My notes give an example of:
3.75 x 10E-4 with the result of -3.425968732
However, when I attempt to run the example through my calculator I end up with 0.60205999....
Since pH = -log[H+] and you have 1x10E-5 moles in 10 L of solution,
your [H+] = (1x10E-5 moles)/10L = 1x10E-6 molar (M),
and therefore your pH = 6.0.
Since Kw = 1x10E-14 = [H+]*[OH-],
your [OH-] = (1x10E-14)/(1x10E-6) = 1x10E-8 M
We can check this since pH + pOH = 14, and the negative log of the [OH-] = 8, so it's right.
I don't know if your ...
Given a certain volume of H+ ions in solution, calculate the pH and [OH-]. The answer is fully explained.