The pH of the solution containing 10 ml of 0.1 N NaOH and 10 ml of 0.05 `NH_(2)SO_(4)` would be

Options

A. 0
B. 1
C. `gt 7`
D. 7

Explanation

To find the pH of the solution containing 10 ml of 0.1 N NaOH and 10 ml of 0.05 N H₂SO₄, we can follow these steps:

### Step 1: Calculate the moles of NaOH
- Normality (N) of NaOH = 0.1 N
- Volume (V) of NaOH = 10 ml = 0.01 L
- Moles of NaOH = Normality × Volume = 0.1 N × 0.01 L = 0.001 moles

### Step 2: Calculate the moles of H₂SO₄
- Normality (N) of H₂SO₄ = 0.05 N
- Volume (V) of H₂SO₄ = 10 ml = 0.01 L
- Moles of H₂SO₄ = Normality × Volume = 0.05 N × 0.01 L = 0.0005 moles

### Step 3: Determine the moles of H⁺ produced by H₂SO₄
- H₂SO₄ is a diprotic acid, meaning it can donate 2 H⁺ ions per molecule.
- Therefore, moles of H⁺ from H₂SO₄ = 2 × moles of H₂SO₄ = 2 × 0.0005 = 0.001 moles

### Step 4: Determine the net moles of OH⁻ and H⁺
- Moles of OH⁻ from NaOH = 0.001 moles
- Moles of H⁺ from H₂SO₄ = 0.001 moles

### Step 5: Calculate the excess moles
- Since the moles of OH⁻ and H⁺ are equal (0.001 moles each), they will neutralize each other completely.
- After neutralization, there are no excess H⁺ or OH⁻ ions left.

### Step 6: Determine the pH of the solution
- Since there are no excess H⁺ or OH⁻ ions, the solution is neutral.
- Therefore, the pH of the solution is 7.

### Final Answer:
The pH of the solution is **7**.