The normality of solution obtained by mixing 100 ml of 0.2 M `H_(2) SO_(4)` with 100 ml of 0.2 M NaOH is

Options

A. 0.05 N
B. 0.1 N
C. 0.15 N
D. 0.2 N

Answer

To find the normality of the solution obtained by mixing 100 mL of 0.2 M \( H_2SO_4 \) with 100 mL of 0.2 M NaOH, we will follow these steps:

### Step 1: Calculate the Normality of \( H_2SO_4 \)
- **Molarity of \( H_2SO_4 \)** = 0.2 M
- **n factor for \( H_2SO_4 \)**: Since \( H_2SO_4 \) can donate 2 protons (H⁺ ions), the n factor is 2.

\[ \text{Normality of } H_2SO_4 = \text{Molarity} \times n \text{ factor} = 0.2 \, \text{M} \times 2 = 0.4 \, \text{N} \]

### Step 2: Calculate the Normality of NaOH
- **Molarity of NaOH** = 0.2 M
- **n factor for NaOH**: Since NaOH can donate 1 hydroxide ion (OH⁻), the n factor is 1.

\[ \text{Normality of NaOH} = \text{Molarity} \times n \text{ factor} = 0.2 \, \text{M} \times 1 = 0.2 \, \text{N} \]

### Step 3: Set Up the Neutralization Reaction
In a neutralization reaction, the total equivalents of acid will equal the total equivalents of base. We can use the formula:

\[ n_1 V_1 - n_2 V_2 = N \times V_{\text{total}} \]

Where:
- \( n_1 \) = Normality of \( H_2SO_4 \) = 0.4 N
- \( V_1 \) = Volume of \( H_2SO_4 \) = 100 mL
- \( n_2 \) = Normality of NaOH = 0.2 N
- \( V_2 \) = Volume of NaOH = 100 mL
- \( V_{\text{total}} \) = Total volume = 100 mL + 100 mL = 200 mL

### Step 4: Substitute Values into the Equation
Substituting the values into the equation:

\[ (0.4 \, \text{N} \times 100 \, \text{mL}) - (0.2 \, \text{N} \times 100 \, \text{mL}) = N \times 200 \, \text{mL} \]

### Step 5: Solve for \( N \)
Calculating the left side:

\[ (40 - 20) = N \times 200 \]

This simplifies to:

\[ 20 = N \times 200 \]

Now, solving for \( N \):

\[ N = \frac{20}{200} = 0.1 \, \text{N} \]

### Conclusion
The normality of the solution obtained by mixing 100 mL of 0.2 M \( H_2SO_4 \) with 100 mL of 0.2 M NaOH is **0.1 N**.

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