`NaHC_(2)O_(4)` is neutralised by NaOH. It can also be oxidised by `KMnO_(4)` (in acidic medium). Equivalent weight is related to molecular weight (M) of `NaHC_(2)O_(4)` in these two reaction as :

Explanation

To determine the equivalent weight of sodium hydrogen oxalate (NaHC₂O₄) in two different reactions (neutralization with NaOH and oxidation with KMnO₄), we will follow these steps:

### Step 1: Determine the Molecular Weight of NaHC₂O₄
The molecular weight (M) of NaHC₂O₄ can be calculated using the atomic weights of its constituent elements:
- Sodium (Na): 23 g/mol
- Hydrogen (H): 1 g/mol
- Carbon (C): 12 g/mol (2 Carbon atoms)
- Oxygen (O): 16 g/mol (4 Oxygen atoms)

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

### Step 2: Neutralization Reaction with NaOH
In the neutralization reaction:
\[ \text{NaHC}_2\text{O}_4 + \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + \text{H}_2\text{O} \]
Here, NaHC₂O₄ acts as an acidic salt and can donate one proton (H⁺). Therefore, the number of equivalents (n) is 1.

### Step 3: Calculate Equivalent Weight in Neutralization
The equivalent weight (E) in the neutralization reaction is given by:
\[ E = \frac{M}{n} = \frac{112 \text{ g/mol}}{1} = 112 \text{ g/equiv} \]

### Step 4: Oxidation Reaction with KMnO₄
In the oxidation reaction:
\[ \text{NaHC}_2\text{O}_4 + \text{KMnO}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{MnSO}_4 + \text{K}_2\text{SO}_4 + \text{CO}_2 + \text{H}_2\text{O} \]
In this reaction, the carbon in NaHC₂O₄ is oxidized. The oxidation state of carbon changes from +3 in NaHC₂O₄ to +4 in CO₂.

### Step 5: Determine Change in Oxidation State
The change in oxidation state for one carbon atom is:
\[ \text{Change} = +4 - (+3) = +1 \]
Since there are 2 carbon atoms, the total change is:
\[ \text{Total Change} = 2 \times 1 = 2 \]
Thus, the number of equivalents (n) for this reaction is 2.

### Step 6: Calculate Equivalent Weight in Oxidation
The equivalent weight (E) in the oxidation reaction is given by:
\[ E = \frac{M}{n} = \frac{112 \text{ g/mol}}{2} = 56 \text{ g/equiv} \]

### Summary of Results
1. In the neutralization reaction with NaOH, the equivalent weight of NaHC₂O₄ is 112 g/equiv.
2. In the oxidation reaction with KMnO₄, the equivalent weight of NaHC₂O₄ is 56 g/equiv.