Diprotic Acid Dissociation Explained

The substance $\text{H}_2\text{X}$ is a diprotic acid, meaning it can donate two protons in successive steps. The ionization constants given are:

• First ionization: $\text{H}_2\text{X} \rightleftharpoons \text{H}^+ + \text{HX}^-$, with $K_{a1} = 2.5 \times 10^{-8}$.
• Second ionization: $\text{HX}^- \rightleftharpoons \text{H}^+ + \text{X}^{2-}$, with $K_{a2} = 1.0 \times 10^{-13}$.

We need to determine the concentration of the fully dissociated species, $\text{X}^{2-}$, in a $0.1\text{ M}$ solution of $\text{H}_2\text{X}$.

Calculating Concentration from First Ionization

We first consider the contribution of the first ionization step. The equilibrium expression is: $K_{a1} = \frac{[\text{H}^+][\text{HX}^-]}{[\text{H}_2\text{X}]}$ Let the initial concentration of $\text{H}_2\text{X}$ be $C = 0.1\text{ M}$. Let $x$ be the concentration of $\text{H}^+$ and $\text{HX}^-$ produced. Assuming $x$ is small compared to $C$: $K_{a1} \approx \frac{x^2}{C}$ $2.5 \times 10^{-8} \approx \frac{x^2}{0.1}$ $x^2 \approx (2.5 \times 10^{-8}) \times 0.1 = 2.5 \times 10^{-9}$ $x = \sqrt{2.5 \times 10^{-9}} \approx 5.0 \times 10^{-5}\text{ M}$

This approximation ($x \ll C$) is valid since $5.0 \times 10^{-5}$ is much smaller than $0.1$. Thus, after the first ionization:

• $[\text{H}^+] \approx 5.0 \times 10^{-5}\text{ M}$
• $[\text{HX}^-] \approx 5.0 \times 10^{-5}\text{ M}$

Calculating Concentration from Second Ionization

Now, we analyze the second ionization step: $\text{HX}^- \rightleftharpoons \text{H}^+ + \text{X}^{2-}$. The equilibrium expression is: $K_{a2} = \frac{[\text{H}^+][\text{X}^{2-}]}{[\text{HX}^-]}$. The concentration of $\text{H}^+$ comes from both steps. Since $K_{a1} \gg K_{a2}$, the $[\text{H}^+]$ from the first step dominates. We approximate the total $[\text{H}^+]$ concentration in the second equilibrium as $[\text{H}^+] \approx 5.0 \times 10^{-5}\text{ M}$. The concentration of $\text{HX}^-$ is also approximately $5.0 \times 10^{-5}\text{ M}$, as the amount dissociating further to form $\text{X}^{2-}$ will be small due to the very low $K_{a2}$. Let $[\text{X}^{2-}] = y$.

Substituting these values into the $K_{a2}$ expression: $K_{a2} \approx \frac{[\text{H}^+]_{\text{from step 1}} \times [\text{X}^{2-}]}{[\text{HX}^-]_{\text{from step 1}}}$ $1.0 \times 10^{-13} \approx \frac{(5.0 \times 10^{-5}) \times y}{(5.0 \times 10^{-5})}$ This simplifies to $y \approx 1.0 \times 10^{-13}\text{ M}$.

The concentration of $\text{X}^{2-}$ is therefore approximately $1.0 \times 10^{-13}\text{ M}$. This result is consistent because the value of $y$ ($1.0 \times 10^{-13}$) is negligible compared to the concentrations of $\text{H}^+$ and $\text{HX}^-$ ($5.0 \times 10^{-5}$).

Final Answer Formatting

The concentration of $\text{X}^{2-}$ is $1.0 \times 10^{-13}\text{ M}$. We need to express this in the format $\text{__________} \times 10^{-15}\text{ M}$. Converting the unit: $1.0 \times 10^{-13}\text{ M} = 1.0 \times 10^2 \times 10^{-15}\text{ M} = 100 \times 10^{-15}\text{ M}$ The nearest integer value is 100.