Which of the following reagents can be used to distinguish between a sulphite and a sulphate in solution?

Options

A. `FeSO_(4)`
B. `Na_(2)[Fe(CN)_(5)NO]`
C. `BaCl_(2)+"dil. "HCl`
D. `Na_(3)[Co(NO_(2))_(6)]`

Answer

To distinguish between a sulfite (SO3^2-) and a sulfate (SO4^2-) in solution, we can use the following step-by-step approach:

### Step-by-Step Solution:

1. **Identify the Reagents**: The reagents provided in the question are:
- Ferrous sulfate
- Na2Fe(CN)5NO
- Barium chloride plus dilute HCl
- Na3Co(NO2)6

2. **Understanding the Reactions**:
- **Sulfite (SO3^2-)**: When sulfite ions are treated with barium chloride (BaCl2), they form barium sulfite (BaSO3), which is a white precipitate. This precipitate is soluble in dilute hydrochloric acid (HCl).
- **Sulfate (SO4^2-)**: When sulfate ions are treated with barium chloride, they form barium sulfate (BaSO4), which is also a white precipitate. However, this precipitate is insoluble in dilute hydrochloric acid.

3. **Perform the Test**:
- Add barium chloride solution to the test solution containing either sulfite or sulfate.
- Observe the formation of a white precipitate.

4. **Add Dilute HCl**:
- To the precipitate formed, add dilute HCl.
- If the precipitate dissolves, it indicates the presence of sulfite (SO3^2-).
- If the precipitate does not dissolve, it indicates the presence of sulfate (SO4^2-).

5. **Conclusion**:
- The reagent that can be used to distinguish between sulfite and sulfate in solution is **barium chloride plus dilute HCl**.

### Final Answer:
The correct reagent to distinguish between a sulfite and a sulfate in solution is **Barium chloride plus dilute HCl**.