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Calculating the Ka from pH and initial concentration

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data
    A 0.0200 mol/L acid solution has a pH of 2.347. Calculate the percent ionization of this acid. Calculate the Ka of this acid

    3. The attempt at a solution

    I first calculated the hydronium ion concentration: 10^-pH = 10^-2.347 = 4.5*10^-3

    I then used this concentration to calculate the percent ionization: 0.0200 / 4.5*10^-3 = 22.5%

    Can anyone confirm this answer. Now the second part of the question asks for the Ka. My prof said that if you got 1.3*10^-3 as the Ka value it is wrong. I keep getting this value and it makes no sense why it would be anything other than this value:

    Since the ratio in the balanced chemical equation is 1:1 , the equilibrium expression is:
    Ka = (4.5*10^-3)(4.5*10^-3) / (0.0200 - 4.5*10^-3)

    solving for x I get 1.3*10^-3, however this isn't suppose to be the right answer.. can anyone tell me where I've gone wrong or maybe my prof is mistaken? Thanks.
     
  2. jcsd
  3. Nov 3, 2012 #2

    symbolipoint

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    Your Ka value is wrong simply because you miscomputed your numeric expression. The numeric expression seems good; Try recomputing and you will, or should find a much smaller Ka value.
     
  4. Nov 4, 2012 #3
    I don't know I've put it into my calculator numerous times and I get the same value,
     
  5. Nov 4, 2012 #4

    symbolipoint

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    You are incorrectly handling this computation on the calculator. Your expression seems to be written well and appears credible, so as a check, try doing the computation the old fashioned, manual way. You SHOULD find something about ?.?? x 10-6
     
  6. Nov 4, 2012 #5

    Borek

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    OK

    No.

    [tex]\frac {0.0200} {4.5\times 10^{-3}} = 4.44[/tex]

    Expression looks OK to me, but first - there is no x to solve for, second - as symbolipoint stated, it evaluates to something completely different.
     
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