20 ml of `0.1 M FeC_(2)O_(4)` solution is titrated with `0.1 M KMnO_(4)` is acidic medium. Volume of `KMnO_(4)` solution required to oxidise `FeC_(2)O_(4)` completely is

Options

A. 20 ml
B. 12 ml
C. 8 ml
D. 4 ml

Answer

To solve the problem of determining the volume of `KMnO4` solution required to completely oxidize `FeC2O4` in an acidic medium, we will follow these steps:

### Step 1: Write the Balanced Chemical Equation
The balanced chemical equation for the reaction between `KMnO4` and `FeC2O4` in acidic medium is:

\[ 6 \text{KMnO}_4 + 10 \text{FeC}_2\text{O}_4 + 24 \text{H}_2\text{SO}_4 \rightarrow 5 \text{Fe}_2(\text{SO}_4)_3 + 20 \text{CO}_2 + 6 \text{MnSO}_4 + 3 \text{K}_2\text{SO}_4 + 24 \text{H}_2\text{O} \]

### Step 2: Calculate Moles of `FeC2O4`
Given:
- Volume of `FeC2O4` solution = 20 mL
- Molarity of `FeC2O4` solution = 0.1 M

To calculate the number of moles of `FeC2O4`, we use the formula:

\[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} \]

First, convert the volume from mL to L:

\[ \text{Volume in L} = \frac{20 \text{ mL}}{1000} = 0.02 \text{ L} \]

Now, calculate the moles:

\[ \text{Moles of } FeC2O4 = 0.1 \text{ M} \times 0.02 \text{ L} = 0.002 \text{ moles} \]

### Step 3: Determine Moles of `KMnO4` Required
From the balanced equation, we see that:

- 10 moles of `FeC2O4` react with 6 moles of `KMnO4`.

To find the moles of `KMnO4` required for 0.002 moles of `FeC2O4`, we set up a proportion:

\[ \frac{6 \text{ moles of } KMnO4}{10 \text{ moles of } FeC2O4} = \frac{x \text{ moles of } KMnO4}{0.002 \text{ moles of } FeC2O4} \]

Cross-multiplying gives:

\[ 10x = 6 \times 0.002 \]

Solving for \(x\):

\[ x = \frac{6 \times 0.002}{10} = 0.0012 \text{ moles of } KMnO4 \]

### Step 4: Calculate Volume of `KMnO4` Required
We know the molarity of `KMnO4` is 0.1 M. We can use the formula:

\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} \]

Substituting the values:

\[ \text{Volume of } KMnO4 = \frac{0.0012 \text{ moles}}{0.1 \text{ M}} = 0.012 \text{ L} \]

Convert this volume to mL:

\[ \text{Volume in mL} = 0.012 \text{ L} \times 1000 = 12 \text{ mL} \]

### Final Answer
The volume of `KMnO4` solution required to completely oxidize `FeC2O4` is **12 mL**.

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