The first and second ionization constants of a weak dibasic acid $$H_2A$$ are $$8.1 \times 10^{-8}$$ and $$1.0 \times 10^{-13}$$ respectively. 0.1 mol of $$H_2A$$ was dissolved in 1 L of 0.1 M HCl solution. The concentration of $$HA^-$$ in the resultant solution is :

The two successive ionisation steps of the dibasic acid are

$$H_2A \;\rightleftharpoons\; H^+ + HA^- \qquad\qquad K_{a1}=8.1\times10^{-8}$$
$$HA^- \;\rightleftharpoons\; H^+ + A^{2-} \qquad\qquad K_{a2}=1.0\times10^{-13}$$

A solution is prepared by dissolving $$0.1\,$$mol of $$H_2A$$ in $$1\,$$L of $$0.1\,$$M $$HCl$$. Hence

$$[\!H^+\!]_{\text{from HCl}} = 0.1\ \text{M},\qquad [\!H_2A]_{\text{initial}} = 0.1\ \text{M}$$

The strong acid ($$HCl$$) fixes the hydrogen-ion concentration at about $$0.1\,$$M. Because this value is much larger than the acid’s own $$K_{a1}$$, the dissociation of $$H_2A$$ will be greatly suppressed. Let

$$x = [HA^-]$$ formed at equilibrium from the first dissociation.

Then
$$[H^+] \approx 0.1 + x \approx 0.1\ \text{M}\quad(\text{since }x\ll0.1)$$
$$[H_2A] \approx 0.1 - x \approx 0.1\ \text{M}$$

Apply the first-step ionisation constant:

$$K_{a1} = \frac{[H^+][HA^-]}{[H_2A]}$$

Substituting the approximate concentrations:

$$8.1\times10^{-8} = \frac{(0.1)\,x}{0.1} = x$$

Thus

$$[HA^-] = 8.1 \times 10^{-8}\ \text{M}$$

The second ionisation (governed by $$K_{a2}=1.0\times10^{-13}$$) is even more strongly suppressed in such an acidic medium, so its contribution to $$[HA^-]$$ is negligible.

Therefore, the concentration of $$HA^-$$ in the solution is $$8.1 \times 10^{-8}\,$$M.

Option C which is: $$8.1 \times 10^{-8}\ \text{M}$$