kishka77
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Equivalence points and titration of polyprotic acid
Would anyone know the answer to this question:
What is the relationship of the successive equivalence-point volumes in the titration of a polyprotic acid?
Thanks!
Cheers,
K
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Magpie
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This has to be found by experiment for each polyprotic acid. The equilibrium constants Ka1, Ka2, etc represent the results. Eg, for an acid H3A
[H3A] <----> [H+][H2A-]
Ka1 = [H+][H2A-]/[H3A]
etc.
Also, for HnA <----> nH+ + A-
K = ([H+]^n)[A-]/[HnA]
then K = (Ka1)(Ka2)...(Kan)
maybe that is what you are looking for. Your textbook should have all this with better notation than I can give.
[Edited on 10-7-2011 by Magpie]
[Edited on 10-7-2011 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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DJF90
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The volumes are equal. If you need x moles for the first neutralisation, then you'll also need x moles for the second neutralisation. (i.e. to reach
the equivalence points). Thus as volume = moles/concentration, and you're using the same standard solution for the titration, it follows that the
volume added between successive equivalence points will always be the same.
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Magpie
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Quote: Originally posted by DJF90  | | ... it follows that the volume added between successive equivalence points will always be the same. |
Duh! That does make perfect sense. 
The single most important condition for a successful synthesis is good mixing - Nicodem
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