An acidic buffer is obtained on mixing:

First, we recall the definition of an acidic buffer. An acidic buffer is a solution that contains a weak acid and its conjugate base (usually provided by a salt of the weak acid). Such a mixture resists changes in pH when small amounts of strong acid or base are added. Mathematically, the Henderson-Hasselbalch equation for an acidic buffer is stated as

$$\text{pH}= \text{p}K_a + \log_{10}\!\left(\dfrac{[\text{salt}]}{[\text{acid}]}\right).$$

Hence, to identify which option forms an acidic buffer, we must look for a final mixture that simultaneously contains a weak acid, for example $$\mathrm{CH_3COOH},$$ and its salt with a strong base, for example $$\mathrm{CH_3COONa}.$$ We now examine each option one by one, carefully calculating moles so that every algebraic step is visible.

Option A. 100 mL of 0.1 M $$\mathrm{CH_3COOH}$$ and 100 mL of 0.1 M $$\mathrm{NaOH}$$.

We determine the number of moles present before mixing:

$$\text{Moles of } \mathrm{CH_3COOH}=0.1\,\text{M}\times0.100\,\text{L}=0.010\,\text{mol}.$$

$$\text{Moles of } \mathrm{NaOH}=0.1\,\text{M}\times0.100\,\text{L}=0.010\,\text{mol}.$$

Because $$\mathrm{NaOH}$$ is a strong base, it completely neutralises the weak acid.

Neutralisation reaction:

$$\mathrm{CH_3COOH}+\mathrm{NaOH}\;\longrightarrow\;\mathrm{CH_3COONa}+\mathrm{H_2O}.$$

Equal moles react, so both 0.010 mol are consumed, leaving

$$0\,\text{mol }\mathrm{CH_3COOH}$$ and $$0.010\,\text{mol }\mathrm{CH_3COONa}.$$

No weak acid remains; only its salt is present. Therefore no acidic buffer is produced in Option A.

Option B. 100 mL of 0.1 M $$\mathrm{HCl}$$ and 200 mL of 0.1 M $$\mathrm{NaCl}$$.

$$\mathrm{HCl}$$ is a strong acid, not a weak one. Furthermore, $$\mathrm{NaCl}$$ is a neutral salt and is not the conjugate base of any weak acid present in the solution. Consequently, the mixture is simply a diluted strong acid solution, not a buffer. Hence Option B does not give an acidic buffer.

Option C. 100 mL of 0.1 M $$\mathrm{CH_3COOH}$$ and 200 mL of 0.1 M $$\mathrm{NaOH}$$.

Again, calculate moles:

$$\text{Moles of } \mathrm{CH_3COOH}=0.1\,\text{M}\times0.100\,\text{L}=0.010\,\text{mol},$$

$$\text{Moles of } \mathrm{NaOH}=0.1\,\text{M}\times0.200\,\text{L}=0.020\,\text{mol}.$$

Neutralisation occurs according to the same reaction used in Option A. The limiting reagent is the acid (0.010 mol). After complete reaction, the leftover base is

$$0.020\,\text{mol}-0.010\,\text{mol}=0.010\,\text{mol }\mathrm{NaOH}.$$

The solution therefore contains excess strong base, making the medium decidedly basic. Since a buffer requires both weak acid and its salt, and here no weak acid is left, Option C also fails to produce an acidic buffer.

Option D. 100 mL of 0.1 M $$\mathrm{HCl}$$ and 200 mL of 0.1 M $$\mathrm{CH_3COONa}$$.

We again compute moles precisely:

$$\text{Moles of } \mathrm{HCl}=0.1\,\text{M}\times0.100\,\text{L}=0.010\,\text{mol},$$

$$\text{Moles of } \mathrm{CH_3COONa}=0.1\,\text{M}\times0.200\,\text{L}=0.020\,\text{mol}.$$

Now the strong acid $$\mathrm{HCl}$$ reacts with the salt of the weak acid as follows:

$$\mathrm{HCl}+\mathrm{CH_3COONa}\;\longrightarrow\;\mathrm{CH_3COOH}+\mathrm{NaCl}.$$

Mole-by-mole subtraction shows the change:

$$$\begin{aligned} \text{Initial }&\;[\mathrm{HCl}]=0.010\,\text{mol},\quad[\mathrm{CH_3COONa}]=0.020\,\text{mol}.\\ \text{Reacted }&\;0.010\,\text{mol HCl with }0.010\,\text{mol }\mathrm{CH_3COONa}.\\ \text{Remaining }&\;[\mathrm{HCl}]=0\,\text{mol},\quad[\mathrm{CH_3COONa}]=0.020-0.010=0.010\,\text{mol}.\\ \text{Formed }&\;[\mathrm{CH_3COOH}]=0.010\,\text{mol}.\\ \end{aligned}$$$

After reaction, the solution now contains 0.010 mol of the weak acid $$\mathrm{CH_3COOH}$$ and 0.010 mol of its conjugate-base salt $$\mathrm{CH_3COONa}$$ in the final total volume

$$V_{\text{final}}=100\,\text{mL}+200\,\text{mL}=300\,\text{mL}=0.300\,\text{L}.$$

Because both the weak acid and its salt are simultaneously present, the mixture satisfies the buffer condition. Specifically, it is an acidic buffer because the pH will be <7 (the weak acid component dominates over the weak base part of the conjugate pair).

Therefore, only Option D yields an acidic buffer.

Hence, the correct answer is Option D.