The percentage hydrolysis in $0.003 M$ aqueous solution of $N a I = O C N$ $(K_{a}$ for $H O C N$ is $3.33 \times 10^{- 4} M)$ is $0.0 X$ %. The value of X is

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Solution

For salt of weak acid- strong base,
Hydrolysis constant $K_{h} = \frac{K_{W}}{K_{a}}$
And degree of hydrolysis $h = \sqrt{\frac{K_{h}}{C}}$
$∴ h = \sqrt{\frac{K_{W}}{K_{a} \cdot C}} = \sqrt{\frac{10^{- 14}}{3.333 \times 10^{- 4} \times 3 \times 10^{- 3}}}$
$⟹ h = 1.0005 \times 10^{- 4}$
$∴$ Percentage hydrolysis $= h \times 100 = 0.01 %$
Hence, $X = 1$

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