Need help with an Acid-Base problem. Please

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Offline CowTipper

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Need help with an Acid-Base problem. Please
« on: 03/12/2008 03:07:43 »
I need help solving this problem:

A solution containing .123 mol Sodium oxalate (NaC2O4) and .017 mol sodium hydrogen oxalate (NaHC2O4) was treated with 23.5 mL of a .766 M HCL solution and diluted to 1.00 L. The solution pOH was found to be 9.31. Determine Ka2 for oxalic acid and Kb1 for the oxalate ion.

I have no idea where to begin.

Do I set up like this?

NaC2O4 + NaHC2O4 + HCL ---> C2O4 + H2C2O4

Or this?

NaC2O4 + NaHC2O4 + HCL + H2O  ---> C2O4 + H2C2O4 + H3O + OH

What would I do next? Is there an easier way?

Any guidance would be really appreciated.
Thank you.

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Offline Chemistry4me

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Need help with an Acid-Base problem. Please
« Reply #1 on: 03/12/2008 08:02:55 »
The trouble with this question is not knowing how the HCl reacts. I'll have a crack though... it's been a while since I 've done these kind of questions. What have you got so far?

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Offline Chemistry4me

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Need help with an Acid-Base problem. Please
« Reply #2 on: 03/12/2008 08:40:53 »
I am assuming that the reactions with water goes like:
For oxalic acid: H2C2O4 + H2O [revarrow] H3O+ + HC2O4-

so then for the Ka of oxalic acid you'll get Ka=[H3O+][HC2O4-]/[H2C2O4]

For the oxalate ion: C2O42- + H2O [revarrow] HC2O4- + OH-

so then the Kb of oxalate ion you'll get: Kb=[HC2O4-][OH-  ]/[C2O42-]

Does the HCl only react with the oxalate ions? Assuming that it does
2H+ + C2O42-→ H2C2O4
There are 0.018001 moles of H+ so n/c(H2C2O4) and n/c(C2O42-) will be 0.0090005 moles/L-1. You know the concentration of hydronium and hydroxide ions. Substitute all these values back into the Ka/Kb. Hang on I knew I forgot something, hmmm... now I'm not quite sure what to do with the other two numbers (0.123 and 0.017). Anyway, I don't even know if my method is correct  [::)] [::)], but its just something else for you to think about.



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Offline Bored chemist

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Need help with an Acid-Base problem. Please
« Reply #3 on: 03/12/2008 19:36:02 »
At the end the pH is still quite alkaline so there's clearly no leftover HCl or oxalic acid.
So the HCl must have converted oxalate into hydrogen oxalate.
You need to calculate how many moles of oxalate; hydrogen oxalate and HCl you started with. Then work out how many moles of oxalate were turned into hydrogen oxalate to give a new pair of concentrations.
Then you need to look at this equation.
http://en.wikipedia.org/wiki/Henderson-Hasselbalch_equation
to see how to calculate a pKa
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