a 250.0ml buffer solution is 0.250M in ascetic acid and 0.250M in sodium acetate.

Question

B). What is pH after addition of 0.0050mol of HCl?

C). What is pH after addition of 0.0050mol of NaOH?

Answer 1

250 mL x 0.250M = 62.5 millimols HAc
250 mL x 0.250M = 62.5 mmols NaAc.
add 5 mmols HAc
add 5 mmols NaOH
............Ac^- + H^+ ==> HAc
initial.....62.5..........62.5
add................5...........
change......-5....-5......+5
equil.....57.5.....0......67.5

Substitute into the Henderson-Hasselbalch equation and solve for pH.
The same process will work the NaOH part.
equil.....