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Topic: acid / base titration question  (Read 3836 times)

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Offline johnj7

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acid / base titration question
« on: February 13, 2008, 01:36:17 AM »
Hello,
I'm struggling with an weak acid / strong base titration of a diprotic acid.. question
How would you get the pH of the solution after the 2nd equivalence point has been reached?

For example:

titrating 100 mL of 1.0 M oxalic acid with 1.0 M NaOH
Acid dissociation constants of oxalic acid are Ka1= 5.36X10^-2 and  Ka2= 5.3X10^-5
-It would take a total of 200 mL of NaOh to reach the equivalence point
What is the pH after 50 mL of NaOH has been added after the 2nd equivalence point has been reached?

Is the only thing I have to consider the excess diluted NaOH?
ie...
- 0.05 L X 1.0 M =  0.05 moles base
- 0.05 / 0.250 L = 0.2 M base
pOH = -log (0.2) = 0.699
pH = 14 - 0.699 = 13.30

Is this correct?

Thanks in advance!

Offline Borek

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Re: acid / base titration question
« Reply #1 on: February 13, 2008, 03:36:13 AM »
http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-simplified

You may assume only 2nd dissociation counts and the pH is governed by the presence of weak (conjugated) base.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline johnj7

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Re: acid / base titration question
« Reply #2 on: February 13, 2008, 07:33:23 PM »
hmm i think i understand how to calculate pH at the equivalence point,
but what about after the eq. point, just adding excess NaOH?

Offline Borek

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Re: acid / base titration question
« Reply #3 on: February 14, 2008, 03:16:48 AM »
Exactly - just calculate pH of the excess NaOH.
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