# Calculation of pH in diluted buffer

I have been preparing $$\pu{150 mM}$$ and $$\pu{30 mM}$$ 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 $$\displaystyle = 8.3 + \log\frac{\pu{0.375 M} }{\pu{1.125 M}} = 7.822$$

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

pH $$\displaystyle = 8.3 + \log \frac{\pu{0.375 M} - \pu{0.125 M}}{\pu{1.125 M} + \pu{0.125 M}} = 7.50$$

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.

[From comments] ... 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.

• 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? Feb 2, 2016 at 0:39
• I'd recommend formatting your equations using MathJax to make it more understandable. Feb 2, 2016 at 1:12

The real question I have is that will there be an effect on pH by the water in the dilute solutions?

There is very little effect on the pH of a buffer when you dilute it with water (as long as the concentration is still high compared to that of hydronium and hydroxide concentrations).

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.

The buffer capacity is lowered by diluting the buffer. If you add the same amount of strong acid to a buffer or a diluted buffer, the diluted buffer will show a larger change in pH.

If I understand the calculations presented in the question correctly, the concentrations are off by a factor 10. Comparing the 150 mM with the 30 mM buffer, the relative change in concentrations upon adding the same amount of HCl is larger for the more diluted buffer, thus the pH change is predicted to be more pronounced. The experimental data supports that prediction.

Here are the updated calculations for row 2, 150 mM:

pH $$\displaystyle = 8.3 + \log\frac{\pu{0.0375 M} }{\pu{0.1125 M}} = 7.822$$

For the addition of .1 mL 1.25 M HCl:

pH $$\displaystyle = 8.3 + \log \frac{\pu{0.0375 M} - \pu{0.0125 M}}{\pu{0.1125 M} + \pu{0.0125 M}} = 7.50$$

For the 30 mM buffer, the buffer concentration is one fifth of the 150 mM buffer, but the amount (and concentration) of HCl added is the same:

pH $$\displaystyle = 8.3 + \log\frac{\pu{0.0075 M} }{\pu{0.0225 M}} = 7.822$$

For the addition of .1 mL 1.25 M HCl, we can try the same approach:

pH $$\displaystyle = 8.3 + \log \frac{\pu{0.0075 M} - \pu{0.0125 M}}{\pu{0.0225 M} + \pu{0.0125 M}} =$$ oops, we ran out of buffer (HCl is in excess of the deprotonated form of the buffer substance).

pH $$= \log (0.0125 - 0.0075) = \log 0.005 = 2.3$$

This matches the experimental result fairly well.

• Hmm, with different charges of respective conjugate acid and base, their activities would change by dilution at different rate. I would have to quantify it, but I am not sure if it is very little for different ionic strengths. OTOH, if the ionic strength is kept constant by added salts, then you are right. Jul 15, 2021 at 6:16
• @Poutnik You are right. At the level of theory presented here, however, you get the same answer.
– Karsten
Jul 15, 2021 at 11:21
• Yes. I use this argument often myself. :-) All depends on the expected level. Jul 15, 2021 at 11:35