"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor")
n-factor is very important in redox as well as non-redox reactions.With the help of n-factor we can predict the molar ratio of the reactant species specis taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants.
In general n-factor of acid/base is number of moles of `H^(+)//OH^(-)` furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant.
Example 1:
(1)In acidic medium : `KMnO_(4) (n=5)to Mn^(2+)`
(2) In neutral medium : `KMnO_(4)(n=3) to Mn^(2+)`
(3) In basic medium : `KMnO_(4)(n=1) to Mn^(6+)`
Example 2 : `FeC_(2)O_(4)to Fe^(3+)+2CO_(2)`
Total number of moles `e^(-)` lost by 1 mole of `FeC_(2)O_(4)`
`=1+1xx2 implies 3`
Consider the following reaction.
`H_(3)PO_(2)+NaOH to NaH_(2)PO_(2)+H_(2)O`
What is the equivalent mass of `H_(3)PO_(2)` ?(mol.Wt.is M)

Options

A. M
B. `M/2
C. M/3
D. None of these

Answer

To find the equivalent mass of \( H_3PO_2 \) in the reaction \( H_3PO_2 + NaOH \rightarrow NaH_2PO_2 + H_2O \), we will follow these steps:

### Step 1: Understand the Concept of Equivalent Mass
The equivalent mass of a substance is calculated using the formula:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{\text{n-factor}} \]
where the n-factor is the number of moles of reactive species (like \( H^+ \) ions for acids) that one mole of the substance can furnish.

### Step 2: Determine the Molar Mass of \( H_3PO_2 \)
The molar mass of \( H_3PO_2 \) 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 × 2 = 32 g/mol

Adding these together gives:
\[ \text{Molar mass of } H_3PO_2 = 3 + 31 + 32 = 66 \text{ g/mol} \]

### Step 3: Determine the n-factor for \( H_3PO_2 \)
In the given reaction, \( H_3PO_2 \) donates one \( H^+ \) ion when it reacts with \( NaOH \). Therefore, the n-factor for \( H_3PO_2 \) is:
\[ \text{n-factor} = 1 \]

### Step 4: Calculate the Equivalent Mass
Now, substituting the values into the equivalent mass formula:
\[ \text{Equivalent mass of } H_3PO_2 = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{66 \text{ g/mol}}{1} = 66 \text{ g/equiv} \]

### Final Answer
The equivalent mass of \( H_3PO_2 \) is **66 g/equiv**.

---