Question:Calculate the equilibrium concentration of Zn²⁺ in a solution initially with 0.150 M Zn²⁺and 2.50 M CN^-.

Let us calculate the equilibrium concentration of Zn²⁺in a solution initially with 0.150 M Zn²⁺and 2.50 M NH₃.CN^-.

The reaction of formation of\({\left( {{\rm{Zn}}{{({\rm{CN}})}_4}} \right)^{2 - }}\)

Since Zn²⁺ is a limiting reagent, 0.150 M Zn²⁺will react with\((0.150{\rm{M}} \cdot 4)\)\(0.6{\rm{MC}}{{\rm{N}}^ - }\), to produce 0.150 M\({\left( {{\rm{Zn}}{{({\rm{CN}})}_4}} \right)^{2 - }}\)

Therefore, we will be left with

\(\begin{array}{*{20}{c}}{\left( {Z{n^{2 + }}} \right) = 0{\rm{M}}}\\{({\rm{CN}}\mid = 2.50{\rm{M}} - 0.6{\rm{M}} = 1.9{\rm{M}}}\\{\left( {{{\left( {{\rm{Zn}}{{({\rm{CN}})}_4}} \right)}^{2 - }}} \right) = 0.150{\rm{M}}}\end{array}\)

The reaction ofdissociation of\({\left( {{\rm{Zn}}{{({\rm{CN}})}_4}} \right)^{2 - }}\)

The dissociation constant of\({\left( {{\rm{Zn}}{{({\rm{CN}})}_4}} \right)^{2 - }}{\rm{\;is\;}}{K_d} = \frac{1}{{{K_f}}} = \frac{1}{{2.1 \cdot {{10}^{19}}}} = 4.76 \cdot {10^{ - 20}}\)

\(\begin{array}{*{20}{c}}{{K_d} = \frac{{\left( {Z{n^{2 + }}} \right) \cdot {{\left( {C{N^ - }} \right)}^4}}}{{\left( {{{\left( {Zn{{(CN)}_4}} \right)}^{2 - }}} \right)}}}\\{4.76 \cdot {{10}^{ - 20}} = \frac{{x \cdot {{(1.9 + 4x)}^4}}}{{0.150 - x}}}\end{array}\)

Since K_d is smaller than 10^-4,

We will assume that\(0.150 - {\rm{x}} \approx 0.150\)

And that\(1.9 + 4{\rm{x}} \approx 1.9\)

\(\begin{array}{*{20}{c}}{4.76 \cdot {{10}^{ - 20}} = \frac{{x \cdot {{(1.9)}^4}}}{{0.150}}}\\{x = 5.48 \cdot {{10}^{ - 22}}{\rm{M}}}\\{\left( {Z{n^{2 + }}} \right) = 5.48 \cdot {{10}^{ - 22}}{\rm{M}}}\end{array}\)