Homework Help: Chem

Posted by Katie on Sunday, March 28, 2010 at 8:03pm.

Aspirin (acetylsalicylic acid, ) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved 2.00 g of aspirin in 0.600 L of water and measured the pH . What was the Ka value calculated by the student if the of the pH of the solution was 2.62?

I know that Ka =[h3O+][a-]/[ha] but i don't know how to solve this.

so far i have
[h3o+]=-log(2.62)=2.4*10^-3
and
[oh-]=4.17*10-12

but i'm not sure what exactly this question is looking for. if you could point me in the right direction that would be great. thank you.

Chem - DrBob222, Sunday, March 28, 2010 at 8:17pm
It tells you it wants you to calculate the Ka for ASA. If we call it HA, then
HA ==> H^+ + A^-

Ka = (H^+)(A^-)/(HA)
You know (H^+) and you have that right.
(A^-) is the same as (H^+). I have used H^+ for H3O^+. For HA, you want to plug in the molarity of the aspirin which is moles/L. Find moles by grams/molar mass and you have it in 0.6 L.
Chem - Katie, Sunday, March 28, 2010 at 8:42pm
I'm still not getting the correct answer:

i have 2.00g/180.15g/mol =1.11*10^-2

1.11*10^-2/.6=1.85*10^-3

and (2.4_10^-3)^2/1.85*10^-3 = 3.11*10^-9.

what am i doing wrong?

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