Question

Aspirin is an acid with a \(\mathrm{p} K_{\mathrm{a}}\) of \(3.5 ;\) its structure includes a carboxyl group. To be absorbed into the bloodstream, it must pass through the membrane lining the stomach and the small intestine. Electrically neutral molecules can pass through a membrane more easily than can charged molecules. Would you expect more aspirin to be absorbed in the stomach, where the pH of gastric juice is about \(1,\) or in the small intestine, where the pH is about \(6 ?\) Explain your answer.

Short answer

More aspirin is expected to be absorbed in the stomach, where the pH is 1.

Step-by-step solution

Understand the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation relates the pH, the pKa of the acid, and the ratio of the concentrations of its dissociated and undissociated forms. It is given by: \[ \text{pH} = \text{pKa} + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \] Here, pKa = 3.5 (of aspirin), [A^-] = concentration of the deprotonated (charged) form of the acid, [HA] = concentration of the protonated (uncharged) form of the acid.

Compare the pH of the Stomach and pKa

In the stomach, the pH is about 1. Plug the pH of the stomach and the pKa of the aspirin into the Henderson-Hasselbalch equation to determine the ratio: \[ 1 = 3.5 + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \] Solving for the ratio: \[ \text{log} \frac{[\text{A}^-]}{[\text{HA}]} = 1 - 3.5 \] \[ \text{log} \frac{[\text{A}^-]}{[\text{HA}]} = -2.5 \] Taking the antilog: \[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{-2.5} \] \[ \frac{[\text{A}^-]}{[\text{HA}]} = 0.0032 \] This means there are many more uncharged molecules (HA) than charged ones (A^-) in the stomach.

Compare the pH of the Small Intestine and pKa

In the small intestine, the pH is about 6. Plug the pH of the small intestine and the pKa of the aspirin into the Henderson-Hasselbalch equation: \[ 6 = 3.5 + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \] Solving for the ratio: \[ \text{log} \frac{[\text{A}^-]}{[\text{HA}]} = 6 - 3.5 \] \[ \text{log} \frac{[\text{A}^-]}{[\text{HA}]} = 2.5 \] Taking the antilog: \[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{2.5} \] \[ \frac{[\text{A}^-]}{[\text{HA}]} = 316.2 \] This indicates there are many more charged molecules (A^-) than uncharged ones (HA) in the small intestine.

Conclusion

Neutral (uncharged) molecules pass through the membrane more easily. In the stomach, where pH=1, the majority of aspirin will be in its protonated (uncharged) form due to the ratio being 0.0032 (more HA). In the small intestine, where pH=6, the majority of aspirin will be in its deprotonated (charged) form as the ratio is 316.2 (more A^-). Therefore, more aspirin is likely to be absorbed in the stomach where it exists more in its uncharged (HA) form.

Key concepts

Henderson-Hasselbalch Equation

To understand why aspirin is absorbed more in the stomach than in the small intestine, we need to use the Henderson-Hasselbalch equation. This equation helps us find the relationship between the pH of a solution and the pKa of an acid. The formula is given by:

\( \text{pH} = \text{pKa} + \text{log} \frac{[\text{A}^{-}]}{[\text{HA}]} \)

Here, pKa is a constant specific to aspirin (which is 3.5), \( [\text{A}^{-}] \) represents the concentration of the charged form of aspirin, and \( [\text{HA}] \) represents the concentration of the uncharged form.

By solving this equation for different pH values, we can determine how much of aspirin is in its neutral form. This is important because neutral (uncharged) molecules easily pass through cell membranes and get absorbed.

So, the Henderson-Hasselbalch equation helps us predict the form aspirin will take in different environments like the stomach or small intestine.

pH and pKa Relationship

The relationship between pH and pKa is crucial in understanding aspirin absorption. Let's break it down.

pKa is a measure of the strength of an acid. For aspirin, pKa is 3.5.

When the pH of the environment is equal to the pKa of the acid, the concentrations of the charged (A^-) and uncharged (HA) forms are equal.

In environments with a pH lower than the pKa, the uncharged form (HA) is predominant.
When the pH is higher than the pKa, the charged form (A^-) is more common.

Let's apply this to aspirin:

• In the stomach (pH=1), which is lower than the pKa of aspirin, the uncharged form (HA) will dominate.


• In the small intestine (pH=6), which is higher than the pKa of aspirin, the charged form (A^-) will dominate.

This means that in the stomach, there will be more aspirin in a form that can easily pass through the membrane (uncharged). In the small intestine, aspirin will mostly be in a charged form, making it less likely to pass through the membrane.

Membrane Permeability

Understanding membrane permeability is key to knowing where aspirin is more likely to be absorbed.

Cell membranes are made up of lipid bilayers that form barriers to charged molecules. Neutral (uncharged) molecules can pass through these membranes much more easily.

Think of it this way:

• Charged molecules (like \( \text{A}^{-} \)) have a harder time getting through the membrane due to their charge.


• Uncharged molecules (like \( \text{HA} \)) can easily slip through the membrane because they aren't hindered by charge.

For aspirin to pass into the bloodstream, it needs to be in its uncharged form. In the low pH of the stomach, aspirin exists mostly in this uncharged form, making it easier to pass through the stomach lining into the bloodstream.

In contrast, in the higher pH environment of the small intestine, aspirin is mainly in its charged form, thus struggling to cross the membrane.

Therefore, more aspirin gets absorbed in the stomach where the conditions favor its uncharged form, maximizing its membrane permeability.