500 ml of 0.2( M ) `H_(2) SO_(4)` are mixed with 250 ml of 0.1 ( M ) `Ba( OH)_(2)` , the normality of resulting solution is -

Options

A. 0.2
B. 0.1
C. 0.5
D. 0.25

Explanation

To find the normality of the resulting solution when 500 ml of 0.2 M H₂SO₄ is mixed with 250 ml of 0.1 M Ba(OH)₂, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between sulfuric acid and barium hydroxide can be represented as follows:
\[ \text{H}_2\text{SO}_4 + \text{Ba(OH)}_2 \rightarrow \text{BaSO}_4 + 2 \text{H}_2\text{O} \]

### Step 2: Determine the n-factor for the reactants
- For H₂SO₄: It can donate 2 H⁺ ions, so its n-factor is 2.
- For Ba(OH)₂: It can donate 2 OH⁻ ions, so its n-factor is also 2.

### Step 3: Calculate the number of equivalents for H₂SO₄
Using the formula:
\[ \text{Equivalents} = \text{Molarity} \times \text{Volume (L)} \times \text{n-factor} \]
For H₂SO₄:
\[ \text{Equivalents of H}_2\text{SO}_4 = 0.2 \, \text{M} \times \frac{500 \, \text{ml}}{1000} \times 2 = 0.2 \, \text{equivalents} \]

### Step 4: Calculate the number of equivalents for Ba(OH)₂
For Ba(OH)₂:
\[ \text{Equivalents of Ba(OH)}_2 = 0.1 \, \text{M} \times \frac{250 \, \text{ml}}{1000} \times 2 = 0.05 \, \text{equivalents} \]

### Step 5: Determine the excess equivalents
Since H₂SO₄ is in excess:
\[ \text{Excess Equivalents} = \text{Equivalents of H}_2\text{SO}_4 - \text{Equivalents of Ba(OH)}_2 = 0.2 - 0.05 = 0.15 \, \text{equivalents} \]

### Step 6: Calculate the total volume of the solution
Total volume = 500 ml + 250 ml = 750 ml = 0.75 L

### Step 7: Calculate the normality of the resulting solution
Normality (N) is given by:
\[ \text{Normality} = \frac{\text{Equivalents}}{\text{Volume (L)}} \]
Thus,
\[ \text{Normality} = \frac{0.15 \, \text{equivalents}}{0.75 \, \text{L}} = 0.2 \, \text{N} \]

### Final Answer
The normality of the resulting solution is **0.2 N**.

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