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 (in mL) 1 M `KMnO_(4)` reacting in acid solution with 5.08 gram of the acid oxalate.

Answer

To solve the problem of calculating the volume of 1 M potassium permanganate (KMnO₄) reacting with 5.08 grams of potassium acid oxalate (K₂C₂O₄·3H₂C₂O₄·4H₂O), we will follow these steps:

### Step 1: Determine the Molar Mass of Potassium Acid Oxalate
The first step is to calculate the molar mass of potassium acid oxalate (K₂C₂O₄·3H₂C₂O₄·4H₂O).

- Potassium (K): 39.1 g/mol × 2 = 78.2 g/mol
- Carbon (C): 12.01 g/mol × 4 = 48.04 g/mol
- Oxygen (O): 16.00 g/mol × 8 = 128.00 g/mol (4 from K₂C₂O₄ and 4 from 3H₂C₂O₄)
- Hydrogen (H): 1.008 g/mol × 6 = 6.048 g/mol (from 3H₂C₂O₄ and 4H₂O)

Adding these together:
Molar mass = 78.2 + 48.04 + 128.00 + 6.048 = 260.288 g/mol

### Step 2: Calculate the Number of Moles of Potassium Acid Oxalate
Using the mass of potassium acid oxalate given (5.08 g), we can find the number of moles.

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{5.08 \text{ g}}{260.288 \text{ g/mol}} \approx 0.0195 \text{ moles} \]

### Step 3: Determine the Equivalent Factor (n-factor)
The n-factor for potassium acid oxalate (K₂C₂O₄) is 2 because it can donate 2 moles of electrons when oxidized.

### Step 4: Calculate the Gram Equivalent of Potassium Acid Oxalate
The gram equivalent can be calculated using the formula:

\[ \text{Gram equivalent} = \text{Number of moles} \times n \]
\[ \text{Gram equivalent} = 0.0195 \text{ moles} \times 2 = 0.0390 \text{ equivalents} \]

### Step 5: Determine the n-factor for KMnO₄
The n-factor for KMnO₄ in acidic medium is 5, as it can accept 5 electrons.

### Step 6: Calculate the Gram Equivalent of KMnO₄
Using the equivalence of the two reactants:

\[ \text{Gram equivalent of KMnO₄} = \text{Gram equivalent of K₂C₂O₄} \]

Let \( V \) be the volume of KMnO₄ in liters.

\[ \text{Normality} = \frac{\text{Gram equivalent}}{\text{Volume in liters}} \]
\[ 1 \text{ M KMnO₄} = 1 \text{ equivalent/L} \]

Thus, we can write:

\[ 5 \cdot V = 0.0390 \text{ equivalents} \]

### Step 7: Solve for Volume \( V \)
\[ V = \frac{0.0390}{5} = 0.0078 \text{ L} \]

### Step 8: Convert Volume to mL
To convert liters to milliliters, multiply by 1000:

\[ V = 0.0078 \text{ L} \times 1000 = 7.8 \text{ mL} \]

### Final Answer
The volume of 1 M KMnO₄ required to react with 5.08 grams of potassium acid oxalate is **7.8 mL**.

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