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Finding pH and pKb

  1. Apr 19, 2007 #1
    HELP! I just need the steps to figured out these 3 problem. I have the answes but I can't seem to figure out how to do it and my notes aren't helping. thanks!

    Determine the pH of a solution obtained by titrating 25.00 mL of a 0.107 M solution of a weak base using 75.25 mL of 0.100 M HCl( aq) as a titrant. (hint: this involves a dilution problem.)

    answer is 1.32

    65.5 mL of a 0.0670 M HCl( aq) solution were required to titrate a 25.00 mL sample of a weak base to its equivalence point. The pH at the equivalence point was 4.668. What is the pK b of the weak base? (assume pK w = 14.00) (Remember, pH at the equivalence point is based on total volume.)

    answer is 5.98

    Determine the pH of a solution obtained by titrating 25.00 mL of a 0.121 M solution of a weak monoprotic acid using 60.25 mL of 0.125 M NaOH( aq) as a titrant. (assume pK w = 14.00) (hint: this involves a dilution problem.)

    answer is 12.72
  2. jcsd
  3. Apr 19, 2007 #2


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    Let me help you with the 2nd exercise.

    You have the titration's equivalence point, so you can find the molarity of the weak base:

    [tex] \[
    65.5ml \times 0.067M = 25ml \times M_b
    \] [/tex]
    ...where the unknown there is molarity of weak base.

    Now you also need formal concentration of the weak conjugate acid which formed, which I like to show as WHCl, with ions WH+ and Cl-.
    For this weak acid, you have dissociation constant equation:
    [tex] \[
    K_a = \frac{{[H][W]}}{{[WHCl]}}
    \] [/tex]
    Please excuse the lack of signs in that Ka formula.

    Since [H] is roughly close enough to [W-], you can also then say:
    [tex] \[
    K_a = \frac{{[H][H]}}{{0.04848 - [H]}}
    \] [/tex]
    ... Do you know where the 0.04848 comes from?

    Now you need to use pH at equivalence, given, to find [H] value.
    Since pH = 4.668, [H]= [tex] \[
    2.148 \times 10^{ - 5}
    \] [/tex]

    Are you with this now? You can now easily find Ka; from which you use Ka*Kb=Kw to find your Kb value.
  4. Apr 19, 2007 #3


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    For the first one, the HCl is in excess, so all of the base will be used up. You need to calculate the total number of moles of the base and the HCl (n/V=M), then subtract the moles of HCl by the moles of the weak base, and use that to calculate the amount of H+ produced. Don't forget to adjust the concentration of HCl to account for the combined volume. The [H+] will equal the remainder of the [HCl], since HCl is a strong acid and completely reacts.

    The last one can be solved the same way as the first one, except you calculate [OH-] instead of [H+] so you will have to solve for p[OH-]. p[OH-] + p[H+] = pKw = 14
    Last edited: Apr 19, 2007
  5. Apr 19, 2007 #4


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    Why do you have [H] = 2.148 x 10^-5?

    pH = -log[H]
    4.668 = -log[H]
    10^-4.668 = [H]

    edit: Oh lol, I should've put that in my calculator before posting.. they're the same thing. I wasn't thinking straight.
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