Chapter 16.4-16.5
Titrations of Weak Acids and Bases
Objectives
◦Separate the titration curve for a weak acid into
regions in which a single equilibrium dominates
◦Estimate the pH of mixtures of a weak acid and
strong base or weak base and strong acid.
◦Calculate the pH in the titration of a weak acid with
strong base.
◦Calculate the pH in the titration of a weak base with
strong acid.
October 13th, 2015
Titration Calculation Strategy
 Break titration into 4 zones:
• pH before any titrant is added.
• pH after titrant is added but before
equivalence point.
• pH at equivalence point.
• pH after equivalence point.
Titration of Weak Acid with Strong Base
Strong base region
Buffer region
pH
Equivalence point
Weak acid
0
5
10
15
20
25
Volume of NaOH added, mL
30
Titration of a Weak Base with a Strong Acid
14
Weak base
12
10
8
pH
Buffer region
6
4
Equivalence point
2
Strong acid region
0
0
5
10
15
20
Volume of HCl added
25
30
In Class Example
 Calculate the pH in the titration of
20.00 mL of 0.500 M formic acid (Ka =
1.8 x 10-4) with 0.500 M NaOH after 0,
10.00, 20.00, and 30.00 mL of base have
been added. Graph the titration curve. The
reaction is:
HCOOH + OH-  HCOO- + H2O
Zone 1: pH before any titrant is added
0 mL 0.500 M NaOH + 20.0 mL 0.500 M HCOOH
Ka = 1.8 x 10-4
No base has been added, so the solution only contains a
weak acid.
Therefore, use our knowledge of how to calculated pH
from Ka values.
H3 O+ [HCOO− ]
Ka =
[HCOOH]
Zone 2: pH after titrant is added but before equivalence point
The solution is a mixture of the weak acid and its
conjugate base – a buffer.
The pH can be determined using a sRfc table and the
Henderson-Hasselbalch equation.
nb
pH = pKa + log
na
sRfc Table for Zone 2
10.00 mL 0.500 M NaOH + 20.0 mL 0.500 M HCOOH
HCOOH + OH-  HCOO- + H2O
s, mmol
R, mmol
f, mmol
c, M
Zone 3: pH at equivalence point
Use the sRfc table to find the equivalence point.
The equivalence point occurs when the mmols of titrant
added equals the initial mmols of HCOOH.
At the equivalence point, only weak base is present in
solution, so the pH can be calculated from our
knowledge of Kb
HCOOH [OH− ]
Kb =
[HCOO− ]
sCrf Table at equivalence point
20.00 mL 0.500 M NaOH + 20.0 mL 0.500 M HCOOH
HCOOH + OH-  HCOO- + H2O
s, mmol
R, mmol
f, mmol
c, M
Zone 4: pH after equivalence point
After the equivalence point there is excess strong base.
There is also weak base around, but in Chapter 15.7 we
learned that the stronger base dominates the pH and the
weak base can be ignored.
Use the sRfc table to determine the amount of strong
base and use our knowledge of pKw to find pH
pKw = 14 = pH + pOH
sCrf Table at equivalence point
30.00 mL 0.500 M NaOH + 20.0 mL 0.500 M HCOOH
HCOOH + OH-  HCOO- + H2O
s, mmol
R, mmol
f, mmol
c, M
Student Example
 Calculate the pH in the titration of 12.00
mL of 0.100 M CH3COO- with 0.200 M HCl
after 0, 3.00, 6.00 and 9.00 mL of base
have been added. Kb = 5.510-10