To standardize a solution of KMnO4 for a redox titration, 0.3926 g of dry oxalic

Question

To standardize a solution of KMnO4 for a redox titration, 0.3926 g of dry oxalic acid ( molar mass = 90.03488 g/mol; ignore buoyancy correction) is added to an Erlenmeyer flask and dissolved in excess 1 M H2SO4. The intensely puruple titrant (KMNO4) reacts with the standard to form a clear solution UNTIL the endpoint at 36.24 mL, where the first hint of purple is persistent in the flask. Another titration using 1 M H2SO4 and no oxalic acid requires 0.34 mL of titrant. What is the molar concentration of the standardized titrant?

2 KMnO4 + 6 H+ + 5 H2C2O4 -> 2 Mn2+ + 8 H2O + 5 CO2

please show work and include any relevant information! thankyou!

Explanation / Answer

Ans. Given,

            I. Mass of oxalic acid = 0.3926 g

            II. Volume of KMnO4 solution consumed for (oxalic acid + H2SO4) = 36.24 mL

            III. Volume of KMnO4 solution consumed for (H2SO4 only) = 0.34 mL

Now,

            Moles of oxalic acid taken = Mass / molar mass           

                                                            = 0.3926 g/ (90.03488 g/mol)

                                                            = 0.004360 mol

            Volume of KMnO4 consumed by oxalic acid alone = reading (II - III)

                                                            = 36.24 mL – 0.34 mL

                                                            = 35.90 mL                                     

Now, coming to stoichiometry of reaction: 2 mol completely reacts with 5 mol H2C3O4.

            5.0 mol H2C2O4 is equivalent to 2.0 mol KMnO4

     Or, 1.0 mol           -           -           -           (2.0 / 5.0) mol KMnO4

Or, 0.004360 mol    -          -          -          (2.0 / 5.0) x 0.004360 mol KMnO4

                                                                        = 0.001744 mol

Thus, required moles of KMnO4 consumed during titration of H2C2O4 alone = 0.001744 mol.

That is,

35.90 mL of KMnO4 solution contains 0.001744 mol KMnO4.

So,

Molarity of KMnO4 solution = Number of moles / Volume of solution in liters

                                                = 0.001744 mol / (0.03590 L)                     ; [1 L = 1000 mL]

                                                = 0.0486 mol/ L

                                                = 0.0486 M

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