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Chemical and Physical Principles

Analytical Chemistry

ANALYTICAL CHEMISTRY
Analytical Chemistry
 Branch of Chemistry that deals with the analysis, identification, separation and
composition of matter
 Involves methods used to identify the substances that are present in a sample
(qualitative analysis) and the exact amount of the identified substances
(quantitative analysis)
Quantitative Methods of Analysis
A
...


Classical methods
a
...
Volumetric method – measurement of the volume of solution necessary to react
completely with the analyte

2
...
Spectroscopic method – measurement of the electromagnetic radiationproduced
by the analyte or its interactions with it
b
...


Other methods – involves the measurement of the properties of the analyte such as
heat of reaction (calorimeter), index of refraction (refractometer), optical activity
(polarimeter) or mass-to-charge ratio (mass spectrometer)

B
...
Selection of an appropriate method
 In the selection of method of analysis, it is necessary to consider the level of
accuracy, complexity and component of the sample, availability of equipment
and trained personnel and the time of analysis
 Standard procedures are usually available from literature such as Chemical
Abstracts, Analytica Chimica Acta, Applied Spectroscopy, Journal of the
Association of Analytical Chemists, etc
...


Obtaining a representative sample
 The American Society for Testing and Materials (ASTM), National Bureau of
Standards (NBS) and Association of Official Analytical Chemists (AOAC) are
such a few organizations that impose standard sampling procedures for analysis
of some samples
 Three steps are generally followed in obtaining samples: obtaining a gross
sample, obtaining a laboratory sample and obtaining an analysis sample
 A gross sample is obtained from a bulk sample and obtained in such a manner
that it is considered a representative of the bulk sample
 A laboratory sample is a fraction of the gross sample weighing several grams
wherein further reduction to few milligrams results into an analysis sample

Chemical and Physical Principles
3
...

Samples that decompose upon heat treatment are analyzed on a wet basis or asreceived basis
 Replicate samples are taken for analysis to ensure accuracy of the method used
and quality of the results
...
Classification of analysis based on sample size
Method
Sample Mass
Sample Volume
macro
more than 100 mg
more than 0
...
050 mL to 0
...
050 mL
ultra-micro
less than 1 mg
–

4
...
Water
...

b
...
In many instances, some portion of the sample will
not dissolve in water and usually the addition of acids render the sample
soluble
...

c
...
For more stubborn samples, hot, concentrated sulfuricacid,
nitric acid and aqua regia are used
...
Hydrofluoric acid is also used for
dissolving silicate ores
...
Fluxing agents
...
Fluxing agents may be
classified as acidic (K2S2O7, KHF2 and B2O3), basic (Na2CO3, K2CO3, NaOH
or KOH) and oxidizing (Na2O2)
...

 Reagents and chemicals used in the laboratory are classified as follows:
a
...
Reagents that undergo superficial
purification and not directly used for analysis
b
...
Reagents used by pharmacists and unfit for analysis
c
...
Reagents that are more refined compared to
technical reagents
d
...
Reagents
analyzed by the manufacturer with the analysis found on the label of the
container
e
...
Chemicals with purity greater than 99
...
The following are most commonly used to express concentrations of
solutions:
a
...


Volume percent – commonly used to specify the concentration of a pure
liquid compound diluted with another liquid
vol  volume of solute
%  
100%
vol  volume of solution
For alcoholic beverages, percentage of alcohol is usually expressed in
terms of proof as follows:
proof   vol 
2 %  
 vol 

c
...


Mole fraction (x) – commonly used in unit operations to express
concentrations of solute present in a stream of gas or liquid
mole of solute
x
mole of solute  mole of solvent

e
...


Molarity (M) – most commonly used in titration and denotes the amount of
solute, in moles, dissolved in a solvent and diluting to a final volume of 1L
in a volumetric flask
mole of solute (mol)
M
volume of solution (L)
Formality (F) – concentration term identical to molarity commonly used for
solutions of ionic salts that do not exist as molecule in solid or in solution

g
...
The number of equivalents is given by thenumber
of moles multiplied by the number of reacting units per molecule or atom
...
Molar equivalents of solutes
Nature of solute
Molar equivalent
acid
number of replaceable H+
base
number of equivalent HO–
salt
net charge of an ion
oxidant
gain of electron
reductant
loss of electron
h
...


Treatment of the sample
 Some samples has to be reduced or oxidized prior to analysis or sometimes
treated to become colored or converted to a form that it can be readily volatilized
 More often, the accuracy of an analysis is affected by the presence of unwanted
components called interferences
 Interferences can be eliminated by converting it into non-interfering form by a
process called masking
...


Measurement of the analyte
 Using classical methods of analysis, results can be accurate up to a few parts per
thousand or better, requires relatively large amount of sample and usually
applied to measurement of major constituents in a sample
 Instrumental methods are generally more sensitive and selective
...


Analytical Chemistry

Chemical and Physical Principles

Table 3
...
00%
semi-micro
0
...
00%
micro
0
...
10%
ultra-micro
less than 0
...


