Water Chemistry: Hardness of water, EDTA Method, Ion Exchange Method, Reverse Osmosis, Chlorination. Corrosion: Definition, Cause and Effects, Types of corrosion, Oxidation corrosion, Electrochemical Corrosion, Factors influencing corrosion, Waterline corrosion, Pitting corrosion, Cathodic protection, Galvanizing

1. MATRUSRI ENGINEERING COLLEGE
DEPARTMENT OF SCIENCES AND HUMANITIES
SUBJECT NAME: CHEMISTRY
FACULTY NAME: VISHNU THUMMA
MATRUSRI
ENGINEERING COLLEGE
TOPIC: WATER CHEMISTRY AND CORROSION

2. CHEMISTRY
COURSE OBJECTIVES:
➢Correlate the properties of materials with their internal structure and use
the for Engineering applications
➢Apply the principles of electrochemistry in storage of electrical energy in
batteries.
➢Gains knowledge in causes of corrosion and its prevention.
➢Attains knowledge about the disadvantages of hard water for domestic
and industrial purposes.
➢Also learns the techniques of softening of hard water and treatment of
water for drinking purpose.
➢Exposed to qualitative and quantitative parameters of chemical fuels.
➢Aware eco-friendly materials and processes.
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ENGINEERING COLLEGE
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3. CHEMISTRY
COURSE OUTCOMES: After completion of course students will be able to
➢Analyze and apply knowledge of electrodics in calculation of cell
potentials of batteries.
➢Identify the different types of hardness and alkalinities in water and
make use of softening methods, analyze and apply the knowledge of
corrosion for its prevention.
➢Discuss different types of polymers based on their end on use and the
need to replace the conventional polymers with polymers of engineering
applications.
➢Identify and analyze different types of chemical fuels for domestic and
automobile applications.
➢Outline the principles of green chemistry for sustainable environment
and preparation of biodiesel from renewable sources.
MATRUSRI
ENGINEERING COLLEGE
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4. UNIT-II WATER CHEMISTRY AND CORROSION
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ENGINEERING COLLEGE
Definition: Corrosion is the process of gradual deterioration of a metal from
its surface due to an unwanted chemical or electrochemical interaction of
metal with its environment.
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OUTCOMES: After completion of course students will be able to Identify the different
types of hardness and alkalinities in water and make use of softening methods, analyze
and apply the knowledge of corrosion for its prevention.
MODULE-I: INTRODUCTION TO CORROSION
Ex: Reddish brown
scale and powder of
rust (Fe2O3.3H2O) on
the surface of iron.
Green film of basic
carbonate
[CaCO3+Cu(OH)2]
on the surface of
copper.

5. Cause of Corrosion
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Most metals (with the exception of noble metals such as Au, Pt, etc.) exist in nature in
combined forms as their oxides, carbonates, hydroxides, sulphides, chlorides and silicates.
These chemically combined states of metal are known as ‘ores’.
These chemically combined states of metal are thermodynamically more stable states for
metal.

6. Cause of Corrosion
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During extraction of metals, considerable amounts of energy are required in metallurgy.
Consequently, isolated pure metals can be regarded in a higher energy state which are
thermodynamically unstable than their corresponding ores.
That is why, metals have a natural tendency to revert back to combined state
(thermodynamically more stable state).

7. Cause of Corrosion
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As a result when metals are put into use, in various forms, they are exposed to environment
such as dry gases, moisture, liquids, etc. the exposed metal surfaces begin to decay and
form more stable compounds of metals like oxides, carbonates, etc.
Thus, corrosion is a process “reverse of extraction of metals”.

