UNIT_3_VOLUMETRIC_TITRIMETRIC_METHODS_OF.pdf

UNIT 3: VOLUMETRIC
(TITRIMETRIC) METHODS OF
ANALYSIS
Analytical Chemistry (CHY3022)
Dr. D. Bennett
1

SELF LEARNING TASKS
REVIEW YOUR LEARNING
TOPICS YOU MUST REVIEW
• Lowry and Bronsted’s definitions of acids and
bases
• Conjugate acid; conjugate base pairs
• Strong acid and bases; weak acids and bases
• Autoionization/protolysis of water
• pH calculations
• Henderson Hasselbalch equation
• Equilibrium constants
• Buffers
2

UNIT CONTENT
 Uses of titrimetry in chemical analyses
 Definition of general terms
 Basic calculations
 Neutralisation titrations
– Acid-base indicators
– Titration curve for strong acid-strong base titrations
– Weak acid-strong base titration curves
– Strong acid – weak base titration curves
3

TITRIMETRIC METHODS OF ANALYSIS
• Classified based on the type of reaction involved
 Acid-base titration – acidic or basic titrant reacts
with an analyte that is a base or an acid.
 Complexometric titrations – involving a metal-
ligand complexation reaction
 Redox titrations- the titrant is an oxidizing or
reducing agent
 Precipitation titrations – analyte and titrant react
to form a precipitate
4

Conducting a
titration
experiment
A buret:
most essential piece
of equipment for an
acid base titration
and is a means of
delivering the
titrant to the
solution containing
the analyte
5

GENERAL TERMS IN TITRIMETRY
 Titration:
 Addition of a solution from a burette
(titrant) to a solution in a flask (titrand)
until there is an indication that the titrant
has reacted stoichiometrically with the
titrand.
 Standards:
 reagents of exactly known concentration
{(x ± y) units} used in volumetric analysis.
6

GENERAL TERMS IN TITRIMETRY
 Primary (1°) standard:
 substance of sufficient purity such that a
standard solution can be prepared from it by
directly weighing a quantity of it, dissolving it
and then diluting to a defined volume of
solution.
 Secondary (2°) standard:
 solution whose concentration has been
determined by comparison against a 1°
standard.
7

THE CRITERIA USED IN SELECTING PRIMARY STANDARDS
A primary standard
Can be weighed accurately either for the purpose of preparing a standard
solution (which then does not have to be standardized) or for the
comparison to a solution with which it reacts for the purpose of
standardizing that solution.
The quality of the primary standard substance is the
ultimately the basis for a successful standardization.
It must meet some special requirements with respect to
purity, etc.,
8

•Primary standard
 It must be 100% pure, or at least its purity must be
known.
 If it is impure the impurity must be inert
 It should be stable at drying oven temperatures
 It should not be hygroscopic- it should not absorb water
when exposed to laboratory air.
 The reaction in which it takes place must be quantitative
and preferably fast
 Have a high formula weight is desirable, so that the
number of significant figures in the calculated result is
not diminished eg: borax (Na2B4O7.10H2O) as a
primary standard for HCl and KHP for NaOH
9

EQUIVALENCE POINTS AND END POINTS
• For a titration to be accurate we must add a stoichiometrically
equivalent amount of titrant to a solution containing the analyte.
• This stoichiometric mixture is the EQUIVALENCE POINT.
 moles titrant reacting with the analyte = Veq × CT
 Knowing the stoichiometry of the titration reaction we can calculate
the moles of analyte.
• Most titrations give no obvious indication that the equivalence point has
been reached. We stop adding titrant when we reach an END POINT
indicated by a colour change of an indicator added to the solution
containing the analyte.
• Titration error = VEP – VEQ
• Selecting an appropriate end point is critical if a titrimetric method is to
give accurate results.
10

An indicator is a substance added to the reaction flask
ahead of time to cause a colour change at or near the
equivalence point of the titration.
The point of a titration at which an indicator changes
colour, the visual indication of the equivalence point,
is called the end point of the titration
Later we will see equivalence
points can be determined in
ways other than the use of an
indicator
11
Basic terms related to volumetric analysis cont’d

