0.01, M, KMnO_{4} solution was added to 20.0, | vedclass

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Question

(A) The balanced chemical equation for the reaction between $KMnO_{4}$ and Mohr's salt $(FeSO_{4} \cdot (NH_{4})_{2}SO_{4} \cdot 6H_{2}O)$ in acidic m

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) The balanced chemical equation for the reaction between $KMnO_{4}$ and Mohr's salt $(FeSO_{4} \cdot (NH_{4})_{2}SO_{4} \cdot 6H_{2}O)$ in acidic medium is:$MnO_{4}^{-} + 5Fe^{2+} + 8H^{+} ightarrow Mn^{2+} + 5Fe^{3+} + 4H_{2}O$Using the equivalence concept:$n_{factor} 	imes M_{1} 	imes V_{1} = n_{factor} 	imes M_{2} 	imes V_{2}$For $KMnO_{4}$,$n_{factor} = 5$ (reduction of $Mn^{+7}$ to $Mn^{+2}$).For Mohr's salt,$n_{factor} = 1$ (oxidation of $Fe^{+2}$ to $Fe^{+3}$).$5 	imes 0.01 	imes V_{1} = 1 	imes 0.05 	imes 20.0$$0.05 	imes V_{1} = 1.0$$V_{1} = \frac{1.0}{0.05} = 20.0\, mL$The volume of $KMnO_{4}$ used is $20.0\, mL$.The volume left in the $50\, mL$ burette is:$V_{left} = 50.0 - 20.0 = 30.0\, mL$.

$0.01\, M\, KMnO_{4}$ solution was added to $20.0\, mL$ of $0.05\, M$ Mohr's salt solution through a burette. The initial reading of $50\, mL$ burette is zero. The volume of $KMnO_{4}$ solution left in the burette after the end point is $.....\, mL$. (nearest integer)

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