8. Effects of Corrosion
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9. • The valuable metallic properties like conductivity, malleability, ductility etc.
are lost due to corrosion and thus loss of efficiency.
• The process of corrosion is not incurred and is responsible for the
enormous wastage of machines, equipment and different types of metallic
products.
• Losses occurring due to corrosion cannot be measured in terms of the cost
of metals alone, but the high cost of fabrication into equipment/machine
tool/ structures should also be considered.
• The approximate estimate of loss of metal due to corrosion, as 2 to 2.5
billion dollars per annum all-over the world.
Effects of Corrosion
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10. The corrosion process proceeds in two types by chemical and electrochemical
attack of environment.
A) Dry or Chemical Corrosion
B) Wet or Electrochemical Corrosion
Types of Corrosion
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11. QUIZ
1. The process of deterioration of a metal form its surface due to unwanted
chemical or electrochemical interaction of the metal with its environment is
called
a) electrolysis b) electrodialysis c) corrosion d) deposition
2. Metals exist in the nature in the form of
a) mineral b) ores c) combined state d) all the above
3. Metals undergo corrosion due to
a) pure form of metal is thermodynamically unstable.
b) pure form of metal is thermodynamically stable.
c) pure form of metal is regarded as low energy state
d) all the above
4. Loss of corrosion is
a) loss of metallic properties d) metallurgy cost c) design cost d) all the above
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12. This type of corrosion occurs by direct chemical reactions between the
environment and the metals and alloys.
Presence of an electrolyte is not at all essential for the corrosion to occur.
Eg: Direct chemical action of environmental gases such as oxygen, halogens,
hydrogen sulphide, sulphur dioxide, nitrogen or anhydrous inorganic liquid
with the metal.
MODULE-2: DRY OR CHEMICAL CORROSION
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13. a) Oxidation corrosion,
b) Corrosion by other gases,
c) Liquid metal corrosion.
There are three main types of chemical corrosion:
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14. Direct action of oxygen at high or low temperatures on metals in the absence of moisture
is called oxidation corrosion.
2 M → 2 Mn+ + 2n e-
n/2 O2 + 2n e- → n O2-
2 M + n/2 O2 → 2 Mn+ + n O2-
Oxidation occur first at the surface of the metal by forming a metal oxide scale which acts
as a barrier between metal surface and environment.
The nature of oxide formed plays an important role in oxidation corrosion process which
decides further action.
Oxidation corrosion
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15. If the oxide film formed is continues and rigidly adhered to the surface of metal is
impervious in nature, and is called stable oxide layer.
It is protective and shields the metal from further corrosion.
Ex: Oxide films on Al, Sn, Pb, Cu etc. acts as a protective coating and further
corrosion is prevented.
If unstable oxide film is formed, it decomposes back into the metal and oxygen.
Consequently, oxidation corrosion is not possible.
Ex: Noble metals like Ag, Au and Pt do not undergo corrosion.
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16. If oxide layer formed is volatile, it is non-protective and more feasible for further attack of
environment.
This causes rapid and continuous corrosion, leading to excessive corrosion.
Ex: Mo forms a volatile oxide layer.
If oxide layer is having pores or cracks, the atmospheric oxygen have access to the
underlying surface of metal, through the pores or cracks of the layer, thereby the
corrosion continues unobstructed till the entire metal is completely converted into its
oxide.
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17. According to it, ‘”greater is the specific volume ratio, lesser is the rate of corrosion.”
If the volume of metal oxide layer is at least as great as the volume of metal from which it
is formed is non-porous and becomes protective layer by tightly adhering to the base
metal.
Ex: The specific volume ratios of W, Cr and Ni are 3.6, 2.0 and 1.6 respectively.
Hence, the rate of corrosion is least in Tungsten (W).
If the volume of metal oxide is less than the volume of the metal, the oxide layer is
porous, non continuous and non-protective and faces strains.
Hence, cracks and pores are developed in the layer, creating access to atmospheric
oxygen to reach the underlying metal. In this case corrosion is continuous and rapidly
increases.
Ex: Li, Na and K.
Pilling-Bedworth rule:
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18. In the absence of moisture a few gases like SO2, CO2, Cl2, H2S and F2 etc. attack the
metal.
The degree of corrosion depends on the formation of protective or non-protective films on
the metal surface.
If the film formed is protective or non-porous, the intensity or extent of attack decreases,
because the film formed protects the metal from further attack.
Ex: AgCl film, resulting from the attack of Cl2 on Ag.
If the film formed is non-protective or porous, the surface of the whole metal is gradually
destroyed.
Ex: Dry Cl2 gas attacks on tin (Sn) forming volatile SnCl4. H2S at high temperature
attacks steel forming a FeS scale in petroleum industry.
Corrosion by other gases:
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19. It is due to chemical action of flowing liquid metal at high temperatures on solid metal or
alloy. Such corrosion occurs in devices used for nuclear power.
The corrosion reaction involves either dissolution of a solid metal by a liquid metal or
internal penetration of the liquid metal into the solid metal.
Both these modes of corrosion cause weakening of the solid metal.
Liquid Metal Corrosion:
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20. QUIZ
1. Which of the following is not dry corrosion?
a) oxidation corrosion b) rusting of iron
c) corrosion by gases d) liquid metal corrosion
2. Direct action of environment on metal surface is called
a) chemical corrosion b) electrochemical corrosion
c) wet corrosion d) none
3. Type of corrosion product layer which prevents further action of environment
a) stable oxide layer b) volatile oxide layer
c) porous oxide layer d) all the above
4. When specific volume ration is high
a) Corrosion is high b)Corrosion is low
c) No change in corrosion rate d) none
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21. Wet corrosion or electrochemical corrosion takes place under wet or moist
conditions through the formation of short circuited tiny electrochemical cells.
Wet corrosion is more common than dry corrosion.
This type of corrosion can be observed
i) When a metal is in contact with conducting liquid (or)
ii) When two dissimilar metals are dipped partially in a solution.
This corrosion occurs due to the existence of separate ‘anodic’ and ‘cathodic’
areas between which current flows through the conducting solution.
MODULE-3: ELECTROCHEMICAL CORROSION
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22. ➢ The formation of anodic and cathodic areas or parts in contact with each
other.
➢ Presence of a conducting medium.
➢ Corrosion of anodic areas only.
➢ Formation of corrosion product somewhere between anodic and cathodic
areas.
Electrochemical Corrosion involves:
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23. Involves flow of electron-current between the anodic and cathodic areas.
The anodic reaction involves in dissolution of metal as corresponding metallic ions with
the liberation of free electrons.
At anodic area: M → Mn+ + n e- (Oxidation)
The cathodic reaction consumes electrons with either by
i) evolution of hydrogen, or
ii) absorption of oxygen
depending on the nature of the corrosive environment.
Mechanism of Electrochemical Corrosion:
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24. It occurs usually in acidic environment.
Ex: Fe → Fe2+ + 2e- (Oxidation)
These electrons flow through the metal, from anode to cathode, where H+ ions are
eliminated as hydrogen gas from acidic solution.
2 H+ + 2 e- → H2  (Reduction)
The overall reaction: Fe + 2 H+ → Fe2+ + H2 
Evolution of Hydrogen:
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25. Thus, this type of corrosion causes “displacement of hydrogen ions from the acidic
solution by metal ions. “
Consequently, all metals above hydrogen in the electrochemical series have a tendency
to get dissolved in acidic solution with simultaneous evolution of hydrogen.
It may be noted that in hydrogen evolution type corrosion the anodes are usually very
large in areas where as the cathodes are small areas.
Evolution of Hydrogen
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26. • Occurs in the presence of atmospheric oxygen.
• Rusting of Iron is a common example of this type of corrosion.
• Usually the surface of iron is coated with a thin film of iron oxide.
• However, if this iron oxide film develops some cracks, anodic areas are
created on the surface while the well-metal parts act as cathodes.
• If follows that the anodic areas are small surface parts while nearly the rest
of the surface of the metal forms large cathodes.
Absorption of Oxygen
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27. At the anodic areas, the metal (iron) dissolves as ferrous ions with liberation of electrons.
Fe → Fe2+ + 2e- (Oxidation)
The liberated electrons flow from anodic to cathodic areas, through iron metal, where
electrons are intercepted by the dissolved oxygen as:
½ O2 + H2O + 2e- → 2 OH- (Reduction)
The Fe2+ ions at anode and OH- ions at cathode diffuse through medium and when they
meet, ferrous hydroxide is precipitated.
Fe2+ + 2 OH- → Fe(OH)2 ↓
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Absorption of Oxygen

28. If enough oxygen is present, ferrous hydroxide is easily oxidized to ferric
hydroxide.
4 Fe(OH)2 + O2 + 2 H2O → 4 Fe(OH)3
This product is called yellow rust, actually corresponds to Fe2O3.3H2O.
If the supply of oxygen is limited the corrosion product may be even black
anhydrous magnetite, Fe3O4.
An increase in oxygen content forces the cathodic reaction to produce more
OH- ions which in turn removes more electrons from anode and accelerates
the corrosion.
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Absorption of Oxygen

29. QUIZ
1. Electrochemical corrosion involves
a) Formation of anodic & cathodic areas b) Corrosion at anode
c) Conducting medium d) All the above
2. Rusting of iron is characterized by
a) absorption of oxygen b) small anodic areas
c) reddish brown scale d) all the above
3. In which of the case corrosion is more rapid
a) small anode & large cathodic areas b) volatile corrosion product
c) autocatalytic action of metal d) all the above
4. When supply of oxygen is limited the corrosion product may be
a) Fe2O3.3H2O b) Fe3O4 c) Fe2O3 d)Fe(OH)2
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30. It is a case of differential aeration corrosion.
More prevalent in cases such as ocean going ships, water storage steel
tanks etc, in which a portion of metal is always under water.
MOLDULE-4: WATERLINE CORROSION
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The water line corrosion takes place due to the formation of differential
oxygen concentration cell.

31. The part of the metal below the water line exposed only to the dissolved oxygen
while the part above the water is exposed to higher concentration of the
atmosphere oxygen.
Thus, part of the metal below the water acts as anode and undergoes corrosion.
Waterline Corrosion
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The part above the water line is free from corrosion.

