[h20h]

Q) Oxalic acid is a diprotic acid (it has 2 hydrogens which can be neutralized).  A 0.0211g. sample of the acid was titrated with 31.2 mL of  .015M NaOH.  Using ONLY THIS GIVEN information, what is the molecular mass of oxalic acid?

A)  I am sort of stuck but not really:  The equation for the reaction is:

H2C204 + 2NaOH= Na2C204+ 2H20

That is all I have at this point?  Is that equation even right?  I would assume if given the mL of the NaOH and the M we could figure out the moles....use the stoichiometry of the reaction being 2 moles of NaOH to 1 mole of acid...divide the moles of NaOH in half to get the moles of the acid...

we would then have moles of the acid and then am stuck?

any pointers???

Thanks

[Yggdrasil]

Everything you have said is completely correct and on the right track.  Now, you have figured out the number of moles of oxalic acid in your 0.0211g sample of oxalic acid.  You just need to do one more (simple!) step to figure out the molar mass (i.e. grams per mole) of oxalic acid.

[h20h]

So I just take the grams given which is .0211g of the acid and divide by the moles of the acid..that will give me grams per mole

so:  31.2mL = .0312L of NaOH

.0312 liters of NaOH (.015M)/1 liter = 4.68x10-4

stoich:  2 moles of NaOH : 1 mole of H2C2O4

4.68x10-4 (1 mole H2C2O4)/2 moles NaOH =  2.34x10-4

molecular mass= grams/mole     .0211 g H2C204/2.34x10-4  =  90.17g/mol H2C204

look ok???

[Borek]

OK, although 90 looks even better (hint: significant digits).

[h20h]

Thanks so much for your *delete me* 

[Yggdrasil]

Note that the actual molecular mass of oxalic acid is 90.03 g/mol which agrees with your answer (since you have two significant digits).