I have been preparing 150 mM and 30 mM Tris Buffers (pka=8.3). Using .15 M solutions of both Tris and Tris-HCL.

Here is the information in which I have calculated theoretical pH values as well as collected experimental data.

I used (Ka or Kb)= x^2 / [concentration]-x to solve for the values in solution 1 and 5 of both the concentrated and diluted buffers. I also used Henderson-Hasselbach for solutions 2-4. For example, solution 2 in 150mM:

pH=8.3+log(.375M unprotonated/1.125 M protonated)=7.822

For the addition of .1 mL 1.25 M HCl I added and subtracted appropriately:

pH=8.3+log(.375-.125M unprotonated/1.125+.125M protonated)=7.50

150 mM Tris Buffer

30 mM Tris Buffer

I am having trouble applying these concepts to a diluted buffer solution. I calculated solution 1 for 30 mM buffers by simply changing the concentration by a factor of 5 (per the dilution factor) while using the ICE table Ka=x^2/ concentration-x.

I need some direction in being able to figure out the pH of the diluted solution and how to calculate the effect of .1 mL 1.25 M HCl addition to each of the dilute solutions.

  • $\begingroup$ Sorry to clarify the 30mM are also 10 mL total. They are diluted with water until 10mL after the addition of the acid and base as laid out in the 2nd graph. The real question I have is that will there be an effect on pH by the water in the dilute solutions? The ratio of acid and base is the same, but I am not sure if the values will be identical to those in the 150mM calculations. Can anybody shed some light on that? $\endgroup$ – jpa317 Feb 2 '16 at 0:39
  • 1
    $\begingroup$ I'd recommend formatting your equations using MathJax to make it more understandable. $\endgroup$ – boxspah Feb 2 '16 at 1:12

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