The added Base is 180mL of .1M NaOH. This is the base which will have .018 moles. Divided by the 380mL or .380 L then we have the value to plug into the HH (approx .04736). The acid concentration we get from the ICE table. HAc <--> H + Ac-

Ka = [H][Ac-]/[HAc]

Ka = x^2/.1-x we ignore the bottom x because the Ka is so small and we get x^2/.1 = 1.8e-5, x^2 = 1.8e-6 and x = 1.34e-3. This is the Molarity in 200mL so we need to compensate for the new volume. 1.34e-3 x .2L = 2.68e-4 moles. Divide this by new volume (.38L) get 7.05e-4 as the Molarity of acid. Plugging this into HH we get pH = p(1.8e-5) + log(.047/7.05e-4) = 6.56 which is too high, the answer given is 5.7. Am I making a math error, a procedural error, or is this just the wrong use of the equation? Why would we use the concentration of acid and the pKa of the acid but only the concentration of the base? What happens if we have strong acid and weak base? Weak and weak? Do we always use this equation?