Question

In $Fe\left(II\right)–Mn{O}_4^{-}$ titration, $HN{O}_3$ is not used because:

• A. It oxidizes $M{n}^{(}2+)$
• B. It reduces $Mn{O}_4^{-}$
• C. It oxidizes $F{e}^{2+}$
• D. It reduces $F{e}^{3+}$ formed

Answer 1

The correct option is C It oxidizes $F{e}^{2+}$

Reaction of Fe(II) with $Mn{O}_4^{-}$ is a redox reaction and is a test for strength of $KMn{O}_4$ against Mohr's salt.
In this titration dil. $H_{2}S{O}_4$ is used as it doesn't oxidise or reduce the elements where $HN{O}_3$ being a very powerful oxidising agent oxidises Fe(II) ion to Fe(III) and a lower amount of $KMn{O}_4$ gets used up, giving an incorrect result in the titration.