UNIT-1

WATER AND ITS TREATMENT

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 Topic - Module

Introduction

 Water is nature’s most wonderful, abundant and useful compound. Water is not
only essential for the lives of animals and plants, but also occupies a unique
position in industries.
 It is widely used in drinking, bathing, sanitary, washing, irrigation, fire-fights,
air-conditioning and also the production of industrial materials.
 The distribution of water on the Earth’s surface is extremely uneven.
 Only one percent of the water on the Earth’s surface is usable by humans, and
99% of the usable quantity is situated underground.

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Types of Impurities present in water:

1. Physical impurities:
 Colour: in water is caused by metallic substances like salts.
 Turbidity: is due to the colloidal, extremely fine suspensions such as insoluble
substances like clay, slit, and micro-organisms.
 Taste: presence of dissolved minerals in water produces taste. Bitter taste can be
due to the presence of Fe, Al, Mn, Sulphates and lime. Soap taste can be due to the
presence of large amount of sodium bicarbonate.
 Odour: In water is undesirable for domestic as well as industrial purpose.

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2. Chemical impurities:

 Inorganic chemicals: Cations (Al+3, Ca+2, Mg+2, Fe+2, Zn+2, Cu+2, Na+, K+), Anions

(Cl−, SO4−2, NO3−, HCO3−, F−, NO2−)

 Organic chemicals: dyes, paints, petroleum products, pesticides, detergents, drugs

textile materials, other organic related materials.

3. Biological Impurities:

 Biological impurities are Algae, pathogenic bacteria, fungi, viruses, pathogens,

parasite-worms.

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Hardness of Water
Hardness of water defined as which prevent the lathering of soap. This is due to
presence of certain salts like Ca+2, Mg+2 and other heavy metals dissolved in water.
Soaps (Sodium or Potassium salts of higher fatty acids) like Stearic acids
(C17H35COONa).
C17H35COONa + H2O → C17H35COONa + NaOH
soap Stearic acid
Hard Water: The water which does not give lather with soap is called hard water.
This is due to presence of certain salts like Ca +2, Mg+2 and other heavy metals
dissolved in water

2C17H35COONa + CaCl2/MgCl2 → (C17H35COO)2 Ca/Mg + 2NaCl

soap (soluble) salts (soluble) insoluble scum

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Causes of Hardness

Hardness of water is due to the presence of Bicarbonates, Chlorides, Sulphates and

Nitrates of Calcium and Magnesium. These soluble salts get mixed with natural water
due to the following reasons:
1. When natural water containing CO2 flows over the rocks of the limestone (CaCO3)
and Dolamite (CaCO3 & MgCO3), they get converted into soluble bicarbonates. Thus,
water gets hardness.

CaCO3 + H2O + CO2 → Ca(HCO3)2

Insoluble Soluble

2. When natural water flows over the rocks containing chlorides and sulphates and
Nitrates of Calcium and magnesium, these salts dissolve in water. Thus, water gets
hardness.
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Disadvantages of Hardness
In Domestic use:
 Washing: Hard water, when used for washing purposes, does not producing lather
freely with soap. As a result, cleaning quality of soap is decreased and a lot of it is
wasted.
 Bathing: Hard water does not lather freely with soap solution, but produces sticky
scum on the bath-tub and body. Thus, the cleaning quality of soap is depressed and
a lot of it is wasted.
 Cooking: The boiling point of water is increased because of presence of salts.
Hence more fuel and time are required for cooking.
 Drinking: Hard water causes bad effects on our digestive system. Moreover, the
possibility of forming calcium oxalate crystals in urinary tracks is increased.

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2. Industrial Use:
 Textile Industry: Hard water causes wastage of soap. Precipitates of calcium and
magnesium soaps adhere to the fabrics and cause problem.
 Sugar Industry: The water which containing sulphates, nitrates, alkali carbonates
are used in sugar refining, cause difficulties in the crystallization of sugar.
 Dyeing Industry: The dissolved salts in hard water may reacts with costly dyes
forming precipitates.
 Paper Industry: Calcium, magnesium, Iron salts in water may affect the quality of
paper.
 Pharmaceutical Industry: Hard water may cause some undesirable products while
preparation of pharmaceutical products.

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3. Steam generation in Boilers:

For steam generation, boilers are almost invariably employed. If the hard water is fed
directly to the boilers, there arise many troubles such as: Scales & sludges formation,
Corrosion, Priming & Foaming and Caustic embrittlement.

