Question

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

Hint:

For any polyprotic acid where successive ionization constants differ significantly (e.g., $K_{a1} \gg K_{a2} \gg K_{a3}$), the concentration of the dianion is simply $K_{a2}$ and the concentration of the trianion is $(K_{a2} \cdot K_{a3})/[H^{+}]$.

Answer 1

Correct Answer: 1

Step 1: Understanding the Concept:
A diprotic acid $H_{2}X$ dissociates in two successive steps.
The first dissociation is: $H_{2}X \rightleftharpoons H^{+} + HX^{-}$ with equilibrium constant $K_{a1}$.
The second dissociation is: $HX^{-} \rightleftharpoons H^{+} + X^{2-}$ with equilibrium constant $K_{a2}$.
In such systems, if $K_{a1} \gg K_{a2}$, the first ionization produces the bulk of $H^{+}$ and $HX^{-}$.

Step 2: Key Formula or Approach:
For a weak diprotic acid, the concentration of the second acid radical ($X^{2-}$) is approximately equal to the second dissociation constant $K_{a2}$.
This holds true because in the expression for $K_{a2}$:
\[ K_{a2} = \frac{[H^{+}][X^{2-}]}{[HX^{-}]} \]
Since $[H^{+}] \approx [HX^{-}]$ from the first dissociation step, they cancel out.

Step 3: Detailed Explanation:
From step 1: $[H^{+}] \approx [HX^{-}] = \sqrt{K_{a1} \cdot C}$.
From step 2:
\[ [X^{2-}] = \frac{K_{a2} \cdot [HX^{-}]}{[H^{+}]} \]
Substituting the approximation $[H^{+}] \approx [HX^{-}]$:
\[ [X^{2-}] \approx K_{a2} \]
Given $K_{a2} = 1.0 \times 10^{-13}$.
Thus, $[X^{2-}] = 1 \times 10^{-13}$ M.

Step 4: Final Answer:
The concentration of $X^{2-}$ is $1 \times 10^{-13}$ M.
The value to be filled in the blank is 1.