The volume of 0.1M `H_(2)SO_(4)` solution required to neutralise 50ml of 0.2M NaOH solution is -

Options

A. 25ml
B. 50ml
C. 75ml
D. 100ml

Answer

To solve the problem of finding the volume of 0.1 M `H₂SO₄` solution required to neutralize 50 mL of 0.2 M NaOH solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The neutralization reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) can be represented as:
\[ H₂SO₄ + 2NaOH \rightarrow Na₂SO₄ + 2H₂O \]
This shows that one mole of H₂SO₄ reacts with two moles of NaOH.

2. **Determine the Moles of NaOH**:
We need to calculate the number of moles of NaOH in the 50 mL of 0.2 M solution.
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume (in L)} \]
Convert 50 mL to liters:
\[ \text{Volume of NaOH} = \frac{50}{1000} = 0.05 \text{ L} \]
Now calculate the moles:
\[ \text{Moles of NaOH} = 0.2 \, \text{mol/L} \times 0.05 \, \text{L} = 0.01 \, \text{mol} \]

3. **Calculate the Moles of H₂SO₄ Required**:
From the balanced equation, we see that 1 mole of H₂SO₄ is needed for every 2 moles of NaOH. Therefore, the moles of H₂SO₄ required can be calculated as:
\[ \text{Moles of H₂SO₄} = \frac{\text{Moles of NaOH}}{2} = \frac{0.01}{2} = 0.005 \, \text{mol} \]

4. **Calculate the Volume of H₂SO₄ Required**:
We know the molarity of H₂SO₄ is 0.1 M. We can use the formula:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} \]
Substitute the values:
\[ \text{Volume of H₂SO₄} = \frac{0.005 \, \text{mol}}{0.1 \, \text{mol/L}} = 0.05 \, \text{L} \]
Convert this volume to milliliters:
\[ \text{Volume of H₂SO₄} = 0.05 \, \text{L} \times 1000 = 50 \, \text{mL} \]

5. **Final Answer**:
The volume of 0.1 M H₂SO₄ required to neutralize 50 mL of 0.2 M NaOH solution is **50 mL**.