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Moles of a conjugate base in a buffer solution

  1. Jul 8, 2015 #1
    1. The problem statement, all variables and given/known data
    You need to produce a buffer solution that has a pH of 5.70. You already have a solution that contains 0.0200 moles of acetic acid. Using the Henderson-Hasselbalch equation calculate the moles of sodium acetate needed to create a buffer with the desired pH? The Ka of acetic acid is 1.8 × 10^-5.

    2. Relevant equations
    pKa = -log (Ka)
    pH= pKa + log (base / acid)
    3. The attempt at a solution
    pKa= -log (1.8 × 10^-5) = 4.74
    5.70 = 4.74 + log (base/acid)
    Unfortunately, I have been staring at this problem for about an hour and am unsure how to find the concentration of the acid to discover the consent ration of the base. I am not sure how to proceed since the problem does not give a volume to find the molarity.
    Any help with this would be greatly appreciated!!
     
  2. jcsd
  3. Jul 9, 2015 #2

    Borek

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    Staff: Mentor

    Hint: you don't need the acid concentration.

    [tex]C_A = \frac {n_A} {V}[/tex]
    [tex]C_B = \frac {n_B} {V}[/tex]

    [tex]\frac {C_A}{C_B} = ...[/tex]
     
  4. Jul 9, 2015 #3
    I appreciate your responce, any responce really. Although, your equation set seems to need a volume as well, and the problem I am to work out does not include one. That is a big area I am being messed up. As well as the problem states to use that particular equation to find the moles of base needed.
     
  5. Jul 9, 2015 #4

    Borek

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    Staff: Mentor

    Plug concentrations into ratio and see what happens. V/V = ?
     
  6. Jul 9, 2015 #5
    Since the problem called for using the henderson-hasselbalch equation, and that entails using the concentration of the acid, I used the volume used in the lab for the titration (25 mL). Not sure if that is right... but I got an answer. I will check with a few of my class mates before lab in the morning. Thank you for your help.
     
  7. Jul 9, 2015 #6

    Borek

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    Staff: Mentor

    You don't need any specific volume.

    V/V = 1 - volume cancels out. Ratio of concentrations (if these are concentrations in the same solution) is identical to the ratio of numbers of moles.
     
  8. Jul 9, 2015 #7
    That very simple answer was probably the best answer you could have given me! Tha k you so very much for having the patience for dealing with slightly dense people like me!
     
  9. Jul 9, 2015 #8

    Borek

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    Staff: Mentor

    I hoped you will see it by yourself. You will learn much more when you put an effort and find such things by yourself, than when they are presented on a silver plate.
     
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