32. A distinct brown line is formed just blow the water line due to the deposition of
rust.
• At anode: Fe → Fe2+ + 2e- (Oxidation)
• At cathode: ½ O2 + H2O + 2e- → 2 OH- (Reduction)
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Waterline Corrosion

33. QUIZ
1. The corrosion which occurs due to the difference in oxygen concentration is
called
a) oxidation corrosion b) differential aeration corrosion
c) galvanic corrosion d) none
2. The part of the metal below the waterline acts as ___ in water line corrosion
a) cathode b) anode c) electrolyte d) none
3. In waterline corrosion the reaction involved at cathode is
a) absorption of oxygen b) evolution of hydrogen
c) evolution of oxygen d) absorption of hydrogen
4. A distinct brown line is formed _______ the waterline due to deposition of
rust.
a) above b) below c) surface of water line d) none
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34. Pitting corrosion is a localized and accelerated corrosion, resulting in
the formation of pits or pin holes around which the metal is relatively
un-attacked.
MODULE-5: PITTING CORROSION
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It is characterized by small anodic and large cathodic areas, resulting in
accelerated corrosion at the anodic area.

35. It is generally initiated by the deposition of extraneous matter such as
sand, scale, water drop, dust etc,
or
Due to the breakdown or cracking of the protective film on metal
surface.
This gives rise to the formation of small anodic and large cathodic
areas.
In the correct environment, this produces corrosion current.
It is an auto catalytic process, with the initially formed pit produces
conditions which are both stimulating and necessary for the
continuing activity of the pit.
PITTING CORROSION
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The metal surface which is covered by the drop has low oxygen
concentration and thus acts as an anode and suffers corrosion.
The uncovered metal surface due to high O2 concentration acts as
cathode.

37. At anode: Fe → Fe2+ + 2e- (Oxidation)
At Cathode: ½ O2 + H2O + 2e- → 2 OH- (Reduction)
Oxidation
Fe2+ + 2OH- → Fe(OH)2 → Fe(OH)3
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Once the corrosion product is formed, it further provides the condition for
differential aeration below the corrosion product and the surrounding metal
parts. The pit grows and ultimately may cause failure of metal.

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➢Pitting is one of the most destructive forms of corrosion.
➢It causes equipment to fail because of perforation with only a small
percent weight loss of the entire structure.
➢It is often difficult to detect pits because of their small size and also
because pits are covered with corrosion products.
➢Pitting is dangerous because it is localized and intense corrosion
and failures often occur with extreme suddenness.
➢As such, it is rather difficult to assess precisely the life of metal
component undergoing pitting corrosion.

39. QUIZ
1. Which of the following is most destructive type of corrosion?
a) galvanic corrosion b) waterline corrosion
c) pitting corrosion d) none
2. Which of the following type corrosion is an autocatalytic process?
a) pitting corrosion b) oxidation corrosion
c) waterline corrosion d) liquid metal corrosion
3. It is difficult to detect the pitting corrosion because
a) pits are small in size b) pits are covered with rust
c) small % of weight loss d) all the above
4. Which of the following corrosion reaction involves absorption of oxygen at
cathode?
a) waterline corrosion b) differential aeration corrosion
c) pitting corrosion d) all the above
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40. 1. Nature of Metal:
i) Position of metal in the Galvanic series
ii) Relative areas of anode and cathodes
iii) Overvoltage
iV) Purity of metal
v) Nature of surface oxide film
2. Nature of Environment:
i) Temperature
ii) Humidity
iii) pH
MODULE-6: FACTORS AFFECTING THE RATE OF CORROSION
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41. NATURE OF METAL
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Position of metal in the Galvanic series: When the metal is higher up in the
galvanic series, greater is the oxidation potential. Thus, greater is its tendency to
become anodic and hence greater is the rate of corrosion. When two metals are
in electrical contact, greater is the difference in their positions in the
electrochemical series, faster is the corrosion of anodic metal.

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NATURE OF METAL
Relative areas of anode and cathodes: Corrosion is more rapid and severe, and
highly localized, if the anodic area is small, because the current density at a
smaller anodic area is much greater, and the demand for electrons by cathodic
areas can be met by smaller anodic areas only by undergoing corrosion more
briskly.

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NATURE OF METAL
Overvoltage: When a metal which occupies a high position in galvanic
series, say zinc is placed in 1N H2SO4, it undergoes corrosion forming a film
and evolving hydrogen gas, the initial rate of reaction is quite slow, because
of high over voltage of zinc metal, which reduces the effective electrode
potential (= 0.70V) to a small value.
However, if a few drops of copper sulphate are added, the corrosion rate of
zinc is accelerated, because some copper gets deposited on the zinc metal,
forming minute cathodes, where the hydrogen overvoltage is only 0.33V.
Thus, reduction in overvoltage of the corroding metal accelerates the
corrosion rate.

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NATURE OF METAL
Purity of metal: Impurities in a metal cause heterogeneity and form tiny
electrochemical cells at the exposed parts, and the anodic parts get corroded.
For example, zinc metal containing impurity such as Pb or Fe undergoes
corrosion due to formation of local electrochemical cells. The rate and extent of
corrosion increase with the increasing exposure and extent of the impurities.
Nature of surface oxide film: In aerated atmosphere, practically all metals get
covered with a thin surface film of metal oxide. The ratio of the volumes of the
metal oxide to the metal is known as specific volume ratio. Greater the specific
volume ratio, lesser is the oxidation corrosion rate.
For example, the specific volume ratios of Ni, Cr and W are 1.6, 2.0 and 3.6
respectively. Consequently, the rate of oxidation of tungsten is least, even at
elevated temperatures.

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NATURE OF ENVIRONMENT
Temperature: The rate of a chemical reaction, in general, increases with rise in
temperature. Corrosion process is one such chemical reaction. Therefore, the
rate of corrosion increases as the temperature increases. Increase in
temperature increases the conductance of the corrosion medium, which also
contributes to the increase in rate of corrosion.
Humidity: The greater is humidity, the greater is the rate and extent of
corrosion. This is due to the fact that the moisture or vapours present in
atmosphere acts as a solvent for O2, H2S, SO2 and NaCl etc. to furnish the
electrolyte essential for setting up an electrochemical cell.
pH: In general, lower the pH of the corrosion medium, higher is the corrosion
rate. However, some metals like Al, Zn undergo fast corrosion in highly alkaline
solution. The pH of the solutions also decides the type of cathodic reaction.

