Find concentration of X^2– at equilibrium in 0.1 MH2 X. Given K_a1 = 2.5 × 10–7, K_2a = 1 × 10^–13

The correct option is 1 × 10^–13.

Step-by-Step Explanation:

For a diprotic acid H₂X, the ionization occurs in two sequential steps:

First Dissociation Step:

${\text{H}}_2\text{X}\rightleftharpoons {\text{H}}^{+}+{\text{HX}}^{-}$

The first dissociation constant, K_a1, is given by:

${\text{K}}_{\text{a1}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{HX}}^{-}\right]}{\left[{\text{H}}_2\text{X}\right]}$

Since K_a1 (2.5 × 10^–7) is very small, the degree of dissociation is extremely low. Thus, the concentration of H⁺ ions produced in the first step is equal to the concentration of intermediate anion species, [HX^–]:

$\left[{\text{H}}^{+}\right]\approx \left[{\text{HX}}^{-}\right]$

Second Dissociation Step:

The intermediate anion further dissociates according to:

${\text{HX}}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{X}}^{2-}$

The second dissociation constant, K_a2, is expressed as:

${\text{K}}_{\text{a2}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{X}}^{2-}\right]}{\left[{\text{HX}}^{-}\right]}$

Since [H⁺] ≈ [HX^–] from the first ionization step, we can substitute this equivalence into the K_a2 expression:

${\text{K}}_{\text{a2}}=\frac{\left[{\text{HX}}^{-}\right]\left[{\text{X}}^{2-}\right]}{\left[{\text{HX}}^{-}\right]}$

Canceling [HX^–] from the numerator and denominator gives:

$\left[{\text{X}}^{2-}\right]={\text{K}}_{\text{a2}}$

Given K_a2 = 1 × 10^–13, the equilibrium concentration of X^2– ions is equal to:

$\left[{\text{X}}^{2-}\right]=1\times {10}^{-13}\text{ M}$