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PH of buffer with addition of conjugate base

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider 1L of a solution that is 0.100 M in acetic acid and 0.100 M in sodium acetate. What will the pH be, when 0.0900 mol of sodium acetate are added?
    Ignore changes in volume.

    acetic acid pka = 1.75 E-5

    2. Relevant equations

    HA --> H+ + A-
    pH=pka +log A-/HA


    3. The attempt at a solution

    I had similar questions that I know how to do with strong acids. The difference in this question is that the added acid/base is a weak one. With the strong base addition I can just add it to A- concentration and subtract it from HA and use Henderson-Hasselbach equation.

    This one I tried that but the numbers I know are off. I also tried doing ka = H+*(0.1+.009)/0.1

    Solving for H+ and then using that to find out concentrations of A- and HA but the H+ number is far too small to make a difference and I'm fairly sure I can't just add the conjugate bases together.

    Can someone please help.
     
  2. jcsd
  3. Feb 10, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    0.009? * instead of log? This is a correct approach, just check your math.
     
  4. Feb 10, 2009 #3
    Oops I made it a mistake, anyways when I did my math I just did 1.75E-5 = H+(0.19)/0.1

    I get 9.2E-6

    but I have no idea what to do with this. If I do it the same way as a strong base question I would add it to the A- concentration and subtract it from HA, but this value is too small to make a difference to a 0.1 M.
     
  5. Feb 10, 2009 #4

    Borek

    User Avatar

    Staff: Mentor

    You have added conjugate base - so change only base concentration, acid concentration is untouched. Simply plug numbers into HH equation.
     
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