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Calculating pH from pKa

Discussion in 'MCAT Study Question Q&A' started by nfg05, Jun 18, 2008.

  1. nfg05

    nfg05 2+ Year Member

    May 19, 2008
    So I got the following question correct, but my method is quite different than the Kaplan explanation, which frankly I don't really follow. Here's the question and how I solved it, would really appreciate if someone could confirm my method is valid so I know I didn't just get it right by blind luck. I'd also like to hear easier/other ways of thinking about the question since these types always give me trouble.

    If the pKa1 value for H2CO3 is 6.4, what is the pH of a 10^-3 M solution of this acid?

    My solution:
    pKa = -log(Ka) = 6.4
    log(Ka) = -6.4
    10^-6.4 = Ka = [H+][HCO3-]/[H2CO3]

    Now the H2CO3, being a weak acid, will dissociate to give some concentration x of H+ ions and that same concentration x of HCO3- ions

    Ka = (x)(x)/(10^-3 - x) = 10^-6.4

    Ignoring the x in the denominator since it is small relative to 10^-3, we have:

    x^2 = (10^-3)(10^-6.4) = 10^-9.4.
    x = (10^-9.4)^(1/2) = 10^-4.7

    Since x = [H+], pH = -log(x) = -log(10^-4.7) = 4.7

    And pH = 4.7 is the correct answer given by Kaplan.

    Kaplan starts off by saying " since the negative log of the dissociation constant is 6.4, the antilog is 4 x 10^-7 " (How the F do you get that exactly w/o a calculator?). But as per above, I got Ka = 10^-6.4 and got the correct answer. WTF...

    I can post the rest of the long, rather confusing Kaplan explanation if people want to hear it, but I'd rather hear a straightforward approach other people have used successfully for these kinds of problems.
    Last edited: Jun 18, 2008
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  3. cwfergus

    cwfergus 7+ Year Member

    May 12, 2008
    Ive always done it your way, and will always do it that way haha
  4. minhaj

    minhaj Awesomeness Incarnate 7+ Year Member

    Dec 3, 2007
    what you did is in fact what kaplan but with a calculator.10^-6.4 = 3.98 x 10^-7. so your method is fine.
  5. 161927

    161927 Guest

    Aug 5, 2007
    your method seems faster than using an ICE (Initial Change Equilibrium) table

    Does your method work for situations where there is not 1 : 1 stoichiometry?
  6. nfg05

    nfg05 2+ Year Member

    May 19, 2008
    Thanks for clearing that up, I try to work all problems w/o a calculator and I'm continually frustrated by some of the Kaplan questions/explanations which expect you to use one.

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