Question

The percentage hydrolysis in $0.003M$ aqueous solution of $NaI=OCN$ $(K_a$ for $HOCN$ is $3.33\times {10}^{-4}M)$ is $0.0X$%. The value of X is

Answer 1

For salt of weak acid- strong base,

Hydrolysis constant $K_h=\frac{K_W}{K_a}$

And degree of hydrolysis $h=\sqrt{\frac{K_h}{C}}$

$\therefore h=\sqrt{\frac{K_W}{K_a\cdot C}}=\sqrt{\frac{{10}^{-14}}{3.333\times {10}^{-4}\times 3\times {10}^{-3}}}$

$⟹h=1.0005\times {10}^{-4}$

$\therefore$ Percentage hydrolysis $=h\times 100=0.01\%$

Hence, $X=1$