[Asoom]

1) If 0.4586g of sodium oxalate,Na2C2O4, requires 34.53 ml of KMnO4 solution to reach the end point , What is the molarity of the KMnO4 solution ?


2) Titration of an oxalate sample gave the following percentage 15.35% , 15.55% and 15.65%. Calculate the average and the standard deviation


3) Why does the solution decolorize on standing after the equivalence point has been reached ?

[Perito]

The reaction is the following. The H+ comes from acid in the solution, usually present in excess. I haven't written the counter ions:

2 MnO4- + 5 H2C2O4 + 6 H+ --> 2Mn2+ + 10 CO2 + 14 H2O

You can add the counterions from the following equation:

2K+ + 6Cl- --> 2KCl + 4Cl-

The excess chlorides match with the Mn2+ to form MnCl2

Try to take it from here. If you have problems, ask.

[Asoom]

Thanks for your helping..

I did't understand what you explain

Na2C2O4 and you write H2C2O4 . Why ?

[Perito]

H2C2O4 is oxalic acid. Na2C2O4 is the sodium salt (sodium oxalate). The anion, C2O4(-2) is the same. The reaction involves acid and since oxalic acid is a weak acid, it will form H2C2O4 in acidic solution

Na2C2O4 + 2H+ = H2C2O4 + 2Na+

[Asoom]

I solve Q1

The result was 0.099mol/L

What about another Q (2,3) , how can I solve it ?

[Perito]

2) Titration of an oxalate sample gave the following percentage 15.35% , 15.55% and 15.65%. Calculate the average and the standard deviation

Wikipedia has a discussion on the standard deviation:

Standard deviation - Wikipedia, the free encyclopedia

this is the most common equation for



Each of the percentages in the problem is given by

3) Why does the solution decolorize on standing after the equivalence point has been reached ?

What is the reaction? KMnO4 is a deep purple. What are the colors of the products? Manganese+2 is a very light purple, but nowhere near as strongly colored as KMnO4.

[Asoom]

Thanks a lot for your helping..