To find the pH of a 0.10 M solution of $\text{H}_2\text{SO}_3$ (sulfurous acid), we need to consider its two dissociation steps:

1. First dissociation: $\text{H}_2\text{SO}_3 \rightleftharpoons \text{H}^+ + \text{HSO}_3^-$ $K_{a1} = 1.3 \times 10^{-2}$

2. Second dissociation: $\text{HSO}_3^- \rightleftharpoons \text{H}^+ + \text{SO}_3^{2-}$ $K_{a2} = 6.3 \times 10^{-8}$

Step 1: Calculate the pH from the first dissociation

Let $x$ be the concentration of $\text{H}^+$ produced from the first dissociation. The expression for $K_{a1}$ is:

[ K_{a1} = \frac{[\text{H}^+][\text{HSO}_3^-]}{[\text{H}_2\text{SO}_3]} ]

At equilibrium:

• $[\text{H}_2\text{SO}_3] = 0.10 - x$
• $[\text{H}^+] = x$
• $[\text{HSO}_3^-] = x$

Substituting into the $K_{a1}$ expression gives:

$1.3 \times 10^{-2} = \frac{x^2}{0.10 - x}$

Assuming $x$ is small compared to 0.10, we can simplify this to:

$1.3 \times 10^{-2} \approx \frac{x^2}{0.10}$ $x^2 = 1.3 \times 10^{-3}$ $x = \sqrt{1.3 \times 10^{-3}} \approx 0.036 \, \text{M}$

Step 2: Calculate the contribution of $\text{H}^+$ from the second dissociation

Now we need to consider the second dissociation, where $\text{H}^+$ from the first dissociation also contributes to this step.

Using $K_{a2}$:

[ K_{a2} = \frac{[\text{H}^+][\text{SO}_3^{2-}]}{[\text{HSO}_3^-]} ]

Let $y$ be the concentration of $\text{H}^+$ from the second dissociation. Then:

$K_{a2} = \frac{(0.036 + y)(y)}{(0.10 - 0.036 - y)}$

For simplicity, assume $y$ is small compared to $0.036$:

$K_{a2} = \frac{(0.036)(y)}{0.064}$ $6.3 \times 10^{-8} = \frac{0.036y}{0.064}$

Solving for $y$:

$y = \frac{6.3 \times 10^{-8} \times 0.064}{0.036} \approx 1.11 \times 10^{-8} \, \text{M}$

Step 3: Total concentration of $\text{H}^+$

Total $[\text{H}^+]$:

$[\text{H}^+]_{\text{total}} = 0.036 + 1.11 \times 10^{-8} \approx 0.036 \, \text{M}$

Step 4: Calculate pH

Finally, calculate the pH:

$\text{pH} = -\log(0.036) \approx 1.44$

Thus, the pH of the 0.10 M $\text{H}_2\text{SO}_3$ solution is approximately 1.44.

Summary

1. The main contribution to $\text{H}^+$ comes from the first dissociation.
2. The second dissociation contributes a negligible amount of $\text{H}^+$.
3. The pH reflects the strong acidity of the solution.

Do you want further details or have any questions? Here are some related questions you might find interesting:

1. How would the pH change if the concentration of $\text{H}_2\text{SO}_3$ were increased?
2. What would the pH be if $\text{H}_2\text{SO}_3$ were a strong acid?
3. How does temperature affect the dissociation constants $K_{a1}$ and $K_{a2}$?
4. What would be the pH of a mixture of $\text{H}_2\text{SO}_3$ and a strong base?
5. Can you explain the significance of $K_{a}$ values in understanding acid strength?

Tip: Always check if the approximation made during calculations is valid, particularly with weak acids and bases!