PH of a acetic acid, which has a dissociation constant?

CH3COOH <-> CH3COO- + H+

Ka = [CH3COO-][H+]/[CH3COOH] = 1.76E-5 m/l

[CH3COOH] = 0.05 m/l
[H+] = [CH3COO-] m/l

let x = [H+] m/l

then

1.76E-5 = x^2/5E-2

solve for x

x = ((1.76E-5)(5E-2))^.5 = (8.8E-7)^.5 = 9.38E-4

[H+] = x = 9.38E-4 m/l

pH = -log[H+] = -log(9.38E-4) ~= -log(1E-3) = 3

Rap