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Buffer question

  1. Jul 14, 2013 #1
    1. The problem statement, all variables and given/known data
    We have the buffer Tricine 0.1M and pKa=8.15 and the pH=8.8


    2. Relevant equations

    what are the molar conc. of tricine+ (the protonated form of tricine) and tricine0 (the deprotonated form of tricine)?

    3. The attempt at a solution
    Well, if you go to henderson-hassalbach equation you get:
    8.8=8.15+log X/0.1-X
    where X is the molar conc. of both H3O+ and tricine0 at equillibrium, further solving gives X=0.082M
    But I don't understand - if the pH is given to me can't I find X, which is also the H3O+ conc. at equillibrium, from 8.8=-log[H3O+]? If I do that, I'll get a very different answer to X... why is that?
     
  2. jcsd
  3. Jul 14, 2013 #2

    epenguin

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    Not all that clearly stated - helps to state what X is and use correct brackets in your (X/(0.1-X)) - and if you wrote stuff more explicitly you would probably get it. Your answer is correct for [tricine0] which I guess you mean by X. Your answer fits the equation for equilibrium between the participants - check it. Then hopefully you will see there was no reason to expect that [tricine0] equals [H3O+]. (In fact they are many orders of magnitude different!)
     
  4. Jul 14, 2013 #3
    What do you mean by "there was no reason to expact that [tricine0] = [H3O+]??
    I'm attaching the solution from the presentation as it was given to us in class. As you can see - X stands for [H+] and [tricine0] at equillibrium, while 0.1-X stands for [tricine+] at equillibrium.
    According to Henderson-Hasselbach: X=0.082M, so how come [tricine0] isn't equal to [H+]??

    And, if you go by my way:

    Ka=10^(-8.15)=7.08*10^-9.

    Ka=[H+]*[tricine0]/[tricine+] --> 7.08*10^-9 = X^2/(0.1-X)---> X=2.66*10^-5

    How does this add up?
     

    Attached Files:

  5. Jul 14, 2013 #4

    epenguin

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    That is OK. In fact I always write that out as it saves me having to remember whether there is + or a - in the HH equation or which way up the fraction should be.

    In the buffering range which most questions will be about, the ratio of protonated/unprotonated forms of the buffer will be in the ballpark of 1, rarely much outside the range 0.1 to 10. So the [H +] will be in the ballpark of Ka.

    As well as ballpark figures which it is always helpful and error-signalling to have in mind, put your exact figures for X which you have managed to calculate and [H +] = 10-8.8 and Ka = 10-8.15 and it is all OK.

    There is just no reason for the x (X?) in the second line of your insert, which is otherwise correct, and no reason for an X^2 term.
     
  6. Jul 14, 2013 #5
    Why not? If x molecules of tricine+ were deprotonated until equillibrium, then x molecules of tricine0 were formed and also x molecules of H3O+ were formed also during this process (as indicated in the attachment I uploaded, which was done by the teacher). How is this different from "normal" weak acid questions?
     
  7. Jul 14, 2013 #6

    epenguin

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    You will deprotonate your tricine as you add e.g. NaOH to the solution but your protons are transferred not to H2O but to OH-.
     
  8. Jul 14, 2013 #7
    So why is it correct to represent the equillibrium conc. like they did in the attachment?
     
  9. Jul 14, 2013 #8

    epenguin

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    I don't know, typo or something?

    Another thought, you talk about deprotonating. Just in case useful, it helps to be a bit material. Deprotonating what? If you're achieving the pH by adding base and deprotonating you'd be starting presumably from the salt tricene.HCl. If you start with tricene base you'd achieve the pH by protonating it adding acid.
     
    Last edited: Jul 14, 2013
  10. Jul 15, 2013 #9
    I don't think it's a typo...
    Can you explain me what happens in this exercise step by step?

    1. We started with 0.1M of the tricine buffer... well what does this mean exactly? why is it correct to conclude from this sentence that tricine+ is 0.1M at first? and what "at first" really means? Did at first we had no tricine0? If so - how can we say that 0.1M of the tricine buffer was present? Don't we need both tricine+ and tricine0 to be present at almost equal conc.?

    2. What does the reaction equation as it's written in the attachment represent? From how it's written down I conclude: We starte with 0.1M of tricine+, xM deprotonated, so xM of H+ and xM of tricine0 were formed - why can I even put 0.1-x and x into the Hendersen-Hasselbacg equation?

    3. Why isn't the given pH of 8.8 represent this xM of H+ that are written down in the reaction equation like it did in other non-buffer problems?

    Thanks a lot!
     
  11. Jul 15, 2013 #10

    epenguin

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    1. When we say e.g. 0.1 M buffer, we mean 0.1 M in the total of all its ionic forms, usually only two. This is what we'd need to calculate how much to weigh or measure out of the tricene which we then dissolve, then add acid to bring to required pH then water to make to volume.Or else of tricene.HCl to which we then add alkali etc., or occasionally if we have both tricene and tricene.HCl mix in calculated amounts and make to volumes.

    When tricine+ and tricine0 are in equal concentrations you have pH = pK it is useful to realise. Otherwise, as I mentioned, in buffering problems they will usually be in the same ballpark, and pH within about 1 unit of pK.

    2. Don't understand your second sentence - you did just that.

    3. If as well as your calculation you also did the check I suggested and everything fits, that should give you confidence in it. Textbooks and handouts do sometimes contain typos and even bloopers. Understanding their main message and argument will see you through them - you do not get stuck reading a novel because of a misprint. True the student will make more mistakes than the book so you will want to be careful or ask. If you insist that [H+] has to be equal to [tricene0] and the pH is around 9 and work your equilibrium out you will run into an absurdity somewhere.
     
    Last edited: Jul 15, 2013
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