A solution of 10 mL of frac{M}{10} FeSO_4 w | vedclass

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(D) The balanced redox reaction in an acidic medium is: $MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$ According to the law of equ

Vedclass · Accepted Answer

$10 \ mL$ of $0.02 \ M$

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(D) The balanced redox reaction in an acidic medium is: $MnO_4^- + 8H^+ + 5Fe^{2+} ightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$ According to the law of equivalence,the number of equivalents of $KMnO_4$ must equal the number of equivalents of $FeSO_4$. Equivalents = $Molarity 	imes n-factor 	imes Volume(L)$. For $FeSO_4$,the $n-factor$ is $1$ $(Fe^{2+} ightarrow Fe^{3+} + e^-)$. For $KMnO_4$,the $n-factor$ is $5$ $(MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O)$. Equivalents of $FeSO_4 = \frac{1}{10} 	imes 1 	imes 10 \ mL = 1 \ meq$. Equivalents of $KMnO_4 = M 	imes 5 	imes V = 1 \ meq$. Checking option $(d)$: $0.02 \ M 	imes 5 	imes 10 \ mL = 1 \ meq$. Thus,the correct option is $(d)$.

$A$ solution of $10 \ mL$ of $\frac{M}{10} \ FeSO_4$ was titrated with $KMnO_4$ solution in an acidic medium. The amount of $KMnO_4$ used will be:

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