Question

Consider the salt $AgCN$ present in acidic medium having $H^{+}$ concentration equal to $cM$.
If $s$ is the solubility of $AgCN$, $K_a$ is the acid dissociation constant of $HCN$ and $K_{sp}$ is the solubility product of $AgCN$.

Find the value of $\frac{K_{sp}}{(K_a{)}_{HCN}}$

Assume $(K_a{)}_{HCN}$ to have a fairly low value.

• A. $\frac{s}{c-s}$
• B. $\frac{s^2}{c-s}$
• C. $\frac{s^2}{c+s}$
• D. $\frac{c-s}{s^2}$

Answer 1

The correct option is B $\frac{s^2}{c-s}$

Solubility of salt in acidic medium :
$AgCN\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+C{N}^{-}\left(aq\right)$
$ss-x$

$C{N}^{-}\left(aq\right)+H^{+}\left(aq\right)\stackrel{\frac{1}{(K_a{)}_{HCN}}}{\rightleftharpoons }HCN\left(aq\right)$
$sc0$
$s-xc-xx$

$HCN$ 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 both the above equations :
$AgCN\left(s\right)+H^{+}\left(aq\right)\stackrel{\frac{K_{sp}}{K_a}}{\rightleftharpoons }A{g}^{+}\left(aq\right)+HCN\left(aq\right)$
$c-sss$
$\frac{K_{sp}}{K_a}=\frac{s^2}{c-s}$