Question

A 25.0-mL solution of 0.100 M acetic acid is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution:

a) 0.0 mL b) 5.0 mL c) 10.0 mL d) 12.5 mL e) 15.0 mL

Answer 1

Ka = 1.8 x 10^-5

pKa = 4.74

millimoles of acid = 25 x 0.10 = 2.5

1) before the titration

pH = 1/2 (pKa -logC)

pH = 1/2 (4.74-log(0.1))

pH = 2.87

2) 5 ml 0.2M NaOH

    NaOH millimoles = 5 x0.2 = 1

CH3COOH + NaOH --------------------> CH3COONa + H2O

2.5 1                                0                       0 ---------------initial

   1.5 0 1 1 ------------equilibrium   

mixture contain weak acid and salt so it forms buffer

For acidic buffer

Henderson-Hasselbalch equation

pH = pKa + log[salt/acid]

pH = 4.74 + log[1/1.5]

pH = 4.56

4) 10 ml NaOH added

millimoles of NaOH = 10 x 0.2 = 2

CH3COOH + NaOH --------------------> CH3COONa   +     H2O

2.5 2                                0                       0 ---------------initial

0.5 0 2 2---------------equilibrium

pH = 4.74 + log[2/0.5]

pH = 5.34

4)12.5

equivalence point

C =   2.5 / (25 + 12.5) = 0.067 M

pH = 8.78