CLASSICAL METHODS IN
TECHNIQUES OF ANALYTICAL
CHEMISTRY: TITRIMETRIC
METHODS OF ANALYSIS
ERT 207 ANALYTICAL CHEMISTRY
SEMESTER 1, ACADEMIC SESSION 2015/16
Overview
2
OVERVIEW OF TITRIMETRY
 GRADES OF CHEMICALS
 TITRATION CALCULATIONS
 PRECIPITATION TITRATIONS
- TITRATION CURVE
 ARGENTOMETRIC TITRATION

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OVERVIEW OF TITRIMETRY
3
Defination:
 Titrations (or titrimetric method) are based on
measuring the amount of a reagent of known
concentration that reacts with the unknown.
 It is the process of determining unknown
concentration by adding the small increments of
standard solution until the reaction is just
complete.
 A general equation can be expressed:
aA + tT → products
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where, A: analyte; T: titrant

OVERVIEW OF TITRIMETRY
4

Revisiting Keywords of Titration
 Equivalence point: The point (e.g., volume of titrant)
in a titration where (theoretically) stoichiometrically
equivalent amounts of analyte and titrant react.
 Indicator: A colored compound whose change in
color signals the (experimental) end point of a
titration.
 End point: The point (e.g., volume of titrant) in a
(experimental) titration where we stop adding
titrant in an experiment.
 Titration error: The determinate error in a titration
due to the difference between the end point and
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the equivalence point.
OVERVIEW OF TITRIMETRY
5
Type of Titrations based on Chemical Reactions:
i. Acid-Base Titrations:
H+ + OH– → H2O
K= 1/Kw
ii. Precipitation Titrations:
Ag+(aq) + Cl–(aq) → AgCl(s)
K=1/Ksp
iii. Redox Titrations:
5 H2O2 + 2 MnO4– + H+ → 5 O2 + 2 Mn2+ +
8H2O
iv. Complexometric Titrations:
EDTA + Ca2+ → (Ca–EDTA)2+
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
OVERVIEW OF TITRIMETRY
Type of Titrations based on Measuring Techniques:
i. Volumetric titrimetry: Measuring the volume of a
solution of a known concentration (e.g., mol/L) that
is needed to react completely with the analyte.
ii. Gravimetric (weight) titrimetry: Measuring the
mass of a solution of a known concentration (e.g.,
mol/kg) that is needed to react completely with the
analyte.
iii. Coulometric titrimetry: Measuring total charge
(current x time) to complete the redox reaction, then
estimating analyte concentration by the moles of
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electron transferred.

6
OVERVIEW OF TITRIMETRY

Type of Titration Curves:
7
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GRADES OF CHEMICALS
Terms and Definitions:
i. Reagent Grade:
 The reagents which meets or surpasses the latest
American Chemical Society specifications.
ii. Primary standard:
 The reagent which is ready to be weighted and
used prepare a solution with known
concentration (standard).
 Requirements of primary reagent are:

8
- Known stoichiometric composition
- High purity
- Nonhygroscopic
- Chemically stable both in solid & solution
- High MW or FW
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GRADES OF CHEMICALS
9
Secondary standard:
 A standard which is standardized against a
primary standard.
 Certified reference materials (CRM):
 A reference material, accompanied by a
certificate, which has been analysed by
different laboratories to determine consensus
levels of the analyte concentration.

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GRADES OF CHEMICALS
10
NIST Standard Reference Material® (SRM):
 A CRM issued by NIST that also meets additional
NIST-specific certification criteria and is issued
with a certificate.
 Standardization:
 The process by which the concentration of a
reagent is determined by reaction with a known
quantity of a second reagent.

