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

Options

A. 10ml
B. 20ml
C. 5ml
D. 15ml

Explanation

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

### 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 reaction, we see that 1 mole of H₂SO₄ reacts with 2 moles of NaOH.

2. **Calculate Normality of H₂SO₄**:
The normality (N) of a solution is given by the formula:
\[ N = M \times n \]
where \( M \) is the molarity and \( n \) is the number of equivalents (basicity in this case). For H₂SO₄, the basicity is 2 (since it can donate 2 protons).
\[ N_{H_2SO_4} = 0.1 \, M \times 2 = 0.2 \, N \]

3. **Identify Normality of NaOH**:
NaOH is a strong base that donates 1 hydroxide ion (OH⁻) per molecule. Therefore, its normality is equal to its molarity:
\[ N_{NaOH} = 0.1 \, N \]

4. **Use the Neutralization Equation**:
According to the neutralization equation:
\[ N_1 V_1 = N_2 V_2 \]
where:
- \( N_1 \) = Normality of H₂SO₄ = 0.2 N
- \( V_1 \) = Volume of H₂SO₄ (unknown)
- \( N_2 \) = Normality of NaOH = 0.1 N
- \( V_2 \) = Volume of NaOH = 10 ml

5. **Substitute the Values**:
Substitute the known values into the equation:
\[ 0.2 \, N \times V_1 = 0.1 \, N \times 10 \, ml \]

6. **Solve for \( V_1 \)**:
Rearranging gives:
\[ V_1 = \frac{0.1 \, N \times 10 \, ml}{0.2 \, N} \]
\[ V_1 = \frac{1}{2} \times 10 \, ml = 5 \, ml \]

### Final Answer:
The volume of 0.1 M H₂SO₄ required to neutralize 10 ml of 0.1 M NaOH solution is **5 ml**.

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