Question

A chemist wishes to prepare 250mL of a buffer that is pH = 4.50. Beginning with 100mL of 0.12 mol L^(-1) acetic acid and a supply of 0.10 mol L^(-1) NaOH, explain how this could be done. How much 0.20 mol L^(-1) NaOH must be added to this buffer to raise the pH to 5.1? If the same amount of 0.20 mol L^(-1) NaOH were added to 250 mL of deionized water, what would the new pH be?

Answer 1

I couldn't guess.
How many millimoles acetic acid (HAc) do you have? That's 100 mL x 0.12M = 12.
How much NaOH must be added? That's 0.1 M * x mL = 0.1x

.........HAc + OH^- ==> Ac^- + H2O
initial..12.....0........0.......0
add............0.1x...............
change.-0.1x..-0.1x......0.1x....0.1x
equil..12-0.1x..0.......0.1x.....0.1x

pH = pKa + log(base)/(acid)
4.50 = pKa + log(0.1x)/(12-0.1x)
Solve for x = mL 0.1M NaOH to be added to form buffer of pH = 4.50.

I will leave the last two parts for you; the last one is strictly a strong base problem. The other one follows this same kind of procedure using the Henderson-Hasselbalch equation.
Post your work if you get stuck.

Answer 2

uno for the pKa would it simply be 4.50 as well?

Answer 3

No. It's the pKa for acetic acid. Ka = 1.8E-5. pKa = -log Ka.