Question

To determine the concentration in terms of molarity of KMnO4 by titrating it against 0.1 molar standard solution of oxalic acid​

Answer 1

The reaction involved in this titration is:

2MnO4− + 5H2C2O4 + 6H+⟶ 2Mn2+ + 10CO2 + 8H2O

So, this defines the stoichiometry of the reaction and for every 2 equivalents of permanganate (MnO4−), 5 oxalic acids (H2C2O4) are consumed. Also note that the color of solutions of permanganate are intensely purple colored while manganese (Mn2+) solutions are a very light pink color.

So, you will need to know the concentration of your permanganate solution (lets say it is 0.001 M). With this solution, titrate the oxalic acid solution until the purple color disappears (a little tricky since manganese solutions are lightly pink). Note what this volume is and multiple the concentration. Lets say the titration took 25 mL (or 0.025 L) of you 0.001 M solution and remember that the ratio of oxalic acid to permanganate is 5/2. The amount of oxalic acid then is:

Oxalic acid (Moles) = vol of permanganate solution * concentration of permanganate * ratio of oxalic acid/permanganate or

Oxalic acid (Moles) = 0.025 L * 0.001 moles/L * 5/2 = 0.0000625 moles.

If you wanted a concentration then you would need to know the volume of the oxalic acid solution. Lets say it was also 25 mL (0.025 L) then the concentration of the oxalic acid would be:

[Oxalic Acid] = (0.0000625 moles)/(0.025 L) = 0.0025 m/L (or M)