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Buffer solutions - relating volume and concentration

  1. Feb 24, 2010 #1
    I am really confused about how to approach the following question. Can someone please help me out?

    A buffer is made by mixing CH3COO- and CH3COOH of the same concentration of 0.5M. If the volume of the former is v1 and the latter is v2, show algebraically that the following will always hold true:

    [CH3COO-]/[CH3COOH] = v1/v2

    where [CH3COO-] and [CH3COOH] are the concentrations at equilibrium.

    Thanks a ton!!!!!
  2. jcsd
  3. Feb 27, 2010 #2


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    Gold Member

    Yes I find that difficult to explain.

    My problem is that's so obvious I don't know how to explain it.

    What's yours? :biggrin:

    Well OK on second thoughts maybe. Forget your substances are what they are. They could be anything and the same result applies. As long as they don't actually do something, like react chemically. Or evaporate. Which yours will a bit so better solve this quickly!
  4. Mar 7, 2010 #3
    Yeah, I got the answer xD I was confused because I didnt know that the number of moles remain constant. I thought some of either reactant could react to produce the other, especially since its asking for the concentrations at equilibrium. Just in case anyone was looking for the answer:

    Concentration of CH3COO- at equilibrium = (0.05M)V1/(V1+V2) = [CH3COO-]
    Concentration of CH3COOH at equilibrium = (0.05M)V2/(V1+V2) = [CH3COOH]

    dividing the two equations,

    (0.05V1)(v1+v2)/ (0.05V2)(v1+v2) = [CH3COO-]/[CH3COOH]
    V1/V2 = [CH3COO-]/[CH3COOH]

    @.@ wasnt at all tht hard but just confusing
  5. Mar 7, 2010 #4


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    You're welcome.

    Nice to hear someone come back.

    I suppose the majority do get to solutions with, without or in spite of our help, but we don't always hear.
  6. Mar 7, 2010 #5
    xD sorry, I completely forgot that I posted this question until I saw the email when I was clearing my Inbox.
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