Making an ICE table with given info

CH3CH2OH (l) + 1/2 O2 (g) ----> CH3COOH (aq) +H2O (l)
I can't figure out how to create an ICE table to solve for the concentration:
"Breathalyzers stop working when the concentration of acetic acid becomes too high (0.15 M). What is the pH of a .15 M solution of scetic acid (Ka=1.8 x 10 ^-5^)

Ka = 1.8x10^-5 = [H+][C2H3O2-] / [HC2H3O2]

ICE chart

[HC2H3O2]..[H+]..[C2H3O2-]
0.15..............0..........0
-x................+x........+x
0.15-x..........x............x

1.8x10^-5 = x^2 / 0.15 - x
x^2 + 1.8x10^-5x - 2.7x10^-6 = 0
x = [H+] = [C2H3O2-] = 2.67x10^-3M
pH = -log[H+] = 2.57