stomach wall has a protective lining of mucous that prevent the acid from penetrating the underlying tissues. using asprin too often can damage the stomach wall. Aspirn is a weak carboxlyliic acid with a Ka of 3.2 x 10^-4.

HC8H7O2CO2(aq) <--> H^+(aq) + C8H7O2CO2^-(aq)

Stomach acid has a pH about 1.5. Given the acidity of the stomach, would the aspirn in the stomach fluid be mostly ionized or un-ionized?

I know that weak acid dissociate about 40%. That's all I know. I don't knwo how to answer the question :S

Hasp ==> H^+ + asp^-

(H^+)(asp^-)/(Hasp) = K
From a qualitative stand point, we can see that making H^+ high (1.5 pH) would shift the ionization equilibrium to the left making fewer ions and more of the unionized material. However, you can calculate the two by substituting the H^+ from pH of 1.5, and calculating the ratio of asp^- to Hasp. I estimated the value of (asp^-)/(Hasp)= about 0.01 (but you need to confirm that) which means that (asp^-) is small and (Hasp) is large so the quantitative data confirms the qualitative suggestion that most of the aspirin is in the unionized form.

Isn't the ratio of Hasp to asp 1 from the balanced chemical equation?

No. Think of acetic acid which you have done many times. If we call acetic acid, HA, (and we fan call aspirin HA), then

HA <==> H^+ + A^-
We know that for every molecule of HA that ionizes, 1 H^+ and `1 A^- are produced but that says nothing about HA and H^+ or HA and A^- being equal or the ratio being 1. It is true that the ratio of H^+ to A^- = 1. But as you can see from the calculation I did that (A^-)/(HA) = 0.01 and you know that must be so (at least that the ratio is a small number) because the H^+ of pH 1.5 will force the equilibrium to the far left.

are we doing the ratio between the number of mols? ;S

Actually, (A^-)/(HA) = 0.01 is the ratio of the concentrations in moles/L or M.

did you do this to find out A-:

3.2 x 10^-4 = (10^-1.5)^2/HA

No.

H^+ is not equal to A^.
HA ==> H^+ + A^-
Ka = 3.2 x 10^-4 = (H^+)(A^-)/(HA)
The problem tells us that the pH of the stomach is about 1.5 so that is the H^+.
1.5 = -log(H^+) and (H^+) = 0.0316 M.
Then (H^+)(A^-)/(HA) = 3.2 x 10^-4
0.0316(A^-)/(HA) = 3.2 x 10^-4
(A^-)/(HA) = 3.2 x 10^-4/0.0316 = about 0.01.
We can't calculate the actual values of (A^-) and (HA) because the (HA) in the stomach is not given; therefore, we calculate the ratio of th two to compare them.

What does the comparison tell us? :S

Does it tell us that there is more acid than aspirn?

(A^-)/(HA) = 0.01 tells us that the ionized portion of aspirin is MUCH less than the unionized portion; i.e. that little ionization has taken place and that most of the aspirin is in its original unionized form. That is that the (A^- or the ionized part) part is very low and that the (HA or the unionized part) is very high which is just another way of stating that the aspirin is mostly unionized when placed in a solution that is pH 1.5.

Ohhh. Should have thought about that. Thanks a lot Dr Bob :D