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Types of Hardness
Hardness of water is mainly two types:
1. Temporary Hardness: Temporary Hardness mainly caused by the presence of
dissolved bicarbonates of Calcium, Magnesium [(Ca(HCO 3)2, Mg(HCO3)2)].
Temporary Hardness can be largely removed by boiling of water.

2. Permanent Hardness: It is due to the presence of dissolved Chlorides, Nitrates and

Sulphates of Calcium, Magnesium, Iron and other metals. Permanent hardness
responsible salts are CaCl2, MgCl2, CaSO4, MgSO4, FeSO4, Al2(SO4)3. Permanent
Hardness cannot be removed by boiling but it can be removed by the use of chemical
agents.
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Expression and Units of Hardness

The expression of hardness producing salts usually expressed in terms of an equivalent

amount of CaCO3. Calcium Carbonate is chosen as a standard because:
i. Its molecular weight (100) and equivalent weight (50) is a whole number, so the
calculations in water analysis can be simplified.
ii. It is the most insoluble salt that can be precipitated in water treatment.

The conversion of the hardness causing salts into CaCO 3 equivalents can be achieved
by using the following formula:

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Units of Hardness:
1. Parts per Million (ppm): The number of parts of calcium carbonate equivalent hardness presents in
106 parts of water.
1ppm = 1 part of CaCO3 eq hardness in 106 parts of water

2. Milligrams per litre (mg/l): The number of milligrams of calcium carbonate equivalent hardness
presents in litre of water.
1 mg/L = 1 mg of CaCO3 eq hardness in 1 litre of water

But one litre of water weights =1 kg =1000g = 1000 x 1000 mg = 106 mg = 1 ppm

3. Clark’s degree (°Cl): The number of parts of calcium carbonate equivalent hardness presents in
70,000 or (7×104) parts of water.
1° Clarke = 1 part of CaCO3 eq hardness per 70,000 parts of water

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4. Degree French (°Fr): The number of parts of calcium carbonate equivalent

hardness presents in 105 parts of water.
1° Fr = 1 part of CaCO3 hardness eq per 105 parts of water

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Determination of Hardness by Complexometric Method / EDTA Method

Hardness present in a given water sample can be determined using the complexometric
method, in which the disodium salt of EDTA is employed (soluble in water) and can be
represented as follows.

EDTA - Ethylene Diamine Tetra Acetic acid

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 EDTA forms complexes with calcium and magnesium ions when the pH is in the
range of around 9.5 to 10.5 and to maintain the pH, a basic buffer solution is used
(NH4OH + NH4Cl).
 The complexes of calcium and magnesium with EDTA are colourless; therefore, an
indicator (In) must be used to locate the endpoint.
 In this titration, Eriochrome black-T (EBT) is used as an indicator, which forms an
unstable wine-red coloured complex with calcium and magnesium.
 Calcium ions are complexed first with EDTA, but the colour change only occurs
once all the magnesium has completely reacted. It is thus possible to determine the
total amount of these metals in the solution and the total hardness can be calculated.
 Calcium itself does not give a satisfactory end point with EBT indicator unless the
solution also contains magnesium.
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 Once EDTA completely removes all the calcium and magnesium ions, the free
indicator is left in the solution, imparting a blue colour. So, the colour change at the
endpoint is wine red to blue.

 During the estimation, one can take 50.0 ml of the hard water sample into a clean
conical flask, followed by adding 2 or 3 ml of buffer solution (pH 9.5 - 10.5) and 2
or 3 drops of EBT indicator.
 Then this mixture is titrated with standard EDTA (0.01 M) solution until the colour
changes from wine-red to blue, which is the reaction's endpoint.

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 So, finally, one can find out that 50.0 ml of hard water samples consumes 'x' ml of
0.01 M EDTA solution, which needs to substitute in the below formula for getting
the hardness of the water sample.
Total hardness of water sample per liter = X x 1000 / 50 = X x 20 mg/litre or ppm

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 Problems on Hardness of water Topic - Module

