1. STK1094 Analytical Chemistry 1
Dayang Norafizan binti Awang Chee
Faculty of Resource Science and Technology
Universiti Malaysia Sarawak
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2. Learning Objectives
At the end of this lesson, students should be able to:
 Describe the difference between an “endpoint” and an
“equivalence point” in an acid-base titration
 Identify the equivalence point in an acid-base titration
from the pH titration curve
 Illustrate titration of a weak acid with a strong base.

3. Defining Terms
• Standard solution: A reagent of a known concentration which used in
the titrimetric analysis
• Titration: This is performed by adding a standard solution from burette
or other liquid-dispensing device to a solution of the analyte until the
point at which the reaction is believe to be completed
• Equivalence point: Occurs in a titration at the point in which the
amount of added titrant is chemically equivalent to the amount of
analyte in a sample
• Back-titration: This is a process in which an excess of the standard
titrant is added, and the amount of the excess is determined by back
titration with a second standard titrant.
In this instance, the equivalent point corresponds with the amount of
initial titrant is chemically equivalent to the amount of analyte plus the
amount of back titrant.

4. Defining Terms
• End point: The point in titration when a physical change occurs that is
associated with the condition of chemical equivalence
• Indicators are used to give an observable physical change (end point)
or at near the equivalence point by adding them to the analyte. The
difference between end point and equivalence point should be very
small and this difference is referred to as titration error.
• Titration error, Et
Et=Vep – Veq
Vep is the actual volume used to get to the end point.
Veq is the theoretical value of reagent required to reach the end point.

5. 5
Acid-Base Titrations
• A quick and accurate method for determining acidic
or basic substances in many samples.
• The titrant is typically a strong acid or base.
• The sample species can be either a strong or weak
acid or base.

6. 6
Acid-Base Titration
Types of acid-base titrations:
1)strong acid – strong base titration
2)weak acid – strong base titration
3)strong acid – weak base titration
4) polyprotic acid – strong base titration
5) polybasic base – strong acid titration

7. 7
Strong Acid - Strong Base Titration
In strong acid – strong base titration, there are three
regions of the titration curve that represent different kinds
of calculations :
- before equivalence point
- at equivalence point
- after equivalence point

8. Titration curve
Strong acid titrated with a strong base:
• The net reaction is
H3O++ OH- → 2H2O
• Before the equivalence point, acid is present in excess
• pH is determined by the concentration of excess HCl
[𝐻3 𝑂+
] =
𝑚𝑚𝑜𝑙𝑒𝑠 𝑎𝑐𝑖𝑑−𝑚𝑚𝑜𝑙𝑒𝑠 𝑏𝑎𝑠𝑒
𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
8

9. Strong acid titrated with a strong base
• At equivalence point, moles of acid and moles of base are
equal.
• At equivalence point,
[H3O+] = [OH- ]
pKw = 14 = pH + pOH
pH = 7
• So, the equivalence point for strong acid/base is always a
pH=7
9

10. Strong acid titrated with a strong base
Overtitration
 Pass the equivalence point, we don’t have any acid remaining. All that we are doing
is diluting our titrant.
[𝑶𝑯−
] =
𝒎𝒎𝒐𝒍𝒆𝒔 𝒆𝒙𝒄𝒆𝒔𝒔
𝒕𝒐𝒕𝒂𝒍 𝒗𝒐𝒍𝒖𝒎𝒆
pH = 14-pOH
Eg. Construct a titration curve for the titration of 100 mL 0.1 M HCl with 0.1 M NaOH
1) Volume of NaOH needed to reach eq. point
Moles HCl = moles NaOH
VNaOH = 100.0 mL
2) Before addition of NaOH
pH = - log [0.1] = 1
3) After addition of 10mL NaOH
[𝐻3 𝑂+] =
100 𝑚𝑙 0.10𝑀 −(10𝑚𝑙)(0.10𝑀)
100 𝑚𝑙+10 𝑚𝑙
= 0.082 M, pH= 1.09
10

11. 11
mL
titrant
Total mL [H3O+] pH
0 100 0.10 1.00
10 110 0.082 1.09
20 120 0.067 1.17
30 130 0.054 1.28
40 140 0.043 1.37
50 150 0.033 1.48
60 160 0.025 1.60
70 170 0.018 1.74
80 180 0.011 1.96
90 190 0.0053 2.28

12. Titration curve
4) At equivalence point
• Equivalence point, moles of
HCl= moles of NaOH
• Since neither is in excess, pH is
determined by Kw
Kw = 1.00 x 10-14 =
[H3O+][OH-] = [H3O+]2
[H3O+] = 1.00 x 10-7
pH= 7
• Note that for the first 90 mL of
titration, pH = 2.28
• At eq. point, the pH value jump
of 4.72 pH unit
12

13. Titration curve
5) Overtitration
• Account for the dilution of titrant
• 10 mL overtitration
[OH-] = moles excess NaOH = MVNaOH- MVHCl
Vtotal
= (0.1M)(110mL)-(0.1M)(100mL)
210 mL
= 0.0048M
pOH = 2.32
pH = 14-2.32 = 11.68
13

14. 14
mL
titrant
Total
volume
[OH-] pH
110 210 0.0048 11.68
120 220 0.0091 11.96
130 230 0.013 12.11
140 240 0.017 12.23
150 250 0.020 12.30
160 260 0.023 12.36
170 270 0.026 12.41
180 280 0.029 12.46
190 290 0.031 12.49
200 300 0.033 12.52

