You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10^{- 5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10^{- 7}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10^{- 9}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

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

$α = \sqrt{\frac{K a}{C}}$

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

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

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

$K a = 10^{- 9}$

Suggest Corrections

Similar questions

Q. The ionization constant of base

{(C_{2} H_{5})}_{3}

6.4 \times 10^{- 5}

. Calculate its degree of dissociation in its

0.1 M

solution when it is mixed with

0.01 M

N a O H

solution.

Q. Consider a weak electrolyte,

A B_{3}

having a concentration of 0.1 M in solution. If the degree of dissociation is 0.2, Calculate the ionisation constant

K_{i}

A B_{3}

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 $7.2 \times 10^{- 10}$ .

Q. Calculate the

p H

0.01

M solution of

N H_{4} C N

. The dissociation constants

K_{a}

for

H C N = 6.2 \times 10^{- 10}

and

K_{b}

for

N H_{3} = 1.6 \times 10^{- 5}

Q. The ionisation constant of Propanoic acid is

1.32 \times 10^{- 5}

. Calculate the degree of ionisation of the acid in its

0.05 M

solution and also its

P H

. What will be its degree of ionisation if the solution is

0.01 M

H C l

also?

Join BYJU'S Learning Program

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 $10^{- 9}$
$% α = 0.01 % = \frac{0.01}{100}$

$α = \sqrt{\frac{K a}{C}}$

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

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

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

$K a = 10^{- 9}$