Bromine in excess is dropped to a 0.01 M `SO_(2)`. All of `SO_(2)` is oxidized to `H_(2)SO_(4)` and the excess `Br_(2)` is removed by flushing with gaseous `N_(2)`. Determine the pH of the resulting solution assuming `K_(a1)` of `H_(2)SO_(4)` vary large & `K_(a2) = 10^(-2)`. Take the value of `log (3.24) = 0.51`.

Answer

To determine the pH of the resulting solution after the oxidation of SO₂ to H₂SO₄ by excess bromine, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactions**:
The oxidation of sulfur dioxide (SO₂) to sulfuric acid (H₂SO₄) can be represented as:
\[ \text{SO}_2 + \text{Br}_2 \rightarrow \text{H}_2\text{SO}_4 + \text{HBr} \]
Given that the concentration of SO₂ is 0.01 M, after the reaction, we will have:
- H₂SO₄ produced: 0.01 M
- HBr produced: 0.02 M (since 2 moles of HBr are produced for every mole of SO₂ oxidized).

2. **Concentration of H⁺ Ions from HBr**:
HBr is a strong acid and will completely dissociate in solution:
\[ \text{HBr} \rightarrow \text{H}^+ + \text{Br}^- \]
Thus, the concentration of H⁺ ions from HBr will be:
\[ [\text{H}^+] = 0.02 \text{ M} \]

3. **Concentration of H⁺ Ions from H₂SO₄**:
H₂SO₄ is a strong acid for its first dissociation:
\[ \text{H}_2\text{SO}_4 \rightarrow \text{H}^+ + \text{HSO}_4^- \]
The first dissociation is complete, contributing 0.01 M of H⁺ ions. The second dissociation is:
\[ \text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-} \]
Given that \( K_{a2} = 10^{-2} \), we can set up the equilibrium expression:
\[ K_{a2} = \frac{[\text{H}^+][\text{SO}_4^{2-}]}{[\text{HSO}_4^-]} \]

4. **Setting Up the Equilibrium**:
Let \( x \) be the amount of H⁺ produced from the dissociation of HSO₄⁻. The concentrations at equilibrium will be:
- \([\text{H}^+] = 0.02 + x\)
- \([\text{HSO}_4^-] = 0.01 - x\)
- \([\text{SO}_4^{2-}] = x\)

Substituting into the \( K_{a2} \) expression:
\[ 10^{-2} = \frac{(0.02 + x)(x)}{(0.01 - x)} \]

5. **Approximating x**:
Assuming \( x \) is small compared to 0.01, we can simplify:
\[ 10^{-2} \approx \frac{(0.02)(x)}{0.01} \]
Solving for \( x \):
\[ 10^{-2} \cdot 0.01 = 0.02x \implies x = \frac{10^{-4}}{0.02} = 0.005 \]

6. **Total H⁺ Concentration**:
Now, we can find the total concentration of H⁺ ions:
\[ [\text{H}^+] = 0.02 + 0.005 = 0.025 \text{ M} \]

7. **Calculating pH**:
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the value:
\[ \text{pH} = -\log(0.025) = -\log(3.24 \times 10^{-2}) = 2 - \log(3.24) \]
Given that \(\log(3.24) = 0.51\),
\[ \text{pH} = 2 - 0.51 = 1.49 \]

### Final Answer:
The pH of the resulting solution is **1.49**.