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Volume of Tris Needed to Prepare Buffer Solution

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate the theoretical volumes of 1.0 M Tris base, 1.0 N HCL or 0.2 NaOH, and water needed to prepare 50 ml of a 0.1 M Tris buffer pH 9.0. pKa of Tris: 8.21


    2. Relevant equations
    Henderson-Hasselbalch: pH=pKa + log[A-]/[HA]


    3. The attempt at a solution
    Honestly, I don't even know how to start.

    I know how to calculate moles of each, ratio of [Tris]/[Tris+] (I got 6.16 by 10^(9-8.21)), etc, but I don't what I actually need to use for the calculation... Help?
     
  2. jcsd
  3. Apr 4, 2012 #2

    Borek

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    Staff: Mentor

    First of all - how much TRIS do you need to prepare 50 mL of 0.1 M solution?

    Then you need to add strong acid (or strong base) to convert part of the TRIS to conjugate acid (or base) - this is a simple stoichiometry.

    Finally you need to fill up to 50 mL.
     
  4. Apr 4, 2012 #3
    I would need .005 mols, (0.1M Tris)x(0.05L)=.005. But then how do I go back to ml from here? And where does the 1.0M Tris base come in, since that is the solution I am diluting to get the 0.1M solution?

    EDIT: I think I figured it out. Would it be (1.0M Tris)(X L)=(0.1M Tris)(.05L), so I would need 5mL of 1.0M Tris?

    I still don't understand the HCl and NaOH part though; what do you mean by simple stoichiometry?
     
    Last edited: Apr 4, 2012
  5. Apr 4, 2012 #4

    Borek

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    Staff: Mentor

    Yes, it is just an application of molar concentration definition (##C = \frac n V##) and mass conservation (amount of substance doesn't change during dilution).

    First things first - for buffer solution you need a pair of conjugate acid and base. Is TRIS a base, or an acid? How can you prepare other substance from the pair?
     
  6. Apr 4, 2012 #5
    It's a base. So I would use the 1.0 N HCl to generate the acid, Tris+?
     
  7. Apr 4, 2012 #6

    Borek

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    Staff: Mentor

    TRISH+ to be exact, but yes, you are no the right track.

    Assume protonation reaction went to completion - that's where the stoichiometry comes into play.
     
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