Question: Calculate the equilibrium concentration of Zn²⁺ in a 3.0 M solution of\(Zn(CN)_4^{2 - }\) .

Let us calculate the equilibrium concentration of Zn²⁺in a 3.0 M solution of\({\rm{Zn}}({\rm{CN}})_4^{2 - }\)

The reaction of dissociation of\({\rm{Zn}}({\rm{CN}})_4^{2 - }\)

• The formation constant of\({\rm{Zn}}({\rm{CN}})_4^{2 - }\)is\({K_f} = 2.1 \cdot {10^{19}}\)
• Hence, the dissociation constant is\({K_d} = \frac{1}{{{K_f}}} = \frac{1}{{2.1 \cdot {{10}^{19}}}} = 4.76 \cdot {10^{ - 20}}\)

calculate the equilibrium concentration of Zn²⁺

\(\begin{array}{*{20}{c}}{{K_d} = \frac{{\left( {Z{n^{2 + }}} \right) \cdot {{\left( {C{N^ - }} \right)}^4}}}{{\left( {Zn(CN)_4^{2 - }} \right)}}}\\{4.76 \cdot {{10}^{ - 20}} = \frac{{x \cdot {{(4x)}^4}}}{{(0.30 - x)}}}\end{array}\)

Since K_d is smaller than 10^-4,

We will assume that\(0.30 - {\rm{x}} \approx 0.30\)

\(\begin{array}{*{20}{c}}{4.76 \cdot {{10}^{ - 20}} = \frac{{x \cdot {{(4x)}^4}}}{{0.30}}}\\{256{x^5} = 1.429 \cdot {{10}^{ - 20}}}\\{x = 3.54 \cdot {{10}^{ - 5}}{\rm{M}}}\\{\left( {Z{n^{2 + }}} \right) = 3.54 \cdot {{10}^{ - 5}}{\rm{M}}}\end{array}\)