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 given reaction, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction provided is:
\[ H_3PO_4 + Ca(OH)_2 \rightarrow CaHPO_4 + 2H_2O \]

### Step 2: Identify the change in oxidation state
In this reaction, \( H_3PO_4 \) (phosphoric acid) is converted to \( CaHPO_4 \) (calcium hydrogen phosphate). The key point here is to determine how many hydrogen ions (H\(^+\)) are released during the reaction.

### Step 3: Determine the number of replaceable hydrogen ions
In \( H_3PO_4 \), there are 3 hydrogen ions. However, in the product \( CaHPO_4 \), only 1 hydrogen ion remains attached to the phosphate group. Therefore, 2 hydrogen ions are released in the reaction.

### Step 4: Calculate the valency factor
The valency factor (n) is the number of moles of H\(^+\) ions that can be replaced or provided by one mole of the acid. Here, since 2 H\(^+\) ions are released, the valency factor is:
\[ n = 2 \]

### Step 5: Calculate the molecular weight of \( H_3PO_4 \)
The molecular weight of \( H_3PO_4 \) can be calculated as follows:
- Hydrogen (H): 3 atoms × 1 g/mol = 3 g/mol
- Phosphorus (P): 1 atom × 31 g/mol = 31 g/mol
- Oxygen (O): 4 atoms × 16 g/mol = 64 g/mol

Adding these together:
\[ \text{Molecular weight of } H_3PO_4 = 3 + 31 + 64 = 98 \text{ g/mol} \]

### Step 6: Calculate the equivalent weight of \( H_3PO_4 \)
The equivalent weight can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n} \]
Substituting the values:
\[ \text{Equivalent weight} = \frac{98 \text{ g/mol}}{2} = 49 \text{ g/equiv} \]

### Final Answer
The equivalent mass of \( H_3PO_4 \) is \( 49 \text{ g/equiv} \).

---