0.5 g of `Ba(OH)_2, 0.01 mol Ba(OH)_(2) ` and 0.01 eq. `Ba(OH)_(2)` were together diluted to one litre. Calculate the normality of basic solution.

Answer

To calculate the normality of the basic solution formed by mixing 0.5 g of Ba(OH)₂, 0.01 mol of Ba(OH)₂, and 0.01 eq of Ba(OH)₂ diluted to one liter, we will follow these steps:

### Step 1: Calculate the equivalent mass of Ba(OH)₂
1. **Molecular mass of Ba(OH)₂**:
- Barium (Ba) = 137 g/mol
- Oxygen (O) = 16 g/mol (2 O atoms = 2 × 16 = 32 g/mol)
- Hydrogen (H) = 1 g/mol (2 H atoms = 2 × 1 = 2 g/mol)
- Therefore, molecular mass of Ba(OH)₂ = 137 + 32 + 2 = 171 g/mol

2. **Equivalent mass of Ba(OH)₂**:
- The acidity (n-factor) of Ba(OH)₂ is 2 (it can donate 2 OH⁻ ions).
- Equivalent mass = Molecular mass / n-factor = 171 g/mol / 2 = 85.5 g/equiv

### Step 2: Calculate equivalents from 0.5 g of Ba(OH)₂
- Equivalents from 0.5 g of Ba(OH)₂:
\[ \text{Equivalents} = \frac{\text{mass (g)}}{\text{equivalent mass (g/equiv)}} = \frac{0.5 \text{ g}}{85.5 \text{ g/equiv}} \approx 0.00585 \text{ equiv} \]

### Step 3: Calculate equivalents from 0.01 mol of Ba(OH)₂
- Since 0.01 mol of Ba(OH)₂ can donate 2 equivalents:
\[ \text{Equivalents} = 0.01 \text{ mol} \times 2 \text{ equiv/mol} = 0.02 \text{ equiv} \]

### Step 4: Calculate equivalents from 0.01 eq of Ba(OH)₂
- The third sample is already given in equivalents:
\[ \text{Equivalents} = 0.01 \text{ equiv} \]

### Step 5: Total equivalents of Ba(OH)₂
- Now, sum all the equivalents:
\[ \text{Total equivalents} = 0.00585 + 0.02 + 0.01 = 0.03585 \text{ equiv} \]

### Step 6: Calculate Normality
- Normality (N) is defined as the number of equivalents per liter of solution:
\[ \text{Normality} = \frac{\text{Total equivalents}}{\text{Volume of solution in liters}} = \frac{0.03585 \text{ equiv}}{1 \text{ L}} = 0.03585 \text{ N} \]

### Final Answer
The normality of the basic solution is approximately **0.03585 N**.

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