Question

Calculate the solubility of $AgCN$ in a buffer solution of $pH=3.00$ assuming no complex formation.
Given : $K_{sp}\left(AgCN\right)=2.2\times {10}^{-16},K_a\left(HCN\right)=6.2\times {10}^{-10}$.
(Given $\sqrt{1.33\times {10}^{-9}}=3.64\times {10}^{-5}$)

• A. $5.9\times {10}^{-2}M$
• B. $1.9\times {10}^{-5}M$
• C. $3.9\times {10}^{-8}M$
• D. $9.4\times {10}^{-8}M$

Answer 1

The correct option is B $1.9\times {10}^{-5}M$

Let "s" be the solubility of $A{g}^{+}$ in the buffer solution.
On dissolution silver remains as $A{g}^{+}$ but $C{N}^{-}$ ions are converted mostly to $HCN$ due to fixed acidity of the buffer.
$K_a=\frac{\left[H^{+}\right]\left[C{N}^{-}\right]}{\left[HCN\right]}Or,\frac{\left[HCN\right]}{\left[C{N}^{-}\right]}=\frac{\left[H^{+}\right]}{K_a}=\frac{{10}^{-3}}{6.2\times {10}^{-10}}=1.6\times {10}^6$
As each $C{N}^{-}$ ion hydrolyses to yield one $HCN.$
$s=\left[A{g}^{+}\right]=\left[C{N}^{-}\right]+\left[HCN\right]$
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 ,
$\left[C{N}^{-}\right]=\frac{s}{1.6\times {10}^6}{K}_{sp}=\left[A{g}^{+}\right]\left[C{N}^{-}\right]$
$2.2\times {10}^{-16}=s\times \frac{s}{1.6\times {10}^6}s=\left[A{g}^{+}\right]=1.87\times {10}^{-5}M$


Alternate solution :
Let "s" be the solubility of $A{g}^{+}$ in the buffer solution.
The equilibrium reaction of $AgCN$ is given below:
$AgCN\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+C{N}^{-}\left(aq\right)..\left(1\right)$
$ss-x$
$C{N}^{-}\left(aq\right)+H^{+}\left(aq\right)\stackrel{\frac{1}{(K_a{)}_{HCN}}}{\rightleftharpoons }HCN\left(aq\right)----\left(2\right)$
$sc0$
$s-xc-xx$
HCN is a weak acid. So, $K_a$ will be very small. Hence, $\frac{1}{(K_a{)}_{HCN}}$ is very high i.e., equilibrium shifts in the forward direction
$\therefore s-x\approx 0\to s\approx x$
Adding equations (1) and (2) :
$AgCN\left(s\right)+H^{+}\left(aq\right)\stackrel{\frac{K_{sp}}{K_a}}{\rightleftharpoons }A{g}^{+}\left(aq\right)+HCN\left(aq\right)$
$c-sss\frac{K_{sp}}{K_a}=\frac{s^2}{c-s}\frac{2\times {10}^{-16}}{6\times {10}^{-10}}=\frac{s^2}{0.001-s}3.33\times {10}^{-7}=\frac{s^2}{0.001-s}(3.33\times {10}^{-10})-(3.33\times {10}^{-7})s=s^2{s}^2+(3.33\times {10}^{-7})s-(3.33\times {10}^{-10})=0s=\frac{-(3.33\times {10}^{-7})\pm \sqrt{(3.33\times {10}^{-7}{)}^2-4\times 1\times (-3.33\times {10}^{-10})}}{2\times 1}s=\frac{-(3.33\times {10}^{-7})+\sqrt{(3.33\times {10}^{-7}{)}^2-4\times 1\times (-3.33\times {10}^{-10})}}{2\times 1}s=\frac{(-3.33\times {10}^{-7})+\sqrt{1.33\times {10}^{-9}}}{2}s=\frac{(-3.33\times {10}^{-7})+3.64\times {10}^{-5}}{2}s=1.9\times {10}^{-5}M.$