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# Molarity of Oxalic Acid?

• watai0102
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4 years 3 months ago #830 by watai0102
watai0102 created the topic: Molarity of Oxalic Acid?
This is a question I have on my titration lab between NaOH and Oxalic Acid.
I am supposed to find the molarity of Oxalic acid (H2C2O4) based on what I know about the base NaOH, but I don't think I have enough info. Here is what I have...

Grams of Acid: .98 g
Molarity of NaOH: .6417 M
Initial Volume of NaOH: .35 mL
Final Volume of NaOH: 28.3 mL

My chem teacher specifically told us the volume wasn't important but all the sites I check say it is.

I can find the # of moles easily but the volume of acid is what gets me... any thoughts?

• watai0102
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4 years 3 months ago #838 by watai0102
watai0102 replied the topic: Molarity of Oxalic Acid?
(28.3 mL - .35 mL) = 27.95 mL = 0.02795 L NaOH used

(0.02795 L) x (.6417 mol/L) = 0.0179 mol NaOH used

Since oxalic acid is a dicarboxylic acid, 1 mole of acid reacts with 2 moles of NaOH. So by actual experiment 0.0179 x (1/2) = 0.00895 mol of acid was present.

(.98 g) / (90.0352 g H2C2O4/mol) = 0.01088 mol H2C2O4 is what was stared with, which compares roughly with what was determined by titration.

If you really need the molarity of the oxalic acid you have to have the volume of the acid solution that contains the 0.00895 mol that you measured.

The data that you have suggest that the point of this experiment was not to measure the molarity of the oxalic acid, but to determine the molarity of the NaOH. In that case, the volume of the acid wouldn't matter. Supposing this is the case:

(.98 g H2C2O4) / (90.0352 g H2C2O4/mol) x (2/1) / (0.02795 L NaOH) =
0.779 M NaOH