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Autoprotolysis is not just the transfer of a H+ ion between an acid and a base, it's the transfer of an H+ between two identical molecules, one that can act as an acid and a base (like H2O). If you then take the equilibrium constant of the equation 2 H2O -> OH- + H30+, you'll get Kw = [OH-][H30+].
Autoprotolysis is the reaction of 2 H2O --> OH - + H3O+. Like any reaction, you can write the K expression (products over reactants), and get Kw = [OH-][H3O+], since H2O is a liquid and so isn't counted in the K expression. Now , you can use the properties of logs, and do logKw = log[OH-] + log[H3O+]. If you make the entire equation negative, you get -logKw=-log[OH-] - log[H3O+]. This will get you pKw = pOH + pH.
Jasleen Kahlon wrote:I understand that autoprotolysis is the transfer of an H+ from an acid to a base in an equation (I think), I'm unclear on how pKW = pH + pOH is derived from it though.
The PkW=Pka+Pkb equation is derived from autoprotolysis because autoprotolysis is the simplest example of acid and conjugate base pairs.
From Kw = [H+][OH-], you can negative log both sides => -log[Kw] = -log([H+][OH-]), and using the log rules, you can get =log[Kw] = -log[H+] + -log[OH-]. Since "p-" in chemistry denotes "-log" (as in pH = -log[H+]), you can get the equation pKw = pH + pOH.
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