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PH of a polyprotic acid using multiple equilibria - is this correct?

  1. Jul 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the pH of 0.02025 M tartaric acid using the method of multiple equilibria given that its Ka1 value is 9.2*10-4 and its Ka2 value is 4.31*10-5.


    2. Relevant equations
    Kaoverall = Ka1 * Ka2


    3. The attempt at a solution
    Please check the Word document.

    Is my method and calculation correct?

    Thank you.
     

    Attached Files:

  2. jcsd
  3. Jul 10, 2014 #2

    Borek

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    Staff: Mentor

    For some reason I can't open the file. .docx is not the best format. Try to post it here using LaTeX, or - at worst - as an image.
     
  4. Jul 10, 2014 #3
    Hm... that's strange.

    I'll upload images files then (LaTeX is too much trouble...)

    They're in order of page number.
     

    Attached Files:

  5. Jul 10, 2014 #4

    Borek

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    Staff: Mentor

    Seems like you have completely ignored presence of HT-.

    pH of the 0.02025 M tartaric acid is 2.41, as long as your calculations don't reproduce this result they are wrong.
     

    Attached Files:

  6. Jul 10, 2014 #5
    The reason I'm asking is because I answered the same question but instead used the step-by-step equilibrium method and I got a different answer.

    I can't see where I went wrong though. Both methods should yield the same answer.

    They're in order of page number.
     

    Attached Files:

  7. Jul 10, 2014 #6
    Hm... but wouldn't it cancel out in the equation?
     
  8. Jul 10, 2014 #7

    Borek

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    Staff: Mentor

    No. Just because it cancels out in a partial system of equations doesn't mean it shouldn't be taken into account. Overall equation doesn't say a word about intermediate steps, which doesn't mean they don't exist.

    Correct derivation is shown on the page I already linked to.
     
  9. Jul 10, 2014 #8
    Ah. You're right. I see where I went wrong.

    It's not working because I'm writing an overall reaction that doesn't exist. And it wouldn't work unless there's a third reaction. Otherwise, as you said, I would miss the intermediate reactions.

    Thank you.
     
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