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Homework Help: H-H equation question

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data

    prepare 40mM TES buffer pH 7.4
    prepare 100mls of it
    mw TES = 229.25
    pka = 7.5

    2. Relevant equations

    ph = pka +log(A-/HA)

    3. The attempt at a solution

    My problem is that I don't know what the MW of the acid or base is for this buffer. Is the given MW the acid, and I subract a proton weight for the base or something? I can do the calculation though:

    7.4 = 7.5 +log(A-/HA)

    -.1 = log(A-/HA)

    A-/HA = .79

    56% base
    44% acid

    .56* 40mM * MW of base *.1L = moles of base
    .44*40mM *MW of acid *.1L = moles of acid

    So how should I figure out the MW of acid and base?
  2. jcsd
  3. Oct 15, 2011 #2


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    Staff: Mentor

    http://en.wikipedia.org/wiki/TES_(buffer [Broken])

    Given molar mass refers to the acid.
    Last edited by a moderator: May 5, 2017
  4. Oct 15, 2011 #3
    Thank you,

    Ok, thats kind of what I thought. So the acid MW is 229.25, and the base would be (229.25 - 1.008) is that correct?
    Last edited by a moderator: May 5, 2017
  5. Oct 15, 2011 #4


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    Staff: Mentor

    You don't need molar mass of the conjugate base. Use enough acid to prepare 40 mM solution, neutralize it with any strong base.
  6. Oct 15, 2011 #5
    so I would go:

    229.25g/mol * .04moles/liter *.1L = .917g TES added

    then to calulate strong base needed to convert 56% of acid to base

    .04moles/liter *.1L = .004 moles TES * .56 = .00224 moles of strong base needed
  7. Oct 15, 2011 #6


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    Staff: Mentor

    Check your math. If more than half of the acid is converted to base pH is higher than pKa.

    But in general you are on the right track.
  8. Oct 15, 2011 #7
    whoops it should be 44% converted to base
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