Question

A student prepared a 0.10 M solutino of acetic acid, CH3COOH. Acetic acid has a Ka of 1.75 x 10^-5. What are the hydronium ion concentration and the pH of the solution?

Answer 1

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.

Answer 2

1.75 x 10^-5 = [x][x]/[0.10-x]?

I'm completely lost...

Answer 3

See my response to your later post above.
Repost if you still don't get it.