find the pH of a 0.10M H2SO3 ka1=1.3x10^-2, Ka2 = 6.3x10^-8

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!

Discover how to calculate the pH of a 0.10 M solution of sulfurous acid (H2SO3) with dissociation constants Ka1=1.3x10^-2 and Ka2=6.3x10^-8. This guide covers acid-base equilibrium concepts, providing a clear solution with detailed steps.

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