Question

A solution of acetic acid, 1.0 mol L–1 CH3COOH, contains sodium acetate, NaCH3COO. The percent ionization of the above solution is 0.25%. The Ka for acetic acid is 1.8 x 10–5. Which of the following is the concentration of NaCH3COO in acetic acid?

So i use Ka = [H+][CH3COO] / [CH3COOH]

after this im guessing

but am i to go 1 - 0.0025 = 0.9975

and then, sub for ka =

1.8 x 10–5. = (0.0025x)^2 / 0.9975x

but when i did it i didn't get the right answer.. please help :)

Answer 1

Your error is in thinking the Ac is the same as HAc. They are equal in a solution of acetic acid but the problem tells you that NaAc has been added. Now the solution has H^+ one number and the Ac is whatever you've added. So make that
Ka = (0.0025)(Ac^-)/(0.9975)
and solve for Ac^-. That will be the concn of the sodium acetate in that solution.

Answer 2

Hi thanks for answering.
so after solving for [CH3OO-] = 7.182x10^-3

but the answer provided is, any ideas?

4.7 x 10–3 mol L–1