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Standard XII

Chemistry

Auto Protolysis of Water

The degree of...

Question

The degree of dissociation of $0.1$ M HCN solution is $0.01 %$ . Its ionisation constant would be:

10^{- 3}

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10^{- 5}

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10^{- 7}

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10^{- 9}

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Solution

The correct option is D $10^{- 9}$
$% α = 0.01 % = \frac{0.01}{100}$

$α = \sqrt{\frac{K a}{C}}$

$\frac{0.01}{100} = \sqrt{\frac{K a}{0.1}}$

$\frac{1 \times 10^{- 4}}{10000} = \frac{K a}{0.1}$

$10^{- 8} \times 0.1 = K a$

$K a = 10^{- 9}$

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The degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionisation constant would be:

The correct option is D 10−9%α=0.01%=0.01100α=√KaC0.01100=√Ka0.11×10−410000=Ka0.110−8×0.1=KaKa=10−9

The degree of dissociation of $0.1$ M HCN solution is $0.01 %$ . Its ionisation constant would be:

The correct option is D $10^{- 9}$
$% α = 0.01 % = \frac{0.01}{100}$

$α = \sqrt{\frac{K a}{C}}$

$\frac{0.01}{100} = \sqrt{\frac{K a}{0.1}}$

$\frac{1 \times 10^{- 4}}{10000} = \frac{K a}{0.1}$

$10^{- 8} \times 0.1 = K a$

$K a = 10^{- 9}$