averaging pKa's

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plzNOCarribbean

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So, if you have a unknown diprotic acid that your are titrating with strong base, I understand that you could find what that unknown acid is by looking at the titration curve plotted, finding the half equivalence point, and seeing that the pH=pKa at the half equivalence point.

but what about the equivalence point. Some people have pointed out that the pH of the solution at the first equivalence point would = (pKa1 + pKa2)/2 , sort of like how we find the isoelectric point for an amino acid.

But what about the second equivalence point pH? or in the case of phosphoric acid, the third equivalence point? I think what I am trying to ask is how to calculate the pH of the LAST equivalence point, given that we are given the pKa values for each proton of the acid.


I know TBR mentioned a useful formula for equivalence points, which was

pH equivlance = (pKa acid + PH titrant) / 2 , so could you use the same formula here?
 
Say you started with 1M H3PO4. At the last equivalence point (that is, after adding 3 equivalents of OH-), you will have one equivalent of PO4 3-. You can then treat this as a base and use the Kb value to calculate the pH at this point.

Don't forget to account for the dilution effect. If you started with 1M H3PO4, you need to convert to moles, then divide by the final volume to calculate the concentration of PO4 3- (which should be less than 1M).
 
Say you started with 1M H3PO4. At the last equivalence point (that is, after adding 3 equivalents of OH-), you will have one equivalent of PO4 3-. You can then treat this as a base and use the Kb value to calculate the pH at this point.

Don't forget to account for the dilution effect. If you started with 1M H3PO4, you need to convert to moles, then divide by the final volume to calculate the concentration of PO4 3- (which should be less than 1M).

I see what your saying. Is that the only way to solve it? it just seems way too calculation heavy. Is there any other way to do it, maybe a shortcut formula or another way graphically?
 
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