46. QUIZ
1. The rate of corrosion is more when
a) Metal is higher up in galvanic series b) Rise in temperature
c) Small anode & large cathodic areas d) All the above
2. Which of the following factor reduce rate of corrosion
a) overvoltage b) stable oxide layer
c) absence of humidity d) all the above
3. Rate of corrosion of anodic region is directly proportional to the
a) cathode area b) anode area
c) product of anode & cathode areas d) sum of anode & cathode areas
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47. The principle involved in this method of protection is to force the metal to be
protected to behave like a cathode, thereby corrosion does not occur.
There are two types of cathodic protection.
a) Sacrificial anode method.
b) Impressed current method.
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MODULE-7: CATHODIC PROTECTION

48. In this method of protection, the metallic structure to be protected is connected
to more anodic metal through a wire.
So that all the corrosion is concentrated at the more anodic metal.
Sacrificial Anodic Protection
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49. The more anodic metal itself gets corroded slowly; while the parent structure is
protected.
The more active metal so employed is called “sacrificial anode”.
The corroded sacrificial anode block is replaced by a fresh one, when
consumed completely.
Metals commonly used as sacrificial anodes are Zn, Al, Mg and their alloys.
Mg is used in high resistivity electrolytes such as soils due to its most negative
potential and it can provide highest current output.
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Sacrificial Anodic Protection

50. Examples:
Ship's steel hulls
Offshore drilling platforms
Oil and gas under sea pipelines
Containers used to store water and other liquids
are protected by this method.
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51. In this method current from an external source is impressed in the opposite
direction to nullify the corrosion current.
Thus, the anodic corroding metal becomes cathodic and protected from
corrosion.
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Impressed Current Method

52. The anode may be either an inert metal or one which deteriorates and will have
to be replaced periodically.
The commonly used anodic materials are graphite, carbon, stainless steel,
scrap iron, high silica iron and platinum.
The anode is buried in back fill such as gypsum to increase the electrical
contact between itself and the surrounding soil.
This protection method is useful when electrolyte resistivity and current
requirements are high.
It is well suited for large structures and long-term applications.
Applications: This protection technique is employed in the case of open water box
coolers, water tanks, buried pipe-lines, marine pipes etc.
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Impressed Current Method

53. QUIZ
1. In the cathodic protection, the metal which is wanted to be protected is
a) forced to behave like an anode b) forced to behave like a cathode
c) forced to behave like a conductor d) none
2. In the sacrificial anodic protection metal the base metal is connected to a
_______ through a wire.
a) more active metal b) less active metal
c) cathodic metal d) none
3. Which of the following metals can provide cathodic protection to Fe?
a) Al & Cu b) Al & Zn c) Zn & Cu d) Cu & Ni
4. The method in which corrosion current is nullified using an external source of
emf is called as
a) impressed current method b) sacrificial anodic protection
c) sacrificial cathodic protection d) none
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54. MODULE-8: HOT DIPPING - GALVANIZING
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Hot dipping is a method of coating a low melting metal such as Zn
(m.p.=419oC), Sn (m.p.=232oC) Pb, Al, etc., on iron, steel and copper which
have relatively higher melting points.

55. HOT DIPPING
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The base metal is dipped in a molten bath of the coating metal which is covered
by a molten flux layer.
Flux cleans the base metal surface and prevents the oxidation of the coating
metal.
For good adhesion of the coating metal on the surface of base metal, the base
metal surface must be very clean.
The most commonly used hot dipping methods are Galvanizing and Tinning.

56. Galvanizing is a process of coating iron or steel sheets with a thin coat of Zn to
prevent them from rusting.
GALVANIZING
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57. The base metal iron or steel sheet is cleaned by acid pickling method with
dil.H2SO4 for 15-20 minutes at 60-90oC.
The sheet is then washed well and dried. It is dipped in a bath of molten zinc
maintained at 425-435oC.
The surface of the bath is kept covered with ammonium chloride flux to prevent
oxide formation.
The sheet is taken out and excess Zn is removed by passing it between a pair
of hot rollers.
Then the sheet is subjected to annealing process at 650oC and cooled slowly.
An alloy of iron and zinc were formed at the junction of the base metal and
coating metal.
GALVANIZING PROCESS
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58. Applications: It is mostly used to protect iron used for roofing sheets, wires,
pipes, nails, bolts, screws, buckets and tubes.
Galvanizing utensils cannot be used for preparing and storing food stuffs
especially acidic in nature, because zinc dissolves to form highly toxic or
poisonous compounds.
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59. QUIZ
1. Coating a low melting metal on a relatively high melting metal surface is
called___
a) Hot dipping b) Cathodic protection
c) Anodic protection d) none
2. The process of coating Fe or Steel with a zinc coating metal is called
a) tinning b) galvanizing
c) cladding d) none
3. The surface of the molten bath is covered with ammonium chloride flux to
prevent______
a) carbonate formation b) oxide formation
c) sulphide formation d) none
4. The galvanizing products cannot be used for
a) roof top b) food storage
c) buckets d) all 59
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60. MODULE-9: WATER CHEMISTRY
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Hardness of water: Water which does not produce lather with soap solution
readily but forms a white curd is called hard water.
Hardness in water is that characteristic, which “prevents the lathering of soap”.
This is due to presence of certain salts of calcium, magnesium and other heavy
metals dissolved in it.
When hard water is treated with soap does not produce lather and forms a white
scum or precipitate due to the formation of insoluble soaps of calcium and
magnesium.
2C17H35COONa + CaCl2 → (C17H35COO)2Ca  + 2NaCl
Sodium stearate Hardness Calcium stearate
Soap (insoluble)
2C17H35COONa + MgSO4 → (C17H35COO)2Mg  + Na2SO4
MATRUSRI
ENGINEERING COLLEGE

61. TYPES OF HARDNESS
Temporary hardness (bicarbonate hardness): It is caused by the presence
of dissolved bicarbonates of calcium, magnesium and other heavy metals
and carbonate of iron.
Temporary hardness can be removed by boiling of water.
Bicarbonates are decomposed into insoluble carbonates or hydroxides, which
are deposited as a crust at the bottom of vessel on boiling the water.
Heat
Ca(HCO3)2 → CaCO3  + H2O + CO2
Calcium bicarbonate Calcium carbonate
(insoluble)
Mg(HCO3)2 → Mg(OH)2  + 2 CO2 
Magnesium bicarbonate Magnesium hydroxide
61
MATRUSRI
ENGINEERING COLLEGE

62. TYPES OF HARDNESS
Permanent hardness (Non-carbonate hardness): It is due to the presence of
chlorides and sulphates of calcium, magnesium, iron and other heavy
metals.
Unlike temporary hardness, permanent hardness is not destroyed on boiling.
62
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63. 63
The concentration of hardness and non-hardness salts is expressed in terms
of equivalent amount of CaCO3.
Since this mode permits the multiplication and division of concentration when
required.
The choice of CaCO3 in particular is due to:
its molecular weight is 100 (equivalent weight = 50)
it is the most insoluble salt that can be precipitated in water treatment.
EQUIVALENTS OF CALCIUM CARBONATE
MATRUSRI
ENGINEERING COLLEGE
OR
The equivalents of CaCO3 = Mass of hardness producing substance
Molecular weight of hardness substance
X 100
The equivalents of CaCO3 = Mass of hardness producing substance
Equivalent weight of hardness substance
X 50

