Question

A monoprotic acid in $1.00\, M$ solution is $0.01\%$ ionised. The dissociation constant of this acid is

• A. $ 1\times 10^{-8}$
• B. $ 1\times 10^{-4}$
• C. $ 1\times 10^{-6}$
• D. $ 1\times 10^{-5}$

Answer 1

Answer: (A)

For a weak electrolyte according to Ostwald's dilution law.
$\alpha =\sqrt{K \cdot V}$
or $K =\alpha^{2} C\,\, \left(V=\frac{1}{C}\right)$
where, $K$ is known as dissociation constant
$C$ is concentration in $mol / L$
$\alpha$ is degree of ionisation
Here, $\alpha =0.01 \%$
$=0.0001=1 \times 10^{-4}$
$C=1.00\, M$
We know, $K_{a} =\alpha^{2} \cdot C$
$K_{a} =\left(1 \times 10^{-4}\right)^{2} \times 1$
$K_{a} =1 \times 10^{-8}$