The percentage hydrolysis in $0.003 M$ aqueous solution of $N a O C N$ $(K_{a}$ for $H O C N$ is $3.33 \times 10^{- 4} M)$ is:

0.1 %

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0.01 %

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0.5 %

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0.8 %

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Solution

The correct option is B $0.01 %$
$O C N^{-} 1 (1 - h) + H_{2} O ⇌ H C N 0 h + {O H}^{-} 0 h$

$h = \sqrt{\frac{K_{H}}{c}} = \sqrt{\frac{K_{w}}{K_{a} \cdot c}} = \sqrt{\frac{10^{- 14}}{3.33 \times 10^{- 4} \times 0.003}} = 10^{- 4}$

$∴ h = 10^{- 2} % = 0.01 %$

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