How to calculate log and pH uncertainties?

[missbrokensmile]

I'm looking at the pKa of an acid, and I've done all the calculations, but I'm a little uncertain as to how to calculate the uncertainties for log and pH. Is there any chance any one could help or point me in a direction with information on it?

For example, I just picked an arbitrary value...

pH = 1.23 (+/- 0.05) --> How would I calculate a percentage uncertainty for this?

pKa = -log(1.23x10^-6) = 13.6 --> How would I calculate a percentage uncertainty for this?

Any help is appreciated!

[x___x]

calculating the percentage uncertainty is quite easy, no need to worry.

here is how it's done:

you take the absolute uncertainty value over the value of the pH then multiply the result by 100 to get the percentage, like this: (0.05 ÷ 1.23) × 100 = 4.065 (4 s.f.).

To calculate the percentage uncertainty for the second one is not necessary (not sure though). Because you calculate pKa with this formula (pKa = -log(Ka)), and since Ka is the acid dissociation constant (meaning that its value doesn't change for a certain acid), the value then can be taken as a literature value, and the errors in those are not required to be propagated. What you have to do is find the ka of the acid you're using, and then just plug it into the equation of the pKa, don't worry about the uncertainty.

Or am i getting things wrong?

[missbrokensmile]

For the log uncertainty, I was searching around, and on this PDF file, on page 28: http://www.cc.ysu.edu/~jeclymer/Uncertainty%20Tutorial.pdf, they have a calculus derived equation (I think?). I kind of understand it (a little anyway), but not enough to make it so I can manipulate it to fit my equation. Any help in that area?

Also, since log/pKa and pH are related, I thought that since there was that complex looking method to do it (on that PDF file), the pH uncertainty would be found using the same method.

Also, wouldn't the method of finding uncertainty for pH and log/pKa be a little different than just (0.05 ÷ 1.23) × 100 = 4.065 (4 s.f.) since it's based on a logarithmic 10-fold scale, as opposed to a scale like mL or cm, where it's linear? Correct me if I'm wrong, but that's the impression I have at the moment.

Edited June 24, 2010 by missbrokensmile

[Tilia]

Also, wouldn't the method of finding uncertainty for pH and log/pKa be a little different than just (0.05 ÷ 1.23) × 100 = 4.065 (4 s.f.) since it's based on a logarithmic 10-fold scale, as opposed to a scale like mL or cm, where it's linear? Correct me if I'm wrong, but that's the impression I have at the moment.

Not sure that matters. It should be done as you usually do, but I can't guarantee that I'm correct...