[missbrokensmile]

How do I calculate uncertainties for pH and log? I'm doing an experiment where I measure the pH, then do a few calculations to get to the pKa value. But for this experiment, I need to include the uncertainties. The uncertainties for things like mL, temperature, etc, are simple enough, and I can do them, but I'm unsure of how to calculate uncertainties for pH and log (as the relative scale is a little different). Is there any chance anyone could help?

For example:

pH = 2.79 (+/- 0.05) --> How would I calculate a percentage uncertainty for this?

pKa = -log(6.58x10^-6) = 5.18 --> How would I calculate a percentage uncertainty for this?

Any help is appreciated!

---

Also, I hope it's okay to include another question in this topic. I don't really understand why pKa is only affected by temperature. Why isn't it affected by concentration or pressure, etc? The way I understand it, it has something to do with equilibrium and Le Chatelier's principle, as it shifts the position of the equilibrium... But what makes temperature so special that it is the only thing that affects pKa? Doesn't temperature also shift the position of the equilibrium as well?

[Woofuls]

If you measured the pH, then the relative uncertainty is your uncertainty divided by your best estimate. Take your relative uncertainty times 100 to get the percentage.

As for the error propagation across, say y=log(x), the relative uncertainty in y is proportional to the relative uncertainty in x by 1/ln(10). So, to get the error in y one takes (relative error in x) times (1/ln(10)). I've been told you shouldn't use percent relative uncertainty with these calculations (I don't know why).

For your last question:pkA, obviously, depends on the Ka. So, what affects Ka will change pkA. Changes in concentration, temperature and pressure will affect the equilibrium position (which your principle will tell you about), but Keq remains the same because of Le Chatelier's principle. If you increase the temperature, heat will be consumed by the system, then shift the equilibrium in the endothermic direction (changes K). Likewise, if you add reactant, some of the added reactant is used and the production concentration increases ultimately leaving no change on K. The actual dependence of K on temperature is thermodynamic in nature, and I'm not sure I can explain it in a better way.

[JadenErius]

*Ignore me, I am impatient*, hello?