TITRATION STRATEGIES CONT’D
• If:
 The titration reaction is too slow
 A suitable indicator is not available
 There is no useful direct titration reaction
• INDIRECT ANALYSIS MAY BE POSSIBLE
• Suppose you wish to determine the concentration of
formaldehyde, H2CO, in an aqueous solution?
– The oxidation of H2CO by I3
-
H2CO (aq) + 3OH- (aq) + I3
-(aq) HCO2
- (aq) + 3I-
(aq) + 2H2O (l)
is a useful reaction but it is too slow for a direct titration.
SO , WHAT DO WE DO?
12

• If we add a known amount of I3
-, such that it is in
excess, we can allow the reaction to go to completion.
• The I3
- remaining can be titrated with thiosulfate, S2O3
2-
.
• I3
- + 2S2O3
2- (aq) S4O6
2- (aq) + 3I-(aq)
• BACK TITRATION
– A titration in which a reagent is added to a solution
containing the analyte, and the excess reagent
remaining after its reaction with the analyte is
determined by a titration.
13

• When a suitable reaction involving the analyte does not exist it
may be possible to generate a species that is easily titrated. For
example, the sulphur content of coal can be determined by using a
combustion reaction to convert sulphur to sulphur dioxide.
S(s) + O2 (g)  SO2 (g)
• Passing the SO2 through an aqueous solution of hydrogen
peroxide, H2O2,
SO2 (g) + H2O2 (aq)  H2SO4 (aq)
• produces sulphuric acid, which we can titrate with NaOH,
H2SO4 (aq) + 2OH- (aq) SO4
2- (aq) + 2H2O (l)
• providing an indirect determination of sulfur.
14

BASIC TERMS USED IN VOLUMETRIC ANALYSIS CONT’D
15

Review of Solution concentration
•MOLARITY (M)
•The number of moles of solute dissolved per liter of solution and may
be calculated by dividing the number of moles dissolved by the number
of liters of solution:
–molarity = moles of solute/liters of
solution
16

•What is the molarity of a solution that has 4.5 mol of solute
dissolved in 300.0 mL of solution?
•Convert mL to liters
• 300.0 mL = 0.3000L
17
M
L
mol
molarity 15
3000
.
0
5
.
4



WHAT IS THE MOLARITY OF A SOLUTION OF NAOH THAT HAS
0.491 G DISSOLVED IN 400 ML OF SOLUTION?
Convert from grams of solute to moles by dividing the grams by
the molar mass.
Convert from mL to L
18
M
L
gmol
g
liters
RMM
grams
molarity 0307
.
0
400
.
0
977
.
39
/
491
.
0
/ 1





•SELF TASK
•How would you prepare 500.0 mL
of a 0.2M solution of NaOH?
–The grams of solute calculated is weighed and placed in the
container. Water is added to dissolve the solute and to dilute the
solution to volume. The solution is then shaken to make it
homogeneous.
19

SOLUTION PREPARATION BY DILUTION- THE
MOLES OF SOLUTE IN A SOLUTION IS UNCHANGED BY DILUTION ONLY THE
AMOUNT OF SOLVENT IS CHANGED
If a solution is prepared by diluting another solution
whose concentration is precisely known the following
dilution equation may be used to calculate the
molarity.
CB×VB = CA×VA
C refers to concentration (any unit but same on both
sides), V refers to volume (any unit but same on both
sides), B refers to before dilution and A after dilution
20

Self Learning Task
•What is the molarity of a solution prepared by diluting 10.00 mL of a
4.281 M solution to 50.00 mL?
•How many milliliters of 12 M HCl are needed to prepare 500.0 mL of a
0.60 M solution?
•A student weighs 5.678 g of KHP and dissolves it in water in a 250. mL
volumetric flask. A 25.06 mL aliquot of this KHP solution is
withdrawn and placed in a 250 mL Erlenmeyer flask to which 5 drops
of the acid-base indicator phenolphthalein is added. This solution is
titrated with a sodium hydroxide solution of unknown molarity. 26.64
mL of sodium hydroxide solution is required to reach the
endpoint. Calculate the molarity of the sodium hydroxide solution.
21

