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Buffer preparation problem

  1. Mar 25, 2004 #1
    I don't know if I can effectively convey what I am doing here, but I am going to try because I need the help. I am trying to determine my ratio of acid to base and subsequently make a .5M Tris-HCl buffer utilizing Tris base, and concentrated HCl.

    By my calculations I get a 19.05 parts acid/1 part base ratio.
    Using this, I divide the total amount of buffer I am looking to make (200mL) by the total number of parts in solution (20.05 or 19.05+1). Given this, I get ~190mL acid to 10 mL base.

    Here is where I was a little unsure of myself. To calculate the amount of base I need, I multiplied the MW of Tris-base (121.4 grams/mole) by the molarity of the solution by the number of mL specified above (10mL).

    To calculate a 190mL of a .5M solution of HCL, I used (C1)(V1)=(C2)(V2), where C1=.5M, V1=.19 Liters, C2=11.6M (molarity of concentrated HCl). From this, I calculated that I needed to add 8.2mL of conc. HCl to 181.8mL dd H20.

    The final step was to add my acid solution to my base solution.


    I tested some of this "buffer" on pH paper and got a pH of just above 1.0. Why?
    The stated pKa value seems to fluxuate between 8.08 and 8.3. I was using 8.08 in application to HH.

    Any input would be very much appreciated.

    Thank you.
  2. jcsd
  3. Mar 25, 2004 #2
    The problem is quite straightforward. You are asked to prepare a 200 mL solution of 0.5 M Tris buffer at pH 8.08. You only calculated to prepare 10 mL of 0.5 M Tris. You need to add 12.11 grams of Tris base (according to my calculations) to have, at the end, 200 mL of 0.5 M Tris buffer at pH 8.08.

    My suggestion? Weigh out 12.11 grams of Tris base. Add to 100 mL of water. Carefully pH with HCl (or NaOH if you get too fancy free with the HCl). Bring up to 200 mL with water. And there you go.

    FYI, be careful with temperature. Tris is temperature dependent when it comes to pH.

    My calculation just so you can take a look: (0.2 L buffer) * (0.5 mol/L) * (121.1 grams/mol) = 12.11 grams Tris base
  4. Mar 26, 2004 #3
    Mike has an easy way to do it, however, it would be better if you just used the Henderson-Hasselbach equation. You could easily pull all your info from there.
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