[sharonO]

I am trying to standardize KMnO4 solution with Na2C2O4.
It will be very much appreciated if you could look over my calculations!

- used approx. 0.04M KMnO4 stock (made from 50mL 0.2M KMnO4 with 200 mL water)
- weighed 1.2g of Na2C2O4 ---> transferred into 100mL water ---> took 25mL aliquot
   ----> added 25mL (3M) sulfuric acid
- end point detected at 29.02mL KMnO4

fw of Na2C2O4 is 134g.
(1.2g Na2C2O4)*(1/134g) = 0.008955 mols Na2C2O4

because it's in 100mL solution
(0.008955 mols)/(0.100L)  = 0.08955M Na2C2O4

Na2C2O4 ----> 2Na+  +  C2O4 2-

therefore 0.08955M = [C2O4 2-]


and then the 25mL aliquot and 25mL sulfuric acid (overall 50mL solution)
dilution factor:   (0.08955M)*(25/50) = 0.044776 M

**** then I get a little confused here*****

since it's from 25mL aliquot, there are (0.044766M)*(0.025L) = 0.001119 mols of oxalate.

(from balanced equation, for every 5 mols of oxalate, there is 2 mols permanganate)
so by stoichiometry:

(0.001119 mols oxalate)*(2 mols permanganate/5 mols oxalate) = 4.46E-4 mols permanagate

and then since 29.02 mL permanganate was used in titration,
(4.46E-4 mols)/(0.02902L) = 0.0154M permanganate....

I must have done something wrong because I should have used 0.04M permanaganate solution..
I should have gotten an answer around 0.04M permanganate, right?? 




 

[Train]

0.08955M = [C2O4 2-]

and then the 25mL aliquot and 25mL sulfuric acid (overall 50mL solution)
dilution factor:   (0.08955M)*(25/50) = 0.044776 M

**** then I get a little confused here*****

since it's from 25mL aliquot, there are (0.044766M)*(0.025L) = 0.001119 mols of oxalate.

You're off by a factor of 2 because the 25 mL aliquot is 0.08955 M.  So to find the moles of oxalate that you titrated you should have either

(0.044766M)*(0.05L) = 0.002238 moles

or

(0.08955M)*(0.025L) = 0.02238 moles