Sciencemadness Discussion Board

Equivalence points and titration of polyprotic acid

kishka77 - 9-7-2011 at 11:49

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

Magpie - 9-7-2011 at 21:46

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]

DJF90 - 10-7-2011 at 10:35

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.

Magpie - 10-7-2011 at 12:26

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. ;)