Calculation of results and reporting of data
 Results of analysis can be expressed depending on the nature of analyte
a
...
Calculations on solid samples are based on mass
...
analyte
%  
100%
wt  wt
...
Concentrations of analyte in solid samples in trace concentrations
Unit
Definiton
Unit
mg
analyte
g analyte
wt  gram analyte
parts per
pt   gram sample 10 3
g sample or kg sample
thousand
wt 
μg analyte mg analyte
parts per
wt  gram analyte
ppm   gram sample 10 6
g sample or kg sample
million
wt 
ng analyte μg analyte
wt  gram analyte
parts per
ppb  gram sample 10 9
g sample or kg sample
billion
wt 
b
...
Similarly, concentrations of solid or liquid analytes in liquid
samples obtained from a macro analysis is usually expressed as % weight by
volume or % volume by volume defined as follows:
wt  gram analyte
vol  volume analyte
%  
100% or %  

100%
vol  mL sample

vol  volume sample

Table 5
...
Types of Gravimetric Analysis
1
...

3
...
The analyte in the sample is obtained using as appropriate
solvent and the residue from the solution, after evaporation of the solvent, is
chemically related to the analyte
Precipitation Method
...
The sample is treated to yield a gas that is passed in an
absorbing medium; the analysis is based upon the change in mass of the medium

B
...


Calculation in gravimetric analysis
To calculate the amount of analyte in the sample…
mass of final form
% analyte 
GF 100%
mass of sample
Gravimetric Factor (GF)
 x mol analyte 
molar mass of analyte

GF  molar mass of final form 
y mol final form molar ratio

2
...
Precipitating agents used in precipitation gravimetry
Species

Final Form

Precipitant

Precipitated

Cl

AgCl

Br

AgBr

I
SO4–2
As
Bi
Cd

AgI
BaSO4
As2O3
Bi2S3
CdSO4

Cu
Sn
Sb

CuO
SnO2
Sb2O3

Mg
Zn

Mg2P2O7
Zn2P2O7

C
...


Properties of precipitates

AgNO3

BaCl2

H2S

(NH4) 2HPO 4

Species

Final

Precipitant

Precipitated

Form

Al

Al2O3

Cr

Cr2O3

Fe
Sn
Ba
Cd
Sr

Fe2O3
SnO2
BaSO4
CdSO4
SrSO4

H2SO4

Ca
Mg
Zn

CaCO3
MgCO3
ZnCO3

(NH4)2C2O4

K
Hg

H2PtCl6
HgS

K2PtCl6
(NH4)2S

NH3

Chemical and Physical Principles

Analytical Chemistry

a
...
Solid particles formed from precipitation may vary accordingly:
 Colloidal – tiny particles with size ranging from 0
...
Appearance
...
Relative supersaturation (von Weimarn ratio)
QS
relative supersaturation 
S
where Q = concentration of the solute as precipitation begins and S = solubility
of the precipitate
 In order to obtain low relative supersaturation and form a crystalline
precipitate, Q must be minimized and S must be maximized
...


Mechanism of precipitation
Precipitation is assumed to occur in two ways:
a
...
Particle growth
 Prevails at low relative supersaturation
 Results in the formation of small number of large particles

3
...
Electrical nature of colloidal suspensions
 Suspensions, which are stable since these particles are either positively or
negatively charged, hence repel each other
 By heating, stirring and addition of electrolyte causes this suspension to
combine together and form a readily filterable solid
 This process of converting a colloidal suspension into a readily filterable
solid is called coagulation or agglomeration
b
...
Factors affecting adsorption
 Common Ion Effect
...
In cases that there is more than one ion
adsorbed, the one having a lower solubility is adsorbed to a greater extent
...
The degree of adsorption
increases as the ionization of the contaminant decreases
 Effect of Concentration
...
Important Terminologies
1
...
Standardization – process of determining the concentration of an unknown
solution
3
...
Secondary standard – compound whose purity was established by a chemical
analysis and serves as reference material for volumetric analysis
5
...
End point – an observable change in a titration process which estimates the
equivalence point
7
...
Conditions for a Volumetric Analysis
1
...
The reaction must be complete and no side reaction occurs
3
...
Characteristics of a Good Primary Standard
1
...
Stable towards air, high temperature and humidity
3
...
Types of Titration
1
...
Back Titration – type of titration where an excess standard solution is added and the
excess is determined by the addition of another standard solution
3
...
Acid-Base Titration
1
...


ion, protonated water or solvated proton
ion

Autoprotolysis or self-ionization reactions
 Involves spontaneous reaction of molecules producing a pair of ions
 Protic solvents have reactive H+ and undergo autoprotolysis
∘

H2O + H2O ⮀ H3O+ + HO–

25 C
pK auto
14
...
8

CH3COOH + CH3COOH ⮀ CH3COOH2+ + CH3COO–

pK 25 C 14
...
7

CH3CH2OH + CH3CH2OH ⮀ CH3CH2OH2+ + CH3CH2O–

25 C
pK auto
19
...
Ion product constants for water
T,C
KW1014
T,C
KW1014
T,C
KW1014
0
0
...
69
40
2
...
19
25
1
...
86
10
0
...
45
50
5
...
46
35
2
...
87
Concentrations are expressed in molarity using density of water at each temperature
...
L
...
U
...
Phys
...
Ref
...

295-304
...


Strength of acids and bases
Strong
Acids

Weak
Bases

Acids

Bases

Analytical Chemistry

Chemical and Physical Principles
HCl
HBr
HI
1
2

HNO3
HClO4
1
H2SO4

LiOH
NaOH
KOH

RbOH
CsOH
2
R4NOH

carboxylic acids
polyprotic acids
metal cations

only the first ionization is complete; dissociation of the second proton has an equilibrium constant of 1
...


