Question

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

Answer 1

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