Hi everyone,
I'm making a calibration with different pH values. It's used to determine the effect of the pH on TOC-analyses. This makes it quite difficult because I will need to convert the diluted solutions to to undiluted solutions thus I can only use rounded numbers of volume for my acid/base to make the calculation eassier.
This is what I'm planning to do:
A) The sample has a pH of 8.15 which I will have to lower/increase to pH of 2, 4, 6, 10 and 12.
B) I will be using a 10% and a 30% solution of HNO3 to lower the pH. I will use a 1M and a 0.01M solution of NaOH to increase the pH.
My question is how do I calculate the needed volumes of acid/base to reach these pH values?
I figured that the concentration of the sample is 10^-8.15 M, but I don't know what to do next...
Thanks in advance!
Hi Mitchel,
Try this pH calculator, it may be useful in your calculations.
"pH Calculator | Calculate pH of a Solution | Sensorex" https://sensorex.com/ph-calculator/
Except for the special case (*) where the solution only contains either a strong base (e.g. NaOH) or a strong monoprotic acid (e.g. HCl), besides, possibly, neutral salts; or just those neutral salts; it would be not possibly to correctly predict how the pH would change just based on the volume of added strong base or strong monoprotic acid acid solution.
In general; the titrated solution would have to be characterized in terms of its analytical composition, so that, quantitatively, the hydrolytic (acid-base) equilibria of the dissolved species (e.g. conjugate bases of weak acids) could be accounted for.
You may find perhaps justifiable to accept that the pH can be mainly determined, based on a single buffer system, by the Henderson-Hasselbalch equation.
On the validity of the Henderson-Hasselbalch equation ― You may want to check my posts elsewhere at this forum:
https://www.researchgate.net/post/PH-calculation-of-a-mixture-of-formic-acid-NaOH-and-water
(*) Worked example ― for the special case mentioned above:
― Determine the excess volume (Vexcess) of some unnamed monoprotic strong acid (HA) aq. sol., of nominal molar concentration (formality to be precise), cHA, that should be added to a pure neutral salt aq. sol.; so that the excess acid leads the former neutral pH of that solution to some (given) acidic pH.
― Note that the said solution would become diluted by the dilution factor f = Vexcess /Vfinal, where Vfinal stands for the final volume. Note also that the neutral salt ― obtained from the titration of a strong base with a strong acid ― is not expected to significantly contribute for final pH. The pH of the final acidic solution would be: pH = -log10(f·cHA/(1 M)).
Hence: Vexcess = Vfinal·f = Vfinal·10- pH/{cHA/(1 M)}.
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