The concentration of $H C N$ and $N a C N$ in a solution is $0.01 M$ each. Calculate the concentration of hydroxyl ions if the dissociation constant of $H C N$ is $7.2 \times 10^{- 10}$ .

K_{a}

for

H C N

5 \times 10^{- 10}

25^{o} C

. For maintaining a constant

p H

of 9, the volume of

5 M K C N

solution required to be added to

10 m L

2 M H C N

solution is

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If a 0.1 M solution of HCN is 0.01% ionised, the ionisation constant for HCN is:

The correct option is B 10−9For a weak acid, α=√KaCα=0.01100=10−4∴10−4=√Ka0.1∴10−8=Ka10−1∴Ka=10−9Therefore, the ionization constant for HCN is 10−9

If a 0.1 M solution of $H C N$ is 0.01% ionised, the ionisation constant for $H C N$ is:

The correct option is B $10^{- 9}$
For a weak acid, $α = \sqrt{\frac{K_{a}}{C}}$
$α = \frac{0.01}{100} = 10^{- 4}$
$∴ 10^{- 4} = \sqrt{\frac{K_{a}}{0.1}}$
$∴ 10^{- 8} = \frac{K_{a}}{10^{- 1}}$
$∴ K_{a} = 10^{- 9}$
Therefore, the ionization constant for $H C N$ is $10^{- 9}$