DISSOCATION CONSTANTS

[A][H]/[AH] = Ka (acid dissociation constant); weak acid

[B][H+]/[BH+] = Ka; weak base


pKa = -logKa, where Ka is in molar units


Every drug, whether a weak acid or weak base, has its own pKa or Ka. This is a measure of the structure of the molecule; how strongly a weak acid wants to donate protons, or how strongly a weak base wants to accept protons. This is a structural parameter that we’re not expected to know.




THE USEFUL FORM OF THE HENDERSON-HASSELBALCH EQUATION:

for weak acid: log [A-]/[AH] = pH – pKa


for weak base: log [B]/[BH+] = pH – pKa


**How to remember the H-H equation: on the right side, we’ll always be talking about the pH relative to the pKa, (It’s always pH minus pKa). On the left, it’s the deprotonated form over the protonated form (the form with the H attached, is a little heavier and that’s why its on the bottom).


These equations are simply an algebraic rearrangement of Ka, which is just an expression of the mass action law.


BUT, these are very important in pharmacology BECAUSE if you have a drug and you know its pKa AND the pH of the body environment that the drug is in, you plug that information into the above equation to get the fraction of the drug that’s in the charged and uncharged forms. Remember, the uncharged form can transverse membranes while the charged form cannot. So in certain compartments where the drug is in the ionized form the drug is trapped and cannot go anywhere, whereas if the drug is in the uncharged form it can readily cross membranes and go all over the place.


We’ll be expected to know (more or less) the following ranges of body tissue pH:

*pH for blood = 7.4

*pH for stomach = 2

*pH for bowel = 8-9


So we need to know what percent of the drug is in each form (charged and uncharged). We can use the H-H equation for this. Usually however, what we need to know isn’t gained from the H-H equation, which only gives us the ratio [A-]/[AH]. What is more useful is something slightly different—which is the percent of total drug in form [A-] or in form [AH]. (Refer to page 5 of Cohen’s handout to see how we get this)


The fraction of total drug in the protonated form AH or BH+ is given by:


[AH] = 1 or [BH+] = 1

[Atotal] 1 + [A-]/[AH] [Btotal] 1 + [B]/[BH+]

*the equation for the protonated form is easier to remember. Thus, even if you’re asked for how much is in the deprotonated form, its just 1 minus the fraction of the protonated form. (see page 5 for the equations used to find the deprotonated form [A-] or [B], if you’d rather work it out that way).


Example: (handout p. 6)

A weak-acid drug has pKa = 5.0 and is in the intestine, where pH = 6.0. What percent of the drug is in the permeant (protonated) form?

(See calculations on p. 6)

The answer = 9.1% of the drug is in its protonated, permeant form.


This calculation only depends on the DIFFERENCE between pH and pKa!!!

So, if in the above example we had pH = 7.4 and pKa = 6.4, so that the difference is still 1, the answer will still be the same. Thus, its always the pH relative to the pKa that matters in all of these problems. It doesn’t matter at all what the pH is if we don’t know the pKa. (in the film that we saw, he was pretty sloppy with this concept, he should have specified if the drug was acidic or basic relative to the pKa that we were talking about, instead of only if the drug was ‘acidic’ or ‘basic’).


PERCENT OF DRUG IN PERMEANT OR IMPERMEANT FORM VS. pH:



When pH = pKa, HALF of the drug is in non-ionized (permeant) form [AH]

As pH increases with respect to pKa, free [H+] is below the value needed to keep half the sites protonated, so [AH] falls below 50% of the total drug (the equilibrium shifts to the right). Conversely, when pH is lower than pKa, free [H+] is high, more sites get protonated, so [AH] rises to more than 50% of total drug (the equilibrium shifts to the left)


pH = pKa + 1 = 9% pH = pKa – 1 = 91%

pH = pKa + 2 = 1% pH = pKa – 2 = 99%

pH = pKa + 3 = 0.1% pH = pKa – 3 = 99.9%

pH = pKa + 4 = 0.01% pH = pKa – 4 = 99.99%

pH = pKa + 5 = 0.001% pH = pKa – 5 = 99.999%

etc…


*most of the change (82%) occurs within 1 pH-unit on either side of pKa, and 98% of all change occurs within 2 pH-units on either side of pKa.


If you’re given a problem that says (for example), pKa=4.2, pH=5.8, ask yourself the diffence and the answer will be close; it’ll be within 1% - 9%. In a multiple choice exam, you can really narrow it down this way.


What if pH – pKa is not an integer? It’s the same procedure, using the same keys on your calculator (see top of handout page 8).


Check to see if your answer makes sense.


For a similar question with respect to the B form of the drug, see graph on p. 8 of handout.

-When pH goes above pKa, the proton concentration is dropping. So the equation shifts to the right. If we’re plotting %B, this will go up. So the curve will be opposite that of the A form.

**Don’t memorize this (graph that weak acids go down to the right, and weak bases go up to the right), because what if we’re asked to plot % of form [BH+], the graph won’t go the same.


To solve a problem, you’ve got to know three things:

1.

whether the drug is a weak acid or weak base
2.

pKa of the drug
3.

pH of the body fluid where the drug happens to be


A summary:

-At pH<>pKa, things deprotonate and go to the A- form or B form, here the B form is permeant and A- is not.

Category: Pharmacology Notes