Given:
\[ K_{a1} = 2.5 \times 10^{-8} \]
\[ K_{a2} = 1.0 \times 10^{-13} \]
Initial concentration of H₂X = 0.1 M
Step 1: First dissociation
\[ H_2X \rightleftharpoons H^+ + HX^- \]
Since Ka₁ is small, weak acid approximation applies:
\[ [H^+] \approx \sqrt{K_{a1} \times C} \]
\[ = \sqrt{2.5 \times 10^{-8} \times 0.1} \]
\[ = \sqrt{2.5 \times 10^{-9}} \]
\[ = 5 \times 10^{-5} \text{ M} \]
Step 2: Second dissociation
\[ HX^- \rightleftharpoons H^+ + X^{2-} \]
\[ K_{a2} = \frac{[H^+][X^{2-}]}{[HX^-]} \]
Rearranging:
\[ [X^{2-}] = \frac{K_{a2}[HX^-]}{[H^+]} \]
From first dissociation:
\[ [HX^-] \approx [H^+] = 5 \times 10^{-5} \]
Therefore:
\[ [X^{2-}] = \frac{1.0 \times 10^{-13} \times 5 \times 10^{-5}}{5 \times 10^{-5}} \]
\[ [X^{2-}] = 1.0 \times 10^{-13} \]
But note:
\[ [H^+] = 5 \times 10^{-5} \]
More accurate relation for diprotic acid:
\[ [X^{2-}] = \frac{K_{a1}K_{a2}C}{[H^+]^2} \]
Substitute values:
\[ = \frac{(2.5 \times 10^{-8})(1.0 \times 10^{-13})(0.1)}{(5 \times 10^{-5})^2} \]
\[ = \frac{2.5 \times 10^{-22}}{25 \times 10^{-10}} \]
\[ = 1.0 \times 10^{-13} \]
Express in form ______ × 10⁻15:
\[ 1.0 \times 10^{-13} = 100 \times 10^{-15} \]
Final Answer = 100