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TITRATION CALCULATIONS
11

Terms and Definitions:
 Blank Titration: Titration procedure is
carried out without analyte (e.g., a distilled
water sample). It is used to correct titration
error.
 Back titration: A titration in which a (known)
excess reagent is added to a solution to
react with the analyte. The excess reagent
remaining after its reaction with the analyte,
is determined by a titration.
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TITRATION CALCULATIONS
12
(i) Standardization
 To standardizing a KMnO4 stock solution, the
primary standard of 9.1129 g Na2C2O4 is
dissolved in 250.0 mL volumetric flask.
 10.00 mL of the Na2C2O4 solution require 48.36
mL of KMnO4 to reach the titration end point.
What is the molarity (M) of MnO4– stock solution?
(FW Na2C2O4 134.0)
 Solution:
5C2O42–(aq) + 2MnO4–(aq) + 16H+(aq) → 10CO2(g) + Mn2+(aq) + 8H2O(l)
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TITRATION CALCULATIONS
13
9.1129 g Na 2C2O 4 1 mol Na 2C2O 4
1 mol C2O 24 
10 mL



1
134.0 g Na 2C2O 4 1 mol Na 2C2O 4 250 mL
2 mol MnO 4
1
1000 mL





0
.
02250
M
MnO
4
2
1L
5 mol C2O 4 48.36 mL
(ii) Unknown Analysis with a Blank Correction
 A 0.2865 g sample of an iron ore is dissolved in
2+
acid, and the iron is converted entirely to Fe .
 To titrate the resulting solution, 0.02653 L of
0.02250 M KMnO4 is required. Also a blank
titration require 0.00008 L of KMnO4 solution.
What is the % Fe (w/w) in the ore? (AW Fe
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55.847)
TITRATION CALCULATIONS
Solution:
MnO4–(aq) + 5Fe2+ + 8H+(aq) → Mn2+(aq) + 5Fe3+ + 4H2O(l)
14
Net titrant vol  0.02653  0.00008 L  0.02645 L
0.02645 L titrant 0.02250 mol MnO4 5 mol Fe 2 




1
1 L titrant
1 mol MnO4
55.847 g Fe
1

 100%  58.01% Fe ( w / w)
2
0.2865 g sample
1 mol Fe
(iii) Back Titration
 The arsenic in 1.010 g sample was pretreated to
H3AsO4(aq) by suitable treatment. The 40.00 mL of
0.06222 M AgNO3 was added to the sample
solution forming Ag3AsO4(s):
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H 3 AsO4( aq)  3 Ag  ( aq)  3H  ( aq)  Ag3 AsO
4( s )
TITRATION CALCULATIONS
(iii) Back Titration
 The arsenic in 1.010 g sample was pretreated to
H3AsO4(aq) by suitable treatment. The 40.00 mL of
0.06222 M AgNO3 was added to the sample
solution forming Ag3AsO4(s):
15
H 3 AsO4( aq)  3 Ag  ( aq)  3H  ( aq)  Ag3 AsO4( s )

The excess Ag+ was titrated with 10.76 mL of
0.1000 M KSCN. The reaction was:
Ag  ( aq)  SCN  ( aq)  AgSCN( s )

Calculate the percent (w/w) As2O3(s) (fw 197.84
g/mol) in the sample.
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Total mmol Ag  added
0.06222 mmol AgNO3
 40.00 mL AgNO3 
 2.4888 mmol
1 mL AgNO3
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TITRATION CALCULATIONS
mmol Ag  consumed by SCN 
0.1000 mmol SCN  1 mmol Ag 

 10.76 mL SCN 


1
.
0760
mmol
Ag
mL SCN 
1 mmol SCN 

mmol Ag  consumed by x mmol H 3 AsO4
3 mmol Ag 
 x mmol H 3 AsO4 
 3 x mmol Ag 
1 mmol H 3 AsO4
2.4888 mmol Ag   1.0760 mmol Ag   3 x mmol Ag 
x mmol H 3 AsO4  0.4709 mmol H 3 AsO4
0.4709 mmol H 3 AsO4
1 mmol As
1 mmol As2O3
1 mol As2O3



