posted by Katie .
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
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.
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.
I'm still not getting the correct answer:
i have 2.00g/180.15g/mol =1.11*10^-2
and (2.4_10^-3)^2/1.85*10^-3 = 3.11*10^-9.
what am i doing wrong?