}^2}}{{{{\left( {A{g^ + }} \right)}^2}}}}}}\\{{K_f} = \frac{{\left( {Ag(CN)_2^ - } \right) \cdot \left( {A{g^ + }} \right)}}{{K_{sp}^2}}}\end{array}\)

• 2 moles of AgCN will dissociate into 2 moles of Ag⁺ and 2 moles of CN^-
• 2 moles of CN^-will react with 1 mole of Ag⁺, to produce one mole ofAg (CN)₂⁻and we will have one mole of Ag⁺left

• Therefore, we\(\left( {A{g^ + }} \right) = \left( {Ag(CN)_2^ - } \right)\)

  \(\begin{array}{*{20}{c}}{{K_f} = \frac{{\left( {A{g^ + }} \right) \cdot \left( {A{g^ + }} \right)}}{{K_{sp}^2}}}\\{{{\left( {A{g^ + }} \right)}^2} = {K_f} \cdot K_{sp}^2}\\{ = 1 \cdot {{10}^{21}} \cdot {{\left( {1.2 \cdot {{10}^{ - 16}}} \right)}^2}}\\{\left( {A{g^ + }} \right) = 3.795 \cdot {{10}^{ - 6}}{\rm{M}}}\\{\left( {Ag(CN)_2^ - } \right) = 3.795 \cdot {{10}^{ - 6}}{\rm{M}}}\end{array}\)

  And the concentration of CN^- is

  \(\begin{array}{*{20}{c}}{\left( {C{N^ - }} \right) = \frac{{{K_{sp}}}}{{\left( {A{g^ + }} \right)}}}\\{ = \frac{{1.2 \cdot {{10}^{ - 16}}}}{{3.795 \cdot {{10}^{ - 6}}}}}\\{ = 3.162 \cdot {{10}^{ - 11}}{\rm{M}}}\end{array}\)