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Lowering Solubility

  1. Aug 21, 2011 #1
    1. The problem statement, all variables and given/known data
    What concentration of ethanoate (acetate) ions is
    expected in a 0.18 mol/L solution of ethanoic
    (acetic) acid? Ka is 1.76 x 10-5 at 25°C.


    2. Relevant equations

    Ka=products/reactants

    3. The attempt at a solution

    I know that Ka=1.76 x 10-5=products/reactants but I really don't know what to do from there. I have tried the ICE box method; initial, change, equilibrium but that doesn't seem to work. My textbook only covers the ICE box method and the basic formula, help is appreciated!
     
  2. jcsd
  3. Aug 22, 2011 #2

    Borek

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    Staff: Mentor

    Show your take with the ICE method, not only it is perfectly suitable here but it also works exactly as expected.
     
  4. Aug 22, 2011 #3
    hydrogen acetate acetic acid
    initial: 0 0 0.18mol/L
    change: - +x -x
    equilibrium: 0 0+x 0.18-x

    1.76x10^-5=(0.18-x)/x
    x=.1799999 mol/L

    but apparently the answer is 1.8x10^-3 mol/L.

    Am I missing something?
     
    Last edited: Aug 22, 2011
  5. Aug 22, 2011 #4

    Borek

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    Staff: Mentor

    Yes, your ICE table is incomplete (what happens to H+ concentration during acid dissociation?) and your Ka expression is umop apisdn.

    To some extent it doesn't matter, but the convention is to write dissociation reaction starting with undissociated substance as reactant, and to make ICE table for such a reaction. You reversed everything.
     
  6. Aug 22, 2011 #5
    so would it be

    hydrogen acetate acetic acid
    initial: 0 0 0.18mol/L
    change: +x +x -2x
    equilibrium: x x 0.18-x

    1.76x10^-5=(o.18-x)/x^2
    that still gives an incorrect 0.18 or -1.76...

    What do you mean the Ka is upside down? Isn't it products/reactants?
    So how should I write the table using the 'normal' convention?
     
  7. Aug 22, 2011 #6

    Borek

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    Staff: Mentor

    What are products of the dissociation reaction?
     
  8. Aug 22, 2011 #7
    hydrogen and acetate are dissociated from the acetic acid.
     
  9. Aug 22, 2011 #8

    Borek

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    Staff: Mentor

    So if hydronium (not hydrogen!) and acetate are products, why do you put them in denominator?
     
  10. Aug 22, 2011 #9
    lol I guess I should have written the equation first. will update soon.
     
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