A basic buffer is a mixture of a weak base and its salt with a strong acid. Let's calculate the milli-moles (mmol) of the components in each option:

1. $100 \text{ mL of } 0.1 \text{ M HCl} + 100 \text{ mL of } 0.1 \text{ M NaOH}$: mmol of HCl = $100 \times 0.1 = 10$ mmol mmol of NaOH = $100 \times 0.1 = 10$ mmol Complete neutralization occurs, forming a neutral salt (NaCl). It is not a buffer.

2. $50 \text{ mL of } 0.1 \text{ M NaOH} + 25 \text{ mL of } 0.1 \text{ M CH}_3\text{COOH}$: mmol of NaOH = $50 \times 0.1 = 5$ mmol mmol of $\text{CH}_3\text{COOH} = 25 \times 0.1 = 2.5$ mmol After the reaction, $2.5$ mmol of $\text{CH}_3\text{COONa}$ is formed and $2.5$ mmol of the strong base NaOH is left. It is not a buffer.

3. $100 \text{ mL of } 0.1 \text{ M CH}_3\text{COOH} + 100 \text{ mL of } 0.1 \text{ M NaOH}$: mmol of $\text{CH}_3\text{COOH} = 100 \times 0.1 = 10$ mmol mmol of NaOH = $100 \times 0.1 = 10$ mmol Complete neutralization occurs, forming a salt of a weak acid and a strong base ($\text{CH}_3\text{COONa}$). It is not a buffer.

4. $100 \text{ mL of } 0.1 \text{ M HCl} + 200 \text{ mL of } 0.1 \text{ M NH}_4\text{OH}$: mmol of HCl = $100 \times 0.1 = 10$ mmol mmol of $\text{NH}_4\text{OH} = 200 \times 0.1 = 20$ mmol After the reaction, $10$ mmol of $\text{NH}_4\text{Cl}$ is formed and $10$ mmol of $\text{NH}_4\text{OH}$ is left unreacted. This is a mixture of a weak base ($\text{NH}_4\text{OH}$) and its salt with a strong acid ($\text{NH}_4\text{Cl}$). Hence, it acts as a basic buffer.