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Introduction and Degree of Hydrolysis

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Question

Calculate the percentage of hydrolysis in $0.003 M$ aqueous solution of $N a O C N$ . $K_{a}$ for $H O C N = 3.33 \times 10^{- 4}$ .

0.001

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0.01

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Solution

The correct option is B 0.01
$N a O C N + H_{2} O ⇌ N a O H + H C N$
$H O C N ⇌ H^{+} + O C N^{-}$
From hydrolysis of salt, we know,
$K_{h} = \frac{K_{w}}{K_{a}}$ and
$D e g r e e o f h y d r o l y s i s, α = \sqrt{(\frac{K_{h}}{C})} = \frac{\sqrt{K_{w}}}{\sqrt{(K_{a} . C)}}$
$= \frac{\sqrt{(10^{- 14})}}{\sqrt{(3.33 \times 10^{- 4} \times 0.003)}}$
$D e g r e e o f h y d r o l y s i s = 10^{- 4}$
$% of hydrolysis$ $= 10^{- 4} \times 100 = 10^{- 2}$

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Similar questions

0.003 M

N a O C N

K_{a}

H O C N = 3.33 \times 10^{- 4}

The percentage of hydrolysis in $0.003 M$ aqueous solution of $N a O C N$ is:

( $K_{a}$ for $H O C N$ = $3.33 \times 10^{- 4} M$ )

0.003 M

N a O C N

(K_{a}

H O C N

3.33 \times 10^{- 4} M)

0.003 M

N a O C N

K_{a}

H O C N = 3.33 \times 10^{- 4}

The percentage of hydrolysis in $0.003 M$ aqueous solution of $N a O C N$ is:

( $K a$ for $H O C N$ = $3.33 \times 10^{- 4} M$ )

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