`CO_(3)^(2-)` and `SO_(3)^(2-)` can be distinguished from each other by the production of `CO_2)` and `SO_(2)` respectively upon using

Options

A. Baryta water
B. Lime water
C. acidified dichromate
D. sulpamic acid.

Explanation

To distinguish between the carbonate ion \( \text{CO}_3^{2-} \) and the sulfite ion \( \text{SO}_3^{2-} \), we can use acidified dichromate solution. Here’s a step-by-step solution:

### Step 1: Understand the ions
- **Carbonate ion**: \( \text{CO}_3^{2-} \)
- **Sulfite ion**: \( \text{SO}_3^{2-} \)

### Step 2: Identify the reagents
We will use **acidified dichromate** (\( \text{Cr}_2\text{O}_7^{2-} \) in acidic medium) as our reagent.

### Step 3: Reaction with sulfite ion
When the sulfite ion \( \text{SO}_3^{2-} \) is treated with acidified dichromate, it undergoes oxidation:
\[ \text{SO}_3^{2-} + \text{Cr}_2\text{O}_7^{2-} + \text{H}^+ \rightarrow \text{SO}_2 + \text{Cr}^{3+} + \text{H}_2\text{O} \]
- **Observation**: The solution changes color as the dichromate ion (\( \text{Cr}_2\text{O}_7^{2-} \)) is reduced to \( \text{Cr}^{3+} \), which is green.

### Step 4: Reaction with carbonate ion
When the carbonate ion \( \text{CO}_3^{2-} \) is treated with acidified dichromate, it does not undergo oxidation:
\[ \text{CO}_3^{2-} + \text{Cr}_2\text{O}_7^{2-} + \text{H}^+ \rightarrow \text{No Reaction} \]
- **Observation**: There is no change in color of the solution.

### Step 5: Conclusion
By observing the color change (or lack thereof), we can distinguish between the two ions:
- **If the solution turns green**: \( \text{SO}_3^{2-} \) is present (producing \( \text{SO}_2 \)).
- **If there is no color change**: \( \text{CO}_3^{2-} \) is present.

### Final Answer
Thus, the correct answer to the question is that \( \text{CO}_3^{2-} \) and \( \text{SO}_3^{2-} \) can be distinguished from each other by the production of \( \text{CO}_2 \) and \( \text{SO}_2 \) respectively upon using **acidified dichromate**.

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