During a volumetric titration utilizing Potassium Permanganate in a neutral or faintly alkaline medium, what is the exact equivalent weight of the utilized? (Let M represent the molar mass of ).

Question

During a volumetric titration utilizing Potassium Permanganate in a neutral or faintly alkaline medium, what is the exact equivalent weight of the $KMnO_4$ utilized? (Let M represent the molar mass of $KMnO_4$).

Prashant Kotian · Accepted Answer

Correct answer: B. The basic "BAN" rule for n-factors: Basic (strongly alkaline to manganate) = 1, Acidic = 5, Neutral/faintly alkaline = 3. In neutral media, reduces to a black precipitate of (), a 3-electron change. Thus, Eq. Wt = .

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