# Chemistry

posted by on .

Calculate the final concentration for the standardization of KMnO4 given this information:

2MnO4- + 5C2O4^-2 + 16H+ -> 2Mn^2+ + 10CO2 + 8H2O [final redox equation]

Volume of KMnO4 titrated with H2SO4 + Na2C2O4 = 293 mL

Volume KMnO4 titrated with H2SO4 alone = 0.2 mL

Amount of Na2C2O4 = 0.103 g

It says that I'd need that information to calculate the concentration but I'm confused since Na2C2O4 isn't even a part of that redox equation.. Please help!

• Chemistry - ,

Oh but it is. Na2C2O4 is where the 5C2O4^2- comes from.
293 mL-0.2 mL = 292.8 mL KMnO4 to titrate the Na2C2O4.
mols Na2C2O4 = grams/molar mass
Now convert mols Na2C2O4 to mols KMnO4 using the coefficients in the balanced equation.
Then M KMnO4 = mols KMnO4/L KMnO4.
Post your work if you get stuck.

• Chemistry - ,

Hi DrBob

0.103 g Na2C2O4 divided by 134 g/mol Na2C2O4 = 0.000768 mol Na2C2O4 = 0.000768 mol KMnO4

0.000768 mol C2O4^-2 X 2 mol MnO4-/5 mol C2O4^-2 = 0.0003072 mols KMnO4

M KMnO4 = 0.0003072 mol/0.293 L = 0.001048 M..? is that right? close? or totally wrong?

• Chemistry - ,

partially right, partially wrong.
My calculator reads 0.0007688 for mol Na2C2O4 so I would round that to 0.000769.I think you threw the last digit away.
Then you must convert mols Na2C2O4 to mols KMnO4.
0.000769 mols Na2C2O4 x (2 mols KMnO4/5 mols Na2C2O4) = 0.000769 x 2/5 = 0.000307 as you have. Note that your previous statement that 0.000769 mol Na2C2O4 = 0.000768 mol KMnO4 is not right and in fact it doesn't belong anywhere in the solution.
Then M = mols/L = 0.000768/.2928 L = 0.001048 which I would round to 0.00105 M.
So except for the rounding your answer is right although there were a few missteps along the way.

• Chemistry - ,

Ohh yeah it was actually my fault DrBob i wrote 293 mL but it was actually 2.93 mL but i still got it after your explanation. THANK YOU!!!!