Given 1.00L 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 0.005 moles NaOH.
1000 mL x 0.338M = 338 millimols HAc
1000 mL x 0.093M = 93 mmols Ac^-
5.00 mmols NaOH added.
..........HAc + OH^- ==>Ac^- + H2O
initial...338...0.......93.....
add............5.00...............
change...-5.00..-5.00...+5.00....
equil...333......0.....98..........
Substitute from the ICE chart into the Henderson-Hasselbalch equation and solve for the new pH.