The pH of a $$0.1$$ molar solution of the acid $$HQ$$ is $$3$$. The value of the ionization constant, $$K_a$$ of this acid is :

The acid dissociation equilibrium is
$$HQ \rightleftharpoons H^+ + Q^-$$

Given pH = 3, therefore
$$[H^+] = 10^{-3}\,\text{M}$$

Let the initial concentration of $$HQ$$ be $$C = 0.1\,\text{M}$$ and the degree of ionisation produce $$x$$ mol L$$^{-1}$$ of $$H^+$$.
From the pH we directly get $$x = [H^+] = 10^{-3}\,\text{M}$$.

The equilibrium concentrations are
• $$[H^+] = x = 10^{-3}$$
• $$[Q^-] = x = 10^{-3}$$
• $$[HQ] = C - x = 0.1 - 10^{-3} = 0.099$$

The acid-ionisation constant is defined as
$$K_a = \frac{[H^+][Q^-]}{[HQ]}$$

Substituting the above values,
$$K_a = \frac{(10^{-3})(10^{-3})}{0.099} = \frac{10^{-6}}{0.099} \approx 1.0 \times 10^{-5}$$

Hence $$K_a \approx 1 \times 10^{-5}$$.

Option C which is: $$1 \times 10^{-5}$$