In the reaction, `H_3PO_4+Ca(OH)_2rarrCaHPO_4+2H_2O,` the equivalent mass of `H_3PO_4` is

Options

A. 49 g
B. 32.7 g
C. 196 g
D. 98 g

Answer

To find the equivalent mass of \( H_3PO_4 \) in the reaction \( H_3PO_4 + Ca(OH)_2 \rightarrow CaHPO_4 + 2H_2O \), we can follow these steps:

### Step 1: Identify the reaction and the species involved
The reaction involves phosphoric acid (\( H_3PO_4 \)) reacting with calcium hydroxide (\( Ca(OH)_2 \)) to produce calcium hydrogen phosphate (\( CaHPO_4 \)) and water (\( H_2O \)).

### Step 2: Determine the basicity of \( H_3PO_4 \)
The basicity of an acid is defined as the number of hydrogen ions (\( H^+ \)) that can be donated by one molecule of the acid. In this case, \( H_3PO_4 \) can donate 2 protons (as it reacts with 2 hydroxide ions from \( Ca(OH)_2 \)). Therefore, the basicity of \( H_3PO_4 \) is 2.

### Step 3: Calculate the molecular weight of \( H_3PO_4 \)
The molecular weight of \( H_3PO_4 \) can be calculated as follows:
- Hydrogen (H): 1 g/mol × 3 = 3 g/mol
- Phosphorus (P): 31 g/mol × 1 = 31 g/mol
- Oxygen (O): 16 g/mol × 4 = 64 g/mol

Adding these together gives:
\[ 3 + 31 + 64 = 98 \, \text{g/mol} \]

### Step 4: Calculate the equivalent mass of \( H_3PO_4 \)
The equivalent mass of an acid can be calculated using the formula:
\[ \text{Equivalent mass} = \frac{\text{Molecular weight}}{\text{Basicity}} \]
Substituting the values we have:
\[ \text{Equivalent mass of } H_3PO_4 = \frac{98 \, \text{g/mol}}{2} = 49 \, \text{g} \]

### Conclusion
The equivalent mass of \( H_3PO_4 \) in this reaction is 49 grams.

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