TITRATION CURVES
Finding the end point
 Monitor some property that has a well defined value at the
equivalence point
 Eqv. Pt for NaOH and HCl is pH = 7.0
 We can find the endpoint , by monitoring the pH with a pH
electrode or by adding an indicator that changes colour at a pH
= 7.0
 Suppose that the only available indicator changes colour at pH
= 6.8. Is this endpoint close enough to the equivalence point
that the titration error may be safely ignored?
 To answer this question we need to know how the pH changes
during the titration
22

TITRATION CURVES
 A titration curve provides us with a visual
picture of how a property, such as pH changes as
we add titrant
23
Titration Curve for Strong Acid with Strong Base
0
1
2
3
4
5
6
7
8
9
10
11
12
13
0 10 20 30 40 50 60 70
Volume Base
pH
Equivalence point

TITRATION CURVES
• We can measure the titration curve experimentally by
suspending a pH electrode in the solution containing
the analyte monitoring the pH as titrant is added.
• We can calculate the expected titration curve by
considering the reactions responsible for the change
in pH.
• We use the curve to evaluate an indicator’s likely
titration error.
24

NEUTRALIZATION TITRATIONS
Acid-base indicators
Titration curves
Examples of methods
25

ACID-BASE INDICATORS
]
[
]
[
]
[
]
[
]
][
[
3
3







In
HIn
K
O
H
HIn
O
H
In
K a
a 26
 Weak organic acid or base.
 The acid/base differs in colour from its conjugate
base/acid.
 They display different colours depending on pH.
HIn + H2O  In- + H3O+
Acid
Colour
Base
Colour

]
[
]
[
]
[
]
[
]
][
[
3
3







In
HIn
K
O
H
HIn
O
H
In
K a
a
27
pKa = -logKa
 Indicator pH range = pKa ± 1
 Examples:
 Phenolphthalein 8.3 – 10.0
 Bromothymol Blue 6.2 – 7.6
 Bromocresol Green 3.8 – 5.4

 pH and the magnitude of the change in pH at
the equivalence point must be known.
 Not necessary for indicator to change colour at
the equivalence point.
 Usually, pH change around equivalence point is
rapid.
 Therefore, an indicator beginning and ending its
colour change anywhere on the portion of the pH
curve where there is a sharp rise (or decline) in
pH, will be suitable.
28

29
Possible
indicators
Equivalence volume
End points
pH = -log[H3O+]

NEUTRALIZATION TITRATIONS
 To select the indicator for a neutralization titration
(acid-base) we must know:
 pH at the equivalence point and
 rate change in pH with titrant volume (ΔpH/ Δvol) that occurs
around (Ve) equivalence point.
 Titration curves give us this information.
 These can be constructed from:
 experimental data OR
 calculated pH values.
30

TITRATION CURVE FOR STRONG ACID-STRONG BASE
TITRATIONS
31
Titrate 50 mL of 0.100 M HCl with 0.100 M NaOH.
• Calculate the volume of NaOH needed to reach the
equivalence point
MaVa = MbVb
• Ve (equivalence volume) = 50.00 mL
• pH before any base has been added:
[H3O+] = 0.10 mol L-1
[H3O+] = 10-0.10 ⇒ pH = 1.0

32
0
1
2
3
4
5
6
7
8
9
10
11
12
13
0 10 20 30 40 50 60 70
Volume Base
pH

TITRATIONS
50 mL of 0.100 M HCl with 0.100 M NaOH
• pH after adding Vt & prior to Ve:
HCl + NaOH  H2O + Na+ + Cl-
# mol base added = # mol acid reacted
[HCl] = # mol HCl remaining after adding NaOH
total volume of soln
33

