Volume of 0.1 M `H_2SO_4` solution required to neutralize 40 ml of 0.1 M NaOH solution is :

Answer

To find the volume of 0.1 M \( H_2SO_4 \) required to neutralize 40 ml of 0.1 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 balanced chemical equation for the neutralization of sulfuric acid with sodium hydroxide is:
\[ H_2SO_4 + 2 NaOH \rightarrow Na_2SO_4 + 2 H_2O \]
From this equation, we see that 1 mole of \( H_2SO_4 \) reacts with 2 moles of \( NaOH \).

2. **Calculate Normality of \( H_2SO_4 \)**:
The normality (N) of an acid is defined as the molarity (M) multiplied by the number of protons (H\(^+\)) it can donate. For \( H_2SO_4 \), the basicity is 2 (it can donate 2 protons).
\[ N_{H_2SO_4} = M_{H_2SO_4} \times \text{Basicity} = 0.1 \, M \times 2 = 0.2 \, N \]

3. **Calculate Normality of \( NaOH \)**:
For \( NaOH \), it can donate 1 hydroxide ion (OH\(^-\)), so its normality is equal to its molarity.
\[ N_{NaOH} = 0.1 \, N \]

4. **Use the Normality-Volume Relationship**:
According to the neutralization reaction, the relationship between the normalities and volumes of the acid and base can be expressed as:
\[ N_1 V_1 = N_2 V_2 \]
Where:
- \( N_1 = 0.2 \, N \) (normality of \( H_2SO_4 \))
- \( V_1 \) = volume of \( H_2SO_4 \) (what we need to find)
- \( N_2 = 0.1 \, N \) (normality of \( NaOH \))
- \( V_2 = 40 \, ml \) (volume of \( NaOH \))

5. **Substitute the Values**:
Plugging in the known values:
\[ 0.2 \, N \times V_1 = 0.1 \, N \times 40 \, ml \]

6. **Solve for \( V_1 \)**:
\[ 0.2 \, V_1 = 0.1 \times 40 \]
\[ 0.2 \, V_1 = 4 \]
\[ V_1 = \frac{4}{0.2} = 20 \, ml \]

### Final Answer:
The volume of 0.1 M \( H_2SO_4 \) required to neutralize 40 ml of 0.1 M \( NaOH \) solution is **20 ml**.