Answer:
1) after 10.0 mL base added the pH is 4.75
2) after 20.0 mL base added the pH is 8.74
3) after 30.0 mL base added the pH is 12.36
Explanation:
CH₃COOH + NaoH --------------> CH₃COONa + H₂O
1) after 10.0 mL base added
millimoles of NaOH = 10.0 x 0.125 = 1.25
2.50 - 1.25 = 1.25 means 50% titration complete.
at half equivalence point pH = pKa
pH = 4.75
2) millimoles of NaOH added = 20.0 x 0.125 = 2.50
equivalence point
[CH₃COONa] = 2.50 / 20 + 25 = 0.055 M
pH = 1/2 [pKw + pKa + log C]
pH = 1/2 [14 + 4.75 + log 0.055]
pH = 8.74
3) millimoles of NaOH added = 30.0 x 0.125 = 3.75
3.75 - 2.50 = 1.25 millimoles NaOH left
[NaOH] = 1.25 / 30 + 25 = 0.023 M
as NaOH is strong base
[OH-] = [NaOH] = 0.023 M
pOH = - log [OH-]
pOH = - log [0.023]
pOH = 1.64
pH = 14 - 1.64
pH = 12.36
