Question

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 1L of 0.1 M HCl solution. The concentration of \( \text{HA}^- \) in the resultant solution is:

• A. 0.1 M
• B. \( 9.53 \times 10^{-6}} \) M
• C. \( 8.1 \times 10^{-8} \) M
• D. \( 6.1 \times 10^{-13} \) M

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The dissociation of the weak acid \(\text{H}_2\text{A}\) occurs in a solution already containing a strong acid (\(\text{HCl}\)).
Due to the common ion effect, the presence of the strong acid heavily suppresses the dissociation of the weak acid.
We can assume the concentration of \(\text{H}^+\) is determined entirely by the \(\text{HCl}\), and the concentration of un-ionized \(\text{H}_2\text{A}\) remains equal to its initial concentration.
Step 2: Key Formula or Approach:
The first dissociation step is: \(\text{H}_2\text{A} \rightleftharpoons \text{H}^+ + \text{HA}^-\).
The equilibrium constant expression is: \(K_{a1} = \frac{[\text{H}^+][\text{HA}^-]}{[\text{H}_2\text{A}]}\).
Given values: \(K_{a1} = 8.1 \times 10^{-8}\), \([\text{HCl}] = 0.1 \text{ M}\), Initial \([\text{H}_2\text{A}] = 0.1 \text{ M}\).
Step 3: Detailed Explanation:
The strong acid completely dissociates:
\(\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-\)
So, the initial concentration of \(\text{H}^+\) is \(0.1 \text{ M}\).
Let \(x\) be the amount of \(\text{H}_2\text{A}\) that dissociates. At equilibrium:
\([\text{H}_2\text{A}] = 0.1 - x \approx 0.1 \text{ M}\) (since \(x\) is extremely small).
\([\text{H}^+] = 0.1 + x \approx 0.1 \text{ M}\) (common ion effect).
\([\text{HA}^-] = x\).
Substitute these equilibrium concentrations into the \(K_{a1}\) expression:
\[ 8.1 \times 10^{-8} = \frac{(0.1) \cdot [\text{HA}^-]}{0.1} \] The \(0.1\) terms cancel out perfectly:
\[ [\text{HA}^-] = 8.1 \times 10^{-8} \text{ M} \] (Note: The second dissociation step involving \(K_{a2}\) consumes a negligible amount of \(\text{HA}^-\) due to its extremely small constant \(1.0 \times 10^{-13}\), so this approximation holds true).
Step 4: Final Answer:
The concentration of \(\text{HA}^-\) is \(8.1 \times 10^{-8} \text{ M}\).