Volume of 0.1 M `H_2SO_4` solution required to neutralize 70 ml of 0.1 M NaOH solution is :

Options

A. 35ml
B. 30ml
C. 20ml
D. None of these

Explanation

To solve the problem of finding the volume of 0.1 M H₂SO₄ required to neutralize 70 ml of 0.1 M NaOH solution, we can use the concept of normality and the neutralization reaction between an acid and a base.

### Step-by-Step Solution:

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

2. **Determine Normality**:
- For H₂SO₄, the basicity (number of H⁺ ions it can donate) is 2.
- Therefore, the normality (N) of H₂SO₄ is calculated as:
\[ N_{H_2SO_4} = Molarity \times Basicity = 0.1 \, M \times 2 = 0.2 \, N \]

3. **Calculate Normality of NaOH**:
- NaOH is a strong base that provides one OH⁻ ion per molecule, so its normality is equal to its molarity:
\[ N_{NaOH} = 0.1 \, N \]

4. **Use the Neutralization Equation**:
According to the neutralization equation:
\[ N_1 \times V_1 = N_2 \times V_2 \]
where:
- \(N_1\) = Normality of H₂SO₄ = 0.2 N
- \(V_1\) = Volume of H₂SO₄ (what we need to find)
- \(N_2\) = Normality of NaOH = 0.1 N
- \(V_2\) = Volume of NaOH = 70 ml

5. **Substitute the Values**:
Substitute the known values into the equation:
\[ 0.2 \, N \times V_1 = 0.1 \, N \times 70 \, ml \]

6. **Solve for \(V_1\)**:
Rearranging the equation to solve for \(V_1\):
\[ V_1 = \frac{0.1 \, N \times 70 \, ml}{0.2 \, N} \]
\[ V_1 = \frac{7 \, ml}{0.2} = 35 \, ml \]

### Final Answer:
The volume of 0.1 M H₂SO₄ required to neutralize 70 ml of 0.1 M NaOH solution is **35 ml**.