Determine the volumes of 1 M acetic acid and 1 M NaOH required to make a 200mL 0.1 M buffer with a pH of 5.5. (Assume the pKa of acetic acid is 4.76).

Answer 1

You want 200 mL x 0.1M buffer which means 20 millimoles. For the buffer to be 01M, that means acid (HAc) + base (Ac^-) = 0.1M
Since mL x M = mmoles, we want to start with 20 mL of 1M base to start with 20 mmoles. For final pH = 5.5, we can calculate the base/acid ratio as
5.5 = 4.76 + log (b/a)
b/a = 5.5 or
Ac = 5.5*HAc
Solve these two equations simultaneously as
B + A = 0.1M
B/A = 5.5
You should confirm but I get approximately 0.015M for acid and 0.085M for base. Convert to mmoles gives
200 x 0.015 = about 3 mmoles HAc
200 x 0.085 = about 17 mmoles for Ac
..............HAc + OH^- ==> Ac^- + H2O
initial.......20.....0.........0......0
add..................x..................
change.........-x....-x.........x.....x
equil..........3......0.........17
Thus x = 17 mmoles and you will need to add 17 mL of 1M NaOH to provide 17 mmoles NaAc and leave 3 mmoles Ac^-. Check all of this work. Note that you will need to add enough water to make the final volume 200 mL.

Answer 2

Rachel--As I posted my response above it occurred to me that I've worked several of these buffer problems for you before the 2012 New Year's holidays. You need to learn how to do these yourself. If you care to go into detail about why you still have trouble with them perhaps I can help you over the tight spots.