Question

Calculate the degree of hydrolysis of $0.10$ solution of $KCN$. Dissociation constant of $HCN=7.2\times {10}^{-10}$ at ${25}^{o}C$ and $K_w=1.0\times {10}^{-4}$

Answer 1

$C{N}^{-}+H_{2}O\rightleftharpoons HCN+{OH}^{-}$

$C$

$C(1-h)$ $Ch$ $Ch$

$K_h=\frac{\left(Ch\right)\left(Ch\right)}{C(1-h)}$

Let '$h<<1$'

So $K_h=C{h}^2$

We know that for a salt of weak acid and strong base $K_h=\frac{K_w}{K_a}$

So $K_h=1.38\times {10}^{-5}$

$C{h}^2=1.38\times {10}^{-5}$

$h=1.18\times {10}^{-2}$