Calculation of CaCO3 equivalent Sl.N Dissolved salt or Molecular Multiplification

o. constituent mass factor
The concentration of hardness is
1 Ca(HCO3)2 162 100/162
usually expressed in terms of
2 Mg(HCO3)2 146 100/146
equivalent amount of CaCO3 since this
3 CaSO4 136 100/136
mode permits the multiplication and
4 MgSO4 120 100/120
division of concentration when
5 FeSO4 152 100/152
required. The selection of CaCO3 in
6 CaCl2 111 100/111
particularly to express hardness of 7 MgCl2 95 100/95
water is due to the following reasons. 8 MgCO3 84 100/84
i) CaCO3 has a molecular mass of 100 9 CaCO3 100 100/100
(or eq. mass of 50) a whole number, 10 Mg(NO3)2 148 100/148
easy to do maths and ii) It is the most 11 Ca(NO3)2 164 100/164
insoluble salt that can be precipitated 12 CO2 44 100/44
in water treatment. 13 Al2(SO4)3 114 100/114
14 FeSO4.7H2O 278 100/278
15 HCO3- 122 100/122
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16 OH- 34 100/34
 Problems on Hardness of water
1. Calculate the temporary, permanent and total hardness of a sample of water containing
the dissolved salts in mg/L. Ca(HCO3)2 = 6.8, Mg(HCO3)2 = 9.0, MgCl2 = 4.2, Na2SO4 = 6.5,
CaSO4 = 12.5, MgSO4 = 5.6

Solution:

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 Salt/ion Conc. In mg/L Multip. factor CaCO3 equivalent in mg/L

Ca(HCO3)2 6.8 100/162 6.8 x 100/162 = 4.197

Mg(HCO3)2 9.0 100/146 9.0 x 100/146 = 6.164
MgCl2 4.2 100/95 4.2 x 100/95 = 4.421
Na2SO4 Do not contribute for hardness of water
CaSO4 12.5 100/136 12.5 x 100/136 = 9.191

MgSO4 5.6 100/120 5.6 x 100/120 = 4.666

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 Problems on Hardness of water

2. Calculate the temporary hardness of a sample of water containing the dissolved salts in
mg/L. Ca(HCO3)2 = 7.2, Mg(HCO3)2 = 8.5, MgCl2 = 4.9, Na2SO4 = 6.5, CaSO4 = 12.1, MgSO4 = 4.8.

Solution:

Temporary hardness = Ca(HCO3)2 + Mg(HCO3)2 = (4.444 + 5.821 ) mg/L = 10.265 ppm

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 Salt/ion Conc. In Multip. CaCO3 equivalent in mg/L
mg/L factor
Ca(HCO3)2 7.2 100/162 7.2 x 100/162 =
4.444
Mg(HCO3)2 8.5 100/146 8.5 x 100/146 =
5.821
MgCl2 4.9 100/95 4.9 x 100/95 =
4.421
Na2SO4 Do not contribute for hardness of water
CaSO4 12.1 100/136 12.1 x 100/136 =
8.897
MgSO
Temporary
4 4.8= Ca(HCO ) +100/120
hardness 4.8 x +100/120
Mg(HCO3)2 = (4.197 6.164 ) mg/L =
3 2
=4.000
10.361 ppm
Permanent hardness = MgCl2 + CaSO4 + MgSO4 = (4.421 + 9.191 + 4.666) mg/L
= 18.278 ppm
Total hardness = Temporary + Permanent = 10.361 ppm + 18.278 ppm = 28.639 ppm

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 Problems on Hardness of water

3. One liter of water from an underground reservoir in Nalgonda Town in Andhra Pradesh the following
analysis. for its contents:: Mg(HCO3)2 = 0.0146 g; Ca(HCO3)2 = 0.0081 g; MgSO4 =0.0012 g; CaSO4 = 0.0136 g,
NaCl = 0.0585 g; Organic impurities = 100 mg; Calculate temporary, permanent and total hardness of this
sample of water in degree French.

Solution:

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 Problems on Hardness of water

4. One liter of water sample collected from a water source in Andhra Pradesh has shown the following
analysis. Mg(HCO3)2 = 14.6 mg, MgSO4 = 12 mg, Ca(HCO3)2 = 16.2 mg, CaCl2 = 22.2 mg, MgCl2 = 9.5 mg
and organic impurities 100 mg. Calculate temporary and permanent hardness in Degree French.

