Question

A salt solution containing ferrous ion was titrated with 9 M KMnO₄ solution. The mean of 3 acceptable, corrected titration values was 1.259 mL. Calculate the moles of ferrous ion reacting in each titration portion.

Answer 1

Reduction reaction$MnO_4^- + 8H^+ + 5 e^- \to Mn^{2+}+ 4H_2O$

Oxidation reaction

$Fe^{3+} + e^- \to Fe^{2+}$

Total Reaction

$MnO^{4-} + 5Fe^{2+} + 8H^+ \to 5Fe^{3+} + Mn^{2+} + 4H_2O$

Therefore, 1 mole of MnO₄^- (the oxidizing agent) reacts with 5 moles of Fe²⁺ (the reducing agent) to form 5 moles of Fe³⁺ and 1 mole of Mn²⁺

1:5 mole ratio with respect to the amounts of MnO₄^- and Fe²⁺

number of moles of Fe²⁺ = concentration of MnO4 × volume of Mno4 used/5

= 9 × 1.259/1000 × 1/5 = 0.00227 moles