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
MB
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