Equivalence point

Discussion in 'MCAT Study Question Q&A' started by chiddler, 04.22.12.

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  1. chiddler

    chiddler 5+ Year Member

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    SDN Members don't see this ad. About the ads.
    I know equivalence point is pKa1 + pKa2 / 2

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

    thanks.
     
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  3. pfaction

    pfaction 5+ Year Member

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    Do you mean pI?
     
  4. gettheleadout

    gettheleadout MS-1 Moderator Emeritus 5+ Year Member

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    I have never seen that equation before.
     
  5. pfaction

    pfaction 5+ Year Member

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    Amino acid pI. Zwitterionic point.
     
  6. chiddler

    chiddler 5+ Year Member

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    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

    pfaction 5+ Year Member

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    At 8.3, the molecule will be 100% HCO3-.
    <6.3: majorly or all H2CO3.
    Above 10.3: Majorly or all CO32-
     
  8. chiddler

    chiddler 5+ Year Member

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    there we go. it is the first equivalence point.

    thanks.

    wait, i still need proof lol
     
  9. typicalindian

    typicalindian 5+ Year Member

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    better go make a trip to your nearest lab then :rolleyes:
     
  10. pfaction

    pfaction 5+ Year Member

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    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

    chiddler 5+ Year Member

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    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

    chiddler 5+ Year Member

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    oh man. i have to break in again?!

    :laugh:
     
  13. pfaction

    pfaction 5+ Year Member

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    You know I think you're right, I may have been translating a monoprotic acid into diprotic acid.
     
  14. typicalindian

    typicalindian 5+ Year Member

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    Rookie mistake. My PI likes me enough to give me the keys to our lab and 24/7 access ID badge :)
     
  15. chiddler

    chiddler 5+ Year Member

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    i am so jealous! you get to go to lab at NIGHT?

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

    typicalindian 5+ Year Member

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    [​IMG]
     
  17. chiddler

    chiddler 5+ Year Member

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  18. rjosh33

    rjosh33 2+ Year Member

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    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

    SaintJude

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    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

    chiddler 5+ Year Member

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    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. gettheleadout

    gettheleadout MS-1 Moderator Emeritus 5+ Year Member

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    Ah, well I know absolutely nothing about that. :laugh: Glad I wasn't just forgetting something...
     
  22. rjosh33

    rjosh33 2+ Year Member

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    Nm
     
    Last edited: 04.22.12
  23. rjosh33

    rjosh33 2+ Year Member

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    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

    chiddler 5+ Year Member

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    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.
     

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