Question

Find the concentration of \( \mathrm{X^{2-}} \) at equilibrium in \(0.1\,\mathrm{M}\) \( \mathrm{H_2X} \). Given: \[ K_{a1} = 2.5 \times 10^{-7}, \qquad K_{a2} = 1 \times 10^{-13} \]

• A. \(2.5 \times 10^{-7}\)
• B. \(1 \times 10^{-13}\)
• C. \(6 \times 10^{-12}\)
• D. \(5 \times 10^{-10}\)

Hint:

For weak diprotic acids:

If \(K_{a2} \ll K_{a1}\), the second dissociation is negligible
In such cases, \( [\mathrm{X^{2-}}] \approx K_{a2} \)
This shortcut saves time in MCQs

Answer 1

The Correct Option is B

Concept: \( \mathrm{H_2X} \) is a diprotic weak acid: \[ \mathrm{H_2X \rightleftharpoons H^+ + HX^-} \quad (K_{a1}) \] \[ \mathrm{HX^- \rightleftharpoons H^+ + X^{2-}} \quad (K_{a2}) \] Since: \[ K_{a2} \ll K_{a1} \] the second dissociation is highly suppressed.
Step 1: Hydrogen ion concentration from first dissociation For the first dissociation: \[ [\mathrm{H^+}] \approx \sqrt{K_{a1} \cdot C} \] \[ [\mathrm{H^+}] \approx \sqrt{(2.5 \times 10^{-7})(0.1)} = \sqrt{2.5 \times 10^{-8}} = 5 \times 10^{-4} \]
Step 2: Expression for \( \mathrm{X^{2-}} \) For the second dissociation: \[ K_{a2} = \frac{[\mathrm{H^+}][\mathrm{X^{2-}}]}{[\mathrm{HX^-}]} \] But from the first dissociation: \[ [\mathrm{HX^-}] \approx [\mathrm{H^+}] \] Hence, \[ [\mathrm{X^{2-}}] \approx K_{a2} \]
Step 3: Substitute value \[ [\mathrm{X^{2-}}] = 1 \times 10^{-13}\,\mathrm{M} \] Final Answer: \[ \boxed{1 \times 10^{-13}} \]