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Introduction Page 1
Chapter 1
Introduction
Qualitative chemical analysis, branch of chemistry
that deals with the identification of elements or grouping of
elements present in a sample. In some cases it is necessary
only to verify the presence of certain elements or groups for
which specific tests applicable directly to the sample
(e.g., flame tests, spot tests) may be available. More often
the sample is a complex mixture, and a systematic analysis
must be made in order that all the constituents may be
identified.
The classical procedure for the complete systematic
analysis of an inorganic sample consists of several parts.
First, a preliminary dry test may be performed, which may
consist of heating the sample to detect the presence of such
constituents as carbon (marked by the appearance of smoke
or char) or water (marked by the appearance of moisture)
Introduction Page 2
or introducing the sample into a flame and noting the
colour produced. Certain elements may be identified by
means of their characteristic flame colours. After
preliminary tests have been performed, the sample is
commonly dissolved in water for later determination of
anionic constituents (i.e., negatively charged elements or
groupings of elements) and cationic constituents
(i.e., positively charged elements or groupings of elements).
The procedure followed is based on the principle of treating
the solution with a succession of reagents so that each
reagent separates a group of constituents. The groups are
then treated successively with reagents that divide a large
group into subgroups or separate the constituents singly.
When a constituent has been separated it is further
examined to confirm its presence and to establish the
amount present (quantitative analysis). Portions of the
material are dissolved separately, and different procedures
are used for each to detect the cationic and anionic
constituents.
Introduction Page 3
PROPERTIES OF AQUEOUS SOLUTIONS
Many chemical reactions and virtually all biological
processes take place in aqueous media. Sum frequently used
terms of the aqueous medium in which reactions take place
are described.
A solution is a homogeneous mixture of two or more
substances. The substance present in smaller proportion is
called the solute, and the substance that is present in a larger
amount is called the solvent. An aqueous solution can contain
more than one kind of solutes. Seawater is
a solution that contains more than sixty different substances.
Electrolytes and Non-electrolytes
All solutes in aqueous solutions can be divided into two
categories: Electrolytes and nonelectrolytes. An electrolyte is
a substance that, when dissolved in water, results in a
solution that can conduct electricity. A nonetectrolyte does not
conduct electricity when dissolved in water.
Introduction Page 4
Ionic compounds such as sodium chloride and potassium
iodide, and certain acids and bases such as hydrochloric acid
and sodium hydroxide are all strong electrolytes. They
share the common characteristic of ionising completely
when dissolved in water:
HCI (aq) → H+ (aq) + Cl
¯ (aq)
In other words, all dissolved HCI molecules give H+ , and Cl
¯
ions in solution. On the other hand, weak electrolytes such
as acetic acid ionise much less. We represent the ionisation
of acetic acid as:
CH3COOH (aq) CH3COO¯ (aq) + H
+(aq).
Degree of Dissociation
Molecules cations + anions
Arrhenius introduced a quantity "α" called the degree
of dissociation, defined as follows.
Introduction Page 5
∝= Number of solute molecules dissociated
number of solute molecules before dissociation
Equilibrium Degree of dissociation “α”
HCl H+
+ Cl¯
HNO3 H+ + NO3
¯
H2SO4 H+ + HSO4
¯
CH3OOH H+ + CH3OO
¯
NaOH Na+ + OH
¯
NH4OH NH4+ + OH
¯
NaCl Na+ + Cl
¯
AgNO3 Ag+ + NO3
¯
HgCl2 Hg2+
+ 2Cl¯
0.92
0.92
0.61
0.013
0.91
0.018
0.86
0.82
0.01
Dynamic Equilibrium
Where equilibrium consists of reaction that constantly
and simultaneously proceeds in opposite directions but with
equal speed.
Law of Mass Action
"The rate of a chemical reaction is directly proportional
to the product of the molar concentration of the reacting
Introduction Page 6
substances each raised to a power equal to the number
of ions or molecules appearing in the balanced equation
for the reaction".
aA + bB cC + dD
In which a b, c and d signify the stoichiometric
numbers of particles of A, B, C and D, respectively,
involved in the completely balanced reaction, the
equilibrium constant may be expressed by the relation.
