Calculate the pH of 0.1M Benzoic Acid (C6H5COOH). Ka = 6.3x10-5 M

Ka = [H⁺][B^-]/[HB]Benzoic acid dissociates to C6H6COO^- and H⁺ - this is a 1:1 ratio so the concentrations are equal hence [H⁺][B^-] = [H⁺]²6.3x10-5 = [H⁺]²/0.16.3x10-6 = [H⁺]² therefore H⁺ concentration is 0.0025MpH = -log([H⁺]) therefore pH = -log(0.0025M) = 2.61