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?