Which of the following mixture in aqueous solution of equimolar concentration acts as a buffer solution.

Options

A. `HNO_(3) + NaOH`
B. `H_(2)SO_(4) + KOH`
C. `NH_(4)OH("excess") + HCl`
D. `CH_(3)COOH + NaOH("large excess")`

Explanation

To determine which of the given mixtures in aqueous solution of equimolar concentration acts as a buffer solution, we need to understand the conditions that define a buffer solution. A buffer solution is typically made from a weak acid and its salt (formed by reacting the weak acid with a strong base) or a weak base and its salt (formed by reacting the weak base with a strong acid).

Let's analyze the options step by step:

### Step 1: Identify the Components of Each Mixture
1. **H₂SO₄ and NaOH**:
- H₂SO₄ is a strong acid.
- NaOH is a strong base.
- When mixed, they will completely neutralize each other, resulting in water and a salt (Na₂SO₄). No weak acid or weak base remains, so it cannot act as a buffer.

2. **H₂SO₄ and KOH**:
- H₂SO₄ is a strong acid.
- KOH is a strong base.
- Similar to the first case, they will completely neutralize each other, leading to water and a salt (K₂SO₄). Again, no weak acid or weak base remains, so it cannot act as a buffer.

3. **NH₄OH and HCl**:
- NH₄OH (ammonium hydroxide) is a weak base.
- HCl is a strong acid.
- If NH₄OH is in excess, it will react with HCl to form NH₄Cl (ammonium chloride). In this case, we have a weak base (NH₄OH) and its salt (NH₄Cl) present in the solution, which means it can act as a buffer solution.

4. **Acetic Acid and NaOH**:
- Acetic acid (CH₃COOH) is a weak acid.
- NaOH is a strong base.
- If NaOH is in large excess, it will react with all the acetic acid to form sodium acetate (CH₃COONa) and water. In this case, there will be no acetic acid left, only the salt, so it cannot act as a buffer solution.

### Step 2: Conclusion
From the analysis above, the only mixture that can act as a buffer solution is the third option: **NH₄OH and HCl**.

### Final Answer
The mixture that acts as a buffer solution is **NH₄OH and HCl**.

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