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(A) The balanced chemical equation for the reaction between $KMnO_4$ and oxalic acid in acidic medium is:$2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \rightarrow

Vedclass · Accepted Answer

$8$

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(A) The balanced chemical equation for the reaction between $KMnO_4$ and oxalic acid in acidic medium is:$2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 ightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$.According to the law of equivalence, the number of equivalents of $KMnO_4$ equals the number of equivalents of oxalic acid:$n_{eq}(KMnO_4) = n_{eq}(H_2C_2O_4)$.Equivalents are calculated as $	ext{Molarity} 	imes 	ext{Volume} 	imes n	ext{-factor}$.For $KMnO_4$ in acidic medium, the $n	ext{-factor} = 5$.For oxalic acid $(H_2C_2O_4)$, the $n	ext{-factor} = 2$.Let the molarity of $KMnO_4$ be $M$.$M 	imes 2 \ mL 	imes 5 = 2 \ M 	imes 20 \ mL 	imes 2$.$10M = 80$.$M = 8 \ M$.

Only $2 \ mL$ of $KMnO_4$ solution of unknown molarity is required to reach the end point of a titration of $20 \ mL$ of oxalic acid $(2 \ M)$ in acidic medium. The molarity of $KMnO_4$ solution is: (in $M$)

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