The first and second ionization constants of $$H_{2}X$$ are $$2.5 \times 10^{-8}$$ and $$1.0 \times 10^{-13}$$ respectively. The concentration of $$X^{2-}$$ in $$0.1 MH_{2} X$$ solution is _________ $$\times 10^{-15}M$$. (Nearest Integer)

The diprotic acid is $$H_2X$$ with step-wise ionization constants

$$H_2X \;\rightleftharpoons\; H^+ + HX^- \qquad K_{a1}=2.5\times10^{-8}$$

$$HX^- \;\rightleftharpoons\; H^+ + X^{2-} \qquad K_{a2}=1.0\times10^{-13}$$

The initial concentration of the acid is $$C = 0.1\,\text{M}$$.

Case 1: First dissociation

Let the degree of first dissociation be $$\alpha$$.
Equilibrium concentrations:

$$[H_2X] = C(1-\alpha), \qquad [HX^-] = C\alpha, \qquad [H^+] \approx C\alpha$$

Using the definition of $$K_{a1}$$,

$$K_{a1}= \frac{[H^+][HX^-]}{[H_2X]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} \approx C\alpha^2$$

since $$\alpha \ll 1$$ for a weak acid.
Therefore

$$\alpha = \sqrt{\frac{K_{a1}}{C}} = \sqrt{\frac{2.5\times10^{-8}}{0.1}} = \sqrt{2.5\times10^{-7}} \approx 5.0\times10^{-4}$$

Hence

$$[H^+] \approx C\alpha = 0.1 \times 5.0\times10^{-4}=5.0\times10^{-5}\,\text{M}$$

Case 2: Second dissociation

Let the fraction of $$HX^-$$ that loses the second proton be $$\beta$$.
Equilibrium concentrations:

$$[HX^-]=C\alpha(1-\beta), \qquad [X^{2-}]=C\alpha\beta, \qquad [H^+] \approx 5.0\times10^{-5} + C\alpha\beta$$

Because $$K_{a2}$$ is extremely small, $$\beta$$ will be very small, so the additional $$H^+$$ produced can be neglected in the denominator. Thus $$[H^+] \approx 5.0\times10^{-5}\,\text{M}$$ remains a good approximation.

Using the expression for $$K_{a2}$$,

$$K_{a2}= \frac{[H^+][X^{2-}]}{[HX^-]} = \frac{\left(5.0\times10^{-5}\right)\left(C\alpha\beta\right)} {C\alpha(1-\beta)} \approx \left(5.0\times10^{-5}\right)\beta$$

Therefore

$$\beta = \frac{K_{a2}}{[H^+]} = \frac{1.0\times10^{-13}}{5.0\times10^{-5}} = 2.0\times10^{-9}$$

Concentration of $$X^{2-}$$

$$[X^{2-}] = C\alpha\beta = 0.1 \times \left(5.0\times10^{-4}\right) \times \left(2.0\times10^{-9}\right) = 1.0\times10^{-13}\,\text{M}$$

Expressing this in the form $$n \times 10^{-15}\,\text{M}$$,

$$1.0\times10^{-13}\,\text{M} = 100 \times 10^{-15}\,\text{M}$$

Nearest integer value of $$n$$ = 100.