The degree of hydrolysis of $0.1 R N H_{3} C l$ solution is $1.0 %$ . If the concentration of $R N H_{3} C l$ is made $0.4 M$ , what is the new degree of hydrolysis (in percentage)?

0.01

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0.001

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0.2

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0.5

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Solution

The correct option is D $0.5$
As we know,

$K_{a} = C α^{2}; α = \sqrt{K_{a} / C}$

$α_{2} / α_{1} = \sqrt{C_{1} / C_{2}} = \sqrt{0.1 / 0.4} = 1 / 2$

$α_{2} = 0.5 \times 1 % = 0.5 %$

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