Question

Find the concentration of \( \text{X}^{2-} \) at equilibrium in 0.1 M \( \text{H}_2\text{X} \). Given \( K_{a1} = 2.5 \times 10^{-7} \) and \( K_{a2} = 1 \times 10^{-13} \).

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

Hint:

In weak acid dissociation, the concentration of the ions can be found using the equilibrium constant expressions. When the constants are very small, simplifying assumptions such as \( x \ll 0.1 \) can be used.

Answer 1

The Correct Option is B

Step 1: Understand the dissociation reactions.
The dissociation of \( \text{H}_2\text{X} \) in water can be represented as two steps: \[ \text{H}_2\text{X} \rightleftharpoons \text{H}_+ + \text{HX}^- \] \[ \text{HX}^- \rightleftharpoons \text{H}^+ + \text{X}^{2-} \] The equilibrium constants for these reactions are \( K_{a1} \) and \( K_{a2} \), respectively.
Step 2: Use the ICE table for dissociation.
We assume that the initial concentration of \( \text{H}_2\text{X} \) is 0.1 M, and the initial concentrations of \( \text{H}_+ \) and \( \text{HX}^- \) are 0. From the first dissociation step, we can write the equilibrium concentrations as: - \( [\text{H}_+] = [\text{HX}^-] = x \) - \( [\text{H}_2\text{X}] = 0.1 - x \) For the first dissociation step, the equilibrium expression is: \[ K_{a1} = \frac{[\text{H}_+][\text{HX}^-]}{[\text{H}_2\text{X}]} = \frac{x^2}{0.1 - x} = 2.5 \times 10^{-7} \]
Step 3: Solve for \( x \).
Since \( K_{a1} \) is relatively small, we can assume \( x \) is much smaller than 0.1, so \( 0.1 - x \approx 0.1 \). Therefore: \[ x^2 = (2.5 \times 10^{-7})(0.1) = 2.5 \times 10^{-8} \] \[ x = \sqrt{2.5 \times 10^{-8}} = 5 \times 10^{-4} \, \text{M} \]
Step 4: Use the second dissociation constant.
The second dissociation step involves \( \text{HX}^- \), and we use \( K_{a2} \) for the equilibrium expression: \[ K_{a2} = \frac{[\text{H}^+][\text{X}^{2-}]}{[\text{HX}^-]} = \frac{x_2 \cdot x}{5 \times 10^{-4}} = 1 \times 10^{-13} \] where \( x_2 \) is the concentration of \( \text{X}^{2-} \). Solving for \( x_2 \): \[ x_2 = \frac{5 \times 10^{-4} \times 10^{-13}}{1} = 1 \times 10^{-13} \, \text{M} \] Thus, the concentration of \( \text{X}^{2-} \) at equilibrium is \( 1 \times 10^{-13} \, \text{M} \).