Volume of 0.1 M `H_2SO_4` solution required to neutralize 40 ml of 0.2 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 40 mL of 0.2 M NaOH solution, we can follow these steps:

### Step 1: Understand the Reaction
Sulfuric acid (\( H_2SO_4 \)) is a diprotic acid, meaning it can donate two protons (\( H^+ \)) per molecule. Sodium hydroxide (\( NaOH \)) is a strong base that provides one hydroxide ion (\( OH^- \)) per molecule. The neutralization reaction can be represented as:
\[ H_2SO_4 + 2 NaOH \rightarrow Na_2SO_4 + 2 H_2O \]

### Step 2: Calculate Normality
Normality (\( N \)) is related to molarity (\( M \)) by the formula:
\[ N = M \times n \]
where \( n \) is the number of protons that can be donated by the acid.

For \( H_2SO_4 \):
- Molarity = 0.1 M
- \( n \) factor = 2 (because it can donate 2 \( H^+ \))

So, the normality of \( H_2SO_4 \) is:
\[ N_1 = 0.1 \, M \times 2 = 0.2 \, N \]

For \( NaOH \):
- Molarity = 0.2 M
- \( n \) factor = 1 (because it can donate 1 \( OH^- \))

So, the normality of \( NaOH \) is:
\[ N_2 = 0.2 \, M \times 1 = 0.2 \, N \]

### Step 3: Use the Neutralization Equation
The neutralization equation can be expressed as:
\[ N_1 \times V_1 = N_2 \times V_2 \]
where:
- \( N_1 \) = normality of \( H_2SO_4 \) = 0.2 N
- \( V_1 \) = volume of \( H_2SO_4 \) (what we need to find)
- \( N_2 \) = normality of \( NaOH \) = 0.2 N
- \( V_2 \) = volume of \( NaOH \) = 40 mL

### Step 4: Substitute the Values
Now, substituting the known values into the equation:
\[ 0.2 \, N \times V_1 = 0.2 \, N \times 40 \, mL \]

### Step 5: Solve for \( V_1 \)
Dividing both sides by \( 0.2 \, N \):
\[ V_1 = 40 \, mL \]

### Conclusion
The volume of 0.1 M \( H_2SO_4 \) required to neutralize 40 mL of 0.2 M NaOH solution is **40 mL**.

---