Question

Calculate the percentage of hydrolysis in $0.003M$ aqueous solution of $NaOCN$. $K_a$ for $HOCN=3.33\times {10}^{-4}$.

• A. 0.001
• B. 0.01
• C. 0.02
• D. 0.1

Answer 1

The correct option is B 0.01

$NaOCN+H_{2}O\rightleftharpoons NaOH+HCN$
$HOCN\rightleftharpoons H^{+}+OC{N}^{-}$
From hydrolysis of salt, we know,
$K_h=\frac{K_w}{K_a}$ and
$Degreeofhydrolysis,\alpha =\sqrt{\left(\frac{K_h}{C}\right)}=\frac{\sqrt{K_w}}{\sqrt{(K_a.C)}}$
$=\frac{\sqrt{\left({10}^{-14}\right)}}{\sqrt{(3.33\times {10}^{-4}\times 0.003)}}$
$Degreeofhydrolysis={10}^{-4}$
$\%\text{of hydrolysis}$$={10}^{-4}\times 100={10}^{-2}$