Animated Solution for Chemistry - Basic Concepts in Chemistry: When $10\text{ mL}$ of an aqueous solution of ${\text{KMnO}}_4$ was titrated in acidic medium, equal volume of $0.1\text{ M}$ of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The strength of ${\text{KMnO}}_4$ in $\text{g/L}$ is ......... $\times 10^{-2}$. (Nearest integer) [Atomic mass of $\text{K}=39$, $\text{Mn}=55$, $\text{O}=16$]

Imagine a classic laboratory setup: a conical flask resting on a white tile, and a burette suspended above it. Inside the flask, we have $10\text{ mL}$ of a $0.1\text{ M}$ aqueous solution of ferrous sulphate (${\text{FeSO}}_4$). From the burette, we are carefully adding an aqueous solution of potassium permanganate (${\text{KMnO}}_4$) drop by drop. The reaction is taking place in an acidic medium, which is crucial for the behavior of the permanganate ion.

As the purple ${\text{KMnO}}_4$ drops into the flask, it reacts with the ${\text{FeSO}}_4$ and becomes colorless. The moment the color is completely discharged and a faint permanent pink color appears, we have reached the equivalence point. At this magical point, the fundamental law of volumetric analysis comes into play: the number of equivalents of the oxidizing agent must exactly equal the number of equivalents of the reducing agent.

Mathematically, this is expressed as:

Here, $n$ represents the n-factor, $M$ is the molarity, and $V$ is the volume.

To use our equivalence equation, we first need to determine the n-factors for both reactants. The n-factor is essentially the number of electrons transferred per molecule during the redox reaction.

For potassium permanganate in an acidic medium, the manganese ion undergoes a dramatic reduction from an oxidation state of $+7$ to $+2$:

This means the n-factor for ${\text{KMnO}}_4$ ($n_1$) is $5$.

For ferrous sulphate, the iron ion is oxidized from $+2$ to $+3$:

Thus, the n-factor for ${\text{FeSO}}_4$ ($n_2$) is $1$.

Now, let's substitute our known values into the equivalence equation. We know the volumes are equal ($10\text{ mL}$ each), and the molarity of ${\text{FeSO}}_4$ is $0.1\text{ M}$. Let the molarity of ${\text{KMnO}}_4$ be $M_1$.

The $10$ on both sides cancels out beautifully, leaving us with:

We have found the molarity, but the question asks for the strength in $\text{g/L}$. To convert molarity to strength, we multiply by the molar mass. The molar mass of ${\text{KMnO}}_4$ is calculated as $39+55+(4\times 16)=158\text{ g/mol}$.

The question requires the answer in the format $x\times 10^{-2}$. We can rewrite $3.16$ as $316\times 10^{-2}$. Therefore, the value of $x$ is $316$.