A 25.0 ml sample of a 0.100 m solution of acetic acid is titrated with a 0.125 m solution of naoh. Calculate the ph of the titration mixture after 10.0

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

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