Question

$K_a$ for $HCN$ is $1.4\times {10}^{-9}$. Calculate for $0.01N$ $KCN$ solution:
(a) degree of hydrolysis
(b) $\left[{OH}^{-}\right]$ and $\left[{CN}^{-}\right]$
(c) $pH$

Answer 1

Let '$h$' be degree of hydrolysis

$C{N}^{-}+H_{2}O\to O{H}^{-}+HCN$

$C$

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

$K_h=\frac{K_w}{K_a}=\frac{\left(Ch\right)\left(Ch\right)}{C(1-h)}=\frac{C{h}^2}{(1-h)}$

Let '$h$'<<1 so $(1-h)=1$

$K_h=C{h}^2=7.1\times {10}^{-6}$

$h=2.68\times {10}^{-3}$

B) $\left[O{H}^{-}\right]=Ch=2.68\times {10}^{-5}M$

$\left[C{N}^{-}\right]=C(1-h)=0.01M$

C) $pOH=-log\left(\right[O{H}^{-}\left]\right)=4.57$

$pH+pOH=14$

$pH=9.42$