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Homework Help: Calclulating pH/pKa's

  1. Jun 12, 2006 #1
    First is this correct?

    The pH of a 150 mL aqueous solution of 2.13 x10^-3 M HCl is
    pH = -Log (2.13 x10^-3)
    pH = 2.672

    and how would you solve this one?

    A .1M aqueous solution of an acid HA has a pH of 4.0 what is the pKa of HA?
  2. jcsd
  3. Jun 12, 2006 #2
    First is, indeed, correct. :approve:

    I cannot help until you express some of your thoughts. Sorry, I must respect the rules. :frown:

    Suggestion: What information gives you the question when she says the pH is 4 to help you find Ka and, then, pKa?

    Check: http://en.wikipedia.org/wiki/PKa
    Last edited by a moderator: Jun 12, 2006
  4. Jun 12, 2006 #3


    User Avatar

    Staff: Mentor

    2.67, as you used more significant digits in your answer than you had in the input data.

    As for the second question - what formula do you use for pH calculation? Can it be rearranged/solved for pKa?

    Check these pH calculation lectures.
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