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Finding pH using ICE chart/quadratic formula on the last step!

  1. Jul 24, 2012 #1
    Calculate the pH of a 0.0253 M aq solution of a monoprotic acid. The Ka for the acid is 1.79 x 10-2

    So a monoprotic acid will have the form HA --> H + A, correct?

    I 0.0253 0 0
    C -x x x
    E (0.0253-x) x x

    EDIT: I just realized the formatting is all messed up on my chart, but it shouldn't be too difficult to discern what I'm doing.

    So we have:

    1.79 x 10-2 = x2/(0.0253 - x)
    0 = -x2 - 1.79x10-2x + 4.53x10-4

    After using the quadratic formula, we find the acceptable value of x (the one that is positive) to be:

    4.10 x 10-2 M

    Now my question is: how do I find the pH? I thought I could use -log10(4.10x10-2), but apparently that's wrong.

    Any help? Thank you!
     
  2. jcsd
  3. Jul 27, 2012 #2

    epenguin

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    Homework Helper
    Gold Member

    Your method and your quadratic equation I think are OK so there must be some mistake somewhere in arithmetic.

    I don't know how you know it is wrong. But I think a way to know without anyone telling you (in other words better than 'apparently', rather the sort of checkup thinking you should habitually do) is that your [H+] result is higher than the total concentration of acid that is there, which is impossible.

    I went though a calc. and if you get pH 1.85 same as me it is probably not coincidence, if you get different check a bit but I also am error-prone.
     
    Last edited: Jul 27, 2012
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