A .015M solution of a weak acid is found to be 1.3% ionized what is its ka?
Answer: 2.6 x 10^-5


Some of my thoughts:
Ka = [H+][A-]/[HA] so... it will be [X][X]/[.15-X]
The ionized part is confusing? Does this mean that:

Kw * .013 = [H+][OH-]
SO... 1 x 10^-14 * .013 / [1 x 10^-7] = {[H+]?

I am confused? Thank you in advance!!

Are you sure that is the answer?
I keep getting 2.6 x 10-6

It's pg. 78 in ACS gen chem. Yeah it's 2.6 x 10^-5, can you show me what your doing though. I am REALLY bad at Eq'm.

the trick to this question is that you can't really use any variables, because there is only one equation (the Ka) equation and you have the Ka as a variable already. So, as you might already know, a strong acid completely ionizes in water and a weak acid only partly ionizes in water. When it completely ionizes it would mean that ALL of the HA dissociates into H+ and A-. When it partially ionizes, would mean that only some of the HA would dissociate into H+ and A-. I guess you have to assume that it is a monoprotic acid because it doesnt state otherwise, but if it was something else like H2A or H3A, the equation would be different. But i guess for this one you have to assume its just HA. So the equation would be:
HA --> H+ + A-
HA --> 0.013(HA) + 0.013(HA)
1.3% dissociation means that 1.3% of HA becomes H+ and 1.3% of HA becomes A-.
so you would set this up a:

Ka = (0.013*0.015)(0.013*0.015)/0.015
= (0.013^2)*0.015
= 0.000169*0.015
=1.7*10^-4 * 1.5*10^-2
=2.6*10^-6

I think 2.6*10^-5 must be a typo.