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Solve pKa of unknown monoprotic Acid

  1. Aug 17, 2009 #1
    I've been trying to figure this one out for a while.

    A 0.1276g sample of a monoprotic acid was dissolved in 25mL H2O, and titrated with a .0633 M NaOH solution. Volume required to reach equivalence point was 18.4 mL.
    It asked me to calculate the molar mass, which i found to be 116g/mol:
    0.0633 x .0184 = .0011 moles NaOH.
    1 mol NaOH = 1 mol Acid = .0011 moles Acid.
    [tex]\frac{.1276g}{.0011}[/tex] = 116 g/mol

    Then it asks to find the Ka of the acid after the addition of 10 mL and pH of 5.87.
    I tried using the formula: pH = pKa + log [tex]\frac{[A-]}{[HA]}[/tex]
    but the answer in the back of the text doesn't match up to what i come up with, so I'm guessing that isn't the right way of doing this.
     
  2. jcsd
  3. Aug 17, 2009 #2

    Borek

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    Show details of your pKa calculation - approach seems to be correct.

    --
     
  4. Aug 17, 2009 #3
    I calculated for moles in 10mL of NaOH to be 6.3*10-4
    1 mol of NaOH = 1 mol of monoprotic acid,
    moles monoprotic acid = 6.3*10-4

    then, using:
    pH = pKa + log[tex]\frac{[A-]}{[HA]}[/tex]
    If my reasoning is correct, log[tex]\frac{[A-]}{[HA]}[/tex] will equal 1 since [A-] and [HA] will equal the moles of HA (6.3*10-4)

    solving for Ka i get:

    -log Ka = 5.87
    10-5.87 = Ka = 1.35*10-6

    answer in the text is 1.6*10-6
     
  5. Aug 17, 2009 #4

    chemisttree

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    Hey Commie! Your moles of NaOH is indeed equal to the number of moles of the monoprotic acid but it isn't 6.3 X 10-4. Do over. Remember that #moles=volumeXconcentration (of NaOH, 0.0633M and 18.4 mL).
     
  6. Aug 17, 2009 #5
    Alright, I just keep repeating my same thought pattern or something, because I keep coming up with:

    .0633 M x .01L = .000633moles NaOH

    Someone save me from my sinful ways!

    edit: woah wait a minute. okay, so why am i using the 18.4 mL of base still? if it's asking for Ka after the addition of 10mL and pH 5.87
     
    Last edited: Aug 17, 2009
  7. Aug 17, 2009 #6

    chemisttree

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    AGHHHH! My bad!

    Yes, you are right, the number of moles of acid that have been converted to A- is 6.33 X 10^-4. What would 0.0117 - 6.3X10^-4 represent?

    6.33 X 10^-4 is not [HA] in your equation... it's A- (moles). Divide the moles of A- by the volume in L to get [A-].

    Remember that HA is not equal to [HA] and A- is not equal to [A-].
     
    Last edited: Aug 17, 2009
  8. Aug 17, 2009 #7
    A-ha!
    Okay, here it is.

    after addition of 10mL final volume is .035L. So 6.3*10^-4 / .035L = .018 M [A-]

    .00116 moles HA - 6.3*10^-4 = 5.3*10^-4 moles (HA left in flask).
    5.3*10^-4 / .035 = .015 M [HA]

    then:
    solving for pKA,
    pKa= 5.87 - log(.018/.015)
    pKa= 5.79
    Ka = 10^-5.79
    Ka = 1.62*10^-6

    Thanks chemisttree. This problem would've been bugging me for the rest of my life.
     
  9. Aug 18, 2009 #8

    Borek

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    Note: you don't need concentrations in this case, number of moles will do. Volume is the same for both nominator and denominator, so it cancels out.

    --
     
  10. Aug 18, 2009 #9

    chemisttree

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    One minor point... significant figures. If significant figures aren't important to you, then nevermind.
     
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