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How is pKa related to pKb?
- How can I find a relationship to pKa and pKb using pK = - log K ?
Alex Baskin baskin@scruznet.com
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Imagine you have an acid dissociation
HA H+ + A-
|
 |
Ka = |
[H+][A-]
[HA] |
followed by hydrolysis of the A- ion,
A- + H2O HA + OH-
|
 |
Kb = |
[OH-][HA]
[A-] |
If you add the two reactions together, the net reaction is the hydrolysis of water,
H2O H+ + OH-
|
Kw = [H+][OH-] = 1.01×10-14 |
Looking at the equilibrium expressions, you can see that the product of Ka and Kb is Kw:
Ka Kb = |
[H+][A-]
[HA] |
× |
[OH-][HA]
[A-] |
= |
[H+][OH-] = Kw |
Take minus the log of both sides of this equation and you get pKa + pKb = pKw.
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