Question

A monoprotic acid in $1.00M$ 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, $\alpha =0.01\%$
$=0.0001=1\times 10^{-4}$
$C=1.00M$
We know, $K_a={\alpha }^2\cdot C$
$K_a={\left(1\times 10^{-4}\right)}^2\times 1$
$K_a=1\times 10^{-8}$