64. 64
Parts per million (ppm): It is the part of calcium carbonate equivalent
hardness per 106 parts of water.
i.e. 1 ppm = 1 part of CaCO3 eq. hardness in 106 parts of water.
Milligrams per litre (mg/L) is the number of milligrams of CaCO3 equivalent
hardness present per litre of water.
1mg/L = 1 mg of CaCO3 eq. hardness in 1 L of water.
But ,
1 L of water weighs = 1kg =1000g =1000 X 1000 mg= 106 mg.
Therefore, 1mg/L = 1mg of CaCO3 eq per 106 mg of water.
= 1 part of CaCO3 eq per 106 parts of water
1mg/L=1ppm
UNITS OF HARDNESS
MATRUSRI
ENGINEERING COLLEGE

65. 65
Q1: A sample of water contains the following impurities: Ca(HCO3)2 = 14.6
mg/L, Mg(HCO3)2 = 30 mg/L, MgCl2 = 19 mg/L, MgSO4 = 36 mg/L. Calculate
temporary and permanent hardness in ppm.
Solution:
Substance Weight (mg/L) GMW Eq. of CaCO3
Ca(HCO3)2 14.6 162
Mg(HCO3)2 30 146
MgCl2 19 95
MgSO4 36 120
Temporary hardness (Ca(HCO3)2 + Mg(HCO3)2) = 9 + 20 = 29 mg/L
Permanent hardness (MgCl2 + MgSO4) = 20 + 30 = 50 mg/L
NUMERICAL PROBLEMS
MATRUSRI
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66. 66
Q2: A sample of water on analysis has been found to contain Mg(HCO3)2 = 5.84
mg/L, Ca(HCO3)2 = 4.86 mg/L, CaSO4 = 6.80 mg/L and MgSO4 = 8.40 mg/L.
NUMERICAL PROBLEMS
MATRUSRI
ENGINEERING COLLEGE
Solution:
Substance Mass (mg/L) GMW GEW CaCO3 equivalents
Mg(HCO3)2 5.84 146 73 = x 50 = 4 mg/L
5.84
73
Ca(HCO3)2 4.86 162 81 = x 50 = 3 mg/L
4.86
81
CaSO4 6.80 136 68 = x 50 = 5 mg/L
6.80
68
MgSO4 8.40 120 60 = x 50 = 7 mg/L
8.40
60

67. 67
NUMERICAL PROBLEMS
MATRUSRI
ENGINEERING COLLEGE
Permanent hardness is due to CaSO4 and MgSO4:
= 5 + 7 = 12 mg/L or ppm
Total hardness = 4 + 3 + 5 +7 = 19 mg/L or ppm
Temporary hardness is due to Mg(HCO3)2 and Ca(HCO3)2:
= 4 + 3 = 7 mg/L or ppm

68. QUIZ
1. The phenomenon of prevention of lathering of soap in water is called
a) softness b) hardness
c) turbidity d) none
2. Hardness in water is mainly caused due to presence of _______
a) salts of Ca & Mg b) Salts of Ca & Na
c) Salts of Mg & Na d) all the above
3. Hardness in water which cannot be removed on boiling is known as
a) temporary hardness b) permanent hardness
c) carbonate hardness d) a&c
4. Degree of hardness can be expressed in terms of
a) equivalents of CaCO3 b) equivalents of MgCO3
d) equivalents of Ca(HCO3)2 d) equivalents of Mg(HCO3)2
68
MATRUSRI
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69. 69
MODULE-10: DETERMINATION OF HARDNESS BY EDTA METHOD
MATRUSRI
ENGINEERING COLLEGE
This is a complexometric titration in which EDTA is used as a complexing agent.
EDTA is a hexadentate ligand and forms stable
complexes with most of the metal ions in the pH
range 10.
PRINCPLE

70. 70
DETERMINATION OF HARDNESS BY EDTA METHOD
MATRUSRI
ENGINEERING COLLEGE
A buffer solution of NH4Cl + NH4OH is used to maintain the pH.
To determine the equivalence point Eriochrome black-T (EBT) is used as an
indicator.
Free EBT in water
(Blue)
EBT in Hard water
( Wine Red)

71. 71
DETERMINATION OF HARDNESS BY EDTA METHOD
MATRUSRI
ENGINEERING COLLEGE
It forms unstable wine red
colored complexes with Ca2+
& Mg2+ ions in the pH range
10.
EDTA combines with the free metal ions
in the beginning and the metal ions of
the indicator complex at the end,
displacing the indicator.
Titration against EDTA
Wine red Blue

72. 72
DETERMINATION OF HARDNESS BY EDTA METHOD
MATRUSRI
ENGINEERING COLLEGE
PREPARATION OF REAGENTS
Preparation of standard hard water: Dissolve 1 g of pure CaCO3 in minimum
quantity of dil.HCl and evaporate it to dryness on a water bath. Dissolve it in
distilled water and make up to 1L. Each mL of this solution thus contains 1 mg of
CaCO3 eq hardness.
Preparation of EDTA solution: Dissolve 4.0 g of pure EDTA crystals and add
0.1 g of MgCl2 to it. Make up to 1L using distilled water.
Preparation of indicator: Dissolve 0.5 g of Eriochrome black-T in 100 ml of
alcohol.
Preparation of buffer solution: Add 67.5 g of NH4Cl to 570 ml of concentrated
ammonia solution and dilute with distilled water to 1L.

73. 73
DETERMINATION OF HARDNESS BY EDTA METHOD
MATRUSRI
ENGINEERING COLLEGE
EXPERIMENTAL PROCEDURE
A. Standardization of EDTA solution: Rinse and fill the burette with EDTA
solution. Pipette out 50 ml of standard hard water in a conical flask. Add
10ml of buffer solution and 3 or 4 drops of indicator, the color changes to
wine red. Titrate it against EDTA till the color changes to blue. Let the volume
of EDTA be V1 ml.
50 ml of std. hard water = V1 ml of EDTA
i.e. 50 mg of CaCO3 eq. = V1 ml of EDTA
Hence, 1 ml of EDTA = mg of CaCO3 eq.
50
V1

74. 74
DETERMINATION OF HARDNESS BY EDTA METHOD
MATRUSRI
ENGINEERING COLLEGE
EXPERIMENTAL PROCEDURE
B. Estimation of Total hardness: Pipette out 50 ml of water sample in a
conical flask. Add 10ml of buffer solution and 3 or 4 drops of indicator to it, the
color changes to wine red. Titrate it against EDTA till the color changes to blue.
Let the volume of EDTA be V2 ml.
50 ml of hard water = V2 ml of EDTA
Hence, 50 ml of hard water = x V2 mg of CaCO3 eq.
50
V1
For 1000 ml of H.W = x V2 mg of CaCO3 eq.
1000
V1
Total hardness = x 1000 ppm
V2
V1

