Which of the following when mixed, will given a solution with `pH gt 7`.

Options

A. `0.1M HCl + 0.1M NaOH`
B. `100 mL of 0.1 M H_(2)SO_(4) + 100 mL of 0.3 M NaOH`
C. `100 mL of 0.1 M HC_(2)H_(3)O_(2) + 100 mL of 0.1 M KOH`
D. `25 mL of 0.1 HNO_(3) +25 mL of 0.1 M NH_(3)`

Explanation

To determine which of the given combinations will produce a solution with a pH greater than 7, we need to analyze each option based on the concept of neutralization and the resulting pH of the solutions. A solution with a pH greater than 7 is basic in nature.

### Step-by-Step Solution:

1. **Understand the Options**:
- **Option 1**: 0.1 M HCl with 0.1 M NaOH
- **Option 2**: 100 mL of 0.1 M H₂SO₄ with 100 mL of 0.3 M NaOH
- **Option 3**: 100 mL of 0.1 M oxalic acid (H₂C₂O₄) with 100 mL of 0.1 M KOH
- **Option 4**: 25 mL of 0.1 M HNO₃ with 25 mL of 0.1 M ammonia (NH₃)

2. **Calculate Gram Equivalents for Each Option**:
- **Option 1**:
- HCl: Normality = 0.1 M * 1 = 0.1 N; Volume = 0.1 L
- NaOH: Normality = 0.1 M * 1 = 0.1 N; Volume = 0.1 L
- Gram equivalents of HCl = 0.1 N * 0.1 L = 0.01
- Gram equivalents of NaOH = 0.1 N * 0.1 L = 0.01
- Since both are equal, the solution is neutral (pH = 7).

- **Option 2**:
- H₂SO₄: Normality = 0.1 M * 2 = 0.2 N; Volume = 0.1 L
- NaOH: Normality = 0.3 M * 1 = 0.3 N; Volume = 0.1 L
- Gram equivalents of H₂SO₄ = 0.2 N * 0.1 L = 0.02
- Gram equivalents of NaOH = 0.3 N * 0.1 L = 0.03
- Since NaOH is in excess, the solution is basic (pH > 7).

- **Option 3**:
- Oxalic acid (H₂C₂O₄): Normality = 0.1 M * 2 = 0.2 N; Volume = 0.1 L
- KOH: Normality = 0.1 M * 1 = 0.1 N; Volume = 0.1 L
- Gram equivalents of oxalic acid = 0.2 N * 0.1 L = 0.02
- Gram equivalents of KOH = 0.1 N * 0.1 L = 0.01
- Since oxalic acid is in excess, the solution is acidic (pH < 7).

- **Option 4**:
- HNO₃: Normality = 0.1 M * 1 = 0.1 N; Volume = 0.025 L
- Ammonia (NH₃): Normality = 0.1 M * 1 = 0.1 N; Volume = 0.025 L
- Gram equivalents of HNO₃ = 0.1 N * 0.025 L = 0.0025
- Gram equivalents of NH₃ = 0.1 N * 0.025 L = 0.0025
- Since both are equal, the solution is neutral (pH = 7).

3. **Conclusion**:
- The only option that results in a solution with a pH greater than 7 is **Option 2** (100 mL of 0.1 M H₂SO₄ with 100 mL of 0.3 M NaOH).

### Final Answer:
**Option 2**: 100 mL of 0.1 M H₂SO₄ with 100 mL of 0.3 M NaOH will give a solution with pH > 7.