Posted by Sue on Friday, January 18, 2008 at 9:04pm.
Let's let CH3COOH be represented by HAc (H is the H of the COOH and Ac stands for the remaining part of the molecule.).
HAc ==> H^+ + Ac^-
Beginning concentrations:
(HAc) = 0.1
(H^+) = 0
(Ac^-) = 0
After ionization:
(H^+) = x
(Ac^-) = x
(HAc) = 0.1 - x
Write the Ka expression and plug the after ionization into it in the appropriate places, then solve for x. Following that, use pH = -log(H^+).
Post your work if you get stuck.
1.75 x 10^-5 = [x][x]/[0.10-x]?
I'm completely lost...
See my response to your later post above.
Repost if you still don't get it.
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