# Preparation of Potassium permanganate solution for Lime determination

= 5C

_{2}O_{4}^{2-}+2MnO^{-}_{4}+ 16H^{+}10CO_{2}+ Mn^{2+ }+ 8H_{2}O
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 MnO4

^{2+}at the reaction stage is therefore 1/5 of the corresponding formula weight. That is,
Equivalent weight = KMnO

_{4}g = 31.681g
5

To prepare 0.1N KMnO

_{4}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 KMnO**

_{4}against Na_{2}C_{2}O_{4}
Weigh 0.25g of pure, dried sodium oxalate into two, 250ml flasks, and dissolve in 100ml of water, add 15 – 20ml of H

_{2}SO_{4}(1+1) to each flask; heat to boiling and immediately titrate while hot, with KMnO_{4}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 H

_{2}SO_{4}(1+1); heat to boiling and immediately titrate with the KMnO_{4}. 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 KMnO

_{4}in terms of Na

_{2}C

_{2}O

_{4}. The mill equivalent of KMnO

_{4}in terms of CaO is therefore given as follows:

Wt. of Na

_{2}C_{2}O_{4}/ml of KMnO_{4}= wt. of CaO/ml of KMnO_{4}
134g 56.08g

**Example**– If the volume of KMnO

_{4}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 Na

_{2}C_{2}O_{4}/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 KMnO

_{4}with respect to Na_{2}C_{2}O_{4}is equal to the constant, 5.5970 x 10^{-5}.
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