millimoles HCl = 8.00 mL x 0.1M=0.8mmols.
mmoles NaOH = 2.50 mL x 0.1M = 0.25 mmols.
mmoles NaOH = 9.50 x 0.1M = 0.95 mmoles.
mmoles HAc = 8.00 mL x 0.1M = 0.8 mmoles.
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..........HCl + NaOH ==> H2O + NaCl
initial...0.8....0........0......0
add.............0.25.................
change...-0.25..-0.25....+0.25..0.25
equil...0.55......0.......0.25..0.25
pH of the soln is determined by the excess HCl since there is no NaOH present and NaCl is not hydrolyzed.
(HCl) = 0.55mmoles/(100+8.00+2.50)mL = ?? and convert to pH.
initial......0.8..0......0.......0
add...............0.95.............
change.....-0.8...-0.8...+0.8..+0.8
equil........0....0.15....0.8..0.8
pH of the soln is determined by the excess NaOH since there is no HCl present and NaCl is not hydrolyzed.
(NaOH) = 0.15mmoles/(100+8+9.5)mL = ?, convert to pOH and pH.
...........HAc + NaOH ==> NaAc + H2O
initial...0.8.....0........0.......0
add..............0.25..............
change...-0.25..-0.25....+0.25...+0.25
equil.....0.55.....0.......0.25...0.25
Here you have a soln in HAc and NaAc so that is a buffer. Use the Henderson-Hasselbalch equation to solve for pH.
pH = pKa + log (0.25/0.55) = ?
............HAc + NaOH ==> NaAc + H2O
initial....0.8.....0........0......0
add...............0.95.............
change...-0.8...-0.8......+0.8..+0.8
equil.......0.....0.15.....0.8....0.8
Here you have a soln in NaOH and NaAc but that isn't a buffer and the pH will be determined by the excess NaOH present. It is true that the Ac^- will hydrolyze but it is small enough to ignore it.
(NaOH) = mmoles NaOH/(100 + 9.5+8)mL = ?, convert to pOH and to pH.
Post your work if you get stuck.