Only 2 mL of KMnO_4 solution of unknown mola | vedclass

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(NONE OF THE ABOVE) The balanced chemical equation for the reaction between $KMnO_4$ and oxalic acid in acidic medium is: $2KMnO_4 + 5H_2C_2O_4 + 3H_2

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(NONE OF THE ABOVE) 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 $Molarity 	imes Volume 	imes n-factor$. For $KMnO_4$ in acidic medium,the $n-factor = 5$. For oxalic acid $(H_2C_2O_4)$,the $n-factor = 2$. 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 should be . . . . . . . . . $M$.

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