Volume of 0.1 M `H_2SO_4` solution required to neutralize 50 ml of 0.1 M NaOH solution is :

Options

A. 30ml
B. 25ml
C. 40ml
D. 10ml

Explanation

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

### Step 1: Understand the Neutralization Reaction
The neutralization reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) can be represented as:
\[ \text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]
From this equation, we see that 1 mole of H₂SO₄ reacts with 2 moles of NaOH.

### Step 2: Calculate the Normality of H₂SO₄
Normality (N) is calculated using the formula:
\[ N = M \times \text{Basicity} \]
For H₂SO₄, the molarity (M) is given as 0.1 M and the basicity (number of H⁺ ions released) is 2. Therefore, the normality (N₁) of H₂SO₄ is:
\[ N_1 = 0.1 \, \text{M} \times 2 = 0.2 \, \text{N} \]

### Step 3: Calculate the Normality of NaOH
For NaOH, the molarity is also given as 0.1 M, and since it releases 1 OH⁻ ion, the normality (N₂) is:
\[ N_2 = 0.1 \, \text{N} \]

### Step 4: Use the Neutralization Formula
The relationship between normality and volume in a neutralization reaction is given by:
\[ N_1 V_1 = N_2 V_2 \]
Where:
- \( N_1 \) = Normality of H₂SO₄
- \( V_1 \) = Volume of H₂SO₄ (what we want to find)
- \( N_2 \) = Normality of NaOH
- \( V_2 \) = Volume of NaOH (given as 50 mL)

### Step 5: Substitute the Known Values
Substituting the known values into the equation:
\[ 0.2 \, \text{N} \times V_1 = 0.1 \, \text{N} \times 50 \, \text{mL} \]

### Step 6: Solve for \( V_1 \)
Now, we can solve for \( V_1 \):
\[ 0.2 \, V_1 = 0.1 \times 50 \]
\[ 0.2 \, V_1 = 5 \]
\[ V_1 = \frac{5}{0.2} = 25 \, \text{mL} \]

### Conclusion
The volume of 0.1 M H₂SO₄ required to neutralize 50 mL of 0.1 M NaOH solution is **25 mL**.

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