Question

For the reaction, $\left[Ag\right(C{N}_2){]}^{-}\rightleftharpoons A{g}^{+}+2C{N}^{-}$ the equilibrium constant at 298 K is $4.0\times {10}^{-10}$. The silver ion concentration in a solution which was originally 0.12 molar n $KCN$ and 0.04 molar in $AgN{O}_3$, is:

• A. $1\times {10}^{-10}M$
• B. $1\times {10}^{-8}M$
• C. $1\times {10}^{-12}M$
• D. $1\times {10}^{-4}M$

Answer 1

The correct option is B $1\times {10}^{-8}M$

${\left[{Ag\left(CN\right)}_2\right]}^{-1}\rightleftharpoons {Ag}^{+}+2{CN}^{-}K=4\times {10}^{-10}$

$K=\frac{\left[{Ag}^{+}\right]{\left[{CN}^{-}\right]}^2}{{\left[{Ag\left(CN\right)}_2\right]}^{-1}}$

$\left[{Ag}^{+}\right]$ from ${AgNO}_3=0.04$

$\left[{CN}^{-}\right]$ from $KCN=0.12$

${AgNO}_3+2KCN\to {\left[Ag{\left(CN\right)}_2\right]}^{-1}+{KNO}_3$ (Almost forward reaction)

$t=0$ $0.04$ $0.12$ $0$ $0$

$t=t_{eq}$ $0$ $0.04$ $0.04$

So, at equilibrium

$K=\frac{\left[{Ag}^{+}\right]{\left[{CN}^{-}\right]}^2}{\left[Ag{\left(CN\right)}_2^{-}\right]}$

$4\times {10}^{-10}=\frac{\left[{Ag}^{+}\right]{\left(0.04\right)}^2}{\left(0.04\right)}$

$\left[{Ag}^{+}\right]=\frac{4\times {10}^{-10}}{4\times {10}^{-2}}=1\times {10}^{-8}M$