Question

If you have 200 ml of a 0.1M solution of acetic acid (pka=4.75),
1. How many ml of a solution of 1.O M NaOH would you need to adjust the pH to 5.2?
2. How many ml of water would you then add to make the solution of 0.05M in total acetic acid (ie., acetic acid + acetate ion)

Answer 1

200 mL x 0.1M = 20 mmoles.
1.
............HAc + OH^- ==> Ac^- H2O
initial.....20.0...0........0
added..............x...........
change.......-x....-x........+x
equil......20-x.....0.........+x
Substitute into the Henderson-Hasselbalch equation and solve for x which will be mmoles NaOH. You can take it from there.

2.
You have 200 mL of 0.1M = 20 mmoles to start. If you want it to be 0.05, wouldn't you dilute to 400 mL?