Calculation of pH
 At 25C, the ion product constant for water, KW is equal to 1
...
Strong acids (SA) and strong bases (SB)
pH logM SA
 HA + H2O ⭢ H3O+ (aq) + A–(aq)
(SA):
 MOH ⭢ M+ (aq) + HO–(aq)
b
...
Hydrolysis of salts

(SB):

pH 14  logM SB

(WA):

pH  12 logK a M WA

(WB):

pH 14  12 logK b M WB

 As a general rule, salts coming weak acids or weak bases hydrolyze in water,
that is, only the strong conjugate hydrolyzes in water
 Acidic salt (AS) is formed from the reaction of a strong acid and weak base
HCl(aq) + NH3(aq) ⭢ NH4+(aq) +
Cl–1(aq)
SA

WB

SCA

SCB

Since +only the strong conjugate hydrolyzes in water…+
+
HO
⮀ NH
NH
+ HO
4 (aq)

2

Kh 

Kb

3(g)

[NH3 ][H3O ]

KW



3

(aq)

1

(AS):

[NH4  ]

M AS 

pH 7  2 log 

K b 

 Basic salt (BS) is formed from the reaction of a strong base and weak acid
NaOH(aq) + HCN(aq) ⭢ Na+(aq)
+ CN–1(aq) +
H2O
SB



WA

WCB

SCA

Since only the strong conjugate hydrolyzes in water…
CN–1(aq) +
H2O
⮀ HCN(g) + HO– 1(aq)
1
M BS 
K W [HCN][HO ]
1
Kh 

(BS):
pH
7

log


Ka
2
[CN1]
K a 
 Neutral salt (NS) is formed from the reaction of a strong base and strong acid
 Salts from weak acid and weak base (WAB) will have the following
hydrolytic
equilibrium
expressed by the equation
NH +
+ CN–1
+
HO
⮀ NH OH +
HCN
4 (aq)

K 
h

(aq)

KW



1

K K
a

2

[HCN][NH4OH]


[CN ][NH ][H O]
b

4

2

4

(WAB): pH  1 log
2

K W K a 



 K 
 b 

 Amphoteric salts (HA–1 or HA–2)ionize as a weak acid and also a Brønsted
base that hydrolyzes
+
H2O
⮀
H3O+
+
HA–2
H3A
–1
+
⮀
+
H2O
H3O
+
H2A
+
H2O
⮀
H3O+
+

Analytical Chemistry

Chemical and Physical Principles
H2A–
1

HA–2
A–3
1 

K
K

K
K
[H
A
]
 W a1
a1 a2
2

Ka1
Ka2
Ka3

1
 1 


1
2 log
 2 log K a1 K a2
Ka1 [H2 A ]




2 

K K  K a2 K a3 [HA ]  1
2
 1   W a2


pH of HA
 log K K
2 log

a2 a3
Ka2 [HA 2 ]
2




1

pH of H2A



Analytical Chemistry

Chemical and Physical Principles
Table 8
...
75 10–5
6
...
20 10–10
6
...
10 10–8
6
...
10 10–2
5
...
75 10–5
4
...
50 10–4
5
...
30 10–4
4
...
20 10–6
5
...
30 10–5
8
...
30 10–2

K a2

K a3

1
...
10 10–5
7
...
60 10–7
1
...
00 10–6

4
...
10 10–8
4
...
Buffer solutions
 Solution that has the ability to resist changes in hydrogen ion concentration
upon the addition of small amounts of acid or base (buffer action)
 Usually consists of a mixture of weak acid (HA) and its conjugate salt (A–1)
or of a weak base (B) and its conjugate salt (BH+)
 Henderson-Hasselbalch equation
[Macidic ]
[H ][Mbasic  H ]
K A 

[Macidic  H ]

pH pKA  log

[M basic

[HO1 ][Macidic  HO1 ]
K B 

[Mbasic  HO1]

]
[Macidic ]

pH pKW  pKB  log

[M basic

]

 Buffer capacity or buffer intensity or buffer index is the number of moles of
strong acid or strong base for a liter of solution to cause a unit change in pH
dC HA  dC B 
β 

dpH  dpH 
where CHA and CB = number of moles per liter of strong base or strong acid,
respectively to cause d[pH]
...
001 M, the buffer capacity is
estimated as
CHACA1
β 2
...
Commercial concentrated acids and bases
Acids

%wt

HAc
HF
HCl
HBr
HI
HNO3
HClO4
H2SO4
H3PO4

99
...
0
37
...
0
47
...
0
70
...
5
85
...
4
28
...
1
8
...
5
15
...
7
18
...
7

NH3
KOH
NaOH

29
...
0
51
...
05
1
...
18
1
...
50
1
...
67
1
...
70

0
...
46
1
...
3
11
...
9

6
...
Acidic substances for standardizing basic solutions
Name

Formula

Molar
Mass

Molar
equivalent

Benzoic acid
Potassium hydrogen bis(iodate)
Potassium hydrogen o-phthalate
Sulfamic Acid

C6H5COOH
KH(IO3)2
C6H4(COOH)(COOK)
HSO3NH2

122
...
915
204
...
09

1
1
1
1

b
...
989
216
...
22
121
...
Indicators for acid-base titration
Common Name

pKa

methyl orange
bromocresol green
methyl red
bromothymol blue
m-cresol purple
phenolphthalein
thymolphthalein
thymol blue

