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JEE Main 2020 · Chemistry · Redox Reactions and Titrations · easy

25.0 mL of 0.10 M ferrous oxalate (FeC2O4) solution is titrated in acidic medium with 0.020 M KMnO4, which is reduced to Mn^2+. Volume of KMnO4…

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25.0 mL25.0\ \mathrm{mL} of 0.10 M0.10\ \mathrm{M} ferrous oxalate (FeC2O4\mathrm{FeC_2O_4}) solution is titrated in acidic medium with 0.020 M KMnO40.020\ \mathrm{M\ KMnO_4}, which is reduced to Mn2+\mathrm{Mn^{2+}}. Volume of KMnO4\mathrm{KMnO_4} required for complete oxidation of both Fe2+\mathrm{Fe^{2+}} and C2O42−\mathrm{C_2O_4^{2-}} is
  1. A. 25 mL25\ \mathrm{mL}
  2. B. 50 mL50\ \mathrm{mL}
  3. C. 75 mL75\ \mathrm{mL}
  4. D. 100 mL100\ \mathrm{mL}
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Answer: (C) 75 mL75\ \mathrm{mL}

Per formula unit: Fe2+→Fe3+\mathrm{Fe^{2+}\to Fe^{3+}} loses 1 e−^- and C2O42−→2CO2\mathrm{C_2O_4^{2-}\to2CO_2} loses 2 e−^-, so nn-factor =3=3. MnO4−\mathrm{MnO_4^-} has nn-factor 55. Equivalents: 25×0.10×3=7.525\times0.10\times3=7.5 meq =V×0.020×5=V\times0.020\times5, giving V=75V=75 mL. Ignoring the oxalate gives 2525 mL.
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