When `BaCl_(2)` is added to an aqueous salt solution, a while precipitate is obtained. The anion among `CO_(3)^(2-), SO_(3)^(2-)` and `SO_(4)^(2-)` that was present in the solubtion can be :

Options

A. `CO_(3)^(2-)` but not any of the other two
B. `SO_(3)^(2-)` but not any of the other two
C. `SO_(4)^(2-)` but not any of the other two
D. None of these

Explanation

To solve the problem, we need to determine which anion among \( \text{CO}_3^{2-} \), \( \text{SO}_3^{2-} \), and \( \text{SO}_4^{2-} \) can form a white precipitate when \( \text{BaCl}_2 \) is added to an aqueous salt solution.

### Step-by-Step Solution:

1. **Identify the Reaction**: When \( \text{BaCl}_2 \) is added to the solution, the barium ions \( \text{Ba}^{2+} \) can react with the anions present in the solution to form precipitates.

2. **Consider Each Anion**:
- **For \( \text{CO}_3^{2-} \)**:
\[ \text{Ba}^{2+} + \text{CO}_3^{2-} \rightarrow \text{BaCO}_3 \, (\text{white precipitate}) \]
- **For \( \text{SO}_3^{2-} \)**:
\[ \text{Ba}^{2+} + \text{SO}_3^{2-} \rightarrow \text{BaSO}_3 \, (\text{white precipitate}) \]
- **For \( \text{SO}_4^{2-} \)**:
\[ \text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4 \, (\text{white precipitate}) \]

3. **Determine the Precipitate Color**: All three reactions produce a white precipitate:
- \( \text{BaCO}_3 \) is a white solid.
- \( \text{BaSO}_3 \) is also a white solid.
- \( \text{BaSO}_4 \) is a well-known white precipitate.

4. **Conclusion**: Since all three anions can form white precipitates with \( \text{Ba}^{2+} \), we conclude that the anion present in the solution could be any of the three: \( \text{CO}_3^{2-} \), \( \text{SO}_3^{2-} \), or \( \text{SO}_4^{2-} \).

5. **Final Answer**: The correct answer is that the anion present in the solution can be \( \text{CO}_3^{2-} \), \( \text{SO}_3^{2-} \), or \( \text{SO}_4^{2-} \) since all can form white precipitates with \( \text{Ba}^{2+} \).