3
...
66
5
...
10
8
...
00
10
...
70
8
...
1-4
...
8-5
...
2-6
...
2-7
...
6-9
...
3-10
...
4-10
...
2-2
...
0-9
...
Applications of acid-base titration
a
...
Jones factor for protein conversion
Animal origin

Vegetable origin

Food

Factor

Eggs
Meat
Milk
In general…
Cereals
Meat Products
Dairy Products

6
...
25
6
...
70
6
...
38

Food

Barley
Corn
Oats
Rice
Rye
Sorghums
Peanuts

Factor

5
...
25
5
...
95
5
...
25
5
...
83
6
...
30
5
...
25

Source: Food and Agriculture Organization of the United Nations - http://www
...
org/docrep/006/y5022e/y5022e03
...
Double indicator method for mixture of bases – Warder Titration
 The presence of hydroxide, carbonate and bicarbonate in water is also referred
to as alkalinity which is a measure of the acid-neutralizing capacity of water
 One method requires titration of the mixture to reach the phenolphthalein

Chemical and Physical Principles

Analytical Chemistry

endpoint with the volume recorded as V0-Ph
...
One sample is treated with
phenolphthalein and the other with methyl red
...

 m-Cresol purple can also be used to detect phenolphthalein alkalinity (P) while
bromocresol green or methyl orange for the total alkalinity (T)
Table 10
...
Acid number or acid value
 Defined as the mass (mg) of KOH that will neutralize the acid produced from
water degradative reaction of one gram of fat or oil
)(56
...
Saponification number or Koettstorfer number
H2C

OOCR1

HC

OOCR2

H2C

OOCR3

H2C
+ 3 KOH

HO

CH

H2C

R1COOK

OH
+

OH

R2COOK

R3COOK

 Defined as the mass (mg) of KOH required to saponify one gram of fat or oil
 Can be used to determine the approximate molar mass of fat or oil
 The sample is refluxed with ethanolic KOH and the resulting solution is
titrated with standard HCl
mL
(V mL  Vwith
samplel )(M HCl )(56
...
Precipitation Titration
1
...
Solubility rules for ionic compounds in water at 25°C
Soluble compounds

Insoluble compounds

All nitrates, bicarbonates, chlorates
and compounds containing alkali
metal ions and ammonium ion
...
Solubility product constant (KSP)
 Consider an aqueous saturated solution of a sparingly soluble salt represented by
the equation:
AxBy (s) ⮀ x A+y (aq) + y B–x(aq)
The equilibrium constant for this reaction would be:
[Ay(aq) ]x [Bx (aq) ]y
Keq 
[Ax By(s) ]
However the concentration of the solid AxBy in the solution will be constant (the
ratio of the number moles of AxBy and the volume of the solid is constant)
...
Solubility product constants at 25C
COMPOUND

KSP

COMPOUND

KSP

AgCl
AgBr
AgI
BaCO3
Ba(IO3)2
Al(OH)3
CaCO3 (calcite)

1
...
00 10–13
8
...
00 10–9
1
...
00 10–34
4
...
10 10–12
6
...
70 10–5
7
...
20 10–18
5
...
70 10–29

4
...
Formation of a secondary colored precipitate – Mohr method
 Developed by Karl Friedrich Mohr (1806-1879) in 1865
Titrant:
AgNO3

Analytical Chemistry

Chemical and Physical Principles
Titration reaction:

Ag+ (aq) + Cl–1 (aq) ⮀ AgCl(s)
white

Analytical Chemistry

Chemical and Physical Principles
Indicator:
Indicator reaction:

K2CrO4
2Ag+(aq) + CrO4–2 (aq) ⮀ Ag2CrO 4(s)
yellow

red

 In practice, the indicator concentration is kept between 0
...
005 M
 Titration is done at a pH of 8 to avoid precipitation of silver as hydroxide
(above pH of 10) and eliminate formation of HCrO4–1 (below pH of 6) which
results to consumption of more titrant
 Usually a low concentration of chromate is desired to detect the end point
clearly since a chromate ion imparts an intense yellow color
b
...
01 M
 For the titration of chloride, the resulting precipitate is filtered off before the
back titration since it reacts with the titrant and is more soluble than AgSCN
 For the titration of iodide, the indicator is not added until all iodide is
precipitated since the dissolved iodide is oxidized by the ferric ion
c
...
DCF–1
excess

Indicator:
Indicator reaction:

white

greenish-yellow

fluorescein, dichlorofluorescein or eosin
Ag+ (aq) + Cl–1 (aq) ⮀ AgCl(s) :Ag+1:DCF–1
excess

white

pink

 Before the equivalence point, chloride anion adsorbs to the precipitate in the
primary adsorption layer and drives the adsorption dye anion away by
electrostatic repulsion and the dye imparts a greenish-yellow color in solution
 As soon as the equivalence point is just exceeded with the presence of excess
silver ion, this ion now adsorbs to the precipitate in the primary adsorption
layer where the oppositely-charged adsorption dye anion adsorbs to the
counter-ion layer and imparts a pink color in solution
 For titration of chlorides, fluorescein may be used at an optimum pH range
between 7-10 while dichlorofluorescein is used in acidic solution of pH greater
than 4
...
Complexation Titration
1
...
Titration methods involving complexes
a
...
Determination of nickel
 An ammoniacal solution of nickel is treated with a measured excess of
standard cyanide solution:
Ni(NH3)6+3 + 4CN–1 + 6H2O ⭢ Ni(CN)4–1 + 6NH4OH
 The excess cyanide is determined according to the Liebig method
c
...
5 grams of
NH4Cl and 142 mL of concentrated NH3 and
 Indicators for EDTA titration
 Eriochrome Black T or Solochrome – used for titrations with pH more than
6
...
Oxidation-Reduction Titration
1
...
Oxidation numbers and balancing oxidation-reduction reactions
a
...
Balancing oxidation-reduction reactions
 A reaction is balanced when the number of atoms of each element and the net
charge on both sides are equal

