`KMnO_4` is prepared from the mineral pyrolusite, `MnO_2` (deep purple colour). It acts as an oxidising agent in the neutral, alkaline as well as acidic medium in acidic medium it is used in volumetric analysis for estimation of `Fe^(2+)`, `Cr_2O_4^(2-)` salts etc. The titrations are carried out in presence of `H_2SO_4`. However, before using it as a titrant, it is first standardised with standard oxalic acid solution or Mohr's salt solution . In one of the experiments on titration 26.8g of dry pure sodium oxalate `(Mw=123gmol^-1)` was dissolved in 1L of distilled water and then 100 mL of `2MH_2SO_4` were added. The solution was cooled. Now to this solution `0.1MKMnO_4` solution was added till a very faint pink colour persisted.
Q. The volume of `KMnO_4` solution that must have been added to obtain the faint pink colour at the end point must be

Options

A. 100 mL
B. 200 mL
C. 400 mL
D. 800 mL

Explanation

To solve the problem, we need to determine the volume of `KMnO_4` solution required to reach the endpoint of the titration with sodium oxalate. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the moles of Sodium Oxalate
We start by calculating the number of moles of sodium oxalate `(Na2C2O4)` used in the reaction.

**Given:**
- Mass of sodium oxalate = 26.8 g
- Molar mass of sodium oxalate = 134 g/mol

**Calculation:**
\[ \text{Moles of Na}_2\text{C}_2\text{O}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{26.8 \, \text{g}}{134 \, \text{g/mol}} \approx 0.2 \, \text{mol} \]

### Step 2: Write the balanced chemical equation
The balanced reaction between `KMnO_4` and sodium oxalate in acidic medium is:
\[ \text{MnO}_4^- + 5 \text{C}_2\text{O}_4^{2-} + 8 \text{H}^+ \rightarrow \text{Mn}^{2+} + 10 \text{CO}_2 + 4 \text{H}_2\text{O} \]

From the balanced equation, we see that:
- 1 mole of `KMnO_4` reacts with 5 moles of `C2O4^{2-}`.

### Step 3: Calculate the moles of `KMnO_4` required
Using the stoichiometry from the balanced equation, we can find the moles of `KMnO_4` needed for the reaction.

**Calculation:**
\[ \text{Moles of KMnO}_4 = \frac{1}{5} \times \text{Moles of Na}_2\text{C}_2\text{O}_4 = \frac{1}{5} \times 0.2 \, \text{mol} = 0.04 \, \text{mol} \]

### Step 4: Calculate the volume of `KMnO_4` solution needed
We know the concentration of the `KMnO_4` solution is 0.1 M. We can use the formula:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Concentration (M)}} \]

**Calculation:**
\[ \text{Volume of KMnO}_4 = \frac{0.04 \, \text{mol}}{0.1 \, \text{mol/L}} = 0.4 \, \text{L} = 400 \, \text{mL} \]

### Final Answer
The volume of `KMnO_4` solution that must have been added to obtain the faint pink color at the endpoint is **400 mL**.

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