[𝐂]𝐜[𝐃]𝐝
[𝐀]𝐚[𝐁]𝐛= 𝐊𝐞𝐪
Regardless of the detailed mechanism of the reaction.
The ionic product of water
Water is a very wash electrolyte which ionizes to a very
limited extent according to the equation
H2O → H+ + OH
¯
Applying the law of mass action,
Introduction Page 7
Keq =[H+][OH
¯] / [H2O] . . .
Since the concentration of water, [H2O], is constant at
constant temperature, or it is said to have unit activity, the
quantity [H2O] can be removed, and the equation may be
written
Kw =[H+][OH
¯]
Where Kw represents the ionic product of water. At 25oC,
Kw = I x 10¯14
Furthermore, since the dissociation of water gives rise to an
equal number of hydrogen and hydroxyl ions, the above
equation may be written
[H+]2=Kw = I x 10¯
14
Or in the other words
[H+]=I x 10¯
7
It follows that in a given aqueous solution, if the [H+] =
[OH¯] = I x10¯7 the solution is described as neutral, while if
Introduction Page 8
[H+] = is more than I x10¯
7 that is I x10¯
6, I x10¯
5, etc, the
solution is said to be acidic, and, if [H+] is less
than l0-7, that is l0
-8, l0
-9 etc. the solution is described as
alkaline.
The pH and pOH values of pure water are each 7.0. In
general, in my aqueous solution:
pH + pOH = 14
For convenience, the acidity or basicity of a solution is
commonly expressed in terms of the hydrogen ion exponent,
or pH units.
The pH of a solution is defined as pH = - log [H+].
Precipitation Reaction and Solubility product
Precipitation of a slightly soluble substance from solution
is one of the principle operations of qualitative analysis. The
equilibrium between a slightly soluble electrolyte and its ions
in solution is one of the important applications of the law of
mass action. Let us consider a solution of the slightly soluble
salt AaBb in which excess solid is present:
Introduction Page 9
AaBb aA+ + bB¯
(Solid) (Solution)
Applying the law of mass action to this system:
[𝐀+]𝐚[𝐁−]𝐛
[𝐀𝐚𝐁𝐛]= 𝐊
Since pure solids are considered to be at unit activity and
thus need not to be shown in equilibrium constant expression
the equilibrium constant for the heterogeneous equilibrium
between the solid AaBb, and its ions is termed the solubility
product constant (KSP).
The solubility product constant of a slightly soluble
electrolyte is the product of the molar concentration of its ions
in a saturated solution, each raised to the appropriate power.
As previously mentioned (see law of mass action) for the
calculation of Ksp the concentration must be expressed in
gram moles per liter (molar concentration).
Example 1
The solubility of sliver chloride is I.05 x 10-5
M. calculate
the solubility product constant?
Introduction Page 10
In a saturated solution each 1 mole of dissociated of AgCl
gives 1 mole of Ag and CI.
AgCl Ag+ + Cl
-
Hence, [Ag+] = 1.05 x 10
-5, and [Cl
-]= 1.05 x 10
-5
Ksp= [Ag+] [Cl
-] = (1.05 x 10
-5)( 1.05 x 10
-5)= 1.1 x 10
-10
Example 2
Calculate the solubility product of silver, given that its
solubility is 2.5 mg% (Mwt. of Ag2CrO4 = 332).
First we convert absolute solubility to molar solubility
𝑚𝑜𝑙𝑎𝑟 𝑠𝑜𝑙𝑢𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝐴𝑔2𝐶𝑟𝑂4 = 2.5 𝑥 10
1000 𝑥 332= 7.5 𝑥 10−5 𝑀
Ag2CrO4 2 Ag+ + CrO4
2-
I mole of Ag2CrO4 gives 2 moles of Ag+ and I mole of
CrO42-
Therefore, Ksp= [Ag+]2 [CrO4
2-]
Introduction Page 11
= (2 x 7.5 x 10-5)2 (7.5 x 10
-5)
= 1.7 x I0-12
Example 3
The solubility product or Lead orthophosphate is 1.5 x 10-32
.
