derive pKW = pH + pOH  [ENDORSED]

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Molly Posta 1H
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derive pKW = pH + pOH

Postby Molly Posta 1H » Sun Jan 24, 2021 8:26 pm

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
pKw = -log[H3O+]+ (-log[OH-])
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
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Re: derive pKW = pH + pOH

Postby Astha Patel 2J » Sun Jan 24, 2021 8:26 pm

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

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Re: derive pKW = pH + pOH  [ENDORSED]

Postby Chem_Mod » Sun Jan 24, 2021 8:28 pm

That's it. :-)

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

Ayesha Aslam-Mir 3C
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Re: derive pKW = pH + pOH

Postby Ayesha Aslam-Mir 3C » Sun Jan 24, 2021 8:31 pm

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.

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