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

To find the concentration of \(X^{2-}\) in a \(0.1 \, \text{M} \, H_2X\) solution, we use the ionization constants \(K_{a1}\) and \(K_{a2}\). The dissociation reactions and their constants are:

\( H_2X \rightleftharpoons H^+ + HX^-\), \(K_{a1} = 2.5 \times 10^{-8}\)
\( HX^- \rightleftharpoons H^+ + X^{2-}\), \(K_{a2} = 1.0 \times 10^{-13}\)

Assuming the degree of ionization is small, the concentration of \(H^+\) ions is mainly from the first ionization: \([H^+] = \sqrt{K_{a1} \cdot C} = \sqrt{2.5 \times 10^{-8} \cdot 0.1} = \sqrt{2.5 \times 10^{-9}}\). Calculating, we have:

\([H^+] = 5 \times 10^{-5} \, \text{M}\).

For the second ionization:

\( [HX^-] \approx [H^+] = 5 \times 10^{-5} \, \text{M} \).

Now, \([X^{2-}]\) is calculated using \(K_{a2}\):

\(K_{a2} = \frac{[H^+][X^{2-}]}{[HX^-]}\) implies \( [X^{2-}] = \frac{K_{a2} \cdot [HX^-]}{[H^+]}\). Substituting the known values:

\( [X^{2-}] = \frac{1.0 \times 10^{-13} \cdot 5 \times 10^{-5}}{5 \times 10^{-5}} = 1.0 \times 10^{-13} \, \text{M} \).

Thus, the concentration of \(X^{2-}\) in \(0.1 \, \text{M} \, H_2X\) solution is \(1 \times 10^{-13}\) M. The final answer is within the range given (1,1).