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 ?

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.

Thanks for your helping..

I did't understand what you explain

Na2C2O4 and you write H2C2O4 . Why ?

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+

I solve Q1

The result was 0.099mol/L

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

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 (http://en.wikipedia.org/wiki/Standard_deviation)

this is the most common equation for

StdDev = \sqrt{\frac{\sum{(Value_i -\ Avg)}}{(N-1)}}

Each of the percentages in the problem is given by Value_i




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.

Thanks a lot for your helping..