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.
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
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