Question: Calculate the change in pH of a 0.400 M C6H5COOH solution when 100.0 ml of a 2.50 M KOH solution is added to 1.000 L of C6H5COOH. For C6H5COOH, Ka = 6.46 X 10^-5. Work Done So Far: We're to determine the change in pH, thus we'll need to find the original pH. Ka = 6.46 X 10^-5 = x^2/(.4-x) ==> x = .0050511 ==> pH = 2.296 What should I do next? I understand that KOH is a strong base so it will tear the excess H off of C6H5COOH leaving C6H5COO-, but do we need to account for the H+ left over from the C6H5COOH there before the KOH was added?