TITRATIONS
50 cm3 of 0.100 M HCl with 0.100 M NaOH
 pH after adding Vt & prior to Ve:
For Vt = 10.00 mL
[HCl] = original # mol HCl - # mol NaOH added
= (0.100 M x 50.00 mL) – (0.100 M x 10.00 mL)
50.00 mL + 10.00 mL
= 6.67 x 10-2 mol L-1  pH = 1.18
34

TITRATION CURVE FOR STRONG ACID-
STRONG BASE TITRATIONS
35
 pH after adding Vt & prior to Ve:
SELF TASK
Calculate pH for Vt < Ve
Vt = 1  50.00 mL

 pH at equivalence volume:
HCl + NaOH  H2O + Na+ + Cl-
  at equivalence point, only H2O, Na+ and Cl- are
present.
 Equilibrium at Ve is: 2H2O  H3O+ + OH-
Kw = [H3O+][OH-] = 1 x 10-14 = [H3O+]2
[H3O+] = 1 x 10-7  pH = 7
36

TITRATIONS
 pH after equivalence volume:
 For Vt = 55.00 mL
 Solution now contains only xs NaOH.
[NaOH] = # mol NaOH added – original # mol HCl
= (0.100 M x 55.00 mL) – (0.100 M x 50.00 mL)
50.00 mL + 55.00 mL
= 4.76 x 10-3 M  pOH = 2.32
 pH = 14 – 2.32 = 11.68
37

TITRATION CURVE FOR STRONG ACID- STRONG
BASE TITRATIONS
0
1
2
3
4
5
6
7
8
9
10
11
12
13
0 10 20 30 40 50 60 70
Volume Base
pH
38
 Compute for Vt < Ve, Vt = Ve and Vt > Ve and plot
graph:

TITRATIONS
Strong Acid - Strong Base
0
1
2
3
4
5
6
7
8
9
10
11
12
13
40 45 50 55 60
Volume Base mL
pH
39
 Summary
 pH before equivalence point:
pH = -log[H3O+]
[H3O+] = mol acid remaining/total volume
 pH at equivalence point = 7
 pH after equivalence point = 14 – pOH
& [OH-] = mol xs NaOH/total volume
C: [HCl] = [NaOH] = 0.0001 mol dm-3
pH 4-10
B: [HCl] = [NaOH] = 0.001 mol dm-3
pH 5-9
A: [HCl] = [NaOH] = 0.01 mol dm-3
pH 6-8
Phenolphthalein (P): 8.3 – 10
Bromothymol Blue (BB): 6.2 – 7.6
Methyl Orange (MO): 3.1 – 4.4
P
BB
MO
A
B
C

 In order to construct a titration curve:
 Need to know which acid and which base are involved –
strong or weak?
 Need to know their concentrations.
 To select the appropriate indicator:
 pH change through the equiv. point?
 Indicator pH range?
 If wrong indicator used: equiv. pt  end point,
determined [analyte] will be wrong!!!
40

Issues of Concern:
 end point vs. equivalence point !!!
 Normally we don’t know the analyte
concentration.
 What is the error associated with indicator
selection and colour change detection?
 Analyte concentration and ΔpH/Δvol.
 When has the colour changed? ± x cm-3.
 Error in addition to the burette reading error (± ?)
{error in burette volume = √(0.052 + 0.052 + x2)} 41

TITRATION CURVE FOR WEAK ACID-STRONG BASE
TITRATIONS
 In weak acid-strong base titrations:
 pH before and at the equivalence point depends on
acid concentrations and Ka (NOTE: small ΔpH/Δvol)
42
Ka (AcOH)
1.75 x 10-5
pKa = 4.76
Ka (H3BO3)
5.81 x 10-10
pKa = 9.24
Kb (H2BO3
-)
1.75 x 10-5
pKb = 4.76
Kb (AcO-)
Kb = ?
pKb = ?

