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