In neutral or faintly alkaline medium,MnO_4^{ | vedclass

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Question

(B) The balanced chemical equation for the reaction in neutral or faintly alkaline medium is: $2 \ KMnO_4 + H_2O + KI \rightarrow 2 \ MnO_2 + 2 \ KOH

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(B) The balanced chemical equation for the reaction in neutral or faintly alkaline medium is: $2 \ KMnO_4 + H_2O + KI ightarrow 2 \ MnO_2 + 2 \ KOH + KIO_3$ From the stoichiometry,$1 \ 	ext{mole}$ of $I^-$ requires $2 \ 	ext{moles}$ of $MnO_4^-$. Number of moles of $KI = 1 \ L 	imes 0.5 \ M = 0.5 \ 	ext{mol}$. Therefore,the number of moles of $MnO_4^-$ required $= 2 	imes 0.5 = 1.0 \ 	ext{mol}$. Volume of $0.02 \ M \ KMnO_4$ solution $= \frac{	ext{moles}}{	ext{Molarity}} = \frac{1.0 \ 	ext{mol}}{0.02 \ M} = 50 \ L$.

In neutral or faintly alkaline medium,$MnO_4^{-}$ oxidizes $I^{-}$ to iodate. What is the volume (in $L$) of $0.02 \ M \ KMnO_4$ required to completely convert $1 \ L$ of $0.5 \ M \ KI$ solution to iodate in neutral or faintly alkaline medium?

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