75. 75
DETERMINATION OF HARDNESS BY EDTA METHOD
MATRUSRI
ENGINEERING COLLEGE
EXPERIMENTAL PROCEDURE
C. Estimation of permanent hardness: Take 250 ml of the water sample in a
large beaker and boil it to one third of its volume, all the bicarbonates are
precipitated as carbonates. Filter, wash the precipitate with distilled water, and
collect the filtrate and washings in 250 ml volumetric flask. Makeup the volume
using distilled water. Then titrate 50 ml of this water as under standardization.
Let the volume of EDTA be V3 ml.
50 ml of boiled water = V3 ml of EDTA
50
V1
For 1000 ml of EDTA = x V3 mg of CaCO3 eq.
1000
V1
Permanent hardness = x 1000 ppm
V3
V1
50 ml of boiled water = x V3 mg of CaCO3 eq.

76. 76
DETERMINATION OF HARDNESS BY EDTA METHOD
MATRUSRI
ENGINEERING COLLEGE
EXPERIMENTAL PROCEDURE
D. Temporary hardness : = [Total – Permanent] hardness
Temporary hardness = x 1000 ppm
V2 - V3
V1

77. QUIZ
1. Determination of hardness by EDTA method is a type of
a) acid base titration b) redox titration
c) complexometric titration d) precipitation titrations
2. The composition of buffer solution used to maintain pH in EDTA method
a) NH4Cl + NaOH b) NH4Cl + NH4OH
c) NaCl + NH4OH d) none
3. The colour of metal – indicator complex in EDTA method is
a) green b) blue
c) red d) wine red
4. The blue colour appears at the end point of EDTA method due to
a) free EBT b) Metal – EBT
c) Metal – EDTA d) free EDTA
77
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78. 78
MATRUSRI
ENGINEERING COLLEGE
NUMERICAL PROBLEMS BASED ON EDTA METHOD
Q1: 20 ml of standard hard water containing 15 g of CaCO3 per liter, required 25 ml
of EDTA solution for end point. 100ml of water sample required 18 ml of EDTA
solution; while same water after boiling required 12ml of EDTA solution. Calculate
carbonate and non carbonate hardness of water.
Solution:
1000ml of standard hard water = 15 g of CaCO3 eq. = 15 mg of CaCO3 eq./ml
Now, 25 ml of EDTA solution = 20 ml of standard hard water
= 20 x 15 mg of CaCO3 eq. = 300 mg of CaCO3 eq.
Therefore, 1 ml of EDTA solution =
300
25
= 12 mg of CaCO3 eq. hardness

79. 79
MATRUSRI
ENGINEERING COLLEGE
NUMERICAL PROBLEMS BASED ON EDTA METHOD
Calculation of Total hardness:
100ml of sample water = 18 ml of EDTA solution
= 18 x 12 mg of CaCO3 eq.
= 216 mg of CaCO3 eq.
For 1000ml of sample water = x 1000
216
100
Total hardness = 2160 ppm
= 2160 mg of CaCOeq. hardness

80. 80
MATRUSRI
ENGINEERING COLLEGE
NUMERICAL PROBLEMS BASED ON EDTA METHOD
Calculation of non carbonate (permanent) hardness:
100ml of boiled water = 12 ml of EDTA solution
= 12 x 12 mg of CaCO3 eq.
= 144 mg of CaCO3 eq.
For 1000ml of boiled water = x 1000 = 1440 mg of CaCOeq. hardness
144
100
Permanent hardness = 1440 ppm
Carbonate (temporary) hardness = [Total – Permanent] hardness
= 2160 – 1440 = 720 ppm

81. 81
MATRUSRI
ENGINEERING COLLEGE
NUMERICAL PROBLEMS BASED ON EDTA METHOD
Q2: 50 ml of a sample water consumed 12 ml of 0.01 M EDTA before boiling and
10 ml of the same EDTA after boiling. Calculate the total hardness, permanent
hardness and temporary hardness.
Solution:
Since, EDTA reacts with Mg & Ca ions in 1:1 ratio;
1ml of 1 M EDTA = 100 mg of CaCO3 eq.
i.e. 1 ml of 0.01 M EDTA = 1 mg of CaCO3 eq.
Calculation of Total hardness:
50 ml of sample water = 12 ml of 0.01 M EDTA
= 12 x 1 = 12 mg of CaCO3 eq.
For 1000 ml of sample water = x 1000 = 240 mg of CaCO3 eq.
12
50
Total hardness = 240 ppm

82. 82
MATRUSRI
ENGINEERING COLLEGE
NUMERICAL PROBLEMS BASED ON EDTA METHOD
Calculation of Permanent hardness:
50 ml of boiled water = 10 ml of 0.01 M EDTA
= 10 x 1 = 10 mg of CaCO3 eq.
For 1000 ml of sample water = x 1000 = 200 mg of CaCO3 eq.
10
50
Permanent hardness = 200 ppm
Temporary hardness = [Total – Permanent] hardness
= 240 – 200 = 40 ppm

83. 83
MODULE-11: ALKALINITY
MATRUSRI
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PRINCPLE
Alkalinity of water is measure of acid-neutralizing ability.
It is attributed to the presence of the caustic alkalinity (OH- and CO3
2- ) and
temporary hardness (HCO3
-).
These can be estimated separately by titration against acid, using
phenolphthalein and methyl orange as indicators.
1. [OH-] + [H+] → H2O
2. [CO3
2- ] + [H+] → [HCO3
-]
3. [HCO3
-] + [H+] → H2O + CO2
P M

84. 84
ALKALINITY
MATRUSRI
ENGINEERING COLLEGE
The possible combinations of ions causing alkalinity in water are:
➢OH- only or
➢CO3
2- only or
➢HCO3
- only or
➢OH- & CO3
2- together or
➢CO3
2- & HCO3
- together
OH- & HCO3
- ions cannot exist together in water.
OH- + HCO3
- → CO3
2- + H2O
On the basis of same reasoning, all the three ions (OH- , CO3
2- and
HCO3
-) cannot exist together.

