derive pKW = pH + pOH [ENDORSED]

[Molly Posta 1H]

One of the objectives for outline 2 states: Describe water autoprotolysis and derive pKW = pH + pOH.

When it says be able to derive pKW = pH + pOH, what exactly does that mean, and to what extent? Do we just need to know that since Kw = [H3O+][OH-], and the autoprotolysis constant of water at 25*C is 1.0 x 10^-14, then:

Kw = [H3O+][OH-] = 1.0 x 10^-14
so:
pKw = -log[H3O+]+ (-log[OH-])
so:
pKw = pH + pOH = 14; pKw = 14

Is this all we need to know, or is there more information that I'm missing?

[Astha Patel 2J]

I'm pretty sure that's all you need to know because that is how you derive that equation.

[Chem_Mod]

That's it. :-)

That is what I did in class, and now you can do it. Nice.

[Ayesha Aslam-Mir 3C]

I would say you hit the nail right on the head; understand how pKa and pKb are derived, how you can observe the autoprotolysis of water, know that pH + pOH = 14, know the conditions of a solution when comparing pH to pKa/pKb, and be able to apply these in context when looking at acid/base dissociation.