TITRATION CURVE FOR WEAK ACID-
Plot the titration curve for:
50 mL of 0.10 mol L-1 acetic acid with 0.10 mol
L-1 NaOH,
Ka = 1.75 x 10-5, pKa = 4.76
a) Before titration starts
b) When 0 < Vt < Ve
c) At Ve
d) After Ve
43

TITRATIONS
50 mL of 0.10 mol L-1 acetic acid with 0.10 mol L-1 NaOH, Ka =
1.75 x 10-5, pKa = 4.76
a) Before titration starts (we have a solution of only HOAc):
The weak acid will partially dissociate establishing the
equilibrium:
CH3COOH + H2O  CH3COO- + H3O+
I 0.10 0 0
C -x x x
E 0.10-x x x
Ka = [H3O+][CH3COO-]/[CH3COOH]
44

TITRATION CURVE FOR WEAK ACID-STRONG
BASE TITRATIONS
50 mL of 0.10 mol L-1 acetic acid with 0.10 mol L-1 NaOH, Ka
= 1.75 x 10-5, pKa = 4.76
Since moles [H3O+] is x
and the number of moles of protons produced = the number of moles of conjugate base
produced then….
Ka = [x]2/[0.10-x]  [x]2 = Ka[0.10]
[x] = √{Ka[0.10]} = √{1.75x10-5(0.10)}
= 1.32 x 10-3 M & pH = 2.88
45

BASE TITRATIONS
50 mL of 0.10 mol L-1 acetic acid with 0.10 mol L-1
NaOH, Ka = 1.75 x 10-5, pKa = 4.76
b) As the titration starts up to before Ve:
this reaction takes place in the flask:
CH3COOH + OH- → CH3COO- + H2O
and then this equilibrium re-establishes itself in
the flask:
46

TITRATION CURVE FOR WEAK ACID-STRONG BASE TITRATIONS
NaOH, Ka = 1.75 x 10-5, pKa = 4.76
Ka = [CH3COO-][H3O+]/[CH3COOH] = 1.75 x 10-5
[H3O+] = Ka[CH3COOH]/[CH3COO-] &  pH.
SPOT the difference: the number of moles of protons
produced are NOT = the number of moles of conjugate base
produced
47

BASE TITRATIONS
NaOH, Ka = 1.75 x 10-5, pKa = 4.76
b) As the titration starts up to before Ve:
Initially 50.0 mL of 0.10 M acetic acid in the flask.
 initial moles acetic acid = (50.0)(0.10)/1000 moles
After the addition of Vt mL base, (Vt)(0.10M)/1000
moles of base have been added and reacted with the
same number of moles of acid.
48

TITRATIONS
50 mL of 0.10 mol L-1 acetic acid with 0.10 mol L-1 NaOH, Ka = 1.75 x
10-5, pKa = 4.76
moles acid remaining = initial moles acetic acid – moles
base added
= (50.0)(0.10)/1000 - (Vt)(0.10M)/1000 moles
& moles of acetate (CH3COO-)produced = moles of base
added
= (Vt)(0.10M)/1000 moles 49

BASE TITRATIONS
50 mL of 0.10 mol L-1 acetic acid with 0.10 mol L-1 NaOH, Ka =
1.75 x 10-5, pKa = 4.76
The solution will contain both acetic acid and its conjugate base
acetate – a buffer is created.
[CH3COOH] = (initial moles acetic acid – moles base added)
total volume
= [(50.0)(0.10) - (Vt)(0.10M)]/50.0 + Vt
[CH3COO-] = moles of base added/total volume
= (Vt)(0.10M)/50.0 + Vt
50

TITRATION CURVE FOR WEAK ACID-STRONG BASE TITRATIONS
NaOH, Ka = 1.75 x 10-5, pKa = 4.76
Ka = [CH3COO-][H3O+]/[CH3COOH] = 1.75 x 10-5
[H3O+] = Ka[CH3COOH]/[CH3COO-]
&  pH.
51

