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Ethanoic acid and NaOH titration

  1. Jul 5, 2016 #1

    TT0

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    1. The problem statement, all variables and given/known data
    Screen Shot 2016-07-05 at 5.12.39 PM.png

    2. Relevant equations


    3. The attempt at a solution
    If 50% of the acid has been neutralised, then there is 0.05 M of acetic acid left.Using Ka
    $$ 1.8*10^-5 = \frac {x^2}{0.05-x} $$
    I get x = 0.00094
    -log(0.00094) =3.02687 =pH
    This is not one of the answers provided. I think I interpreted the question wrong. Can someone explain to me what I did wrong?

    Cheers!
     
  2. jcsd
  3. Jul 5, 2016 #2

    Borek

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    Staff: Mentor

    You can't ignore acetate presence.

    Hint: at half titration [CH3COOH] = [CH3COO-]
     
  4. Jul 6, 2016 #3

    TT0

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    I see, then the equilibrium is:

    $$ 1.8*10^{-5} = \frac {(0.05+x)*x} {0.05-x}$$

    Therefore, pH = 4.74

    Answer is B.

    Thanks again!
     
  5. Jul 6, 2016 #4

    Borek

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    Staff: Mentor

    Much easier to solve using Henderson-Hasselbalch equation. Not to mention the fact you should remember at half titration pH=pK.
     
  6. Jul 8, 2016 #5

    TT0

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    I see, thanks a lot!
     
  7. Jul 12, 2016 #6

    epenguin

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    Homework Helper
    Gold Member

    If you had written the equation for the equilibrium constant out explicitly in terms of chemical species you would probably not gone wrong in the way you did.
     
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