Benzoic acid (C6H5COOH)is a monoprotic weak acid with Ka=6.30*10^-5. What is the pH of a solution of benzoic acid that is 0.559M and has 2.25*10^-2M NaOH added?

Answer 1

By adding NaOH you have formed a buffered solution. Take 1 L soln, so you have
1000 x 0.559 M = 559 millimols PhCOOH.
1000 x 0.0225M = 22.5 mmols NaOH

.......PhCOOH + NaOH ==> PhCOONa + H2O
initial..559................0........0
added...........22.5...................
change....-22.5..-22.5.....22.5
equil....536.5...0..........22.5

Use the Henderson-Hasselbalch equation and solve for pH.