Calculate its molar solubility.
Pb3(PO4)2 3Pb2+
+ 2PO43-
X 3 X + 2X
If x is the molar solubility
[Pb2+
] =3 X and [PO43-] = 2 x
Ksp = [Pb2+
] 3 [PO4
3-]
2
1.5 x 10-32
= [3X]3[2X]
2 = 108 X
5
𝑋 = 1.5 𝑥 10−32
108= 1.7 𝑥 10−7
𝑚𝑜𝑙𝑒
𝐿.
5
Applications of the solubility product relation
The solubility product relation is of great value in qualitative
analysis.
Introduction Page 12
I- With its aid it is possible not only to explain but
also to predict precipitation reactions.
If the experimental conditions are such that the ionic
product is different from the solubility product, then the
system will attempt to adjust itself in such a manner that the
ionic and solubility products are equal in value. Thus, if, for a
given electrolyte, the product of the concentrations of the
ions in solutions arbitrarily made to exceed the solubility
product as, for example by the addition of a saIt with a
common ion, the adjustment of the system results in the
precipitation of the solid salt (provided supersaturation
conditions are excluded). If the ionic product is less than the
solubility product, or can arbitrarily made smaller, as for
example by complex saIt formation or by the formation of
weak electrolytes, then a further quantity of the solute can
pass ,into solution until the solubility product is attained, or if
this is not possible, unfit all the soluble has dissolved.
As an example of the formation of a precipitate, let us
consider the case of sliver chloride.
Introduction Page 13
Ksp= [Ag+] [Cl
-] = (1.05 x 10
-5)( 1.05 x 10
-5)= 1.1 x 10
-10
Let us suppose that to a solution which is 0.1 M in silver
ions we add enough potassium chloride to produce
momentarily a chloride concentration of 0.0 I M. The ionic
product (solubility quotient, QS) is then 0.1 x 0.01 = 1 x 10-3
.
Since 1 x 10-3
> 1 x 10-10
to equilibrium will not exist and
precipitation will take place (Ag+ + Cl
- → AgCl) until the value
of the ionic product has been reduced to that of the solubility
product i,e. until [Ag+] [Cl
-] = 1.5 x 10
-10, at this point the rate
of precipitation is equal to the rate of solution of the
precipitate. If now, with a saturated solution of silver chloride
as the initial solution we add either a soluble chloride salt or a
soluble silver, salt in small quantity, a slight further
precipitation, of silver chloride takes place if, after equilibrium
has been reached the concentration of the respective ions
are determined, we find that, although the concentration of
one ions has increased and that of the other has decreased,
the ionic product or solubility product is roughly constant.
Example 4
If the Ksp of silver chloride is 1.5 x 10 -10
, would a
Introduction Page 14
precipitate formed on mixing I x l0-2 M hydrochloric acid
and 3 x I0-5 M silver nitrate,
Ionic product (or QS) = [Ag+] [CI
-]
= (3 xl0-5)(I x 10
-2) = 3 x l0
-7
ionic product > Ksp
.'. Precipitate is formed.
II- The Ksp, value of slightly soluble electrolyte tells
us the maximum concentration of its ions that can
exist in solution
Example 5
What is the concentration of silver ion remaining in solution
after adding sufficient hydrochloric acid to a solution of Silver
nitrate to make the final chloride ion concentration 0.05 molar
of AgCl = 1.5 x l0-10
Ksp= [Ag+] [Cl
-]
1.5 x l0-10
= [Ag+] x (5 x10
-2)
[Ag+]= 3 x 10
-9 M
Introduction Page 15
III- Knowing the analyte concentration it is possible
to calculate the amount of reagent necessary to:
a. Just initiate precipitation of an ion
b. Nearly complete precipitation of an ion.
c. Fractional precipitation.
