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Calculate the pH of 20.00 mL of 0.20 M KHP

  1. Nov 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Calculate the pH of 20.00 mL of 0.20 M KHP.

    Given: pKa1(H2P)= 2.950 pKa2(H2P)= 5.408

    2. Relevant equations

    k = [H][P]/[HP]

    3. The attempt at a solution

    I used this equation and got:

    3.91e-6 = [x][x]/[.2]

    x^2 = 7.82e-7

    x = 8.8e-4 = [H]

    pH = 3.05

    However, my homework is telling me this is incorrect. Is it just a rounding issue or is it something else? Any help is much appreciated.
     
  2. jcsd
  3. Nov 5, 2009 #2

    Borek

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  4. Nov 5, 2009 #3

    symbolipoint

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    Re: titrations

    Either your pK1 value is wrong OR your K1 value is wrong. Fix the wrong one and you may be able to improve your result. I base my comment on the idea that the first dissociation is far more important than the second dissociation in the solution just containing the dissolved KHP. I may be incorrect in this judgement (while still relearning).
     
    Last edited: Nov 5, 2009
  5. Nov 6, 2009 #4

    Borek

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    Re: titrations

    This is Ka2 and pKa2. w330 tries to base calculations only on the second dissociation step, ignoring hydrolysis. As explained on the linked page this is incorrect approach.

    --
    methods
     
  6. Nov 6, 2009 #5

    symbolipoint

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    Re: titrations

    Borek, the discussion in your hyperlink in the post was interesting, and a little clearer than the discussion in the old analyitical textbook I read yesterday. Something almost like it was also shown in an old General Chemistry textbook. I have not seen much of that type of exercise for a long, long time.
     
  7. Nov 6, 2009 #6

    Borek

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    Re: titrations

    In a way I am like an old analytical chemistry textbook :grumpy:
     
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