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Salt of Weak Acid and Weak Base

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Question

Calculate the degree of hydrolysis and $p H$ of $0.02 M$ ammonium cyanide $(N H_{4} C N)$ at 298 K. $[K_{1}$ of $H C N = 4.99 \times 10^{- 9}, K_{b}$ for $N H_{4} O H = 1.77 \times 10^{- 5}]$

A

8.2

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B

3.2

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C

9.3

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D

3.9

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Solution

The correct option is D 9.3
$K_{h} = \frac{10^{- 14}}{4.99 \times 10^{- 9} \times 1.77 \times 10^{- 15}} = 1.132$

It can be seen that hydrolysis constant $(K_{h})$ is not small, and for calculating $h$ , the equation used is

$h = \frac{\sqrt{K_{h}}}{1 + \sqrt{K_{h}}} = \frac{1.06}{1 + 1.06} = 0.51$

Using the formula
$p H = p K_{a} - l o g h + l o g (1 - h)$
$= - l o g (4.99 \times 10^{- 10}) - l o g (0.51) + l o g (1 - 0.51)$
$= 9.3$

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