Given 2.00L of a buffer containing 0.338 M CH3COOH and 0.093 M CH3COONa. Ka(acetic acid) = 1.8 x 10-5. Calculate the pH after the addition of 0.004 moles NaOH.

Answer 1

millimols HAc = 2000 mL x 0.338 = approx 670
mmols NaAc = 2000 x 0.093 = about 180.
Add 4 mmoles NaOH.

............HAc + OH^- ==> Ac^- + H2O
initial.....670....0........180
add...............4..................
change.....-4.....-4.........+4
equil......666.....0...........f

Now plug the equil amounts into the Henderson-Hasselbalch equation and solve for pH. My number above are approximate only; you should recalculate all of them.