Question

Drug X is a weakly acidic drug that has a pKa of 7.2

a) Calculate the percentage of the ionized form of this drug in the duodenum which has a pH of 5.0? explain the concept and the scientific rationale behind your answer.

b)if the pH in the stomach is 2 and that in the intestine is 8, in a new line write which of these two GIT parts drug X is maximally absorbed from; the stomach or the intestine? (just provide the name of the organ).

Answer 1

Answer a)

pka of drug 7.2 Let the lonized form and neutral form be represented as X and Hx respectively, pH in duodenum & 5.0 HX dissociates as HX H t +X- According to Henderson Hasselbalch equation. pH = pka + log [X-] as 5.0= 7.2 te (Hx] 15.0-722) - [x] [mx] 0.0063 = [x -] [mx] apo oft = 0.0063 +0.0063 ·0063 Xloo 2 0.637 Hence, Y of ionized form in duodenum is (0.63)

Answer b)

In stomach, pH is 2 i.e. acidic. Drug mentioned is a weak acid. Due to abundance of H+ in stomach, the weakly acidic drug will remain completely unionized.

In intestine, pH is 8 i.e. basic. Due to decreased H+ in intestine and pH being more than pKa, the drug will be predominantly in its ionized form.

Absorption of drug is readily in its unionized form because of the tendency of such form to cross the hydrophobic membranes easily.

Hence, drug X will be absorbed from the stomach.

One word answer is STOMACH.