Find out the equivalent weight of `H_(3) PO_(4)`in the reaction:
`Ca(OH)_(2) + H_(3) PO_(4) rarr CaHPO_(4) + 2 H_(2) O`

Explanation

To find the equivalent weight of \( H_3PO_4 \) in the given reaction, we can follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ Ca(OH)_2 + H_3PO_4 \rightarrow CaHPO_4 + 2 H_2O \]

In this reaction, phosphoric acid (\( H_3PO_4 \)) reacts with calcium hydroxide (\( Ca(OH)_2 \)) to produce calcium hydrogen phosphate (\( CaHPO_4 \)) and water.

### Step 2: Determine the Number of Replaceable Hydrogen Ions
Phosphoric acid (\( H_3PO_4 \)) can dissociate to release hydrogen ions. In this reaction, it donates 2 hydrogen ions, as it forms \( CaHPO_4 \) which contains one hydrogen ion.

Thus, the value of \( n \) (the number of hydrogen ions that can react) is 2.

### Step 3: Calculate the Molar Mass of \( H_3PO_4 \)
To find the equivalent weight, we first need 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

Now, add these together:
\[ \text{Molar mass of } H_3PO_4 = 3 + 31 + 64 = 98 \text{ g/mol} \]

### Step 4: Calculate the Equivalent Weight
The equivalent weight is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} \]
Substituting the values we have:
\[ \text{Equivalent weight of } H_3PO_4 = \frac{98 \text{ g/mol}}{2} = 49 \text{ g/equiv} \]

### Final Answer
The equivalent weight of \( H_3PO_4 \) in the given reaction is **49 g/equiv**.

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