Equivalent weight of `H_(3)PO_(4)` in each of the reaction will be respectively - `H_(3)+OH^(-)rarrH_(2)PO_(4)+H_(2)`
`H_(3)PO_(4)+2OH^(-)rarrHPO_(4)^(2-)+2H_(2)O`
`H_(3)PO_(4)+2OH^(-)rarrPO_(4)^(3-)+3H_(2)O`

Options

A. `98,49,32.67`
B. `49,98,32.67`
C. `98,32.67,49`
D. `32.67,49,98`

Answer

To find the equivalent weight of \( H_3PO_4 \) in each of the given reactions, we will follow these steps:

### Step 1: Calculate the Molecular Weight of \( H_3PO_4 \)
The molecular weight of \( H_3PO_4 \) can be calculated using the atomic weights of its constituent elements:
- Hydrogen (H): 1 g/mol
- Phosphorus (P): 31 g/mol
- Oxygen (O): 16 g/mol

Using the formula:
\[ \text{Molecular weight of } H_3PO_4 = (3 \times 1) + (1 \times 31) + (4 \times 16) \]
Calculating this gives:
\[ = 3 + 31 + 64 = 98 \text{ g/mol} \]

### Step 2: Determine the N Factor for Each Reaction
The N factor is defined as the number of moles of reactive species (in this case, H ions) that are replaced or removed in the reaction.

1. **First Reaction**:
\[ H_3PO_4 + OH^- \rightarrow H_2PO_4^- + H_2 \]
Here, one hydrogen ion is removed. Therefore, the N factor is 1.

2. **Second Reaction**:
\[ H_3PO_4 + 2OH^- \rightarrow HPO_4^{2-} + 2H_2O \]
Here, two hydrogen ions are removed. Therefore, the N factor is 2.

3. **Third Reaction**:
\[ H_3PO_4 + 2OH^- \rightarrow PO_4^{3-} + 3H_2O \]
Here, three hydrogen ions are removed. Therefore, the N factor is 3.

### Step 3: Calculate the Equivalent Weight for Each Reaction
The equivalent weight can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{N factor}} \]

1. **For the First Reaction**:
\[ \text{Equivalent weight} = \frac{98 \text{ g/mol}}{1} = 98 \text{ g/equiv} \]

2. **For the Second Reaction**:
\[ \text{Equivalent weight} = \frac{98 \text{ g/mol}}{2} = 49 \text{ g/equiv} \]

3. **For the Third Reaction**:
\[ \text{Equivalent weight} = \frac{98 \text{ g/mol}}{3} \approx 32.67 \text{ g/equiv} \]

### Final Results
The equivalent weights of \( H_3PO_4 \) for the three reactions are:
- First Reaction: 98 g/equiv
- Second Reaction: 49 g/equiv
- Third Reaction: 32.67 g/equiv

### Summary
The equivalent weights of \( H_3PO_4 \) in each of the reactions will be respectively:
1. 98 g/equiv
2. 49 g/equiv
3. 32.67 g/equiv