An acidic buffer is obtained on mixing :

Options

A. 100 mL of 0.1 M HCl and 200 mL of 0.1 M `CH_(3)COONa`
B. 100 mL of 0.1 M HCl and 200 mL of 0.1 M NaCl
C. 100 mL of 0.1 M `CH_(3) COOH` and 100 mL of 0.1 M NaOH
D. 100 mL of 0.1 M `CH_(3)COOH` and 200 mL of 0.1 M NaOH

Explanation

To determine which option results in an acidic buffer, we need to understand that an acidic buffer is formed by mixing a weak acid with its conjugate base. Let's analyze each option step by step.

### Step-by-Step Solution:

1. **Understanding the Components of an Acidic Buffer:**
An acidic buffer consists of a weak acid (HA) and its conjugate base (A⁻). For example, acetic acid (CH₃COOH) is a weak acid, and its conjugate base is acetate ion (CH₃COO⁻).

2. **Analyzing Option A:**
- **Components:** 100 ml of 0.1 M HCl and 200 ml of 0.1 M CH₃COOH.
- **Calculating Millimoles:**
- HCl: \( 100 \, \text{ml} \times 0.1 \, \text{mol/L} = 10 \, \text{mmol} \)
- CH₃COOH: \( 200 \, \text{ml} \times 0.1 \, \text{mol/L} = 20 \, \text{mmol} \)
- **Reaction:** HCl is a strong acid and will not contribute to the buffer. The CH₃COOH will remain as the weak acid.
- **Final Mixture:**
- After mixing, we have 10 mmol of CH₃COOH and 10 mmol of CH₃COO⁻ (from the reaction of HCl with CH₃COO⁻).
- **Conclusion:** This mixture contains both a weak acid (CH₃COOH) and its conjugate base (CH₃COO⁻), thus forming an acidic buffer.

3. **Analyzing Option B:**
- **Components:** 100 ml of 0.1 M HCl and 200 ml of 0.1 M NaCl.
- **Calculating Millimoles:**
- HCl: \( 100 \, \text{ml} \times 0.1 \, \text{mol/L} = 10 \, \text{mmol} \)
- NaCl: \( 200 \, \text{ml} \times 0.1 \, \text{mol/L} = 20 \, \text{mmol} \)
- **Conclusion:** HCl is a strong acid, and NaCl is a neutral salt. This mixture does not contain a weak acid and its conjugate base, so it does not form a buffer.

4. **Analyzing Option C:**
- **Components:** 100 ml of 0.1 M CH₃COOH and 100 ml of 0.1 M NaOH.
- **Calculating Millimoles:**
- CH₃COOH: \( 100 \, \text{ml} \times 0.1 \, \text{mol/L} = 10 \, \text{mmol} \)
- NaOH: \( 100 \, \text{ml} \times 0.1 \, \text{mol/L} = 10 \, \text{mmol} \)
- **Reaction:** CH₃COOH reacts with NaOH to form CH₃COO⁻ and water.
- **Conclusion:** After the reaction, we have 10 mmol of CH₃COO⁻ and no remaining CH₃COOH. This does not form a buffer because we need both components.

5. **Analyzing Option D:**
- **Components:** 100 ml of 0.1 M CH₃COOH and 200 ml of 0.1 M NaOH.
- **Calculating Millimoles:**
- CH₃COOH: \( 100 \, \text{ml} \times 0.1 \, \text{mol/L} = 10 \, \text{mmol} \)
- NaOH: \( 200 \, \text{ml} \times 0.1 \, \text{mol/L} = 20 \, \text{mmol} \)
- **Reaction:** CH₃COOH reacts with NaOH to form CH₃COO⁻ and water.
- **Conclusion:** After the reaction, we have 10 mmol of CH₃COO⁻ and 10 mmol of NaOH left. This does not form a buffer because we need both components.

### Final Answer:
The only option that results in an acidic buffer is **Option A**: 100 ml of 0.1 M HCl and 200 ml of 0.1 M CH₃COOH.