12 ml of 0.25 N `H_(2)SO_(4)` is neutralised with 15 ml of sodium hydroxide solution on titration, then the normality of `NaOH` solution is

Options

A. 0.4 N
B. 0.2 N
C. 0.1 N
D. 0.05 N

Explanation

To find the normality of the sodium hydroxide (NaOH) solution used to neutralize sulfuric acid (H₂SO₄), we can use the formula based on the concept of neutralization, which states that the equivalents of acid must equal the equivalents of base at the point of neutralization.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Volume of H₂SO₄ solution (V₁) = 12 ml
- Normality of H₂SO₄ solution (N₁) = 0.25 N
- Volume of NaOH solution (V₂) = 15 ml
- Normality of NaOH solution (N₂) = ?

2. **Use the Neutralization Equation:**
The equation for neutralization is:
\[ N₁V₁ = N₂V₂ \]
where:
- N₁ = Normality of H₂SO₄
- V₁ = Volume of H₂SO₄
- N₂ = Normality of NaOH
- V₂ = Volume of NaOH

3. **Substitute the Known Values into the Equation:**
\[ 0.25 \, \text{N} \times 12 \, \text{ml} = N₂ \times 15 \, \text{ml} \]

4. **Calculate the Left Side:**
\[ 0.25 \times 12 = 3 \, \text{N ml} \]

5. **Set Up the Equation:**
\[ 3 = N₂ \times 15 \]

6. **Solve for N₂:**
\[ N₂ = \frac{3}{15} = 0.2 \, \text{N} \]

### Conclusion:
The normality of the sodium hydroxide (NaOH) solution is **0.2 N**.