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 absorbed in the stomach (pH = 1).

Step-by-step solution

- Understand the problem

We need to determine where aspirin (an acid with \(\text{p}K_a = 3.5\)) is more likely to be absorbed: in the stomach (pH = 1) or in the small intestine (pH = 6). Remember, electrically neutral molecules pass through membranes more easily than charged molecules.

- Recall the concept of \(\text{p}K_a\)

The \(\text{p}K_a\) value indicates the pH at which half of the acid is in its protonated (uncharged, \( \text{HA} \)) form and the other half in its deprotonated (charged, \(\text{A}^-\)) form.

- Apply the Henderson-Hasselbalch equation

Use the Henderson-Hasselbalch equation: \[ \text{pH} = \text{p}K_a + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \] to find the ratio of deprotonated to protonated aspirin at different pH values.

- Calculate the ratio at stomach pH

For the stomach (pH = 1): \[ 1 = 3.5 + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \] Rearrange to find \(\frac{[\text{A}^-]}{[\text{HA}]},\) \[ \text{log} \frac{[\text{A}^-]}{[\text{HA}]} = 1 - 3.5 = -2.5 \] \[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{-2.5} \]

- Calculate the ratio at small intestine pH

For the small intestine (pH = 6): \[ 6 = 3.5 + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \] Rearrange to find \(\frac{[\text{A}^-]}{[\text{HA}]},\) \[ \text{log} \frac{[\text{A}^-]}{[\text{HA}]} = 6 - 3.5 = 2.5 \] \[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{2.5} \]

- Analyze the results

At pH = 1, \[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{-2.5} = 0.0032, \] indicating mostly uncharged species. At pH = 6, \[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{2.5} = 316, \] indicating mostly charged species.

- Conclusion

Since electrically neutral molecules pass through membranes more easily, more aspirin will be absorbed in the stomach (pH = 1), where it is predominantly in its uncharged form.

Key concepts

pKa

To start understanding aspirin absorption, it's crucial to know about **pKa**. The pKa is a measure of the strength of an acid. Specifically, it's the pH at which half the acid molecules are in their protonated (uncharged) form (HA) and half are in their deprotonated (charged) form (A⁻). For aspirin, the pKa is 3.5. This means at a pH of 3.5, half the aspirin will be in its uncharged form and half in its charged form. This balance shifts with the pH of the environment.
In environments with pH below the pKa, like the stomach (pH = 1), more aspirin will be in its uncharged form. In more basic environments, like the small intestine (pH = 6), aspirin will mostly be in its charged form.
Understanding this balance is essential for predicting where aspirin will be more absorbed.

Henderson-Hasselbalch equation

The **Henderson-Hasselbalch equation** helps to understand how pKa and pH relate to each other. The equation is given by:
\[ \text{pH} = \text{p}K_a + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \]
It can predict the proportion of an acid that is in its charged and uncharged forms at a specific pH.
Let's break it down for aspirin in both the stomach and intestine.
For stomach (pH = 1):
\[ 1 = 3.5 + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \]
Rearranging:
\[ \text{log} \frac{[\text{A}^-]}{[\text{HA}]} = 1 - 3.5 = -2.5 \]
\[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{-2.5} = 0.0032 \]
This indicates most aspirin is in the uncharged form in the stomach.
For the small intestine (pH = 6):
\[ 6 = 3.5 + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \]
Rearranging:
\[ \text{log} \frac{[\text{A}^-]}{[\text{HA}]} = 6 - 3.5 = 2.5 \]
\[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{2.5} = 316 \]
This reveals most aspirin is in the charged form in the small intestine.
The equation thus helps visualize aspirin's state in different pH environments.

pH and membrane permeability

The pH of a solution greatly affects aspirin's **membrane permeability**. Membranes are like barriers that selectively allow certain molecules to pass through. Electrically neutral molecules pass through membranes more easily than charged ones. This means the form aspirin takes (charged vs. uncharged) will determine how well it is absorbed.
In the stomach (pH = 1):
Due to the low pH, most aspirin is in its uncharged form. Membranes are more permeable to this form, so more aspirin gets absorbed here.
In the small intestine (pH = 6):
Because of the higher pH, aspirin is mostly in its charged form. Charged molecules struggle to pass through the membrane, leading to less absorption.
Understanding pH and membrane permeability is key for figuring out where a substance like aspirin will be best absorbed in the body.