Preparation of Potassium permanganate solution for Lime determination

Consider the oxidation of oxalate ion by the permanganate:

  =                   5C₂O₄^2-+2MnO^-₄+ 16H ⁺             10CO₂+ Mn²⁺ + 8H₂O

The change in oxidation number for manganese in the reaction is from     +7 to +2 which is equal to +5. For oxidation-reduction reaction such as        this, the numerical value for the equivalent weight is established by    dividing the formula weight of the substance of interest by the change in      oxidation number associated with its reaction. The equivalent weight for MnO4²⁺ at the reaction stage is therefore 1/5 of the corresponding           formula weight. That is,

Equivalent weight       =      KMnO₄ g       =     31.681g

         5

 To prepare 0.1N KMnO₄solution, dissolve 3.1681g or approximately                        3.2g of the solid reagent in 1L of water, heat to boiling and keep hot for 1                   hour or boil continuously for 15 minutes; cool, cover and let stand                              overnight. Filter through a fine-porosity sintered glass crucible, store in                 a clean black bottle and keep in the dark when not in use.

Standardization of KMnO₄ against Na₂C₂O₄

Weigh 0.25g of pure, dried sodium oxalate into two, 250ml flasks, and                  dissolve in 100ml of water, add 15 – 20ml of H₂SO₄ (1+1) to each flask;                     heat to boiling and immediately titrate while hot, with KMnO₄ solution to the             first perceptible pinkish red. Record the volumes of the two titers and average            their values.

To correct for an end point blank, take 100ml of water in a 250ml flask,                add 20ml of H₂SO₄(1+1); heat to boiling and immediately titrate with                         the KMnO₄. Subtract the volume of titer from the average value for the                       standard and calculate the factor (f) for the standardization.

 Calculation:  Divide the weight of sodium oxalate sample by the true                    titer to obtain the milliliter equivalent of KMnO₄ in terms of Na₂C₂O₄. The                  mill equivalent of KMnO₄ in terms of CaO is therefore given as follows:

Wt. of Na₂C₂O₄/ml of KMnO₄    =     wt. of CaO/ml of KMnO₄

  134g                                                        56.08g

 Example – If the volume of KMnO₄ consumed in the standardization      against           sodium oxalate of gram weight, 0.25g and percent purity, 99% is 33ml and if        the volume used up in the titration of calcium as calcium oxalate is 43.06ml.              Then,

Wt. of Na₂C₂O₄ /33ml      =         wt. of CaO/43.06ml

  134g                                            56.08g

                  

          Therefore,

          Factor, f    =     0.25g x 0.99       =                wt. of CaO

   33ml x 134g                  43.06ml x 56.08g

                  

That is,

                            

                             5.5970 x 10^-5      =          wt. of CaO

                                         43.06ml x 56.08g 

Wt. of CaO    =    43.06ml x 5.5970 x 10^-5 x 56.08g

                  

          Since the sample of the impure calcium carbonate is 0.25g, then the                             percent composition of calcium oxide is given as:

% CaO = 43.06ml x 5.5970 x 10^-5 x 56.08g   x 100%

0.25g

                                                  

                         =       54.06% 

          The factor, f, of standardization of KMnO₄ with respect to Na₂C₂O₄ is     equal           to the constant, 5.5970 x 10^-5.        

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