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Consider the oxidation of oxalate ion by the permanganate:

=                   5C2O42-+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.