Volume of 0.1 M `H_2SO_4` solution required to neutralize 10 ml of 0.2 M NaOH solution is :

Options

A. 10ml
B. 20ml
C. 30ml
D. 40ml

Explanation

To find the volume of 0.1 M H₂SO₄ required to neutralize 10 ml of 0.2 M NaOH solution, we can use the concept of normality and the neutralization reaction between the acid and the base.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The neutralization reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) can be represented as:
\[ H_2SO_4 + 2 NaOH \rightarrow Na_2SO_4 + 2 H_2O \]
From the equation, we see that 1 mole of H₂SO₄ reacts with 2 moles of NaOH.

2. **Calculate Normality**:
- The normality (N) of an acid or base is calculated as:
\[ N = M \times n \]
where \( M \) is the molarity and \( n \) is the number of equivalents.
- For H₂SO₄, since it can donate 2 protons (H⁺), the n-factor is 2.
\[ N_{H_2SO_4} = 0.1 \, M \times 2 = 0.2 \, N \]
- For NaOH, it can donate 1 hydroxide ion (OH⁻), so the n-factor is 1.
\[ N_{NaOH} = 0.2 \, M \times 1 = 0.2 \, N \]

3. **Use the Neutralization Formula**:
The relationship between the normalities and volumes of the acid and base can be expressed as:
\[ N_1 \times V_1 = N_2 \times V_2 \]
where:
- \( N_1 \) = Normality of H₂SO₄ = 0.2 N
- \( V_1 \) = Volume of H₂SO₄ (what we need to find)
- \( N_2 \) = Normality of NaOH = 0.2 N
- \( V_2 \) = Volume of NaOH = 10 ml

4. **Substituting the Values**:
Substitute the known values into the equation:
\[ 0.2 \times V_1 = 0.2 \times 10 \]

5. **Solve for \( V_1 \)**:
\[ 0.2 \times V_1 = 2 \]
\[ V_1 = \frac{2}{0.2} = 10 \, ml \]

### Conclusion:
The volume of 0.1 M H₂SO₄ required to neutralize 10 ml of 0.2 M NaOH solution is **10 ml**.