a 250.0 ml buffer solution is 0.250 M in acetic acid and. 250M in sodium acetate. what is the ph after addition of. 0050 mol of HCL? what is the ph after the addition of. 0050 mol of NaOH?

Answer 1

250 mL x 0.250M HAc(CH3COOH) = 62.5 mmoles to begin.
250 mL x 0.250M NaAc(CH3COONa) = 62.5 mmoles to begin. Write the reaction and make a table. It proves invaluable.
............Ac^- + H^+ ==> HAc
begin......62.5.....0.......62.5
add H^+.............5.........
react......-5.0....-5.......+5.0
final......57.5.....0........67.5
pH = pKa + log [(57.5)/(67.5)] = ??

For the addition of 0.0050 mol (5.0 mmols) base, I do this.
.............HAc + OH^- ==> Ac^- + H2O
begin........62.5...0.......62.5....0
add OH^-............5.0................
react........-5.0..-5.0......+5.0.+5.0
final........57.5...0........67.5...5.0

pH = pKa + log[(67.5/57.5) = ??

Answer 2

Ka = 0.000018