Equivalent Weight And Normality

Question

The equivalent weight of KMnO₄ when it acts as an oxidising agent in acidic medium is: (K = 39, Mn = 55, O = 16)

RankGuru · Accepted Answer

1. **Identify the Half-Reaction**: In acidic medium, $MnO_4^-$ is reduced to $Mn^{2+}$:
   - $MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$
2. **Determine n-factor**: The number of electrons gained per formula unit = **5** (Mn goes from +7 to +2).
3. **Calculate Molar Mass of KMnO₄**:
   - $M = 39 + 55 + 4(16) = 39 + 55 + 64 = 158$ g/mol.
4. **Apply the Formula**:
   - $	ext{Equivalent weight} = \frac{	ext{Molar mass}}{n	ext{-factor}} = \frac{158}{5} = 31.6$ g/eq.
5. **Important Distinction**: The equivalent weight of KMnO₄ changes with the medium. In neutral medium (n-factor = 3), it would be 52.7, and in strongly basic medium (n-factor = 1), it would be 158. Always identify the medium first.

Equivalent Weight And Normality

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