Chemical and Ionic Equilibrium - Result Question 119
####63. For the reaction, $\left[Ag(CN) _2\right]^{-} \rightleftharpoons Ag^{+}+2 CN^{-}$
The equilibrium constant, at $25^{\circ} C$, is $4.0 \times 10^{-19}$. Calculate the silver ion concentration in a solution which was originally $0.10 M$ in $KCN$ and $0.03 M$ in $AgNO _3$. (1994, 3M)
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Solution:
- $Ag^{+}+2 CN^{-} \rightleftharpoons Ag(CN) _2^{-}$
Initial : $\quad \begin{array}{cccc}0.03 & 0.10 & 0\end{array}$
Equilibrium : $x \quad 0.10-0.06 \quad 0.03$
$$ \begin{aligned} & K=\frac{1}{4 \times 10^{-19}}=2.5 \times 10^{18} \ \Rightarrow \quad K & =2.5 \times 10^{18}=\frac{0.03}{(0.04)^{2} x} \ x & =7.50 \times 10^{-18} M Ag^{+} \end{aligned} $$