Analytical Chemistry

Chemical and Physical Principles

 Step 1: Assign oxidation
numbers to each of the –1species in the reaction
–1
NO2

+

+3 –2

⮀

MnO4–1

NO3

+7 –2

+

+5 –2

MnO2
+4

 Step 2: Identify oxidation and reduction reactions and indicate the number of
electrons lost or gained, respectively
⮀

NO2–1

Oxidation:

NO3–1

+3

MnO 4–1

Reduction:

+

2e–

⮀

MnO 2

+5

+

3e–

+7

+4

 Step 3: Balance the reaction by multiplying a factor on both sides of the
reaction so that the numbers of electrons on both reactions are the same
Oxidation 3:

⮀

3 NO2–1

3 NO3–1

+3

Reduction 2:

+

6e–

+5

2 MnO 4–1

+

6e–

⮀

2 MnO2

⮀

3 NO3–1

+

2 MnO2

+7

3 NO2–1

+

+4

2 MnO 4–1

 Step 4: Balance the charges (by adding H+ or HO–) and number of hydrogen
and/or oxygen atoms (by adding H2O) on both sides of the equation
 In acidic medium, add H2O to the oxygen-deficient side and supply H+ to
balance the hydrogen
3 NO2–1 + 2 MnO4–1 + 2 H+ ⮀ 3 NO3–1 + 2 MnO2 + H2O

 In basic medium, balance assuming reaction was in acidic medium
...
Standard electrode potential
 The potential of a half-cell reaction with the standard hydrogen electrode (SHE)
used as anode when the activities of all reactant and products are taken as unity,
that is, 1M concentration and 1 atm partial pressure
 Usually listed as standard reduction potential (red) where a positive value
implies that the electrode was used as a cathode and the SHE as anode
 High value of a reduction potential indicates that the electrode is a good oxidizing
agent
Nernst equation
 Formulated by Walther Hermann Nernst (1864-1941)
 Accounts for the effect of concentration on electrode potentials
For a half-cell reduction reaction…Ox ⮀ Red + ne–
ε ε  RT ln a Red
Red

Red

nF

a Ox

where Red = actual cell potential [V], Red = standard reduction potential [V],
R = 8
...
3399 coul-(mol e–)–1
At 25C and for a given cell…
εcell εcell 

0
...
15 K, = 0 and Q = K)
0
...
Oxidation-reduction titration methods
a
...
Permanganate titration
 Titration is carried out in acidic medium using sulfuric acid
 In the presence of HCl, titrant is consumed to oxidize Cl–1
 Acidic and basic solutions of KMnO4 are less stable than neutral ones and
kept in dark-colored bottles to avoid decomposition
Titrant:
KMnO4
Half-cell reactions:
 acidic medium
MnO4–1 + 8H+ + 5e– ⮀ Mn+2 + 4H2O
 basic medium
MnO4–1 + 2H2O + 3e– ⮀ MnO2 + 4HO–
Primary standards:
 As2O3
H3AsO3 + H2O ⮀ H3AsO4 + 2H+ + 2e–
 Na2C2O4
C2O4–2⮀ 2CO2 + 2e–
 Fe metal
Fe ⮀ Fe+2 + 2e–
 FeSO 4
(en)SO 4
4H 2O
Fe +2 ⮀ Fe +3 + e –
Indicator:
self-indicating
Endpoint:
pale pink color that persists for 30 s
c
...
Cerium (IV) titration
 Titration is carried out in acidic medium using sulfuric acid at concentrations
0
...
Iodimetry: Direct titration with iodine
 Titration is carried out in neutral, weak alkaline or weak acidic solutions
Titrant:
I2 dissolved in concentrated solution of KI
Half-cell reaction:
I3–1 + 2e– ⮀ 3I–1
Primary standard:
As2O3
Half-cell reaction:
H3AsO3 + H2O ⮀ H3AsO4 + 2H+ + 2e–
Indicator:
Starch solution
Endpoint:
Formation of intensely blue-colored complex
f
...
Summary of oxidants and reductants used in titration
Oxidants

Half-cell reaction

ε° (V)

KMnO4 (acidic)
KMnO4 (basic)
MnO2
K2Cr2O7
Ce(SO4)2
I2 in KI
I2 (satd)
I2(aq)
KIO3

MnO4–1 + 8H+ + 5e– ⮀ Mn+2 + 4H2O
MnO4–1 + 2H2O + 3e– ⮀ MnO2 + 4HO–
MnO2 + 4H+ + 2e– ⮀ Mn+2 + 2H2O
Cr2O7–2 + 14H+ + 6e– ⮀ 2Cr+3 + 7H2O
Ce+4 + e– ⮀ Ce+3
I3–1 + 2e– ⮀ 3I–1
I2 + 2e– ⮀ 2I–1
I2 + 2e– ⮀ 2I–1
–1
2IO3 + 12H+ + 10e– ⮀ I2 + 6H2O

