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04-22-2012, 10:52 PM   #19
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Quote:
 Originally Posted by rjosh33 The pH of the equivalence point between the pKa1 and pKa2 of a diprotic acid is actually computed from a pretty complicated formula: [H+] = [(Ka1Ka2F x Ka1Kw) / Ka1 + F]^1/2 You need to know the starting concentration of the acid as well as the base (in order to get the formal concentration, F) along with the Ka's (both 1 and 2) and the water dissociation constant. Was a problem asking you for the pH at the equivalence point of a diprotic acid??
no i was just trying to find a better way of remembering the equation. i thought understanding its derivation would help.

if it is indeed this complicated, then thanks and nevermind. i must have remembered incorrectly that i had known the proof before.

Quote:
 Originally Posted by SaintJude Ok, so, So pH of the 1st equivalence point = pKa1 + pKa2 / 2 . And this equals the pI for an amino acid that has a neutral R group, yeah?
for acidic amino acids. for basics, it is pKa2 + pKa3 / 2.