Volume of 0.1 M `H_2SO_4` solution required to neutralize 20 ml of 0.1 M NaOH solution is :

Options

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

Explanation

To solve the problem of finding the volume of 0.1 M \( H_2SO_4 \) required to neutralize 20 ml of 0.1 M NaOH solution, we can follow these steps:

### Step 1: Understand the Neutralization Reaction
The neutralization reaction between sulfuric acid (\( H_2SO_4 \)) 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 \( H_2SO_4 \) reacts with 2 moles of \( NaOH \).

### Step 2: Calculate the Number of Moles of \( NaOH \)
Given:
- Volume of \( NaOH \) solution = 20 ml = 0.020 L
- Molarity of \( NaOH \) = 0.1 M

Using the formula:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (L)} \]
\[ \text{Number of moles of } NaOH = 0.1 \, \text{mol/L} \times 0.020 \, \text{L} = 0.002 \, \text{mol} \]

### Step 3: Determine the Number of Moles of \( H_2SO_4 \) Required
From the stoichiometry of the reaction, 1 mole of \( H_2SO_4 \) neutralizes 2 moles of \( NaOH \). Therefore, the number of moles of \( H_2SO_4 \) required is:
\[ \text{Number of moles of } H_2SO_4 = \frac{\text{Number of moles of } NaOH}{2} = \frac{0.002 \, \text{mol}}{2} = 0.001 \, \text{mol} \]

### Step 4: Calculate the Volume of \( H_2SO_4 \) Required
Given:
- Molarity of \( H_2SO_4 \) = 0.1 M

Using the formula:
\[ \text{Volume (L)} = \frac{\text{Number of moles}}{\text{Molarity}} \]
\[ \text{Volume of } H_2SO_4 = \frac{0.001 \, \text{mol}}{0.1 \, \text{mol/L}} = 0.01 \, \text{L} \]

Convert to ml:
\[ 0.01 \, \text{L} = 10 \, \text{ml} \]

### Final Answer
The volume of 0.1 M \( H_2SO_4 \) required to neutralize 20 ml of 0.1 M \( NaOH \) solution is **10 ml**.

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