Volume of 0.1 M `H_2SO_4` solution required to neutralize 30 ml of 0.1 M NaOH solution is :

Options

A. 20ml
B. 15ml
C. 30ml
D. 35ml

Explanation

To find the volume of 0.1 M \( H_2SO_4 \) required to neutralize 30 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 one mole of \( H_2SO_4 \) reacts with two moles of \( NaOH \).

### Step 2: Calculate the Normality of \( H_2SO_4 \)
Normality (N) is related to molarity (M) and basicity (number of H+ ions contributed by the acid). For \( H_2SO_4 \):
- Molarity = 0.1 M
- Basicity = 2 (since it can donate 2 H+ ions)

Thus, the normality \( N_1 \) of \( H_2SO_4 \) is calculated as:
\[ N_1 = Molarity \times Basicity = 0.1 \, M \times 2 = 0.2 \, N \]

### Step 3: Calculate the Normality of \( NaOH \)
For \( NaOH \):
- Molarity = 0.1 M
- Basicity = 1 (since it can donate 1 OH- ion)

Thus, the normality \( N_2 \) of \( NaOH \) is:
\[ N_2 = Molarity \times Basicity = 0.1 \, M \times 1 = 0.1 \, N \]

### Step 4: Use the Neutralization Equation
According to the neutralization equation:
\[ N_1 \times V_1 = N_2 \times V_2 \]
Where:
- \( N_1 = 0.2 \, N \) (normality of \( H_2SO_4 \))
- \( V_1 \) = volume of \( H_2SO_4 \) (what we need to find)
- \( N_2 = 0.1 \, N \) (normality of \( NaOH \))
- \( V_2 = 30 \, ml \) (volume of \( NaOH \))

### Step 5: Substitute the Values
Now we can substitute the values into the equation:
\[ 0.2 \times V_1 = 0.1 \times 30 \]

### Step 6: Solve for \( V_1 \)
Calculating the right side:
\[ 0.2 \times V_1 = 3 \]
Now, divide both sides by 0.2 to find \( V_1 \):
\[ V_1 = \frac{3}{0.2} = 15 \, ml \]

### Conclusion
The volume of 0.1 M \( H_2SO_4 \) required to neutralize 30 ml of 0.1 M NaOH solution is **15 ml**.

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