Calculate the equivalent weight of $ ext{KMnO}_4$ (molar mass $158\ ext{g mol}^{-1}$) when it acts as an oxidant in (a) strongly acidic medium ($n = 5$), (b) neutral or faintly alkaline medium ($n = 3$), and (c) strongly alkaline medium ($n = 1$).

Question

Calculate the equivalent weight of $	ext{KMnO}_4$ (molar mass $158\ 	ext{g mol}^{-1}$) when it acts as an oxidant in (a) strongly acidic medium ($n = 5$), (b) neutral or faintly alkaline medium ($n = 3$), and (c) strongly alkaline medium ($n = 1$).

Accepted Answer

Concept understanding — Redox Titration

Redox Titration: The Intuition First

Imagine you have a dark room and you want to know exactly how much water is in a bucket. You can't see the water level directly. But you have a measuring cup of ink — and you know that each drop of ink turns a fixed amount of water completely black. You add ink drop by drop, stirring, until the water just turns black. The number of drops tells you exactly how much water was there.

Redox titration works on the same principle — except instead of ink and water, we use an oxidising agent and a reducing agent. One of them is the "unknown" (the water), the other is the "known solution" (the ink). They react with each other in a fixed, predictable ratio. We add the known solution until the reaction is just complete, and that tells us the amount of the unknown.

The Precise Statement

Redox titration is a volumetric analysis technique where a solution of unknown concentration (the analyte) is reacted with a standard solution of known concentration (the titrant) in a redox reaction — one substance gets oxidised, the other gets reduced — until the equivalence point is reached. The volume of titrant used allows calculation of the unknown concentration.

The key difference from acid-base titration: here, electrons are transferred, not protons.

How It Actually Works

You have a flask containing the analyte — say, a solution of ferrous ions ( $\text{Fe}^{2+}$ ). You don't know its concentration. You fill a burette with a standard solution of potassium permanganate ( $\text{KMnO}_4$ ), which is a strong oxidising agent. You know its concentration exactly.

You add the permanganate drop by drop. Each drop reacts with $\text{Fe}^{2+}$ :

\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}

The purple permanganate gets consumed as it reacts. As long as any $\text{Fe}^{2+}$ remains, the purple colour disappears. The moment all $\text{Fe}^{2+}$ is used up, the next drop of permanganate stays purple — the solution turns pink. That's your end point.

Note

In this case, the titrant itself acts as the indicator — no separate indicator needed. This is called a self-indicating titration. Not all redox titrations are self-indicating; some need a separate redox indicator (like starch for iodine titrations).

The Core Idea in One Sentence

You measure the volume of a known oxidising (or reducing) agent needed to completely react with an unknown reducing (or oxidising) agent, and from that volume you calculate the unknown concentration.

The Calculation (Simple Version)

Suppose you titrate 25.0 mL of $\text{Fe}^{2+}$ solution with 0.0200 M $\text{KMnO}_4$ . You use 15.0 mL of permanganate to reach the end point.

From the balanced equation: 1 mole $\text{MnO}_4^-$ reacts with 5 moles $\text{Fe}^{2+}$ .

Moles of $\text{KMnO}_4$ used = $0.0200 \times 0.0150 = 3.00 \times 10^{-4}$ mol

Moles of $\text{Fe}^{2+}$ present = $5 \times 3.00 \times 10^{-4} = 1.50 \times 10^{-3}$ mol

Concentration of $\text{Fe}^{2+}$ = $\frac{1.50 \times 10^{-3}}{0.0250} = 0.0600$ M

C_{\text{analyte}} = \frac{n \times M_{\text{titrant}} \times V_{\text{titrant}}}{V_{\text{analyte}}}

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Equivalent weight is always molar mass divided by n-factor, and the n-factor of $\text{KMnO}_4$ itself depends on how far manganese is reduced, which in turn depends on the medium. (a) In strongly acidic medium, $\text{MnO}_4^{-}$ is reduced all the way to $\text{Mn}^{2+}$ ( $+7\to+2$ , a 5-electron change), so $n=5$ and equivalent weight $=\dfrac{158}{5}=31.6\ \text{g equiv}^{-1}$ . (b) In neutral or faintly alkaline medium, $\text{MnO}_4^{-}$ is reduced only to $\text{MnO}_2$ ( $+7\to+4$ , a 3-electron change), so $n=3$ and equivalent weight $=\dfrac{158}{3}\approx52.67\ \text{g equiv}^{-1}$ . (c) In strongly alkaline medium, $\text{MnO}_4^{-}$ is reduced only as far as manganate, $\text{MnO}_4^{2-}$ ( $+7\to+6$ , a 1-electron change), so $n=1$ and equivalent weight $=\dfrac{158}{1}=158\ \text{g equ …

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Q.Calculate the equivalent weight of KMnO4\text{KMnO}_4KMnO4​ (molar mass 158 g mol−1158\ \text{g mol}^{-1}158 g mol−1) when it acts as an oxidant in

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Q.Calculate the equivalent weight of $\text{KMnO}_4$ (molar mass $158\ \text{g mol}^{-1}$ ) when it acts as an oxidant in

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