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Topic: Finding the Ka value of oxalic acid  (Read 16179 times)

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Offline LilyFFC

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Re: Finding the Ka value of oxalic acid
« Reply #15 on: February 07, 2011, 03:13:42 PM »
Okay so the Ka2 = [H+][C2O42-] / [HC2O4-]

I know [H+] = 10-8.2

Would [HC2O4-] = [H2C2O4] ? And I'm not sure what [C2O42-] is equal to.

Offline DevaDevil

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Re: Finding the Ka value of oxalic acid
« Reply #16 on: February 07, 2011, 04:05:43 PM »
from the values of your experiment you can calculate how much OH- reacted with protons from the acid.

for each molecule of oxalic acid between 1 and 2 protons have reacted. The first proton will have reacted from each of the oxalic acid molecules, the second from only a portion; you can calculate exactly from how many using the amount of OH- that reacted

so: how many mole of OH- has reacted?
how many moles of oxalic acid has reacted?
so from how many moles of oxalic acid did the second proton react as well?

Offline LilyFFC

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Re: Finding the Ka value of oxalic acid
« Reply #17 on: February 07, 2011, 04:23:35 PM »
I still don't follow - how am I supposed to calculate how much OH- reacted and how much I started with?

Offline DevaDevil

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Re: Finding the Ka value of oxalic acid
« Reply #18 on: February 07, 2011, 04:28:54 PM »
well, you know with how many moles of OH you started, right? (the number of moles of NaOH)

you know the pH at the end, and thus the concentration H+ at the end, right?

Do you remember the relation between [H+] and [OH-]? If you don't, then look it up and remember it!

Offline Borek

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Re: Finding the Ka value of oxalic acid
« Reply #19 on: February 07, 2011, 06:00:07 PM »
How much you had is simple - NaOH is fully dissociated. Just assume it it was a limiting reagent for teh neutralization reaction. Amount that is present in the solution at the equivalence point is so small it doesn't matter. Besides, you can't assume it is from NaOH, some of it comes from water autodisociation and oxalate anion hydrolysis.
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