Example 6
Calculate in gram per liter the maximum concentration
of cadmium -ions and manganous ions which will remain
in solution after precipitation by excess saturated aqueous
solution of hydrogen sulphide in 0.3 M hydrochloric acid
(Ksp CdS = 5.5 x 10-25
; MnS =1.4 x 10-15
, atomic weight of
Cd and Mn are 112.4 and 54.9, respectively).
The sulphide ion concentration [S2-] in 0.3 M HCl
saturated with H2S is
[S2-] [H
+]2 = 1.1 x10
-23
[S2-] = 1.1 x 10
-23/(0.3)
2 = 1.2 x 10
-22
[Cd2+
][S2-
] = K=sp = 5.5 x 10-25
[Cd2+
]= = 5.5 x 10-25
/1.2 x 10-22
= 4.6 x 10-3M = 0.5 g/L
[Mn2+
][S2-] =Ksp = 1.4 x10
-15
[Mn2+
] = 1.4 x10-15
/ 1.2 x 10-22
1.2 x10-3M = 6.6 x 10
8 g/L.
Introduction Page 16
This figure clearly show that no MnS will be precipitated by
H2S in 0.3 M HCI. However, if the original concentration of
Cd2+
is I0 g/L. The percentage of precipitate Cd2+
is 95%.
If we used to precipitate MnS we must increase the [S2-
] by
alkalization.
Example 7
A 0.1 M-solution of potassium sulphate is added gradually
to a solution containing Ba2+
(0.l M) and Sr2+
(0.l M);
Calculate:
i) whether barium sulphate or strontium sulphate will
be precipitated first
ii) the concentration of barium ion in solution when
strontium sulphate begins to precipitate and,
iii) the fraction of the original barium ion concentration
remaining in solution when strontium sulphate
commences to precipitate (Ksp, of ,BaSO4 and
SrSO4 are 9.2 x l0-11
and 2.8 x l0-7
,respectively).
(i) Ksp BaSO4 = 9.2 x l0-11
= [Ba+2
] [SO42-
]
= (0.1) [SO42-
]
Introduction Page 17
[SO42-
] = 9.2 X 10-10
Ksp, SrSO4 = 2.8 X 10-7
= [Sr+2
] [SO42-
]
= (0.1) [SO42-
]
[SO42-
] = 2.8 X 10-6
A smaller concentration of [SO42-
] is required to precipitate
BaSO4 and hence it-will precipitates first.
(ii) The SO42-
concentration necessary to begin precipitation
of SrSO4is 2.8 x 10-6
M; at this concentration of SO42-
the
[Ba+2
] will precipitate
[Ba+2
] = Ksp/ [SO42-
] = 9.2 X 10-10
/ 2.8 X 10-6
= 3.3 x 10-5
M
(iii) The fraction of the original Ba+2
remaining in solution
when SrSO4 commences to precipitate
= 3.3xl0-5
/0.1 = X 100 =0.033%
The fraction of precipitated barium=100 - 0.033 = 99.967%.
Introduction Page 18
Dissolution of precipitates or preparation of solutions
In order to dissolve a precipitate or to prepare a solution of
a slightly soluble substance in water, the process of
precipitation must be reversed, i.e. the action of the reagent
must be such as to reduce the concentration of one or both-
of the ions of the slightly soluble substance.
The concentration of an ion can he reduced by two general
methods;
1- Dilution with more solvent
2- Through chemical reaction; some of the more
common chemical reactions for dissolution of
precipitates are described below:
(a) Formation of a weak electrolyte.
(i) Formation of weak acids or bases:
This method is applicable to electrolytes which are
derivatives of weak acids or bases, such as barium
sulphite, calcium oxalate and calcium carbonate which are
soluble in dilute hydrochloric acid because of the formation
of the weak acids H2SO3, H2C2O4 and H2CO3. These weak
Introduction Page 19
acids can exist in equilibrium with only very small
concentration of their ions: and the anion concentration is
diminished still further by the presence of hydrogen .ion
from the completely dissociated hydrochloric acid.