50 mL of 0.10 mol L-1 acetic acid with 0.10 mol L-1 NaOH, Ka = 1.75 x 10-5,
pKa = 4.76
Ka = [CH3COO-][H3O+]/[CH3COOH]
[H3O+] = Ka[CH3COOH]/[CH3COO-]
Hence, at Vt = 5 mL
[H3O+] =
1.75 x 10-5(0.10 x 50.0 – 0.10 x 5.00)/(0.10 x 5.00)
= 1.58 x 10-4 mol L-1  pH = 3.80
OR USING THE HENDERSON HASSELBALCH EQUATION
pH = pKa + log [CH3COO-]/[CH3COOH]
= 4.76 + log[(0.10 x 5.00)/(0.10 x 50.0 – 0.10 x 5.00)]
= 3.80 52

NaOH, Ka = 1.75 x 10-5, pKa = 4.76
SELF TASK
Calculate for 1, 10, 15, 25, 35, 40, 45, 49, 49.9 mL
Note: When Vt = ½ Ve, the point of half neutralization:
[CH3COOH] = [CH3COO-]
[H3O+] = Ka and pH = pKa
pH = pKa at 0.5Ve
53

50 mL of 0.10 mol L-1 acetic acid with 0.10 mol L-1 NaOH, Ka = 1.75 x
10-5, pKa = 4.76
c) At Ve:
All the acetic acid has been neutralized to leave the
conjugate base and the equilibrium:
CH3COO- + H2O  OH- + CH3COOH
Ka x Kb = Kw
Kb= Kw/Ka
Kb = [CH3COOH][OH-]/[CH3COO-]= Kw/(1.75 x 10-5)
= 5.71 x 10-10 54

50 mL of 0.10 mol L-1 acetic acid with 0.10 mol
L-1 NaOH, Ka = 1.75 x 10-5, pKa = 4.76
c) At Ve:
[CH3COOH] = [OH-] = x
[OH-]2 = Kb[CH3COO-]
[OH-] = √{5.71 x 10-10 x (0.10/2)}
= 5.34 x 10-6 M
pOH = 5.27 & pH = 14 – 5.27 = 8.73
d) pH after equivalence point is calculated as
for strong acid strong base titration.
55

56
0.1M Acetic acid titrated with 0.1M NaOH
0
2
4
6
8
10
12
14
0 10 20 30 40 50 60 70 80
ml base
pH
Ve
pH at Ve
pH =
pKa at
0.5Ve
Acetic acid titrated with NaOH
0
2
4
6
8
10
12
14
0 10 20 30 40 50 60 70 80
mL base
pH
0.1 M
0.01 M
0.001 M

57
Acetic acid titrated with NaOH
0
2
4
6
8
10
12
14
0 10 20 30 40 50 60 70 80
mL base
pH
The effect of concentration
The buffering region
Ka = [CH3COO-
][H3O+
]/[CH3COOH]
= [H3O+] x {[CH3COO-]/[CH3COOH]}
The maximum buffering capacity when
[base] = [acid]
Choice of indicators:
as for strong acid –
strong base
titrations

58
The effect of acid strength -
Ka

STRONG ACID – WEAK BASE TITRATION
CURVES
• Completely analogous to the above case, but the
titration curves are the reverse of those for a weak
acid versus a strong base
• SELF TASK
• Sketch the titration curve for 100 mL of 0.1 M NH3
titrated with 0.1 M HCl at 10%, 20% 50% 100% and
110% 120%.
• The neutralizing reaction is:
– NH3 + HCl  NH4Cl
59

SUMMARY
• Basic Solution Calculations
• Terms used in titrimetry
• Neutralization Titrations
– Acid-base indicators
– Titration curve for strong acid-strong
base titrations
– Titration curve for weak acid-strong
base titrations
60

• NEXT TOPICs
• Neutralization titrations (contd.)
– Kjeldahl Method
– Introduction to:
– CO3
2-/HCO3
- titrations
– Polyfunctional acids and bases
– Non aqueous titrations
– Complexation titration
61

REFERENCES
 Lecture notes were prepared by Dr. K. Wilson and Dr.
D. Gordon-Smith and modified by Dr. Deon Bennett
62

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