Question

Potassium permanganate (KMnO4) is sometimes added to water as it enters a drinking water treatment plant to assist in the removal of Fe2+ and Mn2+ ions. The KMnO4 oxidizes those cations by acquiring electrons according to the following reaction: KMnO4 + 4H+ + 3e ↔ K+ + MnO2(s) + 2H2O a) What is the (electron) equivalent weight of Mn, based on this reaction? b) Calculate the mass (g) of KMnO4 in 2 L of a 0.15 M KMnO4 solution. c) If 1.0 mL of 0.15 M KMnO4 is added to each liter of the drinking water, and all the KMnO4 undergoes the reaction shown above, what concentration of KMnO4 will be generated by the reaction, in ppm?

Answer 1

Solution:

a) The reaction is  KMnO4 + 4H+ + 3e ↔ K+ + MnO2(s) + 2H2O

Oxidation number of Mn in KMnO4 is +7

Oxidation number of Mn in MnO2 is +4

Equivalent mass of KMnO4 in this reaction = molar mass/change in oxidation number = 158/3 = 52.67

atomic mass of Mn = 54.93 g/mole

equivalent mass of Mn = 54.93/3 = 18.31

b). molar mass of KMnO4 = 158 g/mol

2 L 0.15 M KMnO4

since molarity is the number of moles present in 1 L, 2 L solution contains 0.15 x 2 moles of KMnO4 = 0.3 moles

mass of KMnO4 = number of moles x molar mass = 0.3 x 158 = 47.4 g

c) Since all the KMnO4 underwent reaction and no KMnO4 will be there in water (I feel the question is wrong)