CaC2O4 Ca2+
+ C2O42-
2 HCl 2Cl- + 2H
+
H2C2O4
In the same manner, manganous, zinc and magnesium
hydroxides dissolve in ammonium chloride solution owing to
the formation of the weak base, ammonia. The presence of
high concentration of ammonium ion reduces the hydroxyl
ion concentration to so low a value that the Ksp of the above
mentioned hydroxides cannot be attained and they
consequently pass into solution.
Mg (OH)2 Mg2+
+ 2 OH-
2 NH4Cl 2Cl- + 2 NH4
+
NH4OH
Introduction Page 20
(ii) Formation of weakly ionised salt:
For example, the dissolution of lead sulphate in
saturated ammonium acetate solution, which is due to
formation of the weakly ionised lead acetate.
PbSO4 + 2CH3COO- Pb(CH3COO)2 + SO4
2-
(b) Formation of complex ion:
Many ions form stable complex ions through the
combination with other ions or molecules, For example, the
Ag+
units with two NH4+ to form the Ag(NH3)2
+ complex
(weak electrolyte); this yield so small a silver ion
concentration, particularly in the presence of excess of
ammonia solution, that the KSP of AgCl is not attained.
AgCl Cl- + Ag
+
2 NH4OH 2H2O + 2 NH3
Ag(NH3)2+
Introduction Page 21
(c) Double decomposition:
In this method the product of the reaction is either
another insoluble substance or a substance can be easily
removed (e.g. decomposed or volatilized). The procedure is
exemplified when strontium sulphate (acid insoluble) is
heated: with a large excess of a saturated solution of
sodium carbonate
SrSO4 + CO32- SrCO3 + SO4
2-
After filtration the formed strontium carbonate may be
dissolved in dilute hydrochloric acid, and the solution
containing the cation is obtained.
(d) Oxidation or reduction:
The oxidation or reduction of ion(s) of a slightly soluble
substance results in decreasing its concentration in solution
that the solubility product of this substance is no attained
and hence passes into solution for example the dissolution
of cupric, cadmium and bismuth sulphide in nitric acid. The
Introduction Page 22
sulphide ion is oxidised to free sulphur, the [S2-
] is thereby
reduced below the Ksp, of the sulphide salt and hence the
latter passes into solution.
3CuS 3 Cu2+
+ 3 S2-
+ 2 NO3-
+8 H
+
3 So + 2NO +4 H2OH
Complex ion and complex formation
A complex ion is .formed by the union of a simple ion with
either other ions of opposite charge or with neutral
molecules. Complex ions are formed in accordance with the
theory which postulates that; certain have two types of
valence:
I- Principal (ionizable) valence which results in the
formation of simple salts.
Example :
AgCl Ag+ + Cl
-
Introduction Page 23
2- Auxiliary valence which gives the metal a definite
capacity for combining with, or attaching to itself other ions
or neutral molecules (these ions or molecules are termed
ligands). lt may be utilized after the principal has been
satisfied.
Example:
AgCl + NH3 [Ag(NH3)2]+Cl
-
Each metal ion has a fixed number of auxiliary valences,
resulting in a fixed coordination number. The coordination
number is frequently, but not always, twice the principal
valence, for example coordination number of Ag+ = 2,
coordination number of Cu2+
, Cd2+
= 4, coordination number
of Fe3+
, Co3+
= 6.
The auxiliary valence may be the result of electric poles
or tilting of energetically suitable electrons orbitals of metal
ions with electron pairs of ligands. Consequently, the bond
by which the ligand is attached to the central cation may be:
Introduction Page 24
I- Ionic bond: which is present in complexes of metal ions
with electronic structures similar to those of inert gases
for example AlF63+
complex.
The metal and the ligands are held together by
electrostatic forces. The coordination number in ionic
complexes is affected by two factors:
1. The space available around the cation and the relative
size of the ligands.
