Question

If 0.1130 g of sodium oxalate, Na2C2O4, requiresof a 35.83 mL KMnO4 solution to reach the end point, what is the molarity of the  KMnO4  solution?

Answer 1

Moles of sodium oxalate(Na2C2O4) = mass/molar mass = 0.1130g/133.99g/mol = 8.4334 ×10^-4

When sodium oxalate and potassium permanganate reacts,following half reaction occurs-

C2O42- ----> 2CO2 + 2e-

MnO4- + 8H+ +5e- -----> Mn2+ + 4H2O

So,multiplying chromium reaction by 5 and Mn reaction by 2 so that electrons can cancel off when we add the half reactions-

5C2O42- + 2MnO4- + 8H+ ----> 10CO2 + 2Mn2+ + 8H2O

So,5 moles of C2O4- reacts with 2 mol of MnO4-

So 8.4334×10^-4 moles of C2O4-reacts with = 8.4334×10^-4 × 2/5 = 3.373 ×10^-4 moles of MnO4-

Volume in L = 35.83 ml = 35.83 /1000 = 0.03583

Molarity of KMnO4 = moles of KMnO4/Volume in L

= 3.373 ×10^-4 /0.03583= 9.415 × 10^-3 M