For the reaction: $[ ext{Ag(CN)}_2]^- ightleftharpoons ext{Ag}^+ + 2 ext{CN}^-$, the equilibri

Question

For the reaction: $[	ext{Ag(CN)}_2]^- ightleftharpoons 	ext{Ag}^+ + 2	ext{CN}^-$, the equilibrium constant at $25^\circ	ext{C}$ is $4.0 	imes 10^{-19}$. Calculate the silver ion concentration (in $	ext{M}$) in a solution which was originally $0.10	ext{ molar}$ in $	ext{KCN}$ and $0.03	ext{ molar}$ in $	ext{AgNO}_3$. Express your answer as $X 	imes 10^{-18}$ and provide the value of $X$.

Professor PC Thomas Classes · Answer

Initially, $	ext{Ag}^+$ reacts with $	ext{CN}^-$ to form the complex ion $[	ext{Ag(CN)}_2]^-$.
$	ext{Ag}^+ + 2	ext{CN}^- ightarrow [	ext{Ag(CN)}_2]^-$
Initial concentrations: $[	ext{Ag}^+] = 0.03	ext{ M}$, $[	ext{CN}^-] = 0.10	ext{ M}$
$0.03	ext{ M}$ of $	ext{Ag}^+$ requires $2 	imes 0.03 = 0.06	ext{ M}$ of $	ext{CN}^-$ to form $0.03	ext{ M}$ of $[	ext{Ag(CN)}_2]^-$.
Since $0.10	ext{ M} > 0.06	ext{ M}$, $	ext{CN}^-$ is in excess. The reaction will go almost to completion.
Concentration of complex formed, $[[	ext{Ag(CN)}_2]^-] \approx 0.03	ext{ M}$.
Concentration of free $	ext{CN}^-$ remaining = $0.10 - 0.06 = 0.04	ext{ M}$.
Now consider the dissociation equilibrium:
$[	ext{Ag(CN)}_2]^- ightleftharpoons 	ext{Ag}^+ + 2	ext{CN}^-$
Let $[	ext{Ag}^+] = x$ at equilibrium.
$K = \frac{[	ext{Ag}^+][	ext{CN}^-]^2}{[[	ext{Ag(CN)}_2]^-]} = 4.0 	imes 10^{-19}$
Substituting the values (assuming $x$ is very small compared to $0.04$):
$\frac{(x)(0.04)^2}{0.03} = 4.0 	imes 10^{-19}$
$x = \frac{4.0 	imes 10^{-19} 	imes 0.03}{1.6 	imes 10^{-3}} = \frac{12 	imes 10^{-21}}{1.6 	imes 10^{-3}} = 7.5 	imes 10^{-18}	ext{ M}$
Thus, $X = 7.5$.