Combining
ratio

1
...
695
1
...
33
1
...
5355
0
...
6197
1
...
559
0
...
440
–0
...
08

4
2
2
1
1

As2O3
Na2C2O4
Fe metal
FeSO4
Na2S2O3

Analytical Chemistry

Chemical and Physical Principles

5
...
Iodine number of oils and fats
 Measure of the degree of unsaturation of fats or oils
 Expressed as the number of centigrams of iodine absorbed by 100 grams of fat
or oil
 Sample is dispersed in chloroform, treated with solution of iodine
monochloride in glacial acetic acid (Wij’s solution) and allowed to react in the
dark for 30 min
C

C

+ IBr (excess) ⭢

C

C

I

Br

+ IBr (unreacted)

 KI is added to liberate the unreacted iodine and titrated with standard
Na2S2O3 solution
IBr (unreacted) + KI ⭢ I2 + KBr
Iodine Number 

V

mL,sample
Na 2S2O3



mL, blank
 V Na
M Na S O
SO
2 2

3

2 2

3



1 I2
2 Na 2S2O3

253
...
Peroxide value of oils and fats
 Measure of the extent of oxidative rancidity of fats and oils during storage
 Expressed as the number of milliequivalent or millimole of peroxide per
kilogram of sample
 Sample is dissolved in a mixture of chloroform and acetic acid (2:3), bubbled
with nitrogen gas to remove remaining oxygen and treated with excess KI to
liberate iodine
2 I–1 + RO2H + H2O ⭢ ROH + 2 HO–1 + I2

V

Peroxide Value 

mL,sample
Na 2S 2O3



blank
 V mL,
M
Na S O
2 2

3



1 meq Na 2S 2 O 3
Na 2S 2O3 1 mmol Na 2S2O3

mass sample (kg)
c
...
Chemical oxygen demand (COD)
 Measure of the amount of oxygen necessary to oxidize all the organic material
in a water sample
 Expressed as milligrams of oxygen required for oxidation per liter of sample
 Sample is refluxed in the presence of HgSO4, Ag2SO4/H2SO4 solution and a
known excess amount of standard K2Cr2O7 solution and back titrated with

Chemical and Physical Principles
standard (NH4)2Fe(SO4)2 solution

Analytical Chemistry

Analytical Chemistry

Chemical and Physical Principles

V

COD 

mL,blank
Fe2



 V mL,sample
M
2
Fe



1 mmol K 2 Cr2 O 7

Fe2

6 mmol Fe2



6 mmol O 2
4 mmol K 2Cr2O7

 32 mg O 2

1 mmol O2

volume sample (L)
Molecular Absorption Spectrometry
A
...


Quantitative Analysis

1
...

 If each of these solutions were assumed to obey Beer’s law, the absorbance (AS)of
each solution is described by:
As 

εbVSCS
VT



εbVXCX

kV C  kV C mC  b

VT

where k = b/VT, m = kVS and b = kCXVX

S S

X

X

S

Chemical and Physical Principles

Analytical Chemistry

2
...
Photometric titration
 Plot of absorbance versus volume of titrant where the curves consist of two
straight line regions with different slopes
 The end point is the intersection of the extrapolated linear regions

REVIEW QUESTIONS AND PROBLEMS
1
...
CaSO4
b
...
P2O5
d
...


Analytical methods classified as micro analysis uses sample mass ranging from
a
...
1-10 mg
c
...
> 100 mg

3
...

a
...
certified reagent
b
...
all of these

4
...
TC
b
...
In
d
...


What proportion by weight of Na 2C 2O 4 (134) to KHC 2O 4
H 2C 2O 4 (218
...
0
...
0
...
1
...
3
...


Platinum crucibles can be used for the following processes without significant loss
except
a
...
Evaporation with hydrofluoric acid
c
...
Heating with sulfides

7
...
955 M HCl solution?
a
...
9 mL
b
...
8 mL
c
...
0 mL
d
...
1 mL

8
...
Grade 1
b
...
Grade 3
d
...
45) at 25C has a density of 1
...

Express the amount of solute in this solution as follows:
9
...
17

b
...
21

d
...
molarity (M)
a
...
98

b
...
96

c
...
92

d
...
88

11
...
3
...
5
...
6
...
8
...
The following describes colloidal suspensions formed during precipitation except
a
...
They settle readily from a given solution
c
...
none of the these
13
...
5176 g sample of a CaCO3 was dissolved in an acidic solution
...
11) and the ignited precipitate at 230C was found
to weigh 0
...
What is the percentage of CaO (56
...
20
...
23
...
41
...
47
...

14
...
Pb+2
b
...
S–2

d
...
The charge of this layer is
a
...
positive

d
...


16
...
Pb+2
b
...
neutral
c
...


d
...
A mixture containing FeCl3 (162
...
33) only weighs 750
...
The
chlorides were precipitated using ammonia and ignited to Fe2O3 (159
...
96), respectively
...
3 mg
...
98) in the sample
...
15
...
41
...
43
...
58
...


