What is the pH at each of the following points for the titra

What is the pH at each of the following points for the titration of 25.00mL of 0.100M CH3COOH (K 1.8 x 10-5) with 0.100M NaOH? a) Before the addition of NaOH b) After the addition of 10.00mL NaOH c) After the addition of 12.50mL NaOH d) After the addition of 25.00mL NaOH e) After the addition of 26.00mL NaOH

Solution

Ka = 1.8 x 10^-5

pKa = 4.74

millimoles of acid = 25 x 0.10 = 2.5

a) before the titration

pH = 1/2 (pKa -logC)

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

pH = 2.87

b) 10.00 mL NaOH :

mmoles of NaOH = 10 x 0.1 = 1

mmoles of acid = 25 x 0.1 = 2.5

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

2.5                        1                                                0

1.5                         0                                           1

pH = pKa + log [salt / acid]

    = 4.74 + log [1 / 1.5]

pH = 4.56

c) half -equivalence point :

at half equivalence point moles of acid = moles of salt

here pH = pKa

[salt/acid] = 1

pH = pKa + log[salt/acid]

pH = 4.74 + log(1)

pH = 4.74

d) 25 mL of NaoH :

Equivalence point . here salt onl y remains .

salt concentration = 2.5 / 50 = 0.05 M

pH = 7 + 1/2 (pKa + log C)

    = 7 + 1/2 (4.74 + log 0.05)

pH = 8.72

e) 26 ml NaOH added

mmoles of NaOH = 2.6

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

2.5              2.6                              0                       0 ---------------initial

0                  0.1                               2.5                 2.5 ----------------equilibrium

here strong base NaOH remianed in the mixture . so pH can be decided by strong base

[OH-] concentration = millimoles /total volume

                                  = 0.1/(25+26)

                                   = 1.96 x 10^-3 M

pOH = -log[OH-] = -log(1.96 x 10^-3) = 2.71

pH = 14-pOH

pH = 14 - 2.71

pH = 11.29