Which of the following is correct ?

Options

A. `AgCl(s)+Na_(3)AsO_(3) to` colourless solution
B. `FeCl_(3)(aq)+K_(4)Fe(CN)_(6) to` brown precipitate
C. `FeCl_(3)(aq)+K_(3)Fe(CN)_(6) to` brown colouration
D. `CuSO_(4)(aq)+KCN("excess")to` blue colouration.

Answer

To determine which compounds can be used to distinguish between Na2SO4 (sodium sulfate) and Na2S (sodium sulfide), we will analyze the reactions of both compounds with the given reagents.

### Step-by-Step Solution:

1. **Using Dilute H2SO4:**
- When Na2S reacts with dilute H2SO4, it produces H2S gas and Na+ ions:
\[ \text{Na}_2\text{S} + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S} + 2 \text{Na}^+ \]
- Na2SO4 does not react with dilute H2SO4:
\[ \text{Na}_2\text{SO}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{No Reaction} \]
- **Conclusion:** Dilute H2SO4 can distinguish Na2SO4 from Na2S.

2. **Using Acidified KMnO4:**
- KMnO4 (potassium permanganate) is a pink solution. When it reacts with Na2S in an acidic medium, it is reduced, and the pink color disappears:
\[ \text{Na}_2\text{S} + \text{KMnO}_4 + \text{H}^+ \rightarrow \text{Colorless Mn}^{2+} + \text{H}_2\text{S} + \text{Na}^+ \]
- Na2SO4 does not react with KMnO4, and the pink color remains:
\[ \text{Na}_2\text{SO}_4 + \text{KMnO}_4 \rightarrow \text{No Color Change} \]
- **Conclusion:** Acidified KMnO4 can distinguish Na2SO4 from Na2S.

3. **Using Sodium Nitroprusside Solution:**
- Sodium nitroprusside reacts with sulfide ions (S2-) from Na2S to form a purple-violet complex:
\[ \text{Na}_2\text{S} + \text{Na}_2[\text{Fe(CN)}_5\text{NO}] \rightarrow \text{Purple-Violet Complex} \]
- Na2SO4 does not produce any color change:
\[ \text{Na}_2\text{SO}_4 + \text{Sodium Nitroprusside} \rightarrow \text{No Color Change} \]
- **Conclusion:** Sodium nitroprusside can distinguish Na2SO4 from Na2S.

4. **Using Cadmium Acetate Solution:**
- When Na2S reacts with cadmium acetate, it forms a yellow precipitate of cadmium sulfide (CdS):
\[ \text{Na}_2\text{S} + \text{Cd(CH}_3\text{COO)}_2 \rightarrow \text{CdS (s)} + 2 \text{Na}^+ + 2 \text{CH}_3\text{COO}^- \]
- Na2SO4 does not form a precipitate with cadmium acetate since CdSO4 is soluble:
\[ \text{Na}_2\text{SO}_4 + \text{Cd(CH}_3\text{COO)}_2 \rightarrow \text{No Precipitate} \]
- **Conclusion:** Cadmium acetate can distinguish Na2SO4 from Na2S.

### Final Conclusion:
All four compounds (dilute H2SO4, acidified KMnO4, sodium nitroprusside solution, and cadmium acetate solution) can be used to distinguish Na2SO4 from Na2S.