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Finding Buffer Volume with Little Information

  1. May 31, 2013 #1
    Sorry. The title should read finding volume with little information. Humongous typo :)

    1. The problem statement, all variables and given/known data
    Suppose you are given 0.50 liter of 0.500 M acetic acid, and 0.50 liter of 0.250 liter sodium acetate. What is the maximum volume of buffer solution that you can make if the buffer must have a pH of 4.58? Ka for acetic acid is 1.8 x 10-5


    2. Relevant equations
    pH=pKa + log([A-]/[HA])


    3. The attempt at a solution
    I've gotten up to this point so far,

    0.69= [A-]/[HA]

    0.69= (0.125/x)/(0.25/x)



    What I don't know is how I should go about from here to get the volume(x).
     
  2. jcsd
  3. May 31, 2013 #2
    You went wrong somewhere as the last line. What you wrote simplifies to 0.69 = 0.125/0.25 which is just not true.

    You're on the right track by considering the HH equation but I believe you manipulated it improperly. Take the -log of the Ka to get the pKa. Then plug the pH and pKa into the equation and solve for something like this:

    # = log (A/HA)

    Then get the above expression into A/HA = # (different from the # above of course).

    This will tell you the ratio of moles of each needed to get a buffer to that pH. You can rearrange this expression to tell you something like: for each mole of A I need # moles of HA.

    Next step would be to work in moles instead of molarity, so use the concentration and volume information given to figure out how many moles of each you actually have in the separate containers.

    At this point it becomes analogous to a limiting reagent problem (remember those?). Once you figure out you're limiting reagent you'll need to figure out how much volume of each solution to deliver to get that mole ratio in your final mixed buffer solution.
     
  4. May 31, 2013 #3

    Borek

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    Staff: Mentor

    Sadly, question is slightly ambiguous. It doesn't say anything about diluting - and it is quite possible to prepare much larger volume of the buffer of a requested pH, just by using correct amounts of reagents and adding copious amount of water.

    Actually it makes the problem much more interesting :wink:
     
  5. May 31, 2013 #4
    Borek you are such a pro that it always amazes me how you envision these things. I'm not so far removed from these types of problems that I already know what it is they (meaning the question writes) are looking for and just go and solve the question. You, however, always look at the actual "situation," (if you get my point). This gets me back to my notion that tests and HW problems, many times, can actually be detrimental to truly learning an area.

    My point is...I want to be just like you when I grow up :)
     
  6. Jun 1, 2013 #5

    Borek

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    Staff: Mentor

    Thanks :blushing:

    Actually "Questions that should be never asked" is a pet peeve of mine. I have quite a collection, perhaps one day I will use them to write a paper for the Journal of Chemical Education.
     
  7. Jun 1, 2013 #6

    epenguin

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    Homework Helper
    Gold Member


    I suppose that should have been !"0.50 liter of 0.250 M sodium acetate."

    And for next step what is [A-] equal to, almost exactly, by electroneutrality? Then volumes needed are in a fairly simple relation with concentrations.

    As far as I can see JH is right except the last line.

    In defence of the problem setters, here we mostly deal with always about the same three or four questions about pH, buffers etc. So to not make it mechanically mindless, but acutally realistic and flexible as is needed in practice they have to try and vary and add tricks.

    They might answer you were not given any water to play with in that problem.
     
    Last edited: Jun 1, 2013
  8. Jun 6, 2013 #7
    Thanks for the clarification everyone. I think I can do this now:


    0.69= [A-]/[HA]


    So, for each mole of HA, I need 0.69 moles of A-.


    I have 0.125 moles A- and 0.250 moles HA.
    The limiting reagent is A-.

    So, in order to keep the 1/0.69 proportion that is required, there must be 0.125 moles A- and 0.1811 moles HA.

    The volume of sodium acetate to get 0.125 moles A- is 0.5 liters. The volume of acetic acid to get 0.1811 moles HA is 0.36 liters, for a total of 0.86 L.


    That was the correct answer. Hmmmm.

    It seems like this problem is asking us to create a buffer that is already at equilibrium. B/c otherwise we could've have used all of the HA(the whole 0.50 liters instead of just 0.36 liters) and let it dissociate, adding the sodium acetate as it is needed until we reach 1/0.69 ratio, right?
     
  9. Jun 6, 2013 #8

    Borek

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    Staff: Mentor

    When doing buffer questions we usually assume neither acid nor conjugate base dissociate/hydrolyze. This is not entirely true, but usually pH shift due to these reactions is completely negligible.
     
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