When equal volumes of $0.2 M {AgNO}_{3}$ and $1 MKCN$ olutions were mixed then at equilibrium, concentration of ${Ag}^{+}$ was found to be $10^{- 6}$ M. While when equal volumes of $0.2 M Zn (NO 3) 2$ solution and of $1 M KCN$ solution were mixed then at equilibrium, concentration of ${Zn}^{2 +}$ ion was found to be $10^{- 12}$ M. Then find the equilibrium constant of following
$\begin{array}{l} 2 {[Ag (CN)_{2}]}^{-} (a q) + {Zn}^{2 +} (a q) ⇌ {[Zn (CN)_{4}]}^{2 -} (a q) + 2 {Ag}^{+} (a q) \end{array}$

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answer is 810.

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Detailed Solution

The reaction given in the problem is given as;

$\begin{array}{lcc} {Ag}^{+} & + 2 {CN}^{-} & ⇌ {Ag}_{2} ({CN}_{2}^{-} \\ 0.2 V & 1 \times V \\ ~ 0 = 10^{- 6} & V - 0.4 V & 0.2 V \end{array}$

$\begin{array}{l} K_{f_{1}} = \frac{(\frac{0.2 V}{2 V})}{10^{- 6} \times {(\frac{0.6 V}{2 V})}^{2}} = \frac{10^{7}}{9} \\ \begin{array}{clc} {Zn}^{2 +} + 4 {CN}^{-} & ⇌ Zn ({CN}_{4}^{2 -} \\ 0.2 V & 1 \times V & - \\ ~ 0 & V - 0.8 V & 0.2 V \\ = 10^{- 12} & = 0.2 V \end{array} \end{array}$

$K_{f_{2}} = \frac{(\frac{0.2 V}{2})}{10^{- 12} \times {(\frac{0.2 V}{2 V})}^{4}} = 10^{15} K_{f (question)} = \frac{K f_{2}}{{(K f_{1})}^{2}} = \frac{10^{15}}{10^{14}} \times 81 = 810$