The normality of orthophosphoric acid having purity of 70% by weight and specific gravity 1.54 is :

Options

A. 11 N
B. 22 N
C. 33 N
D. 44 N

Explanation

To find the normality of orthophosphoric acid (H₃PO₄) with a purity of 70% by weight and a specific gravity of 1.54, we can follow these steps:

### Step 1: Calculate the density of orthophosphoric acid
The density of a substance can be calculated using its specific gravity. The density of water is approximately 1 g/mL. Thus, the density of orthophosphoric acid can be calculated as follows:

\[ \text{Density of H}_3\text{PO}_4 = \text{Specific Gravity} \times \text{Density of Water} = 1.54 \times 1 \, \text{g/mL} = 1.54 \, \text{g/mL} \]

### Step 2: Calculate the weight of orthophosphoric acid in 1 liter of solution
Since the density is 1.54 g/mL, we can find the weight of 1 liter (1000 mL) of the solution:

\[ \text{Weight of H}_3\text{PO}_4 = \text{Density} \times \text{Volume} = 1.54 \, \text{g/mL} \times 1000 \, \text{mL} = 1540 \, \text{g} \]

### Step 3: Calculate the weight of pure orthophosphoric acid
Given that the purity of the acid is 70% by weight, we can calculate the weight of pure H₃PO₄:

\[ \text{Weight of pure H}_3\text{PO}_4 = \text{Weight of solution} \times \text{Purity} = 1540 \, \text{g} \times 0.70 = 1078 \, \text{g} \]

### Step 4: Calculate the molar mass and equivalent weight of orthophosphoric acid
The molar mass of H₃PO₄ is approximately 98 g/mol. The equivalent weight can be calculated using the formula:

\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n} \]

where \( n \) is the number of ionizable hydrogen ions. For H₃PO₄, \( n = 3 \):

\[ \text{Equivalent Weight} = \frac{98 \, \text{g/mol}}{3} \approx 32.66 \, \text{g/equiv} \]

### Step 5: Calculate the number of equivalents of orthophosphoric acid
Now, we can calculate the number of equivalents of H₃PO₄ in the solution:

\[ \text{Number of equivalents} = \frac{\text{Weight of pure H}_3\text{PO}_4}{\text{Equivalent Weight}} = \frac{1078 \, \text{g}}{32.66 \, \text{g/equiv}} \approx 33 \, \text{equivalents} \]

### Step 6: Calculate the normality of the solution
Normality (N) is defined as the number of equivalents of solute per liter of solution. Since we have calculated the number of equivalents in 1 liter of solution:

\[ \text{Normality} = \frac{\text{Number of equivalents}}{\text{Volume of solution in liters}} = \frac{33 \, \text{equivalents}}{1 \, \text{L}} = 33 \, \text{N} \]

Thus, the normality of orthophosphoric acid is **33 N**.

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