At what initial concentration would a solution of acetic acid (ka=1.8x10^-5) be 2% ionized?

Question

Thanks

Answer 1

Let's call acetic acid HAc. Then the ionization is
HAc ==> H^+ + Ac^-

We will call the initial concn of HAc x. If it is 2% ionized, then (H^+) = 0.02*x
(Ac^-) = 0.02*x
and (HAc) = x-(0.02*x) = 0.98*x
Now write the Ka expression and plug in those concns, then solve for x. Post your work if you get stuck. The answer is approximately 0.05 M

Answer 2

what is the difference between hydration and hydrogenation reactions?

Answer 3

How did you get 0.98? Why were you allowed to subtract 0.02 to get that? I'm still confused as to how to plug this in and solve, just because of the 0.98. Thanks