1
1 mmol H 3 AsO4
2 mmol As 1000 mmol As2O3

197.84 g As2O3
1

 100%  4.612% As2O3 (w/w)
1 mol As2O3
1.010 g sample
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TITRATION CALCULATIONS
17
(iv) Kjeldahl Analysis for Total Nitrogen (TN)
(a) KD description:
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TITRATION CALCULATIONS
18
Example 1:
 A typical meat protein contains 16.2% (w/w)
nitrogen.
 A 0.500 mL aliquot of protein solution was
digested, and the liberated NH3 was distilled into
10.00 mL of 0.02140 M HCl. The unreacted HCl
required 3.26 mL of 0.0198 M NaOH for
complete titration.
 Find the concentration of protein (mg protein/mL)
in the original sample.
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TITRATION CALCULATIONS
19
Example 2:
 A 0.7121g sample of wheat flour was analyzed
by the Kjeldahl method.
 The ammonia formed by addition of concentrated
base after digestion with H2SO4 was distilled into
25.00 ml of 0.04977 M HCl.
 The excess HCl was then back-titrated with 3.97
mL of 0.04012 M NaOH.
 Calculate the percent protein in the flour using the
5.70 factor for cereal.
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TITRATION CALCULATIONS
20
(v) Titration of a Mixture
 A solid mixture weighing 1.372 g containing only
sodium carbonate (Na2CO3, FW 105.99) and
sodium bicarbonate (NaHCO3, FW 84.01) require
29.11 mL of 0.7344 M HCl for complete titration:
Na2CO3  2 HCl  2 NaCl( aq)  H 2O  CO 2
NaHCO3  HCl  NaCl( aq)  H 2O  CO 2

Find the mass of each component of the mixture.
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TITRATION CALCULATIONS
21
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PRECIPITATION TITRATIONS
22

A titration in which the reaction between the analyte
and titrant involves a precipitation.
(I) Titration curve:
 Guidance in precipitation titration calculation
 Find Ve (volume of titrant at equivalence point)
 Find y-axis values:
- At beginning
- Before Ve
- At Ve
- After Ve
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PRECIPITATION TITRATIONS
23
EXAMPLE 3:
–
 For the titration of 50.0 mL of 0.0500 M Cl with
0.100 M Ag+.
 The reaction is:
Ag+(aq) + Cl–(aq)  AgCl(s)
K = 1/Ksp = 1/(1.8×10–10) = 5.6 x 109
 Find pAg and pCl of Ag+ solution added
(a) 0 mL (b) 10.0 mL (c) 25.0 mL (d) 35.0 mL
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PRECIPITATION TITRATIONS
24
(ii) Construct a titration curve:
–
 Example: Titration of 50.0 mL of 0.0500 M Cl
with 0.100 M Ag+
pCl
pAg
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PRECIPITATION TITRATIONS
25
(iii) End point determination
dy/dx
d2y/dx2
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PRECIPITATION TITRATIONS
(iv) Diluting effect
of the titration
curves
26
25.00 mL 0.1000 M I–
titrated with 0.05000 M Ag+
25.00 mL 0.01000 M I–
titrated with 0.005000 M Ag+
25.00 mL 0.001000 M I–
titrated with 0.0005000 M
Ag+
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PRECIPITATION TITRATIONS
EXAMPLE 6
–
–
 A 25.00 mL solution containing Br and Cl was
titrated with 0.03333 M AgNO3.
 Ksp(AgBr)=5x10–13, Ksp(AgCl)=1.8x10–10.
(a) Which analyte is precipitated first?
(b) The first end point was observed at 15.55 mL.
Find the concentration of the first that precipitated
(Br– or Cl–?).
(c) The second end point was observed at 42.23 mL.
Find the concentration of the second that
precipitated (Br– or Cl–?).
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ARGENTOMETRIC TITRATION
28
General information:
 Define Argentometric Titration: A precipitation
titration in which Ag+ is the titrant.
 Argentometric Titration classified by types of
End-point detection:
i. Volhard method: A colored complex (back
titration)
ii. Fajans method: An adsorbed/colored indicator
iii. Mohr method: A colored precipitate

