Volume of 0.1 M `H_2SO_4` solution required to neutralize 60 ml of 0.1 M NaOH solution is :

Options

A. 10ml
B. 20ml
C. 30ml
D. None of these

Explanation

To determine the volume of 0.1 M H₂SO₄ required to neutralize 60 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 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 this equation, we see that 1 mole of H₂SO₄ reacts with 2 moles of NaOH.

2. **Calculate Normality of H₂SO₄**:
The normality (N) of an acid is calculated as:
\[ N = \text{Molarity} \times \text{Basicity} \]
For H₂SO₄, the basicity is 2 (since it can donate 2 H⁺ ions). Given that the molarity of H₂SO₄ is 0.1 M:
\[ N_{H_2SO_4} = 0.1 \, \text{M} \times 2 = 0.2 \, \text{N} \]

3. **Identify Normality of NaOH**:
The normality of NaOH is equal to its molarity because it releases one hydroxide ion (OH⁻) per molecule:
\[ N_{NaOH} = 0.1 \, \text{M} = 0.1 \, \text{N} \]

4. **Set Up the Equation**:
Using the formula for neutralization:
\[ N_1 V_1 = N_2 V_2 \]
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 = 60 mL

Plugging in the values:
\[ 0.2 \, V_1 = 0.1 \times 60 \]

5. **Solve for \(V_1\)**:
\[ 0.2 \, V_1 = 6 \]
\[ V_1 = \frac{6}{0.2} = 30 \, \text{mL} \]

### Final Answer:
The volume of 0.1 M H₂SO₄ required to neutralize 60 mL of 0.1 M NaOH solution is **30 mL**.

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