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 Problems on Hardness of water

5. How many mg. Of FeSO4 dissolved per litre gives 200 ppm of hardness?

Solution : 1 mole of FeSO4 + 1 mole of CaCO3 (from mole concept)

i.e FeSO4 = CaCO3

(56+16+64) = 136 g = 100 g

Therefore, 100 ppm of hardness = 136 ppm of FeSO4

Or 200 ppm of hardness = 136 x 200/100 = 272 mg/l

Hence, 272 mg of FeSO4 dissolved per litre gives 200 ppm hardness

6. How many grams of MgCO3 dissolved per litre gives 100ppm of hardness?

Solution: 1 mole MgCO3 = 1 mole CaCO3

(24 + 12+ 16x3) = 84 g = 100 g

Therefore, 100 ppm of hardness = 84 ppm of MgCO3

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 Problems on Hardness of water

7. A sample of hardwater contains 408 mg of CaSO 4 per litre. Calculate the hardness in terms of CaCO 3 equivalents.

Salt/ion Conc. In mg/L Multip. CaCO3 equivalent in mg/L

factor
CaSO4 408 100/136 408 x 100/136 = 300
Therefore, Hardness = 300 ppm of CaCO3

8. Calculate the temporary hardness and total hardness of a sample of water containing Mg(HCO 3)2 = 14.6 mg/L,
Ca(HCO3)2 = 32.4 mg/L , Mg Cl2 = 19.0 mg/L; CaSO4 = 27.2 mg/L.
Salt/ion Conc. Multip CaCO3
In . equivalent in
mg/L factor mg/L
Ca(HCO3) 32.4 100/1 32.4 x 100/162
2
62 = 20
Mg(HCO3) 14.6 100/1 14.6 x 100/146
2
46 = 10
CaSO4 27.2100/1 27.2 x 100/136
36 = 20
MgCl2 19.0 100/9 19.0 x 100/95
5
Temporary hardness = Ca(HCO3)2 + Mg(HCO = 20
3)2 = (20 + 10 ) mg/L = 30 ppm

Permanent hardness = MgCl2 + CaSO4 = (20 + 20) mg/L = 40 ppm

Total hardness = Temporary + Permanent = 30 ppm + 40 ppm = 70 ppm

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 Potable water and its specifications

Definition: “Potable water is defined as the water that can

be considered safe for drinking and can be used for food
preparation”

Essential requirements/Specifications of potable water

 It should be sparking clear and odourless
 It should be pleasant in taste
 It should be perfectly cool
 Its turbidity should be less than 10 ppm
 It should be free from objectionable dissolved gases like H2S
 It should be free from objectionable minerals such as Pb, As, Cr and Mn salts
 Its alkalinity should be high. Its pH should be about 8.0.
 It should be reasonably soft, total hardness should be < 500 ppm.
 It should be free from microorganisms
 Cl-, F- & SO42- contents should be less than 250 ppm, 105ppm and 250ppm

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 Various stages of potable water
treatment

PART A PART B.
Removal of impurities Disinfection

1. Screening- Floating matter like, leaves, wood pieces etc are

1. Chlorination
removed 2. Ozonisation
2. Sedimentation-Suspended matter like sand, clay etc are removed In both the cases
3. Sedimentation with coagulation-Fine suspended inorganic Pathogenic
impurity bacteria are
removed
4. Filtration-Microorganisms and colloidal matter are removed

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 A. Removal of impurities
1. Screening
Screening is the process of removing floating materials like wood pieces, dead animal and leaves from
water. Raw water is allowed to pass through a screen having a large number of holes which removes the
small and large floating matter.

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 A. Removal of impurities
2. Sedimentation
Sedimentation is the process of removing suspended impurities by allowing the water to stand undisturbed
for 2-6 hours in big tanks, about 6M deep. Due to force of gravity, most of the suspended particles settled
down at the bottom and they are removed. Sedimentation removes only 70-75% of the suspended matter.

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 A. Removal of impurities
3. Sedimentation with coagulation
“Sedimentation with coagulation is a process of removing fine suspended and colloidal impurities by the
addition of requisite amount of certain chemicals called Coagulants”.

The coagulants when added to water form an insoluble gelatinous, Chemical reactions of coagulants with water
flocculent precipitates, which descends through the water and mixes-
up into very fine suspended impurities forming bigger impurities
called flocs, which easily settles down. Example for coagulants: (a) Alum, K2SO4∙Al2SO4∙24H2O for water with pH > 7.0
alum, ferrous sulphate, sodium aluminate

(b) Sodium luminate, NaAlO2 for water with pH < 7.0

MgSO4 + 2NaOH → Mg(OH)2 + Na2SO4

(c) Ferrous sulphate, FeSO4∙7H2O for water with pH > 8.5

4Fe(OH)2 + O2 + 2H2O → 4 Fe(OH)3, ferric hydroxide, a heavy floc.

