What is the pH after 5 mL of 0.1M NaOH has been added to 25 mL of 0.2M acetic acid?
millimoles NaOH = 5 mL x 0.1M = 0.5
millimoles HAc = 25 mL x 0.2M = 5
...........NaOH + HAc ==> NaAc + H2O
initial....0.5.....5.0.....0.......0
change....-0.5....-0.5...+0.5...+0.5
equil.......0.....4.5......0.5....0.5
Substitute into the Henderson-Hasselbalch equation and solve for pH.