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Titrations and starting pH

  1. Jan 11, 2016 #1
    1. The problem statement, all variables and given/known data
    What was the concentration of acid S before the titration?

    2. Relevant equations
    MaVa=MbVb

    3. The attempt at a solution
    So for this problem, the solution manual suggests that we use the formula above to solve for the pH of acid S before the titration. But why do we use the concentration of the base before we even add it? Wouldn't it be easier to locate the pH on the graph (which is approximately 3.8) and take 10^-3.8 = 1.6 x 10^-4?

    The correct answer is B. Thanks in advance.
     

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  3. Jan 11, 2016 #2

    BvU

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    pH doesn't help for the concentration if the acid isn't completely dissociated into H+ and Ac-.
    What concentration would you find for acid S from the pH ?
    What would you find for acid T ? pH and concentration ?
     
  4. Jan 11, 2016 #3

    Borek

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    We don't care about initial pH at all, just about the stoichiometry of the reaction. Do you know how the titration works? Why is part of the curve so steep?
     
  5. Jan 11, 2016 #4
    Yes, I understand how it works. The steepest part of the graph is where the equivalence point i and where the pH rises the fastest. Contrarily, the half equivalence point (the horizontal portion) is where the pH changes the least with increased base. But how are we to determine the stoichiometry of the reaction when we know nothing about the acid?
     
  6. Jan 11, 2016 #5
    That is where I am confused. The directions say that we start with 50 mL of acid S and the titrant is 0.1 M NaOH. With MA as the unknown, I came up with:
    MAVA=MBVB
    (0.050L)(MA) = (0.1)(VB)

    But why incorporate any base when the question asks "what is the concentration of acid S before the titration"?
     
  7. Jan 11, 2016 #6

    BvU

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    Don't want to spoil your exercise. Try to move on to 814 and perhaps you'll see. (and note Borek's post #3)
     
  8. Jan 11, 2016 #7
    I think I see what is going on here. The concentration of S is unknown and we have to use the volume of the NaOH added at the equivalence point to determine that.
    So our MA= what we are solving for
    VA=(50 mL)
    MB=0.1 M
    VB=50 mL)
    MAVA=MBVB
    MA=0.1 M

    The question was a bit misleading but I think I get it now.
     
  9. Jan 11, 2016 #8
    I understand 814. We can find the Ka from the pKa at the half eq. point. So, if pH = 4 at half eq. point, then ka = 10^-pH = 1x10^-4.
     
  10. Jan 11, 2016 #9

    BvU

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    Why at the half-eq point ? They ask for the Ka of the acid ...
     
  11. Jan 11, 2016 #10
    Right. Because that is what is unknown. Haven't done too many titrations before but I think I get it. We can only calculate the unknown starting concentration of S when we reach the equivalence point.
     
  12. Jan 11, 2016 #11
    Right. So I used Henderson Hasselbach equation and converted pKa to Ka. Have to be quick on timed exams. Is there another way to calculate the Ka without using HHB?
     
  13. Jan 11, 2016 #12

    Borek

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    Reading pKa from a half-equivalence seems to me the best approach here.
     
  14. Jan 11, 2016 #13

    BvU

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    OK, sorry. Out of my league. Found HH interesting !
     
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