85. 85
ALKALINITY
MATRUSRI
ENGINEERING COLLEGE
COLOUR PROFILE OF INDICATORS
PHENALPTHALEIN
pH = 0 – 8.2 pH = 8.2 - 10
METHYL ORANGE
pH = 2 – 4.4 pH = above 4.4

86. 86
ALKALINITY
MATRUSRI
ENGINEERING COLLEGE
Experimental Procedure
Pipette out 100ml water sample in a clean conical flask. Add 2 to 3 drops of a
phenolphthalein indicator to it. Run N/50 HCl from a burette, till the pink colour
is disappeared. Then to the same solution, add 2 to 3 drops of methyl orange,
continue the titration, till the color changes from yellow to orange pink.
N/50 HCl
Sample water
Burette
Conical Flask

87. 87
ALKALINITY
MATRUSRI
ENGINEERING COLLEGE
0 ml
P
M
When P=0, both OH- & CO3
2- ions are absent, and alkalinity in that case due to
HCO3
- only.
When P=½M, only CO3
2- ion is present, since half of carbonate neutralization
reaction i.e. [CO3
2- ] + [H+] → [HCO3
-] takes place with phenolphthalein indicator;
while complete carbonate neutralization reaction i.e. [HCO3
-] + [H+] → H2O + CO2
occurs when methyl orange indicator used. Thus, alkalinity due to CO3
2- = 2P.
When P=M, only OH- is present, because neither CO3
2- nor HCO3
- is present, thus
alkalinity due to OH- = M.
When P > ½M, in this case, besides CO3
2- , OH- ions are also present. Now half of
CO3
2- equal to M-P; so alkalinity due to complete CO3
2- =2(M-P)
Therefore alkalinity due to OH- = M - 2(M-P) = 2P – M.
When P < ½ M, ;in this case, besides CO3
2- , HCO3
- ions are also present now
alkalinity due to CO3
2- = 2P. Therefore, Alkalinity due to HCO3
- = (M-2P).

88. 88
ALKALINITY
MATRUSRI
ENGINEERING COLLEGE
Alkalinity OH- ppm CO3
2- ppm HCO3
- ppm
P=0 0 0 M
P=½M 0 2P 0
P=M M 0 0
P > ½M 2P-M 2(M-P) 0
P < ½ M 0 2P M-2P

89. 89
QUIZ
MATRUSRI
ENGINEERING COLLEGE
1. Which of the following situation never arises with respect to the constituents causing
alkalinity in water ?
a) CO3
2- and HCO3
– together b) HCO3
– and OH– together
c) OH– only d) OH– and CO3
2- together
2. Which of the following indicator is pink in basic medium?
a) Methyl orange b) Phenolphthalein
c) Starch d) Litmus paper
3. The alkalinity due to hydroxide ion when P > 1/2M will be ____________
a) M-2P b) 2(M-P)
c) 2P-M d) Nil
4. The alkalinity due to carbonate ion is 2P when?
a) P = M b) P > 1/2M
c) P = 1/2M d) P < 1/2M
5. The alkalinity due to bicarbonate ion when P < M/2 will be ____________
a) M-2P b) 2(M-P)
c) d) 2P-M d) Nil

90. MODULE-12: ION EXCHANGE METHOD
Removal of all ions present in water is called demineralization.
Ion Exchange Resin:
•Ion exchange resins are insoluble, cross linked, long chain organic
polymers with a micro-porous structure.
•The ion exchange property of these polymers is due to mainly the
functional groups attached to them.
•These functional groups may be acidic or basic.
•Based on functional groups the resins may be classified as:
a) Cation exchange resins b) Anion exchange resins.
90
MATRUSRI
ENGINEERING COLLEGE

91. Cation Exchange Resins (RH+)
Styrene-di vinyl benzene copolymers, which on sulphonation or
carboxylation, -SO3H or –COOH groups are introduced to polymers.
They become capable to exchange their H+ ions with the cation in
water.
91
MATRUSRI
ENGINEERING COLLEGE

92. Anion Exchange Resins (R’OH-)
Styrene-di vinyl benzene or amino formaldehyde copolymers, which
contain amino or quaternary ammonium or quaternary phosphonium
groups as an integral part of the resin matrix.
These, after treatment with dil.NaOH solution, become capable to
exchange their OH- ion with the anions in water.
92
MATRUSRI
ENGINEERING COLLEGE

93. 93
MATRUSRI
ENGINEERING COLLEGE

94. In Cation Exchanger:
• 2 RH+ + Ca2+ → R2Ca2+ + 2 H+
• 2 RH+ + Mg2+ → R2Mg2+ + 2 H+
In Anion Exchanger:
• R’OH- + Cl- → R’Cl- + OH-
• 2R’OH- + SO4
2- → R’2SO4
2- + 2 OH-
• 2R’OH- + CO3
2- → R’2CO3
2- + 2 OH-
• The H+ and OH- ions released from both the column get
combined to produce water molecule.
H+ + OH- → H2O
94
MATRUSRI
ENGINEERING COLLEGE

95. Regeneration of Ion Exchangers
The exhausted cation exchange resin is regenerated by passing a
solution of dil.HCl or dil.H2SO4.
R2Ca2+ + 2 H+ → 2 RH+ + Ca2+ (washings)
R2Mg2+ + 2 H+ → 2 RH+ + Mg2+ (washings)
The exhausted anion exchange resin is regenerated by passing a
solution of dil.NaOH.
R’2SO4
2- + 2 OH- → 2R’OH- + SO4
2- (washings)
R’2CO3
2- + 2 OH- → 2R’OH- + CO3
2- (washings)
R’Cl- + OH- → R’OH-- + Cl- (washings)
The columns are washed with deionized water and washings which
contain Ca2+, Mg2+ , SO4
2- , Cl- ions are passed to sink or drain.
95
MATRUSRI
ENGINEERING COLLEGE

96. Advantages of Ion Exchange Method
• The process can be used to soften highly acidic or basic waters.
• It produces water of very low hardness (2ppm), so it is very good
for treating water for use in high pressure boilers.
96
MATRUSRI
ENGINEERING COLLEGE

97. Disadvantages of Ion Exchange Method
• The equipment is costly and more expensive chemicals are needed.
• If water contains turbidity, then the output of the process is reduced.
• The turbidity must be below 10ppm. If it is more it has to be removed
first by coagulation and filtration.
97
Matrusri Engineering College
MATRUSRI
ENGINEERING COLLEGE

98. QUIZ
1. Which method are used for preparing of demineralized water?
a) Gas Chromatography b) Batch method ( ion exchange)
c) Mass spectroscopy d) Complexometric Titration
2. Ion-exchange resin is
a) Linear b) Low molecular weight
c) Organic polymer with porous structure d) Soluble polymer
3. Which of the following ion get released from the anion exchange column?
a) CO3
-2 b) OH–
c) Cl– d) SO4
-2
4. The residual hardness after the treatment of water is about __________
a) 1 ppm b) Less than 1ppm
c) about 2 ppm d) 3ppm
98
MATRUSRI
ENGINEERING COLLEGE

99. MODULE-13: REVERSE OSMOSIS
Osmosis: When two solutions of unequal concentrations are separated
by a semi permeable membrane flow of solvent takes place from
dilute to concentrated sides.
99
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ENGINEERING COLLEGE
Semi permeable membrane: does
not permit the ions, atoms,
molecules etc.

