At what pH is acetic acid 30% ionized?

Answer 1

Call acetic acid HAc
...............HAc---> H^+ + Ac^-
initial.........x......0......0
change.........-0.3x..+0.3x...+0.3x
equil........x-0.3x..... 0.3x.....0.3x
Ka = (H^+)(Ac^-)/(HAc)
Substitute equil conditions into Ka expression and solve for (H^+), then convert to pH.

Answer 2

I don't get how you put 30% ionization into te problem. So far I have: pH = 4.76 + log[A^-]/[HAc^-]