The volume of `0.1M H_(2)SO_(4)` required to neutralise completely `40mL` of `0.2M NaOH` solution is

Options

A. `10mL`
B. `40mL`
C. `20mL`
D. `80mL`

Explanation

To find the volume of `0.1M H₂SO₄` required to neutralize `40 mL` of `0.2M NaOH`, we can follow these steps:

### Step 1: Write the balanced chemical equation for the reaction.
The balanced equation for the neutralization reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) is:

\[ H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O \]

This equation shows that one mole of H₂SO₄ reacts with two moles of NaOH.

### Step 2: Calculate the number of moles of NaOH.
Using the formula:

\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in L)} \]

Given:
- Molarity of NaOH = `0.2 M`
- Volume of NaOH = `40 mL = 0.040 L`

Calculating the moles of NaOH:

\[ \text{Number of moles of NaOH} = 0.2 \, \text{mol/L} \times 0.040 \, \text{L} = 0.008 \, \text{mol} \]

### Step 3: Determine the number of moles of H₂SO₄ required.
From the balanced equation, we know that 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Therefore, the moles of H₂SO₄ required can be calculated as follows:

\[ \text{Number of moles of H₂SO₄} = \frac{\text{Number of moles of NaOH}}{2} = \frac{0.008 \, \text{mol}}{2} = 0.004 \, \text{mol} \]

### Step 4: Calculate the volume of H₂SO₄ required.
Using the formula:

\[ \text{Volume (in L)} = \frac{\text{Number of moles}}{\text{Molarity}} \]

Given:
- Molarity of H₂SO₄ = `0.1 M`

Calculating the volume of H₂SO₄:

\[ \text{Volume of H₂SO₄} = \frac{0.004 \, \text{mol}}{0.1 \, \text{mol/L}} = 0.04 \, \text{L} = 40 \, \text{mL} \]

### Final Answer:
The volume of `0.1M H₂SO₄` required to neutralize completely `40 mL` of `0.2M NaOH` solution is **40 mL**.

---