1. Condition for an acidic buffer

An acidic buffer is a mixture of:

• a weak acid, and
• its salt with a strong base (i.e. its conjugate base)

A typical example is $CH_3COOH + CH_3COONa$.

So we check which mixture finally contains both:

• weak acid $CH_3COOH$
• salt $CH_3COONa$

2. Check option A

Given:

• $100\,\text{mL}$ of $0.1\,M$ $HCl$
• $200\,\text{mL}$ of $0.1\,M$ $CH_3COONa$

Moles of $HCl$: $0.100 \times 0.1 = 0.010\,\text{mol}$

Moles of $CH_3COONa$: $0.200 \times 0.1 = 0.020\,\text{mol}$

Reaction: $HCl + CH_3COO^- \rightarrow CH_3COOH + Cl^-$

$HCl$ is limiting, so it reacts with $0.010$ mol acetate.

After reaction:

• acetate left = $0.020 - 0.010 = 0.010$ mol
• acetic acid formed = $0.010$ mol

So final solution contains:

• $CH_3COOH$ (weak acid)
• $CH_3COONa$ / $CH_3COO^-$ (its salt/conjugate base)

Hence an acidic buffer is formed.

3. Check option B

Given:

• $HCl$ and $NaCl$

These are both related to a strong acid system. There is no weak acid/conjugate base pair.

So not a buffer.

4. Check option C

Given:

• $100\,\text{mL}$ of $0.1\,M$ $CH_3COOH$
• $100\,\text{mL}$ of $0.1\,M$ $NaOH$

Moles of each: $0.100 \times 0.1 = 0.010\,\text{mol}$

Reaction: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

Equal moles react completely.

Final solution contains only $CH_3COONa$, not $CH_3COOH + CH_3COONa$ together.

So not a buffer.

5. Check option D

Given:

• $100\,\text{mL}$ of $0.1\,M$ $CH_3COOH$ $\Rightarrow 0.010$ mol
• $200\,\text{mL}$ of $0.1\,M$ $NaOH$ $\Rightarrow 0.020$ mol

NaOH is in excess.

After neutralization, all acetic acid is consumed and excess strong base remains.

So not a buffer.

6. Conclusion

Only Option A gives a mixture of a weak acid and its conjugate base, hence forms an acidic buffer.

$\boxed{\text{A}}$