2. The repulsive force between negative ligands e.g.;
SiO44-
and SiF62-
. The stability of ionic complexes is
largely proportional to the charge density value (charge
density is the ratio of charge to ionic radius). Aluminum
ion has a high charge density, it form stable complexes;
while sodium ion which has small charge density rarely
form complex ions. Comparing the stability of different
aluminum halide complexes, we find that AlF63+
is most
stable (F has the highest charge density) and All63-
is the
stable.
3. Coordinate bond: which is present in complexes of ions
of transition metals e.g. Ag+, Cu
2+, Zn
2+ Co
2+, and Ni
2+.
Both of the shared electrons are donated by the ligand.
Introduction Page 25
There is a great difference in the tendency of different
anions to donate their unshared electrons. Nitrate ion
are very weak electron donors and hence, rarely appear
in complex ions.
On the other hand, cyanide ions have strong tendency to
donate electrons, and hence form complex ions with many
cations. Moreover, a large number of neutral molecules,
such ammonia have a strong tendency to donate their
unshared electrons to cations and, accordingly form stable
complexes.
Formation and structure of coordinate complex
For a cation to accept electrons, it must have
incompletely filled valence shells. Untitled (d) orbital are of
particular importance. Unfilled shells give the necessary
spaces (orbital) in which the donated electrons can be
shared.
The structure of the ferrocyanide complex is represented
as example.
Introduction Page 26
The arrangement of the electrons in the valence Shell of
iron (3d, 45, 4p) is:
Iron, Fe 1s22s22p63s23p63d64s2 [Ar]3d 4s 4p
Since this ion can provide a total of six vacant orbitals (two
3d, one 4s and three 4p), so its coordination number is 6.
Each CN- can donate
one pair of unshared
(nitrogen) electrons;
therefore one ferrous ion
combines with six
cyanide ion.
Since the complex ion was formed by coordination in two
(3d) orbitals, one (4s) orbitals and three (4p) orbitals, it is
said to have a d2sp
3 structure.
Introduction Page 27
Homoatomic Ions
A homoatomic complex ion is a complex formed by
coordination of atoms or molecules of an electronegative
element with an anion of that same element. e
Example:
I- + I2 I3
- (triiodide ion)
S2-
+ (X – 1)So Sx
2- (polysnlphide ion)
Chelate Compounds
The more common compounds described above are
formed in such a manner that the coordinated group is
attached to the central atom at only one point. However, in
many compounds of analytical importance, the complexing
group (ligand) is capable of donating more than one pair of
electron; in this case th ligand is called chelating agent and
the complex is called chelate which has a ring structure.
The nickel dimethylglyoxime is a representative example of
a chelate.
Introduction Page 28
Ionisation of complex ions
Complex ions ionise to a greater or lesser degree giving
ions or molecules from which they are formed.
Generally, the dissociation of complex is represented by the
general equation:
MLn M + nL
(M = metal ion, L = ligand and MLn = complex).
Since this equation represents true equilibrium reaction, its
equilibrium constant can be represented in the conventional
manner
Introduction Page 29
𝐊𝐢𝐧𝐬𝐨𝐥=
𝐌 [𝐋]𝐧
[𝐌𝐋𝐧]
Here the equilibrium constant is called instability
constant. For a stable complex ion the value of Kinsol will be
small, because stable complex ion is slightly ionised so the
quantities in the numerator will be very small and the
quantity in the denominator will be relatively large.
Vice versa, unstable complex ion has relatively large Kinsol,
value. Therefore, the Kinsol value is a measure of the
instability of the complex ion.
