What was the original concentration ( in M) of the acetic acid solution?

Question

A solution of acetic acid is 2.0% dissociated at 25.0 oC. What was the original concentration ( in M) of the acetic acid solution? The Ka at 25.0 oC for acetic acid is 1.8 × 10-5.

Answer 1

In order to minimize typing I use the general formula for monoprotic acids HA

.. .. .. .. .. .. HA <=> H+ +A-
Initial .. .. .. C
Dissoc. .. .. x
Produce .. .. .. .. .. .. x .. ..x
At Equil .. C-x .. .. .. x .. .. x

Ka= [H+][A-] / [HA] =x^2/(C-x) (1)

By definition the degree of dissociation is
a= dissociated / initial = x/C =>x=aC (2)

Substitute x in (1) using (2) and you get

Ka = a^2 * C^2 / (C-aC) =>

Ka = a^2*C / (1-a) (This is a general formula that you can use for example if they were asking you to calculate a)

so C= Ka*(1-a) / a^2 = (1.8*10^-5)*(1-0.02) / 0.02^2 = 0.0441 M