Dear OP,
drug must be non-ionic,
non-charged to be absorbed; nonpolar lipophilic even better . HA (zero charge) will be absorbed, A-(conjugate-base, negative charged) will not be absorbed.
HA <---> [A-] + [H
+]
ka = [A-][H+] / [HA]
-log Ka = -log ([H+][A-]/[HA])
pKa = pH - log [A-]/[HA]
http://chemistry.about.com/od/acidsbase1/a/hendersonhasselbalch.htm
notice:
eg. pKa of 3.6, pH of 5.6
pKa = pH - log [A-]/[HA]
3.6 = 5.6 - log [100]/[1]
3.6 = 5.6 - 2
mathematically, there must be
100 fold more A- than HA.
but why? -- because actually, physically, in the real world, when conditions are basic (high OH-, low H+), you can see with the naked eye the reaction actually drives toward the right to replenish the low H+
eg.
[HA] <--> [A-] + [H+]
under basic conditions, will want to drive toward regenerating more H+
youll end up with:
[HA] --->
[A-] + [H+]
(100 fold more A- than HA)
Conversely,
eg. pH environment of 1.6, pKa 3.6
pKa = pH - log [A-]/[HA]
3.6 = 1.6 - log [1]/[100]
3.6 = 1.6 - (- 2)
mathematically you will find
100 fold more HA than A-.
in the real world, physically, chemically, visually, you will see the reaction actually driving toward the
left. Excess H+
HA <--> A- +
H+
drives to
HA <---
A- +
H+
In a more acidic environment than pKa (excess of H+), reaction will drive toward reactant; left side.
-----
For review:
[HA] --> [A-] + [H+]
you want to be in a low pH (excess H+) to drive the reaction toward the protonated uncharged HA.
for drugs like:
[H20] +
--> [BH+] + [OH-]
you actually want the pH to be high (more OH-) to drive the reaction toward the unprotonated, uncharged B.
eg. NH3 <---- NH4+. You want there to be a surplus of OH, deficit of H+ so the H+ will break off the NH4.
---
Therefore
You can think in terms of raw numbers and use the mathematics of Henderson-Hasselbach, or you can think physically in terms of La Chatelier.
Personally, I prefer the physical mode because I can actually see the reactions with my naked eye.
Best of Luck!
