The equivalent weight of `NaHC_(2)O_(4)` in reaction with NaOH is :-

Options

A. 84
B. 56
C. 224
D. 112

Answer

To find the equivalent weight of NaHC₂O₄ in its reaction with NaOH, we can follow these steps:

### Step 1: Write the Reaction
The reaction between sodium hydrogen oxalate (NaHC₂O₄) and sodium hydroxide (NaOH) can be written as:
\[ \text{NaHC}_2\text{O}_4 + \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + \text{H}_2\text{O} \]

### Step 2: Determine the n-factor
The n-factor of a compound is defined as the number of moles of reactive species (H⁺, OH⁻, electrons, etc.) that one mole of the compound can provide or consume in a reaction. In this case, when NaHC₂O₄ reacts with NaOH, only one hydrogen ion (H⁺) is released to form water (H₂O). Therefore, the n-factor for NaHC₂O₄ in this reaction is:
\[ n = 1 \]

### Step 3: Calculate the Molecular Weight of NaHC₂O₄
To find the equivalent weight, we first need the molecular weight of NaHC₂O₄. The molecular weight can be calculated as follows:
- Sodium (Na): 23 g/mol
- Hydrogen (H): 1 g/mol
- Carbon (C): 12 g/mol (2 Carbons)
- Oxygen (O): 16 g/mol (4 Oxygens)

Calculating the total:
\[ \text{Molecular weight} = 23 + 1 + (2 \times 12) + (4 \times 16) \]
\[ = 23 + 1 + 24 + 64 \]
\[ = 112 \, \text{g/mol} \]

### Step 4: Calculate the Equivalent Weight
The equivalent weight is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} \]
Substituting the values:
\[ \text{Equivalent weight} = \frac{112 \, \text{g/mol}}{1} = 112 \, \text{g} \]

### Final Answer
Thus, the equivalent weight of NaHC₂O₄ in reaction with NaOH is:
\[ \text{Equivalent weight} = 112 \, \text{g} \]

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