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 A. Removal of impurities
4. Filtration
“It is the process of removing colloidal matter and most of the bacteria, micro-organisms etc, by passing
water through a bed of fine sand and other proper-sized granular materials. Filtration is carried out by
using Sand-filter”.
Process:
Water is first passed through a fine sand-bed. The
suspended particles, microorganisms etc are first collected
and gets clogged in fine sand and then passed through
coarse sand-bed. Here colloidal matter gets collected and
the water is free from sediments. After passing coarse sand
bed the water enters through a bed of medium sized stones
called gravel or gravel-bed. This is the last bed where water
is filtered completely and the filtered water is collected at
the filter outlet.
Regeneration of the filter bed
During the filtration process the fine pores of sand bed gets seized and clogged. In order to continue the
filtration process, 2-3 cm of fine sand at the top is scrapped, replaced and leveled with makeup sand.
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 B. Removal of microorganisms or Disinfection or sterilization
1. Chlorination
Chlorination is the process of purifying the drinking water by producing a powerful Germicide like
Hypochlorous acid. When this Chlorine is mixed with water it produces Hypochlorous acid which kills
the germs present in water.
H2O + Cl2 → HOCl + HCl
Hypochlorous acid
(Powerful Germicide)

Chlorinator
Chlorinator is an apparatus, which is used to purify the water by
chlorination process. It is a large tower which contains number
of baffle plates. Raw water and concentrated chlorine solution
are introduced at the top. These two get thoroughly mixed
during their passage through the tower. Disinfected water is
taken out from the inlet at the bottom of chlorinator.

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 B. Removal of microorganisms or Disinfection or sterilization

Chlorination
Advantages
 More effective and economical
 It is stable, requires small space for storage and
doesn’t deteriorate on storage
 It can be used at high as well as low temperatures
 It doesn’t introduce any impurity in treated water

Disadvantages
 Excess chlorine if added produces a characteristic,
unpleasant taste and odour
 Its excess produces an irritation on mucus membrane
 Its concentration should not exceed 0.1-0.2 ppm
 It is more effective below pH 6.5 and less effective at
higher pH values.
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 B. Removal of microorganisms or Disinfection or sterilization

Break-point Chlorination
When liquid chlorine is added or chlorine gas is passed through water, the
consumption of chlorine makes the available chlorine less and after some
time the available chlorine increases. This is due to the consumption of
chlorine in oxidizing some oxidizable organic matter present in water at first.
Then available chlorine again increases due to the formation of chloro-organic
and chloroamine compounds i.e the combined chlorine increases in water.
Then it decreases due to the consumption of this chlorine for decomposition
of chloro-organic compounds. Once all the impurities are reacted or oxidized
further addition of chlorine results in appearance of residual chlorine. This
free chlorine is responsible for destruction of pathogenic bacteria in water.
The point (c in figure) at which the residual chlorine begins to appear is
known as break point and at this point water is devoid of bad taste and
odours and is bacteria free.

Advantages of break point chlorination

• Organic matters present in water are completely oxidized leaving behind odour-free water and sometimes color-free water.
• Pathogenic bacteria are destroyed completely. • It prevents the growth of weed in water in future.

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 B. Removal of microorganisms or Disinfection or sterilization
2. Sterilization by Ozonization
Ozone is an unstable gas and decomposes to give nascent oxygen. This nascent oxygen kills bacteria and oxidizes
the organic matter present in the water. Ozone is an excellent disinfectant and produced by passing silent electric
discharge through cold and dry oxygen.

Electric
3O2 2 O3
Discharge
Advantages
 Sterilization, bleaching, decolorization and deodorization take place at the same time.
 Ozone does not impart any unpleasant taste or odour to the treated water and does not change its pH
appreciably as it simply decomposes into oxygen. So it does not cause any irritation to the mucus membrane
as is the case with chlorine or bleaching powder treatment.
 Time of contact is only 10-15 minutes and dose strength is only 2-3 ppm.
Disadvantage
 The process is comparatively costly.

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 Topic - Module

ION EXCHANGE PROCESS:

∙ Ion exchange process is also known as the demineralization or de-

ionization process.
∙ Ion- Exchange resins are insoluble. Cross linked long chain organic
polymers with a microporous structure, and the “functional Groups”
attached to the chains are responsible for the ion-exchanging properties.
∙ In the De-ionization process all the ions present in water are eliminated by
using ion-exchange resins.
∙ Basically, resins with acidic functional groups are capable of exchanging
H+ ions with other cations.
∙ Resins with functional groups are capable of exchanging OH- ions with
other anions.
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Anion Exchange Resins

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Cation Exchange Resins: These are mainly styrene divinyl benzene co-polymers,
which are on sulphonation or carboxylation. These can exchange their hydrogen ions with
cations in water.

