Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: kevinklee2003 on October 20, 2014, 11:59:22 AM
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So pKa refers to the equilibrium constant for the reaction:
Reaction 1: HA + H2O -><- A- + H3O+
and pKb refers to the reaction Reaction 2: A- + H2O -> <- HA + OH-
So pKb does NOT refer to the equilbrium constant for the reverse reaction of reaction 1 right? But rather to the equilibrium constant for the forward reaction for the conjugate base?
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Also a more fundamental question, I know the relationship between an acid and its conjugate base is
pKa + pkB = 14
My question is why do we use this relationship? Is it because the equilbrium constant for the autoionization of water is 10-14M2. So no matter what acids or bases are added (which wil change the concentration of OH- and H3O+ in solution), after everything equilibriates, H3O+ and OH- must still equilibriate to having the relationship of [H3O+]*[OH-] = 10-14 = Keq for water?
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So pKb does NOT refer to the equilbrium constant for the reverse reaction of reaction 1 right? But rather to the equilibrium constant for the forward reaction for the conjugate base?
Correct. Hence follows the second part, since
Ka = [H3O+][A-]/[HA]; Kb = [HA][OH-]/[A-]; so KaKb = [H3O+][OH-] = Kw
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So pKb does NOT refer to the equilbrium constant for the reverse reaction of reaction 1 right?
No.
But rather to the equilibrium constant for the forward reaction for the conjugate base?
Yes.
Also a more fundamental question, I know the relationship between an acid and its conjugate base is
pKa + pkB = 14
My question is why do we use this relationship?
Because it exists? I am not sure I understand what your question really is. It can be easily proven that pKa + pKb = pKw, once it is proven we can use it, don't we?
http://www.chembuddy.com/?left=pH-calculation&right=bronsted-lowry-theory