Example 8
Calculate the silver ion concentration in 0.1 M [Ag(NH3)2]+
solution in which (a) excess of NH3 is absent and (b) the
NH3 concentration is 3 M (Kinsol = 6.8 x 10-8
).
a) [Ag(NH3)2]+
Ag+ + 2NH3
X X + 2 X
Since the dissociation of the complex ion is small, we may
assume [Ag (NH3)2]+
= 0.1
Introduction Page 30
Kinsol =
Ag+ [NH3]2
[Ag(NH3)2]+
𝟔. 𝟖 𝐱 𝟏𝟎−𝟖 = 𝐱 (𝟐𝐗)𝟐
𝟎. 𝟏
X = 6.0 x 10-8
/4 = 1.2 x 10-3
[Ag+] = 1.2 x 10
-3
b)
Kinsol =
Ag+ [NH3]2
[Ag(NH3)2]+
6.8 x 10−8 = Ag+ (3)2
0.1
[Ag+] = 7.6 x 10
-10 M
Example 9:
In a 0.1 M solution of [C(NH3)4]2+
the concentration of
uncomplexed Cu2+
is 4.5 x 10-4
. Calculate the Kinsol of the
complex.
[C(NH3)4]2+
Cu2+
+ 4 NH3
X X + 4X
Introduction Page 31
Kinsol =
Cu2+ [NH3]4
[Cu(NH3)4]2+
Kinsol = (4.5 x 10
-4 )(4 x 4.5 x10
-4 )
4/ 0.1 = 4.7 x 10
-14
Role of complex formation in qualitative analysis:
1. The specific test of certain ions is a complex formation
reaction.
Examples
- The red colour produced in the reaction between
ferric and thiocyanate ion, forming the complex
ferricthiocyanate ion is used for the detection of either
ion.
- The rose red precipitate produced in the reaction
between nickel ion and dimethylglyoxime (in solution
just alkaline with ammonia) is due to the formation of
the chelate nickel dimethylglyoxime.
2- When testing for specific ion with a reagent,
interferences may occur owing to the presence of other ions
Introduction Page 32
is the solution, -which also react with the reagent. In some
eases it is possible to prevent this interference by the
addition of reagents “masking agent” which form stable
complexes with interfering ions.
Example:
To test Cd2+
in presence of Cu2+
, cyanide ion is used
as masking agent to form the stable curpous cyano
complex and the much weak cadmium cyano complex.
When adding H2S only CdS will be precipitated
because copper was masked through formation of its
cyano complex.
3- Many cations form soluble amine complexes while other
are precipitated. The following are two examples for the
utilization of amine complexes in cation analysis.
A) The formation of amine complexes is the basis for the
separation of AgCl from Hg2Cl2 in group 1 analysis.
Silver chloride is soluble in excess ammonia forming
soluble amine complex while Hg2Cl2 gives with
ammonia insoluble product.
Introduction Page 33
AgCl + 2 NH3 [Ag(NH3)2]+ + Cl
-
Soluble complex
Hg2Cl2 + 2 NH3 HgNH2Cl- + Hg
o + NH4
+ + Cl
-
B) Addition of ammonia to, a mixture of Bi3+
; Cu2+
and
Cd2+
precipitates, Bi(OH)3 while Cu2+
remain soluble as
amine complexes [(Cu(NH3)4]2+
, [Cd(NH3)4]2+
. This is
used in the analysis of group IIA.
Amphoterism
Amphoteric Hydroxides:
The hydroxides of typical metals (e.g.. Na, K. Ca, ...etc.) are
known to bases. The hydroxides of non-metals (e.g,.. S, N,
C, ...etc.) and certain less typical metals, such as
chromium and manganese (in their highest oxidation
states, which have smaller radii exhibit opposite chemical
properties and belong to another class of compounds:
acids. But there are hydroxides that are capable of
Introduction Page 34
functioning as both acids and bases; They are called
ampholytes or are said to be amphiprotic or amphoteric,
and the phenomenon itself is known as amphoterism.
We shall encounter the phenomenon of amphoterism in
studying the cations of groups I - III; and it is of great
importance in analysis.
To illustrate, let us consider zinc hydroxide, Zn(OH)2. Like
other bases. This compound when reacted with acids
dissolves to form the corresponding salts; for example:
Zn(OH)2 + 2 H+ Zn
2+ + 2 H2O
But Zn(OH)2 dissolves in bases too, forming zincates. In
this reaction zinc hydroxide behaves as an acid; this
becomes especially clear if we write its formula in the same
Way as the formulas of acids are usually written:
H2ZnO2 + 2 OH- ZnO2
2+ + 2 H2O
Thus Zn(OH)2 exhibits the properties or both acids and
bases. i.e.. is a typical amphoteric hydroxide. Like Zn(OH)2.