Analytical Chemistry

Which of the following does not describe the correct way to wash precipitates?
a
...
Wash with small portions of washing liquid
c
...
Gelatinous precipitates requires more washing than crystalline precipitates
19
...
94) in the
ore would be twice the mass of Mn3O4 (228
...
19
...
38
...
57
...
76
...
Process by which an agglomerated colloid return to it dispersed state during washing
due to leaching of electrolyte responsible for its coagulation
a
...
coagulation
c
...
peptization
21
...
488-gram sample containing MgCl2 (95
...
45) was dissolved in
sufficient to give 1L solution
...
462 gram AgCl
...
2610 gram of Mg2P2O7 (222
...
Determine the percentage of NaCl in the sample
...
12
...
23
...
36
...
40
...
It is the expressed as the volume of a solution chemically equivalent to a mass of a
solid reagent
a
...
aliquot
c
...
ppm
23
...
5 M H2SO4 at 25C?
a
...
2%
c
...
8%
24
...
NH3
b
...
NH4Cl
d
...
Which of the following solutions at 25C will have the lowest pH value?
a
...
15 M Na2SO4 (KA2 of H2SO4 = 1
...
5
...
0
...
2 10–10)
d
...
01 M NH4Cl (KA of NH4+ = 5
...
Which of the following acid-base pairs will result in the formation of a buffer solution
when titration is done before the equivalence point?
a
...
NH3 – HBr
b
...
all of these
27
...
1025 M HCl must be added to 15
...
0956 M NH3 to
produce a solution of pH = 9
...
5
...
7
...
9
...
11
...
Which of the following statements is not correct?
a
...
The larger the buffer capacity, the more resistant the solution is to pH change
c
...
The acid buffering capacity is maximum at pH equal to pKa

Analytical Chemistry

Chemical and Physical Principles

29
...
50 and a buffer capacity of 0
...
0
...
0
...
0
...
0
...
Approximately how many grams of NH4Cl (53
...
125 F NH3 to reduce the concentration of hydroxide ions to one-thousandth of its
original value?
a
...
1 g
b
...
5 g
c
...
7 g
d
...
91 g
31
...
HPO4–2
b
...
H3PO4

d
...
0 mL of 0
...
100 M NaOH:
32
...
1
...
2
...
4
...
5
...
15 mL
a
...
57

b
...
71

c
...
86

d
...
04

34
...
1
...
2
...
5
...
6
...
49
...
4
...
7
...
8
...
12
...
02 10–10
...
0
...
3
...
5
...
6
...
8
...
0
...
070 M CH3COONa
a
...
44
b
...
05
c
...
22

d
...
82

38
...
035 M CH3COONa
a
...
97

d
...
38

b
...
32

c
...
03

39
...
to eliminate the reaction product, carbon dioxide and carbonic acid
b
...
to achieve a sharper endpoint with methyl red indicator due to the large
decrease in pH
d
...
Calculate the molarity of NaOH solution if 12
...
2615 gram
of primary standard KHP
...
0
...
0
...
0
...
0
...
What is the best basis for choosing the right indicator for a given acid – base titration
from among the following?
a
...
pH at equivalence point
b
...
molarity of the acid or base
42
...
431 grams of KHP, the analyst uses
35
...
25 mL of acid (1mL = 10
...
What is the molarity of the NaOH solution?
a
...
2118 M
b
...
2044 M
c
...
7831 M
d
...
2598 M
43
...
bromocresol green
c
...
methyl orange
d
...
A 0
...
18 ml of 0
...
44 ml of 0
...

Calculate the percentage purity of calcite in terms of % wt/wt CaCO3 (100)
...
17
...
35
...
53
...
71
...
Process of determining the nitrogen content of organic materials by mixing the sample
with powdered copper (II) oxide and ignited to a combustion tube giving CO2, H2O,
N2 and small amounts of nitrogen oxides
...
Kjeldahl Method
c
...
Dumas Method
d
...
A 640 mg sample of P2O5 (141
...
995) impurity
...
867 M NaOH
requiring 20
...
Calculate the percentage of
impurity in the sample
...
0
...
1
...
2
...
3
...


In the titration of phosphoric acid, which of the following statements is true?
a
...
Phenolphthalein indicator is used to detect the second end point
...
It can be treated as a monoprotic or diprotic acid during titration
d
...
1025 M
HCl via double indicator method
...
27
0
...
12
6
...
63
V0-MR (mL)
10
...
19
10
...
38
9
...
Which of the following mixtures contains NaHCO3?
a
...
Mixtures 3 and 4
b
...
Mixtures 1 and 2
49
...


Analytical Chemistry

Chemical and Physical Principles
a
...
82%

b
...
22%

c
...
64%

d
...
13%

50
...
1
...
5
...
10
...
11
...
In the analysis of nitrogen using Kjeldahl Method, which of the following is added to
decompose organic matrices present in the sample?
a
...
HgO
c
...
H2O2
52
...
279-gram sample of meat was analyzed for its nitrogen content using Kjeldahl
Method
...
855 M
H3BO3
...
25 mL of 0
...
Determine the % protein in the sample using 6
...

a
...
98%
b
...
24%
c
...
69%
d
...
14%
53
...
HgO
b
...
HSO3NH2
d
...
05 gram sample of butter is refluxed with ethanolic KOH and required 10
...
1875 M HCl to reach the phenolphthalein end point
...
7 mL of the same standard acid,
54
...
45
b
...
125

d
...
Assuming butter comprised mainly of fat, what is its molar mass?
a
...
970
c
...
1940
56
...
In Volhard method, AgCl is more soluble than AgSCN thus requiring filtration
of AgCl prior to back titration
b
...
In Mohr titration, the concentration of the chromate ion in more basic
solutions is too low to produce the precipitate near the equivalence point
d
...
In the determination of chloride using Mohr method, what should be the theoretical
molar concentration of the chromate ion indicator given that KSP of Ag2CrO4 is 1
...
0
...
0
...
0
...
0
...
Which of the following is not a correct analytical method–titrant pair?
a
...
Volhard – AgNO3
b
...
Liebig – AgNO3
59
...
500-gram sample of impure aluminum chloride was dissolved in water and treated
with 45
...
1000 M AgNO3 using K2CrO4 as indicator
...
33)
...
40
...
13
...
4
...
27
...
In Volhard Method, why is it necessary to carry out titration in acidic solution?
a
...
To prevent formation of AgSCN precipitate
c
...
To prevent precipitation of silver as hydrated as hydrated oxide

