# Equivalence point

Discussion in 'MCAT Study Question Q&A' started by chiddler, Apr 22, 2012.

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1. ### chiddler 5+ Year Member

Apr 6, 2010
I know equivalence point is pKa1 + pKa2 / 2

I don't remember why this is though. Can I please have a proof?

thanks.

3. ### pfaction 7+ Year Member

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Mar 14, 2010
WV
MDApps:
Do you mean pI?

4. ### gettheleadoutMS-2Moderator Emeritus 5+ Year Member

11,821
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Jun 22, 2010
I have never seen that equation before.

5. ### pfaction 7+ Year Member

2,195
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Mar 14, 2010
WV
MDApps:
Amino acid pI. Zwitterionic point.

6. ### chiddler 5+ Year Member

Apr 6, 2010
yes it does mean pI. but i meant equivalence point! is that incorrect?

i'm trying to translate this concept to, say, carbonic acid a diprotic buffer.

what does pKa1 + pKa2 / 2 tell us about carbonic acid? to help visualize, pKa 1 is 6.3, pKa2 is 10.3. both / 2 = 8.3

7. ### pfaction 7+ Year Member

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Mar 14, 2010
WV
MDApps:
At 8.3, the molecule will be 100% HCO3-.
<6.3: majorly or all H2CO3.
Above 10.3: Majorly or all CO32-

8. ### chiddler 5+ Year Member

Apr 6, 2010
there we go. it is the first equivalence point.

thanks.

wait, i still need proof lol

9. ### typicalindian 5+ Year Member

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Dec 25, 2010
better go make a trip to your nearest lab then

10. ### pfaction 7+ Year Member

2,195
35
Mar 14, 2010
WV
MDApps:
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].

11. ### chiddler 5+ Year Member

Apr 6, 2010
when it's 50/50, then it's pH = pKa1 which is 6.3. we're looking at 8.3.

equivalence point is when you add one equivalent of base to the acid. since carbonic acid is weak, it becomes slightly basic which is observed with 8.3.

12. ### chiddler 5+ Year Member

Apr 6, 2010
oh man. i have to break in again?!

13. ### pfaction 7+ Year Member

2,195
35
Mar 14, 2010
WV
MDApps:
You know I think you're right, I may have been translating a monoprotic acid into diprotic acid.

14. ### typicalindian 5+ Year Member

2,345
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Dec 25, 2010
Rookie mistake. My PI likes me enough to give me the keys to our lab and 24/7 access ID badge

15. ### chiddler 5+ Year Member

Apr 6, 2010
i am so jealous! you get to go to lab at NIGHT?

all the fun i'm missing out on! *swoon*

2,345
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Dec 25, 2010

Apr 6, 2010
lol

18. ### rjosh33 5+ Year Member

240
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Jul 28, 2011
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??

19. ### SaintJude

Jan 4, 2012
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?

20. ### chiddler 5+ Year Member

Apr 6, 2010
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.

21. ### gettheleadoutMS-2Moderator Emeritus 5+ Year Member

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Jun 22, 2010
Ah, well I know absolutely nothing about that. Glad I wasn't just forgetting something...

22. ### rjosh33 5+ Year Member

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Jul 28, 2011
Nm

Last edited: Apr 22, 2012
23. ### rjosh33 5+ Year Member

240
4
Jul 28, 2011
Ok, I just went back and looked it up, and you're right. Turns out the equation I wrote earlier can be further simplified to what you said, pH = pKa1 + pKa2 / 2. Sorry about the confusion.

24. ### chiddler 5+ Year Member

Apr 6, 2010
that equation is complicated enough that it won't really help, anyway.

is ok. thanks for the responses. discussion about the equation will have helped memorizing it so i'm satisfied.