calculate (a) the H+ ion concentration,(b) the pH, &(c)the percent ionization of a 0.04 M solution of HC2H3O2?

Equation:
Ka = [H+] * [CH3COO-] / [CH3COOH]
We know that [H+] = [CH3COO-] and because dissociation is small, we can take [CH3COOH] = 0.04
Then:
1.8*10^-5 = [H+]²/0.04
[H+]² = (1.8*10^-5)* 0.04
[H+]² = 7.2*10^-7
[H+] = 8.48*10^-4 that is answer to (a)

pH = -log [H+]
pH = -log ( 8.48*10^-4)
pH = 3.07 answer to (b)

% ionisation = (8.48*10^-4) / 0.04 *100 = 2.12% answer to part (c).