Potassium acid oxalate `K_(2)C_(2)O_(4). 3H_(2)C_(2)O_(4). 4H_(2)O ` can be oxidized by `MnO_(4)^(-)` in acid medium. Calculate the volume of 0.1 M `KMnO_(4)` reacting in acid solution with one gram of the acid oxalate

Answer

To solve the problem of calculating the volume of 0.1 M KMnO4 reacting with 1 gram of potassium acid oxalate (K2C2O4·3H2C2O4·4H2O) in an acidic medium, follow these steps:

### Step 1: Determine the n-factor for KMnO4
In acidic medium, the reaction of MnO4^- (permanganate ion) reduces to Mn^2+. The change in oxidation state is as follows:
- Mn in MnO4^-: +7
- Mn in Mn^2+: +2

The n-factor for KMnO4 is calculated as:
\[ \text{n-factor} = \text{Change in oxidation state} = 7 - 2 = 5 \]

**Hint:** Remember that the n-factor is the total number of electrons transferred per formula unit during the redox reaction.

### Step 2: Determine the n-factor for potassium acid oxalate
The potassium acid oxalate (K2C2O4·3H2C2O4·4H2O) can be simplified to C2O4^2- (oxalate ion) in acidic medium. The oxidation state of carbon in oxalate is +3 and it changes to +4 when oxidized to CO2. The change in oxidation state for 2 carbon atoms is:
- From +3 to +4 for each carbon (2 carbons): \( (4 - 3) \times 2 = 2 \)

Thus, the n-factor for potassium acid oxalate is:
\[ \text{n-factor} = 2 \]

**Hint:** The n-factor can also be determined by the number of moles of electrons transferred in the reaction.

### Step 3: Calculate the equivalent weight of potassium acid oxalate
The molecular weight of potassium acid oxalate (K2C2O4·3H2C2O4·4H2O) is given as 505 g. The equivalent weight can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} \]
\[ \text{Equivalent weight} = \frac{505 \, \text{g}}{2} = 252.5 \, \text{g/equiv} \]

**Hint:** Equivalent weight is the mass of the substance that reacts with or produces 1 mole of hydrogen ions (H+) or electrons.

### Step 4: Calculate the equivalents of potassium acid oxalate
Given that 1 gram of potassium acid oxalate is used, the number of equivalents can be calculated as:
\[ \text{Equivalents} = \frac{\text{Weight}}{\text{Equivalent weight}} \]
\[ \text{Equivalents} = \frac{1 \, \text{g}}{252.5 \, \text{g/equiv}} \approx 0.00396 \, \text{equiv} \]

**Hint:** Always ensure that the units are consistent when calculating equivalents.

### Step 5: Set up the equivalence equation
Since the equivalents of KMnO4 will equal the equivalents of potassium acid oxalate:
\[ \text{Equivalents of KMnO4} = \text{Equivalents of potassium acid oxalate} \]

### Step 6: Calculate the volume of KMnO4 solution
Using the molarity of KMnO4 (0.1 M), we can express the equivalents of KMnO4 in terms of volume:
\[ \text{Equivalents} = \text{Molarity} \times \text{Volume (L)} \]
Rearranging gives:
\[ \text{Volume (L)} = \frac{\text{Equivalents}}{\text{Molarity}} \]
Substituting the known values:
\[ \text{Volume (L)} = \frac{0.00396 \, \text{equiv}}{0.1 \, \text{M}} = 0.0396 \, \text{L} = 39.6 \, \text{mL} \]

**Hint:** Always convert the final volume to the desired unit (mL or L) based on the context of the problem.

### Final Answer
The volume of 0.1 M KMnO4 reacting with 1 gram of potassium acid oxalate in acid solution is approximately **39.6 mL**.