The pKa of acetyl salicylic acid (aspirin) is 3.5. The pH of gastric juice in the human stomach is about 2 - 3 and the pH in the small intestine is about 8. Aspirin will be: A. Unionised in the small intestine and in the stomach B. Completely ionised in the small intestine and in the stomach C. Ionised in the stomach and almost unionised in the small intestine D. Ionised in the small intestine and almost unionised in the stomach

To determine the ionization state of aspirin (acetyl salicylic acid) in different pH environments, we can compare the pH of the environment to the pKa of aspirin.

The pKa of aspirin is 3.5. When the pH of the environment is equal to the pKa of the compound, half of the molecules will be ionized and half will be in the unionized form.

Let's analyze the options:

A. Unionized in the small intestine and in the stomach: The pH of the small intestine is around 8, which is significantly higher than the pKa of aspirin (3.5). So, in the small intestine, aspirin will predominantly exist in its ionized form. In the stomach, the pH is around 2-3, which is lower than the pKa, so it will mostly be in the unionized form. This option is incorrect.

B. Completely ionized in the small intestine and in the stomach: This is incorrect because in the stomach, the pH is below the pKa, so not all of the aspirin molecules will be ionized.

C. Ionized in the stomach and almost unionized in the small intestine: This is also incorrect. In the stomach, where the pH is around 2-3 (lower than the pKa), aspirin will be mostly in its unionized form. In the small intestine, where the pH is around 8 (higher than the pKa), it will be mostly in its ionized form.

D. Ionized in the small intestine and almost unionized in the stomach: This is the correct option. In the small intestine, where the pH is around 8 (higher than the pKa), aspirin will be mostly in its ionized form. In the stomach, where the pH is around 2-3 (lower than the pKa), it will be mostly in its unionized form.

So, the correct answer is option D.