Question

A solution of NaIO was standardized with KMnO4. If 20.0 ml of NaIO required 18.7 ml of .118M KMnO4, calculate the molarity and normality of the solution..

IO^- ----> IO₃^-

MnO₄^- -----> Mn²⁺

Please help example this.

Answer 1

IO^-(aq) -----> IO₃^-(aq) + 4e^- ......(1)

MnO₄^-(aq)+ 5e^- -----> Mn²⁺(aq) .....(2)

Overall balanced reaction :-

5IO^-(aq) + 4MnO₄^-(aq) -----> 5IO₃^-(aq) + 4Mn²⁺(aq)

Moles of MnO₄^- reacting = molarity*volume of solution in litres = 0.118*0.0187 = 0.0022

Thus, moles of NaIO reacting = (5/4)*moles of KMnO₄ = 0.00276

Thus, molarity = moles of NaIO/Volume of solution in litres = 0.138 M

Normality = molarity*number of electron transfer taking place = 0.138*4 = 0.552 N