An equal volume of a reducing agent is titrated separately with 1 M KMnO 4 in acid, neutral and alkaline medium. The volumes of KMnO 4 required are 20 mL in acid, 33.3 mL in neutral and 100 mL in alkaline media. Find out the volume (mL) of 1M K 2 Cr 2 O 7 consumed, if the same volume of the reducing agent is titrated in acid medium. Report your answer by rounding it up to nearest whole number.

Step-by-step solution

Let equal volumes of R.A in each case be X Meqs MnO 4 - + 8 H + + 5 e - → Mn 2 + + 4 H 2 O Acid Z = 5 MnO 4 - + 2 H 2 O + 3 e - → MnO 2 + 4 OH - Neutral Z = 3 MnO 4 - + e - → MnO 4 2 - Alkaline Z = 1 X milli ​ eqs . of R .A = 20×1 ×5 = 33 .3×1 ×3 = 100×1 ×1 =100 milli eqs 100 milli eqs. of R.A = 100 milli eqs. of K 2 Cr 2 O 7 Cr 2 O 7 2 - + 1 4 H + + 6 e - → 2 Cr 3 + + 7 H 2 O 100 milli eqs. of Cr 2 O 7 2 - = 6 × 1 × Volume mL, Vol = 16.7 mL