Redox and Equivalent ConceptEasy

Question

Calculate equivalent weight of KHC2O4 (potassium acid oxalate) when :-
(a) It is neutralized by KOH
(b) It is oxidised by MnO4- in basic medium

Options

A.128, 128
B.64, 128
C.64, 64
D.128, 64

Solution

(a) KHC2O4 + KOH →  K2C2O4 + H2O
∴  eq.ωt = M/1 = 128
(b) KHC2O4 + MnO4- → MnO2 + 2CO2
    ↑               ↑               ↑             ↑
  +6              +7           +4           +8
∴ eq. ωt of KHC2O4 = M/2 = 64

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