15. Titration curve
Titration of a strong base with a
strong acid:
• If we plot pOH rather than pH,
the result still look identical.
• Typically, we still plot pH verses
mL titrant, so the curve is
inverted.
15

16. Titration of weak acids & weak bases with
strong titrant
• Must concerned with conjugate acid/base pairs & their
equilibrium
16

17. Titration of weak acids & weak bases with
strong titrant
Before titration:
• If the sample is weak acid, then use
• 𝐾𝐴=
𝐻3 𝑂+ [𝐴−]
[𝐻𝐴]
• [H3O+]=[A-]
• Calculate the pH value
17
• If the sample is weak base, then use
• 𝐾 𝐵=
𝑂𝐻− [𝐻𝐴−]
[𝐴−]
• [OH-]=[HA]
• Calculate the pH value = 14 - pOH

18. Titration of weak acids & weak bases
with strong titrant
Before equivalence point:
• Equilibrium expression used is the Henderson-
Hasselbalch equation
• Starting with an acid
• pH= 𝑝𝐾 𝐴 + log
[𝐴−]
[𝐻𝐴]
• Starting with base
• pH= 14 − (𝑝𝐾 𝐵 + log
[𝐻𝐴]
[𝐴−]
)
18

19. Titration of weak acids & weak bases
with strong titrant
At equivalence point
• All sample is converted to its conjugate form
• If the sample was an acid-solve the pH using KB
relationship
• If the sample was a base-solve the pH using KA
relationship
𝐾 𝐵+ 𝐾𝐴= 14
19

20. Titration of weak acids & weak bases with
strong titrant
Overtitration:
• Identical to strong acid/strong base example.
• Need to account for the amount of excess titrant & how much it
has been diluted.
• Eg. 100 mL solution of 0.1 M benzoic acid is titrated with 0.1 M
NaOH. Construct a titration curve.
For benzoic acid
Ka=6.31 x 10-5
pKa=4.20
1) Volume of NaOH needed to reach eq. point
Moles C5H6COOH = moles NaOH
VNaOH = 100.0 mL
20

21. Titration of weak acids & weak bases with
strong titrant
2) Before titration:
• 𝐾𝐴=
𝐻3 𝑂+ [𝐴−]
[𝐻𝐴]
• [𝐻3 𝑂+]= [𝐴−]
• Assume [A-] is negligible compared to [HA]
𝐾𝐴= 6.31 × 10−5=
𝑥2
0.10
= 6.31 × 10−5 (0.10)
= 0.025 M
pH= 2.60
21

22. Titration of weak acids & weak bases with
strong titrant
3) After addition of 10mL NaOH
Henderson-Hasselbalch equation
pH = pKa + log [C5H6COO-]
[C5H6COOH]
[C5H6COOH] = moles unreacted C5H6COOH = MVC5H6COOH- MVNaOH
Vtotal Vtotal
= (0.1M)(100mL)-(0.1M)(10mL)
110 mL
= 0.082M
[C5H6COO-] = moles NaOH added = MVNaOH
Vtotal Vtotal
= (0.1M)(10mL)
110 mL
= 0.009M
pH = 4.2 + log (0.009/0.082) = 3.24
• Calculate other point by repeating this process
22

23. Titration of weak acids & weak bases with
strong titrant
23
mL
titrant
pH
0 2.60
10 3.24
20 3.60
30 3.83
40 4.02
50 4.20
60 4.38
70 4.57
80 4.80
90 5.15
mL titrant

24. Titration of weak acids & weak bases with
strong titrant
4) At equivalence point
100mL titrant:
• All acid has been converted to its conjugate base – benzoate
• Use KB relationship.
• 𝐾 𝐵=
[𝑂𝐻−][𝐻𝐴]
[𝐴−]
• 𝐾 𝐵= 𝐾 𝑊/𝐾𝐴= 1.58 × 10−10
n benzoic acid = n NaOH
• Predominate ion in solution is A-, which is a weak base
[A-] = moles acid/ total volume = 0.05M
• We have diluted the sample & the total volume at this point is 200 mL.
• We can assume that [benzoic acid] is negligible compared to [benzoate].
24

25. Titration of weak acids & weak bases with
strong titrant
4) At equivalence point
C5H6COO- (aq) + H2O (l) OH- (aq) + C5H6COOH
𝐾𝑏= 1.58 × 10−10
=
𝑥2
0.050
𝑥 = (1.58 × 10−10)(0.050)
25
mL titrant

26. Titration of weak acids or weak bases with
strong titrant
5) Overtitration
• Need to account for the dilution of titrant.
• Eg: 10 mL excess.
26

27. Titration of weak acids or weak bases with
strong titrant
27
mL
titrant
Total
volume
[OH-] pH
110 210 0.0048 11.68
120 220 0.0091 11.96
130 230 0.013 12.11
140 240 0.017 12.23
150 250 0.020 12.30

28. © 2009, Prentice-Hall, Inc.
Titrations of Polyprotic Acids
When one
titrates a
polyprotic acid
with a base there
is an equivalence
point for each
dissociation.
Titration curve for the reaction of
50.0 mL of 0.10 M H3PO3 with
0.10 M NaOH

29. Titrations of Polybasic Base
8-29

30. Self-Reflection
• What is the difference between end-point and
equivalence point?
• How to build the titration curve for strong
acid/strong base with weak acid/weak base
and vice versa?