A weak monobasic acid is $4\%$ dissociated in $0.05$ M solution. What is percent dissociation in $0.1$ M solution ?

For a weak monobasic acid, the dissociation constant $K_a$ is given by Ostwald's dilution law:

$K_a = C \alpha^2$

where $C$ is the concentration and $\alpha$ is the degree of dissociation.

Given $C_1 = 0.05$ M and $\alpha_1 = 4\% = 0.04$:

$K_a = 0.05 \times (0.04)^2 = 0.05 \times 0.0016 = 8 \times 10^{-5}$

For the new concentration $C_2 = 0.1$ M, let the degree of dissociation be $\alpha_2$:

$K_a = C_2 \alpha_2^2$

$8 \times 10^{-5} = 0.1 \times \alpha_2^2$

$\alpha_2^2 = \dfrac{8 \times 10^{-5}}{0.1} = 8 \times 10^{-4}$

$\alpha_2 = \sqrt{8 \times 10^{-4}} = 2.828 \times 10^{-2}$

The percent dissociation in $0.1$ M solution is $2.828\%$.

Answer: $2.828\%$