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.

Introduction Page 36

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

-

Introduction Page 37

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+

Introduction Page 38

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