I know equivalence point is pKa1 + pKa2 / 2
I don't remember why this is though. Can I please have a proof?
thanks.
I don't remember why this is though. Can I please have a proof?
thanks.
I have never seen that equation before.
Do you mean pI?
At 8.3, the molecule will be 100% HCO3-.
<6.3: majorly or all H2CO3.
Above 10.3: Majorly or all CO32-
there we go. it is the first equivalence point.
thanks.
wait, i still need proof lol
No, it's not the first equivalence point, that's H2CO3->HCO3- at 50/50. pH = pKa1 at that point. Equivalence point is when [H]=[OH].
better go make a trip to your nearest lab then![]()
oh man. i have to break in again?!
![]()
Rookie mistake. My PI likes me enough to give me the keys to our lab and 24/7 access ID badge![]()
i am so jealous! you get to go to lab at NIGHT?
all the fun i'm missing out on! *swoon*
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??
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?
Ah, well I know absolutely nothing about that.Amino acid pI. Zwitterionic point.
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.
for acidic amino acids. for basics, it is pKa2 + pKa3 / 2.