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. 0.1
B. 0.2
C. 0.5
D. 0.3

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 can follow these steps:

### Step 1: Calculate the Normality of \( H_2SO_4 \)
1. **Identify the Molarity and n-factor**:
- Molarity of \( H_2SO_4 \) = 0.2 M
- The n-factor for \( H_2SO_4 \) is 2 (because it can donate 2 protons, \( H^+ \)).

2. **Calculate Normality**:
\[ \text{Normality} (N) = \text{Molarity} (M) \times \text{n-factor} \]
\[ N_{H_2SO_4} = 0.2 \times 2 = 0.4 \, \text{N} \]

### Step 2: Calculate the Normality of NaOH
1. **Identify the Molarity and n-factor**:
- Molarity of NaOH = 0.2 M
- The n-factor for NaOH is 1 (because it can donate 1 hydroxide ion, \( OH^- \)).

2. **Calculate Normality**:
\[ N_{NaOH} = 0.2 \times 1 = 0.2 \, \text{N} \]

### Step 3: Set Up the Neutralization Reaction
The neutralization reaction between \( H_2SO_4 \) and NaOH can be represented as:
\[ H_2SO_4 + 2 NaOH \rightarrow Na_2SO_4 + 2 H_2O \]
This shows that 1 mole of \( H_2SO_4 \) reacts with 2 moles of NaOH.

### Step 4: Use the Neutralization Formula
We can use the formula:
\[ N_1 V_1 - N_2 V_2 = N_f (V_1 + V_2) \]
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
- \( N_f \) = Normality of the final solution

### Step 5: Substitute Values into the Formula
\[ N_f = \frac{N_1 V_1 - N_2 V_2}{V_1 + V_2} \]
Substituting the values:
\[ N_f = \frac{(0.4 \times 100) - (0.2 \times 100)}{100 + 100} \]
\[ N_f = \frac{40 - 20}{200} = \frac{20}{200} = 0.1 \, \text{N} \]

### 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**.