Posted by **Melba** on Tuesday, September 26, 2006 at 4:00pm.

What volume of a 0.2 N NaoH solution do you expect to use in order to titrate 25ml (sample out of 250ml) of a 0.200 N oxalic acid solution?

heeeelp!

Equation is

H2C2O4 + 2NaOH==> 2H2O + Na2C2O4

I assume that is 25.00 mL H2C2O4.

milliequivalents(meq) NaOH = meq H2C2O4

25 mL x 0.2 = ??mL x 0.200

Do you suppose 25.00 mL.

Most schools don't teach normality now, but notice how easy it is to use normality. There are no mols and then convert mols H2C2O4 to mols NaOH, then mols/L = M for NaOH. It is all done for us by using normality. Neat!!

wow...thanks DrBob, I'll try it!

The problem will come when the question does NOT provide normality but provides molarity. Then you must convert M to N yourself OR work it the long way using mols. But note that whereas mL x N = mL x N (ALWAYS, ALWAYS, ALWAYS), mL x M /=/ (is not equal) mL x M EXCEPT when you want to work a dilution problem. For a stoichiometry problem, however, mL x M = mL x M is true ONLY when the reacting ratios are 1:1 (or 2:2 or 3:3 etc).

Thanks...I didn't know that

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