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Triprotic Acid Titration Exact Calculation

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data Calculate the analytical concentrations C2[KOH] and C2[H3PO4] in a solution where H3PO4 (originally of concentration 0.2 moldm-3 and volume 0.250 dm3) is titrated with KOH (originally of concentration 0.8 moldm-3, unknown volume used) until the pH reaches 7.8.

    2. Relevant equations

    Mass balance: C2[H3PO4]=[H3PO4]+[H2PO4-]+[HPO42-]+[PO43-]
    C2[KOH]=[K+]
    Charge balance: [H+]+[K+]=[H2PO4-]+2*[HPO42-]+3*[PO43-]+[OH-]
    Equilibria:
    Kw=[H+]*[OH-]
    Ka1=[H2PO4-]*[H+]/[H3PO4]
    Ka2=[HPO42-]*[H+]/[H2PO4-]
    Ka3=[PO43-]*[H+]/[HPO42-]

    3. The attempt at a solution

    The unknowns are: [OH-], [PO43-], [HPO42-], [H2PO4-], [H3PO4] and [K+]. These number 6; we have 6 equations. So a solution exists. [OH-]=Kw/[H+], but what about the rest?

    My problem is that C2[H3PO4] and C2[KOH] are unknown of themselves. They represent the analytical concentrations of H3PO4 and KOH in the final solution (of pH=7.90) after they have been added together. These are unfortunately the two values we need to work out in the end.
     
  2. jcsd
  3. Mar 25, 2013 #2
    Assume V as the final volume of the solution. Then C2[H3PO4] would be equal to n/V where n is the initial number of moles of H3PO4.

    Now write the anion conc. in terms of the dissociation constants and [H3PO4] plug these in the mass equation to get [H3PO4] in terms of V.

    Now in the charge balance equation, [K+] is equal to 0.8*(V-0.25)/V, the anion concentrations are also in terms of V and [H+] is known. Thus V can be found and hence the analytical concentrations.
     
    Last edited: Mar 25, 2013
  4. Mar 25, 2013 #3
    Can you demonstrate how to write the anion concentrations in terms of dissociation constants? It's this step I always have trouble with for this type of calculation (i.e. the tough step is combining the mass balance and charge balance equations; I can do it sometimes, but not usually for triprotic/diprotic acids).
     
  5. Mar 25, 2013 #4

    epenguin

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    Amounts of [H+] and [OH-] are utterly negligible compared to other things in these conditions so just leave them out of eq. 2. [K+] is the same thing as [KOH]. OK after you have done your main calculation with molarities you finally express in terms of a volume.

    Getting the requested equation is just a matter of applying the equilibrium equations you have written. You have to write your [H2PO4-] in terms [H+] and Ka1, you practically have the expression for that bit, then you have to do it successively for the others and include them - you will get an [H+]2 and [H+]3 term.
     
    Last edited: Mar 26, 2013
  6. Mar 29, 2013 #5
    Thanks, but in trying to get my [H2PO4-] in terms of [H+] and Ka1 how do I get make [H3PO4] cancel down?
     
  7. Mar 30, 2013 #6
    Check out post #2.
     
  8. Mar 30, 2013 #7

    epenguin

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    Don't.
    Don't worry for the moment and if we are taking it step by step don't ask me what to do after the step you haven't shown.
     
  9. Mar 30, 2013 #8
    According to your post this is the first step? You wrote: "You have to write your [H2PO4-] in terms [H+] and Ka1, you practically have the expression for that bit". In any case the problem still applies if I am to find [H3PO4] first: how do I get rid of [H2PO4-] from the expression?

    Perhaps if you could demonstrate for one of them, I will carry it on successively for the others.
     
  10. Mar 30, 2013 #9

    epenguin

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    OK what I'm trying to get you to do is write all the species' concentrations in terms of [H3PO4] and at the end you will apply your mass balance equation. So just use your Ka1 equation to write [H2PO4-] =
    Very simple.
     
  11. Mar 30, 2013 #10
    OK, so by substituting this successively we reach: [PO43-]=[H3PO4]*Ka1*Ka2*Ka3/([H+]3). Now what?
     
  12. Mar 30, 2013 #11

    epenguin

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    OK you've got the last one - have you got the first one that I asked? And the second one?
    (for [HPO42-] = ?)

    Write them down too please.
     
  13. Mar 31, 2013 #12
    [H2PO4-]=[H3PO4]*Ka1/[H+]

    [HPO42-]=[H3PO4]*Ka1*Ka2/[H+]2

    [PO43-]=[H3PO4]*Ka1*Ka2*Ka3/([H+]3).

    I'm beginning to see where this is going. Next we're going to substitute in for each term in the mass balance?
     
  14. Mar 31, 2013 #13

    epenguin

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    Yes. But give us the result. :smile:
     
  15. Mar 31, 2013 #14
    C2[H3PO4]*[H+]3=[H3PO4]*([H+]3+Ka1*[H+]2+Ka1*Ka2*[H+]+Ka1*Ka2*Ka3)

    And now what? We have neither C2[H3PO4] and [H3PO4].
     
  16. Mar 31, 2013 #15

    epenguin

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    You do have C2[H3PO4], it's in the second line of the thread.

    [H3PO4] is what you are trying to find the formula for. You have everything else and could now write

    [H3PO4] = ...

