Standardization of nitrite solution (TITRATION, STUMPED)

[LogicX]

Homework Statement

Potassium permanganate (KMnO4) is used to determine the actual concentration of standard solutions containing the nitrite ion. The permanganate ion reacts with nitrous acid according to the following reaction.

2 MnO4- + 5 HNO2 + 5 H+ = 2Mn2+ + 5 NO3- + 3 H2O

Pipet in order 50.00 mL of standard KMnO4 solution, 5 mL concentrated H2SO4 and 50.00 mL of NaNO2 solution into a glass-stoppered flask. Making sure that the pipet tip is submerged below the surface of the permanganate-acid solution while delivering the nitrite stock solution. Shake gently and warm on a hot plate to about 70oC. Discharge the permanganate color by successive additions of 25.00-mL aliquots of the 0.025 M oxalate solution. Back-titrate the excess oxalate with the standard permanganate solution to the faint pink end point. Run a water blank and make the necessary corrections. Calculate the concentration of NO2- in the stock solution. Calculate NO2 in stock solution.

And here is the formula for KMnO4 reacting with sodium oxalate (N2C2O4):

2 MnO4- + 5 H2C2O4 + 6 H+ = 2Mn2+ + 10 CO2 (g) + 8 H2O

Homework Equations

I should be all stoichiometry...

The Attempt at a Solution

My two partners and I have been attempting this for like the past two hours. We have calculated it a ton of different ways and always get a different answer.

So basically, first you have a solution with excess KMnO4. Then you add oxalate so that you have excess oxalate. We added 100mL of .025M sodium oxalate. Then you titrate the excess oxalate away with KMnO4.

So I know how much excess oxalate was there was based on how much KMnO4 was used to titrate at the end. Now what?

Please help, we have been working on this for so long and we can't continue with the lab until we figure out the concentration of the stock solution.

Numbers aren't really important, I can do those on my own.

Here is a thought I had... since KMnO4 reacts with nitrite, that means that the original KMnO4 in the solution reacted with the nitrite. So there was less KMnO4 than if there was no nitrite. So:

Original KMnO4 - Nitrite = excess KMnO4

So Mol Nitrite= Original KMnO4(we know this) - excess KMnO4 (after multiplying by 5mol NO2/2mol KMnO4)

My partner seemed to think that the excess KMnO4 was equal to:

Total oxalate- excess oxalate = corrected oxalate= excess KMnO4 (again, after applying stiochiometric ratios)

Using this, we got a number that was sort of close to the expected value. We got like .47g/L when it was supposed to be like .41g/L. That still seems wrong to me, but it could be right. I just don't get how to come up with the value of excess permanganate.

PLEASE HELP, we are falling behind.

Thanks!

EDIT: I think I may get his reasoning. When the last titration is done, you have all of the oxalate in equivalence with the excess permanganate plus the permanganate you added during the titration. So if you get rid of the excess oxalate you now have corrected oxalate in equivalence with the excess permanganate (excess as in the permanganate that did not react with nitrite). So convert corrected oxalate moles to moles of permanganate and that is the moles of excess permanagate. Then use my nifty formula:

NO2= Original permanganate - excess permanganate

So we did do it correctly?

Eureka! (I hope)

 

[Borek]

Perhaps this will help. At the end of the procedure:

Equivalents of oxalate + equivalents of nitrite = equivalents of permanganate

Or, if you have never dealt with equivalents, bastardized formula:

Electrons given by oxalate + Electrons given by nitrite = Electrons consumed by permanganate