Find the equivalent mass of `H_3PO_4` in the following reaction :
`Ca(OH)_(2) + H_(3)PO_(4) to CaHPO_(4) + 2H_(2)O`

Explanation

To find the equivalent mass of \( H_3PO_4 \) in the reaction:

\[ Ca(OH)_2 + H_3PO_4 \rightarrow CaHPO_4 + 2H_2O \]

we can follow these steps:

### Step 1: Determine the Molar Mass of \( H_3PO_4 \)

To calculate the molar mass of \( H_3PO_4 \):

- 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:

\[ \text{Molar mass of } H_3PO_4 = 3 + 31 + 64 = 98 \text{ g/mol} \]

### Step 2: Determine the Number of Replaceable Hydrogens

In the reaction, \( H_3PO_4 \) acts as an acid and can donate 2 protons (H\(^+\)). Therefore, the number of replaceable hydrogen ions in this reaction is 2.

### Step 3: Calculate the Equivalent Mass of \( H_3PO_4 \)

The equivalent mass of an acid can be calculated using the formula:

\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{\text{Number of replaceable hydrogens}} \]

Substituting the values we have:

\[ \text{Equivalent mass of } H_3PO_4 = \frac{98 \text{ g/mol}}{2} = 49 \text{ g/equiv} \]

### Final Answer

The equivalent mass of \( H_3PO_4 \) in the given reaction is **49 g/equiv**.

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