Question

Equal volume of $0.02M$ $AgN{O}_3$ and $0.02MHCN$ were mixed. Calculate $\left[A{g}^{+}\right]$ at equilibrium assuming no cynao-complex formation.
Given : $K_{sp}\left(AgCN\right)=2.2\times {10}^{-16},K_a\left(HCN\right)=6.2\times {10}^{-10}.$

• A. $2.64\times {10}^{-4}M$
• B. $3.69\times {10}^{-8}M$
• C. $5.9\times {10}^{-5}M$
• D. $4.64\times {10}^{-2}M$

Answer 1

The correct option is C $5.9\times {10}^{-5}M$

In precipitation reaction we first assume complete precipitation.
$A{g}^{+}+HCN\to AgCN+H^{+}$
Since equal volume of $AgN{O}_{3}andHCN$ are mixed.
Let , v be the volume of $AgN{O}_{3}andHCN$
So , $\left[H^{+}\right]=\frac{0.02\times v}{2v}=0.01M$
Now for the equilibria
$AgCN\rightleftharpoons A{g}^{+}+C{N}^{-};K_{sp}=\left[A{g}^{+}\right]\left[C{N}^{-}\right]=2.2\times {10}^{-16}HCN\rightleftharpoons H^{+}+C{N}^{-};K_a=\frac{\left[H^{+}\right]\left[C{N}^{-}\right]}{\left[HCN\right]}=6.2\times {10}^{-10}$


Let "s" be the solubility of $AgCN.$
The equilibrium reaction of $AgCN$ is given below:
$AgCN\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+C{N}^{-}\left(aq\right)..\left(1\right)$
$c-sss-x$
$C{N}^{-}\left(aq\right)+H^{+}\left(aq\right)\stackrel{\frac{1}{(K_a{)}_{HCN}}}{\rightleftharpoons }HCN\left(aq\right)----\left(2\right)$
$sc0$
$s-xc-xx$
$\left[A{g}^{+}\right]=s\left[A{g}^{+}\right]=(s-x)+x=\left[C{N}^{-}\right]+\left[HCN\right]$
From the ratio of $\left[HCN\right]/\left[C{N}^{-}\right]$ we see , $\left[C{N}^{-}\right]<<\left[HCN\right].So,s=\left[A{g}^{+}\right]=\left[HCN\right]$
Thus from the equation (1) we have ,
$\left[A{g}^{+}\right]=\left[C{N}^{-}\right]+\left[HCN\right]\therefore \frac{2.2\times {10}^{-16}}{\left[C{N}^{-}\right]}=\left[C{N}^{-}\right]+\frac{\left[H^{+}\right]\left[C{N}^{-}\right]}{6.2\times {10}^{-10}}Or,\frac{2.2\times {10}^{-16}}{\left[C{N}^{-}\right]}=\left[C{N}^{-}\right]+\frac{\left(0.01\right)\left[C{N}^{-}\right]}{6.2\times {10}^{-10}}\therefore [C{N}^{-}<<<\frac{\left(0.01\right)\left[C{N}^{-}\right]}{6.2\times {10}^{-10}}Or,\frac{2.2\times {10}^{-16}}{\left[C{N}^{-}\right]}=\frac{\left(0.01\right)\left[C{N}^{-}\right]}{6.2\times {10}^{-10}}[C{N}^{-}{]}^2=\frac{2.2\times {10}^{-16}\times 6.2\times {10}^{-10}}{0.01}[C{N}^{-}{]}^2=13.64\times {10}^{-24}[C{N}^{-}]=3.7\times {10}^{-12}M[A{g}^{+}]=\frac{K_{sp}}{\left[C{N}^{-}\right]}[A{g}^{+}]=\frac{2.2\times {10}^{-16}}{3.7\times {10}^{-12}}[A{g}^{+}]=5.9\times {10}^{-5}M$