Preparation of Potassium permanganate solution for Lime determination
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 KMnO4
solution, 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,
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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