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ARGENTOMETRIC TITRATION

29
Volhard method: A colored complex (back
titration). Analysing Cl– for example:
Step 1: Adding excess Ag+ into sample
Ag+ + Cl– → AgCl(s) + left Ag+
Step 2: Removing AgCl(s) by filtration/washing
Step 3: Adding Fe3+ into filtrate (i.e., the left Ag+)
Step 4: Titrating the left Ag+ by SCN–:
Ag+ + SCN– → AgSCN(s)
Step 5: End point determination by red colored Fe(SCN)2+
complex. (when all Ag+ has been consumed, SCN–
reacts with Fe3+)
SCN– + Fe3+ → Fe(SCN)2+(aq)
Total mol Ag+ = (mol Ag+ consumed by Cl–)
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–)
+ (mol Ag+ consumed by SCN
ARGENTOMETRIC TITRATION
Fajans Method: An adsorbed/colored indicator.
–
 Titrating Cl and adding dichlorofluoroscein:

30
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ARGENTOMETRIC TITRATION
Mohr Method: A colored precipitate formed by
Ag+ with anion, other than analyte, once the Ve
reached.
–
2–
 Analysing Cl and adding CrO4 :

31


Precipitating Cl–:
Ag+ + Cl– → AgCl(s)
Ksp = 1.8 x 10–10
End point determination by red colored
precipitate, Ag2CrO4(s):
2Ag+ + CrO42– → Ag2CrO4(s)
Ksp = 1.2 x 10–12
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EXAMPLE 1
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EXAMPLE 2
mmol
Amount HCl = 25.00mL HClx 0.04977
mLHCl
= 1.2443 mmol
mmol
Amount NaOH = 3.97mL NaOHx 0.04012
mLNaOH
= 0.1593 mmol
Amount N =Amount HCl – Amount NaOH
= (1.2443 – 0.1593) mmol
= 1.0850 mmol
33
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EXAMPLE 2
0.014007 g N
1.0850mmol Nx
% N =
mmol N
x100%
0.7121g sample
= 2.1342
%
protein = 2.1342% N x 5.70% protein
= 12.16
%N
34
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EXAMPLE 3
(a)0 mL Ag+ added (At beginning)
[Ag+] = 0, pAg can not be calculated.
[Cl–] = 0.0500, pCl = 1.30
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EXAMPLE 3
(b) 10 mL Ag+ added (Before Ve)
mmol Cl  Left  original mmol Cl   precipitat ed mmol Cl 







0
.
0500
mmol
Cl
0
.
100
mmol
Ag
1
mmol
Cl


  10.0 mL Ag 

  50.0 mL Cl 



 
1 mL Cl
1 mL Ag
1 mmol Ag 

 
 1.50 mmol Cl 
Vtotal  50.0 mL  10.0 mL  60.0 mL
1.5 mmol Cl 
[ Cl ] 
 2.50 10  2 M
60.0 mL

10
1
.
8

10
9
[ Ag  ] 


7
.
2

10
M

2
[ Cl ] 2.50 10
pCl  1.60
pAg  8.14
K sp
36
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EXAMPLE 3
(c) 25 mL Ag+ added (At Ve)
AgCl(s)  Ag+(aq) + Cl–(aq)
Ksp = 1.8×10–10
s = [Ag+]=[Cl–]
Ksp = 1.8×10–10 = s2
[Ag+]=[Cl–]=1.35x10–5
pAg = 4.89
pCl = 4.89
37
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EXAMPLE 3
(d) 35 mL Ag+ added (At Ve)
mmol Ag  Left  original mmol Ag   precipitat ed mmol Ag  withVe of Ag 






0
.
100
mmol
Ag
0
.
100
mmol
Ag


   25.0 mL Ag 

  35.0 mL Ag 


1 mL Ag
1 mL Ag

 

 1.00 mmol Ag 
Vtotal  50.0 mL  35.0 mL  85.0 mL
1.00 mmol Ag 
[ Ag ] 
 1.18 10  2 M
85.0 mL
K sp
1.8 10 10

8
[Cl ] 


1
.
53

10
M

2
[ Ag ] 1.18 10
pCl  7.82
pAg  1.93

38
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EXAMPLE 4
(a)
 Ag+(aq) + Br–(aq)  AgBr(s)
K = 1/Ksp(AgBr) = 2x1012
 Ag+(aq) + Cl–(aq)  AgCl(s)
K = 1/Ksp(AgCl) = 5.6x109
Ans: AgBr precipitated first
39
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EXAMPLE 4
40
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