Concentration of Monohydrogen Anion in Dibasic Acid Solution

To determine the concentration of $\text{HA}^-$ in the resultant solution, we consider the ionization equilibria of the weak dibasic acid $\text{H}_2\text{A}$ in the presence of a strong acid, $\text{HCl}$.

$\text{HCl}$ is a strong acid and fully dissociates in aqueous solution: $\text{HCl(aq)} \rightarrow \text{H}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$ Therefore, the contribution of $\text{H}^+$ from $0.1\text{ M HCl}$ is: $[\text{H}^+]_{\text{HCl}} = 0.1\text{ M}$

The initial concentration of the weak acid $\text{H}_2\text{A}$ is: $[\text{H}_2\text{A}]_0 = \frac{0.1\text{ mol}}{1\text{ L}} = 0.1\text{ M}$

The first ionization step of $\text{H}_2\text{A}$ is: $\text{H}_2\text{A(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{HA}^-\text{(aq)}$

Let $x$ be the molar concentration of $\text{H}_2\text{A}$ that ionizes at equilibrium.

Setting up the equilibrium concentration table:

• Equilibrium concentration of $\text{H}_2\text{A}$: $[\text{H}_2\text{A}] = 0.1 - x \approx 0.1\text{ M}$ (since $x \ll 0.1$)
• Equilibrium concentration of $\text{H}^+$: $[\text{H}^+] = 0.1 + x \approx 0.1\text{ M}$ (due to the common ion effect from $\text{HCl}$)
• Equilibrium concentration of $\text{HA}^-$: $[\text{HA}^-] = x$

The expression for the first ionization constant $K_{a1}$ is: $K_{a1} = \frac{[\text{H}^+][\text{HA}^-]}{[\text{H}_2\text{A}]}$

Substituting the equilibrium values into the $K_{a1}$ expression: $8.1 \times 10^{-8} = \frac{(0.1)(x)}{0.1}$

Solving for $x$: $x = 8.1 \times 10^{-8}\text{ M}$

Since the second ionization constant $K_{a2} = 1.0 \times 10^{-13}$ is extremely small compared to $K_{a1}$, the further ionization of $\text{HA}^-$ into $\text{A}^{2-}$ is negligible and does not appreciably affect the concentration of $\text{HA}^-$.

Thus, the equilibrium concentration of $\text{HA}^-$ in the solution is: $[\text{HA}^-] = 8.1 \times 10^{-8}\text{ M}$

Correct Option: C