Anion Exchange Resins: Anion exchange resins are styrene-divinyl benzene or amine-
formaldehyde copolymers, which contain amino, quaternary ammonium or quaternary
phosphonium or tertiary sulphonium groups as an internal part of the resin matrix. These are
after treatment with dilute NaOH solution. Become capable of exchanging their OH- ions with
anions in water

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Ion-exchange process:

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• Hard water is allowed to pass through cation exchange

resins, which remove Ca+2 and Mg+2 ions and
exchange equivalent amount of H+ ions.
• Anions exchange resins remove bicarbonates, chlorides,
and sulphates from water exchange equivalent amount
of OH- ions.
• Thus, by passing hard water through cation hardness is
observed by the following reactions.

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Ion-Exchange resins

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REGENERATION:

When cation exchanger loses capacity of producing H+ ions and

exchanger loses capacity of producing OH- ions, they are said to be
exhausted. The exhausted cation exchanger is regenerated by passing it
through dilute sulphuric acid.

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The exhausted anion exchanger is regenerated by passing a dilute solution of NaOH

Advantages: 1. The process can be used to soften highly acidic or

alkaline water.
2. It produces water of very low hardness (2 ppm).

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DEMERITS OF ION-EXCHANGE PROCESS:

• The equipment is costly and more expensive chemicals are needed.

• If water contains turbidity, the output of the process is reduced. The turbidity must be
below 10ppm; else it must be removed by coagulation and filtration.

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DESALINATION OF WATER -REVERSE

OSMOSIS
Definition: The process of removing common salt (Sodium
Chloride) from the water is known as desalination.
The water containing dissolved salts with a salty or brackish taste
is called brackish water.

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Reverse Osmosis

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Reverse Osmosis:
Reverse Osmosis is a process in which pressure greater than the osmotic pressure is applied on the
high concentration side of the membrane, the flow of solvent moves from concentrated side to dilute
side across the membrane.
Osmosis is the phenomenon by virtue of which flow of solvent takes place from a region of low
concentration to high concentration when two solutions of different concentrations are separated by a
semi-permeable membrane.
In this process pure water is separated from salt water. 15-40 kg/cm 2 pressure is applied for
separating the water from its contaminants. The membranes used are cellulose acetate, polymethyl
acrylate and polyamide polymers. The process is also known as super or hyper filtration.

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Advantages:
1. It is simple and reliable process & Capital and operating expenses are low.
2. The life of the semi-permeable membrane is about two years, and it can be
easily replaced within a few minutes, thereby nearly uninterrupted water
supply can be provided.

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 Topic - Module
INTERNAL TREATMENT OF WATER

BOILER TROUBLES

A boiler is a closed vessel in which water under pressure is transformed into steam by
the application of heat. The steam generated is used in industries and generation of
power. In modern pressure boilers and laboratories, the water required is used pure
than the distilled water.
A boiler feed water should correspond with the following composition:
⮚ Its hardness should be below 0.2ppm.
⮚ Its caustic alkalinity (due to OH-) should lie between 0.15ppm to 0.45ppm.
⮚ It should be free from dissolved gases like O2, CO2, to prevent boiler corrosion.

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SLUDGES:

Definition: Sludge is a soft, loose, and slimy precipitate formed within the boiler. It is formed at
comparatively colder portions of the boiler and collects in the area where the flow rate is slow.
Ex: MgCO3, MgCl2, CaCl2, MgSO4.

Reasons for formation of sludges:

The dissolved salts whose solubility is more in hot water and less in cold water produce sludges.
Disadvantages of sludges:
1. Sludges are bad conductors of heat and result in the wastage of heat and fuel.
2. Excessive sludge formation leads to the settling of sludge in slow circulation areas such as pipe
connections, plug openings, gauge–glass connections leading to the choking of the pipes.

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Prevention of sludge formation:

1. Using soft water, which is free from dissolved salts like MgCO3,
MgCl2, CaCl2 and MgSO4 can prevent sludge formation.
2. By blow down operation carried out frequently can prevent sludge
formation.

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58 Committed to excellence since 1980

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59 Committed to excellence since 1980

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60 Committed to excellence since 1980

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61 Committed to excellence since 1980

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62 Committed to excellence since 1980