100. Reverse Osmosis
Principle: If, however, a hydrostatic pressure in excess of osmotic
pressure is applied on the concentrated side, the solvent flow
reverses, i.e. solvent is forced to move from concentrated side to
dilute side across the membrane.
100
MATRUSRI
ENGINEERING COLLEGE

101. Thus, in reverse osmosis process methods pure water is separated from
its contaminants, rather than removing contaminants from the water.
This membrane filtration is sometimes also called “Super-filtration” or
“Hyper filtration”.
101
MATRUSRI
ENGINEERING COLLEGE

102. Process:
In this process, pressure of the order 15 to 40 kg cm2- is applied to the sea
water/ impure water to force its pure water out through the semi
permeable membranes; leaving behind the dissolved solids.
102
MATRUSRI
ENGINEERING COLLEGE

103. Semi-permeable Membrane
The membrane consists of very thin film of cellulose acetate, affixed to
either side of a perforated tube.
More recently superior membranes made of polymethacrylate and
polyamide polymers have come into use.
103
MATRUSRI
ENGINEERING COLLEGE

104. Advantages of Reverse Osmosis:
Reverse osmosis process is a distinct advantage of removing ionic as
well as non ionic, colloidal and high molecular weight organic
matter.
It removes colloidal silica, which is not removed by demineralization.
The maintenance cost is almost entirely on the replacement of the
semi permeable membrane.
The life time of membrane is quite high, about 2 years.
The membrane can be replaced within a few minutes, there providing
uninterrupted water supply.
104
MATRUSRI
ENGINEERING COLLEGE

105. Advantages of Reverse Osmosis
Due to low capital cost, simplicity, low operating cost and high reliability,
the reverse osmosis is gaining ground at present for converting sea
water into drinking water and for obtaining water for very high
pressure boilers.
105
MATRUSRI
ENGINEERING COLLEGE

106. QUIZ
1. The phenomenon of reverse osmosis involve
a) Osmotic pressure is greater than the hydrostatic pressure
b) Osmotic pressure is equal to the hydrostatic pressure
c) Hydrostatic pressure is greater than the osmotic pressure
d) Diffusion
2. Semi-permeable membrane is selective membrane which permits the passage of
________ particles.
a) Solvent b) Solute
c) Anhydrous d) Saturated
3. Which of the following is used as semi-permeable membrane?
a) Polymethyl acrylate b) Cellulose acetate
c) Polyamide polymer d) all the above
4. Select the incorrect statement about reverse osmosis from the following option.
a) It operates at a high temperature
b) Semipermeable membrane can be easily replaced within a few minutes
c) It is simple and reliable process
d) It is relatively energy efficient
106
MATRUSRI
ENGINEERING COLLEGE

107. MODULE-14:SPECIFICATIONS OF POTABLE WATER
It should be
• Sparkling clear and odorless.
• Pleasant in taste.
• Perfectly cool.
• Turbidity should not exceed 10 ppm.
• It should be free from objectionable minerals such as Pb, As, Cr, Mn
salts.
• It should be free from objectionable gases like H2S.
• pH should be in range of 7.0 – 8.0. Alkalinity should not be high.
• Dissolved solid should be less than 500 ppm.
• It should be soft and free from disease causing micro-organisms.
• Fluoride content should be less 1.5 ppm. And Chloride, Sulphate
contents should be less than 250 ppm.
107
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ENGINEERING COLLEGE

108. STERILIZATION BY CHLORINATION
Chlorination is the most commonly used disinfectant in water treatment
throughout world.
It can be employed directly as gas or in the form of concentrated solution
in water.
It produces hypochlorous acid, which is a powerful germicide.
Cl2 + H2O → HOCl + HCl
Bacteria + HOCl → Bacteria are killed
108
MATRUSRI
ENGINEERING COLLEGE

109. CHLORINTOR (Dosage:0.3 to 0.5ppm)
109
MATRUSRI
ENGINEERING COLLEGE

110. Factors affecting efficiency of chlorine
Temperature of water: The rate of reaction with enzymes increases with
temperature. Consequently, death rate of micro-organisms by chlorine
increases with rise in temperature.
Time of contact: Death rate of micro-organisms by chlorine is proportional
to the number of micro-organisms remaining alive. Initially, the death
rate is maximum and with time, it goes on decrease.
pH of water: at lower pH values (between 5 – 6.5), a small contact is
required to kill organisms.
110
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ENGINEERING COLLEGE

111. Advantages of Chlorination
• It is effective and economical.
• It is stable, requires small space for storage, and does not
deteriorate on keeping.
• It can be used at high as well as low temperatures.
• It does not introduce any impurity in water.
• It is most ideal disinfectant.
111
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112. Disadvantages of Chlorination
• Excess of chlorine, produces bad taste and disagreeable odor.
• Excess chlorine produces irritation on mucous membrane.
• The quantity of free chlorine in treated water should not exceed 0.1 to
0.2 ppm.
• It is more effective below pH 6.5 and less effective at higher pH vales.
112
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ENGINEERING COLLEGE

113. QUIZ
1.The pH for the potable water should be in the range of
a) 5-6 b) 7-8.5
c) 9-10 d) 12-13
2.The total hardness of the potable water should be less than
a) 500 ppm b) 700 ppm
c) 900 ppm d) 1000 ppm
3.What is the chemical formula of bleaching powder?
a) Ca(OCl)2 b) Ca(OCl)
c) Ca(OCl)3 d) CaCl2
4.The commonly used chemicals to treat municipal water
a) Chlorine b) bleaching powder
c) Ozone d) all the above
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ENGINEERING COLLEGE

114. MODULE-15 :BREAK-POINT CHLORINATION
It means that chlorination of water to such an extent that living organisms
as well as other organic impurities in water are destroyed.
It involves in addition of sufficient amount of chlorine to oxidize organic
matter, reducing substances and free ammonia in raw water,
Leaving behind mainly free chlorine which possesses disinfecting action
against pathogenic bacteria’s.
It is also known as free-residual chlorination.
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115. Break-Point Chlorination
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116. Advantages
• It ensures complete destruction of organic matter which impart color,
bad odor and unpleasant taste to water.
• It completely destroys all the disease causing bacteria.
• It prevents the growth of any weeds in water.
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117. QUIZ
1.The permissible limit of free residual chlorine is ___________
a) 0.02ppm b) 0.2ppm
c) 1ppm d) 2ppm
2.Chlorine which gets consumed in the oxidation of impurities before disinfection is
a) Free chlorine b) Residual chlorine
c) Chlorine demand d) Residual demand
3.The normal dose of chlorine during break point chlorination is ____________
a) 0.5-1ppm b) 0.1-0.2ppm
c) 3-7ppm d) 1-2ppm
4. The point at which chlorine demand has been totally satisfied, i.e the chlorine has
reacted with all reducing agents, organics, and ammonia in the water is
a) Residual Point b) Break point chlorination
c) chlorine demand point d) none of the above
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118. MATRUSRI
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