Other compounds such as Pb(OH)2, Al(OH)3, Cr(OH)3,
Introduction Page 35
Sb(OH)3, Sn(OH)2, Sn(OH)2, are also amphoteric.
This simplified interpretation of the composition of the
hydroxides of elements, whose oxidation number equals or
exceeds 3. does not correspond to reality. These
hydroxides are in fact hydrous oxides. e.g.. aluminium
hydroxide (Al2O3. xH2O). The dissolution of this hydroxide
in excess alkali may be illustrated by:
Al2O3. xH2O + 2 OH- 2 [Al(OH)4]
- + (X – 3) H2O
The dissolution of zinc hydroxide in an excess of alkali
proceeds according to the equation
Zn(OH)2 + 2 OH- [Zn(OH)4]
2-
This process may be regarded as a complex-formation
reaction: when amphoteric hydroxides are dissolved in
alkalis, their molecules combine with the OH- ions of alkali
to form the corresponding complex ions [Al(OH)4]-,
[Zn(OH)4]2-
...etc. There are actually no. AlO2- or ZnO2
2- ions
in solutions.
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It should be noted that the ions [Al(OH)4]- and [Zn(OH)4]
2-
differ AlO2- and ZnO2
2- in their water content only, and are in
fact the hydrated ions AlO2- or ZnO2
2- , For example,
AlO2- + 2 H2O [Al(OH)4]
-
ZnO22-
+ 2 H2O [Zn(OH)4]2-
Ionizations are equal. The isoelectric point of an amphoteric
hydroxides is very important in analytical chemistry,
because it is the optimum pH for their precipitation.
Amphoteric Sulphides:
Amphoteric sulphides behave in a similar manner to
amphotcric hydroxides.They dissolve in the presence of
both hydrogen ions (H+) and hydroxyl ions (OH
-). In
addition, they also dissolve in hydrosulphide ions (HS-).
Hydrosulphide ions result from the hydrolysis of sulphide
ions:
S2-
+ H2O SH- + OH
-
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The amphoteric sulphides of analytical interest are those of
antimony and tin: Sb2S3, Sb2S5 and SnS2. It is a known fact
that for elements with more than one oxidation state; those
of higher oxidation states are smaller in size and therefore
more acidic. Thus stannous sulphide, SnS, is practically
insoluble in alkali hydroxide solutions and ammonium
sulphide, but oxidation of tin from Sn(II) to Sn(lV) increases
the acidic character of the sulphide. SnS2 sufficiently to
permit it to dissolve in alkali hydroxide and ammonium
sulphide solutions.
Using Sb2S3 as an example, the dissolution of amphoteric
sulphides in acids, alkalis and sulphide (or more accurately,
hydrosulphide) ions may be illustrated by the following
scheme:
H2O + H2S
+ H+
OH- + SH+ + Sb3+ Sb(SH)3 + Sb (OH)3 (basic ionization)
Sb2S3 + H2O
(acidic ionization) HSbS2 + HSbO2 SbS2 SbO2 + H+
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This can be summarized as follows:
(a) Addition of a strong acid dissolves antimony (Ill)
sulphide by converting it into Sb3+
ions and H2S gas.
Sb2S3 + 6 H+ 2 Sb
3+ + 3 H2S
(b) Addition of an alkali solution dissolves antimony (Ill)
suphide by convening it into antimonite, SbO2-, and
thioantimonite, SbS2- ions.
2Sb2S3 + 4 OH- 3 SbS2
- + SbO2
- + 2 H2O
(c) Addition of an excess of hydrogen sulphide water
dissolves antimony (Ill) sulphide by converting it into
thioantimonite, SbS2- ions and H2S gas.
2Sb2S3 + 4 SH- 4 SbS2
- + 2 H2S