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

To find the concentration of \( \mathrm{X^{2-}} \) at equilibrium in a \(0.1\,\mathrm{M}\) \( \mathrm{H_2X} \) solution, we need to understand the dissociation process of the diprotic acid \( \mathrm{H_2X} \). The dissociation occurs in two steps:

1. First dissociation:  

\[\mathrm{H_2X} \rightleftharpoons \mathrm{H^+} + \mathrm{HX^-}\]

1. with dissociation constant \( K_{a1} = 2.5 \times 10^{-7} \).
2. Second dissociation: 

\[\mathrm{HX^-} \rightleftharpoons \mathrm{H^+} + \mathrm{X^{2-}}\]

1. with dissociation constant \( K_{a2} = 1 \times 10^{-13} \).

Given the equilibrium constants, typically for weak acids, the second dissociation is much weaker than the first. This allows us to assume that the concentration of the intermediate species \(\mathrm{HX^-}\) will be significantly higher than that of \(\mathrm{X^{2-}}\).

For such weak acids and their salts, the concentration of \(\mathrm{X^{2-}}\) at equilibrium is approximately equal to the second dissociation constant \( K_{a2} \) due to the much lower concentration of \(\mathrm{H^+} \) influencing the second dissociation:

The justification follows: in the second dissociation step, the concentration of \(\mathrm{HX^-}\) is approximately equal to the initial concentration of \(\mathrm{H_2X}\), assuming that very little \(\mathrm{H_2X}\) is dissociated to form \(\mathrm{HX^-}\). Therefore, the concentration of \(\mathrm{X^{2-}}\) is approximately equal to the second dissociation constant, as this determines the extent of the second dissociation.

Therefore, the concentration of \(\mathrm{X^{2-}}\) at equilibrium is:

\[\mathrm{Concentration \,of\, X^{2-}} = K_{a2} = 1 \times 10^{-13}\,\mathrm{M}\]

Thus, the correct answer is: \(1 \times 10^{-13}\).