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
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
I have another question related to this one. The un-ionized aspirn penetrates the area of the stomach were there is less acidity thus causing irritation. How would Le Chatelier's principle explain why the irritation occurs?
I don't know that it does. As I understand it, the acid breaks through the membrane that is there to protect the stomach lining. Perhaps the membrane becomes thinner because of age or something we eat, but whatever, the acidity gets to the flesh part of the body and that is where the acid does its damage. Its the same thing as placing a drop of HCl on your arm. Of course that will burn because it is damaging the skin. And if allowed to stand there several minutes, it can cause severe damage to the outer layers as well as the inner layers of the skin cells below the skin. What happens in the burning process is more related to biochemistry, I think, and less to Le Chatelier's Principle. But then I'm neither an MD nor a biochemist so I'm guessing.
Le Chatelier's principle involves equillibriums. I don't understand the equillibrium involved in thsi situation?
The question asked why the irritation occur specifically in the area of lower acidity.
I don't know. Perhaps GK or Dr Russ will take a look.
Dr Bob, I think I read the questions wrong. It said that un-ionized aspirn molecules can readily penetrate the stomach lining into a regio of less adicity. This is where the stomach irritation associated with aspirn occurs. Use le Chateliers principle to explain why irritation occurs in this area. Sorry :(