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Aicds & Bases, finding pH problem

Discussion in 'MCAT Study Question Q&A' started by TheJourney, May 12, 2014.

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  1. TheJourney

    TheJourney

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    May 9, 2013
    What would be the pH of a 0.05M solution of acetic acid?
    (From the passage & table you know that the Ka for acetic acid is approximated to be 2 x 10^-5)

    A. 0.05
    B. 2.0
    C. 3.0
    4. 6.0

    First question, why can't you simply plug in 0.05 into the pH = -log [H] equation? Is it because 0.05 is not the H concentration?

    Second, if I wanted to use the "shortcut" TBR equation for weak acids: pH = pka/2 - (log [HA])/2 how would I go about finding pKa knowing pKa = - log Ka.

    Third, the other method of solving this question would be through the dissociation table to get, .00002 = x^2/.05, can someone work this out, I can't seem to get the correct answer unless I use a calc.

    Lastly, are the methods listed the above the only way of solving this question or more importantly these types of questions? Again, thank you in advance!
     
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  3. TheJourney

    TheJourney

    74
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    May 9, 2013
    For the second question, is this correct: pKa = - log 2 x 10^-5 --> pKa = 5 - log 2 ---> pka = 4.7
    Then substituting into the shortcut equation: pH = 4.7/2 - (log .05) /2 --> pH = 2.35 - (log 5 x 10^-2) /2 ---> pH = 2.35 - ((2 - log 5) /2); Assuming log 5 to be > than .5 then ---> pH = 2.35 - (1.4/2)
    ---> pH = 2.35 - .7 = 1.65.

    The answer is answer choice C, therefore I should be adding .7, but I don't see my mistake
     
  4. Chrisz

    Chrisz 2+ Year Member

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    Feb 18, 2014
    I would strongly suggest you to go back to TBR and understand what is the difference between strong acid and weak acid. Since acetic acid is a weak acid, so we can use the second formula. The first formula is for strong acid. Your calculation mistake is this: pH = 2.35 - ((2 - log 5) /2). it is actually 2.35-(log5-2)/2
     
    Teleologist likes this.
  5. Sammy1024

    Sammy1024

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    Dec 12, 2013
    1. You can't plug in 0.05 because acetic acid is not a strong acid. That equation is for strong acids only.

    2. pKa = - log (Ka) so -log (2 x 10^-5) = - (-5 + (.2)) = 5 - .2 = 4.8 (roughly). You're supposed to first just bring down the 5, because the log and 10 cancel out and then subtract 1 - .2 = 0.8. (Shortcut)

    3. Try not to use the table, unless you have to.
    .00002 = x^2/.05 => 2 x 10^-5 = X^2/5 x 10^-2
    (2 x 10^-5) (5 x 10^-2) = x^2
    10 x 10^(-5+ -2) = x^2
    1 x 10^-6 = x^2

    and then it gets complicated because you can't really take the square root of -6. (Maybe I did it wrong).

    4. If it was a strong acid you would just plug and chug in the first equation. When it's a non-strong acid then you have to go through the method I did above.
     
  6. TheJourney

    TheJourney

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    May 9, 2013
    Ahhh, I see. the method I was using for figuring out pKa by dropping the exponent in front and then subtracting the log is also for strong acids. That was causing problems in this questions.
    Thank you both for clarifying the pH = - log [H] as well.
     
  7. Czarcasm

    Czarcasm Hakuna matata, no worries. 2+ Year Member

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    Crypts of Lieberk├╝hn
    No, you were doing it right. You take the square root of 1x10^-6 which is the same thing as (1x10^-6)^1/2 or 6/2 ... so it would equal 1x10^-3. The pH of that is just 3.
     
  8. edgerock24

    edgerock24 2+ Year Member

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    Jul 28, 2012
    Yep. Another helpful hint: if you ever see "Ka" anywhere in the problem, you can assume you are dealing with a weak acid. Ka is only necessary for weak acids. Same with Kb. If you see Kb, then you can assume you're working with a weak base.
     
  9. Czarcasm

    Czarcasm Hakuna matata, no worries. 2+ Year Member

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    Crypts of Lieberk├╝hn
    Well you have to be careful, because on my MCAT (probably shouldn't be saying this), but there was a weak base with the Ka of it's conjugate acid. It's not always clear cut. They really expect you to understand the relationships, and not just plug and chug. Sucks.
     
  10. edgerock24

    edgerock24 2+ Year Member

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    Jul 28, 2012
    Yeah, very true. I'm guessing you simply had to use the Kw=Ka*Kb relationship before you moved forward w/ the problem? Just another annoying step to trick test takers :/
     
  11. TheJourney

    TheJourney

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    May 9, 2013
    Lol funny thing is they probably look at forums and see what topics test takers are having trouble with and what techniques are being employed by test takers to overcome those. What they do with that information, I'm not sure I want to know.
     
  12. Teleologist

    Teleologist 2+ Year Member

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    Jul 7, 2013
    on your 6
    Figure out ionization extent of the acetic acid. Hint: there is a formula for ionization extent.
     
  13. BerkReviewTeach

    BerkReviewTeach Company Rep & Bad Singer Exhibitor 7+ Year Member

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    May 25, 2007
    SDN Exhibitor
    Nice catch Chrisz! You've already explained it well, but I'm going to put some math out there as well.

    For converting from Ka to pKa, take -(power of 10) - log mantissa = -(-5) - log 2 = 5 - log 2 = 4.7

    The BR shortcut equation for the pH of a weak acid is pH = 0.5(pKa) - 0.5(log [HA])
    = 0.5(4.7) - 0.5[log (5 x 10^-2)]
    = 0.5(4.7) + 0.5[-log (5 x 10^-2)]
    = 0.5(4.7) + 0.5(2 - log 5)
    = 0.5(4.7) + 0.5)(1.3)
    = 0.5(6.0) = 3.0​
     

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