The compound whose `0.1 M` solution is basic is

Options

A. Ammonium chloride
B. Ammonium acetate
C. Ammonium sulphate
D. Sodium acetate

Explanation

To determine which compound has a 0.1 M solution that is basic, we can analyze the hydrolysis of each given salt. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds given are:
- Ammonium chloride (NH4Cl)
- Ammonium acetate (CH3COONH4)
- Ammonium sulfate ((NH4)2SO4)
- Sodium acetate (CH3COONa)

2. **Understand Salt Hydrolysis**: Salts can undergo hydrolysis in water, which can affect the pH of the solution. The nature of the solution (acidic, neutral, or basic) depends on the strength of the acids and bases produced during hydrolysis.

3. **Analyze Ammonium Chloride (NH4Cl)**:
- Hydrolysis: NH4Cl → NH4+ + Cl-
- NH4+ (from a weak base NH3) will produce H+ ions, making the solution acidic.
- **Conclusion**: NH4Cl solution is acidic.

4. **Analyze Ammonium Acetate (CH3COONH4)**:
- Hydrolysis: CH3COONH4 → CH3COO- + NH4+
- CH3COO- (from a weak acid CH3COOH) and NH4+ (from a weak base NH3) can create a solution that could be either acidic or basic depending on the concentrations.
- **Conclusion**: CH3COONH4 solution can be neutral, acidic, or basic.

5. **Analyze Ammonium Sulfate ((NH4)2SO4)**:
- Hydrolysis: (NH4)2SO4 → 2NH4+ + SO4^2-
- NH4+ will again produce H+ ions, leading to an acidic solution.
- **Conclusion**: (NH4)2SO4 solution is acidic.

6. **Analyze Sodium Acetate (CH3COONa)**:
- Hydrolysis: CH3COONa → CH3COO- + Na+
- CH3COO- will react with water to produce OH- ions, making the solution basic.
- **Conclusion**: CH3COONa solution is basic.

7. **Final Conclusion**: Among the given options, sodium acetate (CH3COONa) is the compound whose 0.1 M solution is basic.

### Answer:
The compound whose 0.1 M solution is basic is **Sodium Acetate (CH3COONa)**.

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