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Acid/base buffer reactions

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Describe the preparation of 40 L of 0.0500M phosphate buffer, pH 6.9 starting with 1M solutions of KH2PO4 and K2HPO4. pKa= 7.2


    2. Relevant equations

    pH= pKa + log[A-]/[HA]


    3. The attempt at a solution[/b

    6.9=7.2 + log[A-]/[HA]
    -0.3= log[A-]/[HA]
    [A-]/[HA]= 0.50

    0.5[HA]= [A-]
    0.5[HA]+1[A-] = 1.5

    0.5/1.5 = 0.33 = 33% [HA]


    0.05 mol/L x 40L = 2 mol HA
    2 mol x 1L/mol = 2L HA


    2 mol x 0.33= 0.66 mol A-
    0.66 mol x 1L/mol = 0.66 L A-


    2L HA + 0.66L A- = 2.66 <---Edit: 2.66 L of acid/base

    40L-2.66L= add 37.34 L water to make the 40L total.



    I was told by someone that this solution is incorrect. Any suggestions why?
     
    Last edited: Sep 13, 2009
  2. jcsd
  3. Sep 13, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    Earlier you wrote you need 2 moles, here you have 2.66.

    Not that I am sure I understand precisely your working in other places.
     
    Last edited by a moderator: Aug 13, 2013
  4. Jul 17, 2010 #3
    In the equation pH=pKa+log[A]/[HA], the ratio of the log value must be 10^-3=.5. Therefore, [HA] must be double [A].
     
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