Chemical and Ionic Equilibrium - Result Question 63-1
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 $0.10$ M in $ KCN$ and $0.03$ M in $\mathrm{AgNO}_3 $.
$(1994$, 3M $)$
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Solution:
- $\quad \quad\quad \quad \mathrm{Ag}^{+} \quad + \quad 2 \mathrm{CN}^{-} \quad \rightleftharpoons \quad \quad \mathrm{Ag}(\mathrm{CN})_2^{-}$
$\text {Initial }: \quad \quad \quad 0.03 \quad \quad \quad 0.10 \quad \quad \quad \quad 0 $
$\text { Equilibrium }: \quad x \quad \quad 0.10-0.06 \quad \quad 0.03$
$\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} $
BETA
