An equal volume of a reducing agent is titrated separately with $1 ext{ M KMnO}_4$ in acidic, neutr

Question

An equal volume of a reducing agent is titrated separately with $1	ext{ M KMnO}_4$ in acidic, neutral, and strongly alkaline media. The volumes of $1	ext{ M KMnO}_4$ required are $20	ext{ mL}$, $33.3	ext{ mL}$, and $100	ext{ mL}$ respectively. If the same volume of the reducing agent is titrated in an acidic medium with $1	ext{ M K}_2	ext{Cr}_2	ext{O}_7$, the volume of $	ext{K}_2	ext{Cr}_2	ext{O}_7$ consumed is $\frac{100}{x}	ext{ mL}$. Find the value of $x$.

Professor PC Thomas Classes · Answer

Let equivalents of reducing agent be $E$. For $	ext{KMnO}_4$, $E = M 	imes V 	imes n	ext{-factor}$. In acidic medium ($n=5$), $E = 1 	imes 20 	imes 5 = 100	ext{ meq}$. (Matches neutral $33.3 	imes 3 \approx 100$ and alkaline $100 	imes 1 = 100$). For $	ext{K}_2	ext{Cr}_2	ext{O}_7$ in acid, $n	ext{-factor} = 6$. So, $1 	imes V 	imes 6 = 100 \implies V = \frac{100}{6}	ext{ mL}$. Thus $x = 6$.