Question

Calculate the percentage hydrolysis of 0.003 M aqueous solution of $NaOCN.K_a$ for $HOCN=3.3\times {10}^{-4}$.

If the percentage hydrolysis is ${10}^{-x}$, then what is the value of $x$?

Answer 1

$NaOCN+H_{2}O\rightleftharpoons NaOH+HCN$

$h=\sqrt{\left(\frac{K_h}{C}\right)}=\sqrt{\left(\frac{K_w}{K_a\cdot C}\right)}$

$=\sqrt{\left(\frac{{10}^{-14}}{3.33\times {10}^{-4}\times 0.003}\right)}$

$h={10}^{-4}$

$\therefore \%$ hydrolysis $={10}^{-4}\times 100={10}^{-2}$

Hence, the value of $x$ is $2$.