I NEED HELP ASAP

Question

calculate the ph of a 0.5 m solution of acetic acid. acetic acid has a dissociation constant of 1.8x10^-5

Answer 1

Let acetic acid be represented by HAc

HAc ==> H⁺ + Ac^-

Ka = 1.810^-5 = [H⁺][Ac^-]/[HAc] = (x)(x) / 0.5

x² = 9x10^-6

x = 3x10^-3 M = [H⁺]

pH = -log [H⁺]

pH = -log 3x10^-3

pH = 2.52