Chemical and Physical Principles

Analytical Chemistry

61
...
845) and BaBr2 (297
...
00mL
of 0
...
76 mL of 0
...
What is the percentage of BaBr2 in
the sample?
a
...
95%
b
...
05%
c
...
62%
d
...
38%
62
...
The indicator is usually kept at a concentration of 0
...
5 M so as not to
obscure the red precipitate color
b
...
At high pH, silver is precipitated as silver hydroxide thus produces error in the
amount of titrant added
d
...
A 750
...
Its
ammoniacal solution was treated with 50 mL of 0
...
25 mL of 0
...
Determine % Ni (58
...

a
...
86%
b
...
69%
c
...
53%
d
...
72%
64
...
The titration process requires an indicator to signal the end of titration
b
...
A red complex of silver with cyanide is formed which signals the end of the
titration process
d
...
A 500-mg sample containing NaCN required 23
...
1255 M AgNO3 to
obtain a permanent faint turbidity
...

a
...
34%
b
...
01%
c
...
25%
d
...
67%
66
...
Ca+2
b
...
Mg+2
d
...
Which of the following affects the stability of metal complexes?
a
...
chelate effect
b
...
all of these
68
...
It was
standardized using 0
...
50 mL of the solution
...
57 mL of the EDTA solution
...

a
...
316 ppm
c
...
269 ppm
69
...
Multidentate chelating agents form weaker complexes with metal ions
b
...
Eriochrome black T gives a sharp endpoint for the titration of calcium
d
...
The 300 mg sample of impure Na2SO4 (142
...
00 mL of 0
...
The
precipitate was removed by filtration and the remaining BaCl2 consumed 6
...
2467 M EDTA for titration to the Calmagite endpoint
...

a
...
85%
c
...
95%
71
...
8521 gram sample of an alloy was found to contain Cu (63
...
41)
with small amounts of Pb (207
...
59)
...
A 10 mL aliquot was treated with KI to mask the Hg and
the resulting solution required 7
...
0348 M EDTA solution
...
00 to reduce Hg+2
and the metallic Hg was removed from the solution
...
58 mL for titration
...
11 mL for titration
...

a
...
53% Cu and 7% Hg
b
...
56% Cu and 5% Hg
72
...
Prereduction is then necessary before titration
...
This reductor is known as

...
Lindlars catalyst
a
...
Devarda Alloy
d
...
At pH = 7 and a pressure of 1 bar, the potential for the half reaction, 2H+ (aq) + 2e– ⭢
H2 (g) is

...
0 V
b
...
414 V
c
...
828 V
d
...
255 V
74
...
Standard iodine solutions have low smaller electrode potential
b
...
Iodine is very soluble in water and losses are minimal
d
...
What is the molarity of a KMnO4 solution standardized against 1
...
1 mL of the solution in acidic medium?
a
...
161 M
b
...
403 M
c
...
008 M
d
...
856 M
76
...
KMnO4
b
...
K2Cr2O7

d
...
A sample of iron ore weighing 385
...
The resulting solution 52
...
01436 M K2Cr2O7 for titration to
the diphenylamine sulfonic acid endpoint
...
55 g/mol) in the ore
sample
...
15
...
45
...
90
...
67
...
A sample of pyrolusite weighing 0
...
The iodine
liberated required 46
...
1105 M Na2S2O3 solution
...
94)

Analytical Chemistry

Chemical and Physical Principles
in the sample
...
46
...
30
...
92
...
76
...
Which of the following metal is not reduced by Walden reductor?
a
...
Fe
c
...
V
80
...
00 gram sample of cooked-ham was pureed with 200 mL of water, filtered and
the resulting solution containing dissolved potassium nitrite was acidified
...
00 mL of 0
...
97 mL of 0
...
Calculate the amount of nitrite (46
...

a
...
900
c
...
1800
81
...
sodium bismuthate
c
...
ammonium peroxydisulfate
d
...
Which of the following statements is true on the determination of chemical oxygen
demand?
a
...
Chemical oxygen demand accounts for the oxygen demand of both
biodegradable and non-biodegradable oxidizable pollutants in a body of water
c
...
all of these
83
...
If 7
...
0124 M
Na2S2O3 was used for titration, determine ppm O2 in the water sample
...
2
...
3
...
6
...
7
...
The following volumes of standard solutions containing 20 ppm Fe+2 were added to
10-mL aliquot samples of waste water in 500-mL volumetric flasks
...
00-cm cell:
Volume of standards, ppm
0
5
10
15
20
25
Absorbance
0
...
281 0
...
628 0
...
972
Calculate the concentration of Fe+2 in the waste water
...
5
...
8
...
11
...
16
...
A 500 mg sample of a solution containing cobalt (58
...
69) was
dissolved and diluted to 50 mL
...
The following
data were obtained for the simultaneous determination in a 1
...
815
0
...
1172 ppm
b
...
2142 ppm

d

---