Question

A lipophilic drug with a pK = 6.0 is taken orally and dissolves in the gastric environment of the stomach (pH = 1.0).

What is the degree of ionization of the drug in the stomach and the blood plasma (pH = 7.4)?

In what diretion will this drug tend to passively diffuse between the two compartments?

Answer 1

PH in the stomach = 1
conc = 10^-PH = 10^-1 = 0.1 M
at equilibrium conc will be = (0.1-X)
Pka = 6 ; Pka = -logKa ; Ka = 10^-Pka = 10^-6
K_a = [H⁺] [A^-] / [HA] = X² / (0.1 -X)= 10^-6
X² = 1*10^-7 - 10^-6X
X² + 10^-6X -1*10^-7 = 0
solve the above quadratic expression for X,
X = 3.15728*10^-4 M
degree of ionzation in stomach = (3.15728*10^-4 / 0.1)*100 = 0.3157

PH in blood plasma = 7.4
conc = 10^-7.4 = 3.98107*10^-8 M
conc at equilibrium = (3.98107*10^-8 -X)
10^-6 = X² / (3.98107*10^-8 -X)
X² - 3.98107*10^-14 +10^-6X = 0
X = 3.83406*10^-8
degree of ionzation = [3.83406*10^-8 / 3.98107*10^-8] *100 = 96.3072 %

We can observe that the PH in stomach is less and not ionized compared to the PH in blood plasma.
So, the drug will tend to passively diffuse from gastric environment of stomach to or across the blood
plasma and dissociate here.