    Maybe you don't recognise what an answer would be, and in general it would be

    [H3PO4] = C2[H3PO4] X (function of the K's and [H+])

    and that is the important one. In this case C2[H3PO4] can be given a number that they want you to work out.
     
    Last edited: Mar 31, 2013
  17. Apr 1, 2013 #16
    No, we have C1[H3PO4] (the original analytical concentration). C2[H3PO4] is the analytical concentration of H3PO4 once H3PO4 was combined with a certain unknown volume of NaOH to produce the solution of pH = 7.8. How do we find either C2[H3PO4] or [H3PO4]?
     
  18. Apr 1, 2013 #17

    epenguin

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    Now you tell me! I'm sorry, this is the first time this C1 has been mentioned and you have asked a question without ever defining what the terms you ask about meant (!) so I just went on assumption and how suchlike questions are treated always.

    Never mind, it is always much the same thing and you can expect to meet similar questions again and again and they are always dealt with by the same routine. And you do have the information for the calc.

    So I'll just write the various concentrations in rational form - molarities - what need is there for C1, C2 etc? There is a total of all forms of the phosphate which you are given the information to calculate, this will usually be known or knowable, and even when it isn't you still formulate the same equations in terms of it algebraically. So here I'll call it [P]tot, actually it does equal the total molarity of phosphorus in the solution so the tot is just added mind label.

    So I'll rewrite the last line of my last post as

    [H3PO4] = [P]tot X (function of the K's and [H+])

    You can start by writing [P]tot = function given by the mass balance equation with your equilibrium equations enabling you to write in terms on one concentration.
    By which you can calculate [H3PO4] if you know (as you do) [P]tot, [H+] and the K's.

    Likewise you can get an equation for and calculate the concentrations of all the four forms of phosphate at any given pH.
     
    Last edited: Apr 2, 2013
  19. Apr 1, 2013 #18

    epenguin

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    I have to answer this in more than one post because of connection and seeing screen probs.

    Things you should realise:

    The LHS of your mass balance equation should be, in my terms [P]tot, which is what I assumed you meant.

    Before you do the calculation of [H3PO4] I hope you'll realise that it's going to be quite tiny - of the order of less than 10-5 of the main components at pH 7.8. [PO43-] will be likewise negligible.

    To solve the rest of your problem you are going to have to calculate [H2PO4-] and [HPO42-] preferably as fractions of [P]tot, similar kind of calculation to that for [H3PO4].

    The other part of your problem was to calculate [K+] . How are you going to do that? Well think, as well as mass balance what other kind of balance is there here? :wink In fact you already have an equation for [K+].

    I remind you we said [H+] and [OH-] are negligible. I think what throws students is the fact that these seem all important, the questions are about them, and yet at the same time they are negligible though some of the equations involve them vitally, in others they are ignored. Think of them as reporters on the state of the system and on the balance between the big players there in mass.

    In fact there are only two forms of phosphate that matter in this question H2PO4- and HPO42-

    Therefore actually you can solve it using pretty much the theory you will have already done with monobasic acids. And in any case even the full theory for tri- or multi- protic acids is scarcely different from that you have already done for mono-, the only difference is that there are more terms. We often like to write the more general equations first, and then home down on any particular simplifications for particular cases. In the case that the pK's were closer together we'd have to keep the more general ones.

    And I have to say you are taking days to grind out equations that people write in fewer minutes. Conservation equations and equilibrium equations + any simplifications, that's it. In chemistry and biochemistry you'll meet any number of equations like that for multiple equilibria, you'll see them just written down with minimal explanation, maybe definition of the eq. constants by equations, maybe a conservation equation or two, maybe not because considered obvious. You should see what all the terms and the pattern are saying.

    I nay not have time next days to continue this, anyone is welcome to take over if nec.

    Edit: other obscure terminology - I don't know what is meant by 'analytical' :confused: concentrations, unnecessary adjective?
     
    Last edited: Apr 2, 2013
  20. Apr 3, 2013 #19

    epenguin

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    In my (rational and standard :tongue2:) terminology your eq. of #14

    [P]tot[H+]3=[H3PO4]*([H+]3+Ka1*[H+]2+Ka1*Ka2*[H+]+Ka1*Ka2*Ka3)

    appears correct.

    You just need to express that as
    [H3PO4] = ...

    and answer the rest of the question.
     
    Last edited: Apr 3, 2013
  21. Apr 5, 2013 #20
    I see now, thank you. :)

    Since this method involves fairly long computation I was wondering if we could now work on an approximate shortcut. When I asked this elsewhere (not specifying the need for an exact solution) I was told that we need only consider the two species of the acid involved in a pKa value which is within 3 units either way of the pH (assuming no other pKa value is also within 3 units of the pH, which is true for phosphoric acid - pKa2 is the only pKa within 3 units of the pH). So we need only consider [HPO42-] and [H2PO4-] - makes sense so far. I was then told the first step is to calculate the [HPO42-]/[H2PO4-] ratio. This is fairly easy: [HPO42-]/[H2PO4-]=Ka2/[H+]. But do you have any idea how we might proceed from here?

    Perhaps I should clarify what we're looking for. C1 of any acid or base is the total initial concentration of that acid or base in the solution it was originally in (e.g. we're given 0.80 M for the base). C2 of any acid or base is the "initial" (and by this I mean the "total" - this is the collected concentration of all the forms of the acid in the solution) concentration of that acid or base, in the new solution into which they have been placed. With that in mind, the values we now seek are C2 for both the acid and base.
     
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