Baza said:Hi there, what else do you know about the acid you are trying to calculate? Do you know the [HA] concentration and [H+] concentration?
Originally Posted by dr_jo
Hey.
So:
-logKa=2.91
logKa=-2.91 (multiply both sides by -1)
(then take ten to the power of each side)
Ka=10^-2.91=0.00123.....
Hope that helps.
(Assuming your log is to the base 10. If you're not sure, then it'll be 10)
Ormorof said:i have this equation:
[H3O+]2/1 (which is [H3O+]2, 1 being 1mol/dm3 for the conc of the acid) and i have the pH of the acid which is 2.91
the acid in question is C9H15COOH
the ionic equation being
C9H15COOH <> H3O+ + C9H15COO-
thanks for any help btw
edit, someone on general forums seems to have cracked it, thanks for help anyway (needed to use the antilog, and answer looked right)
pixieblue said:Brainy people Scare the hell outta me...