Bharati Vidyapeeth College of Engineering Navi Mumbai

Engineering Chemistry-I

Unit- 6

Water

Prof. Jadhav M R
Department of Applied Science

Unit-6 Water MRJ

 Water
1)Like air, water is one of the few basic materials which is of
prime importance for the preservation of life on this earth.

2) All are aware of the uses of water for drinking, cooking,

bathing & for farming etc.

3) But few know the importance of water as an engineering

matrial.

4)As an engineering material water is used for producing

steam, in boilers to generate hydro-electric power, furnishing
steam for engines, for construction of concrete structures for
manufacturing purposes & as a solvent in chemical process.

Unit-6 Water MRJ

 Sources of Water:
The Main Sources Of Water Are:

1) Rain water
2) River water
3) Spring water and well water
4)Sea water

1). Rain water: Rain water is the purest form of

natural water. However, it dissolves considerable
amount of gases (CO2 – SO2 - NO – NO2 …etc) and
suspended solid particles from atmosphere, during its
journey through it and becomes polluted.

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 2) River water: River are formed by rain and spring waters. During
its flow over the surfacee of land, it dissolves minerals of the soil such as
chlorides, sulfates, bicarbonates of sodium, calcium, magnesium ions
etc..

3) Spring or well water or Lake water: it contains cosnstant

chemical composition. The minerals present in the lake water in the
form dissolved form and high quantity of organic matter.

4) Sea water: It is the most impure form of natural water. It contains

larger percentage of the dissolved salts (above 3.5%) out of which about
2.6% is NaCl. The NaCl which is present in the dissolved form in sea
water will come out as NaCl crystals due to evaporation of sea water.

The other salts present in the sea water are sulphates of sodium,
bicarbonates of potassium, magnesium, calcium, bromides of potassium,
magnesium etc.

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 Underground water: Spring & well waters are the underground water
sources. They are in general clearer in appearance due to the filtering
action of the soil.
They contain more of the dissolved salts generally, underground water
is of high organic purity.

Types of impurities in water:

The impurities present in water are classified as:

1). Dissolved impurities: dissolved impurities may organic or inorganic.

Inorganic impurities: the carbonates, bicarbonates, sulphates,
chlorides of calcium, magnesium, iron potassium and aluminium.
Organic impurities: Organic water products, amino acids, proteins, etc.
Gases: O2 , CO2 , Oxides of nitrogen and sulphur, H2S etc.

Unit-6 Water MRJ

 2) Suspended impurities: It is of two types:

1. Inorganic - sand & clay;

2. Organic – vegetable and animal matter.

3) Biological Impurities: Micro-Organisms like

Pathogenic bacteria,
fungi, algae etc

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 DISADVANTAGES OF HARDWATER / CAUSES OF HARDNESS:

The following are the disadvantages when hard water is used for
various purpose:

(i) DOMESTIC USE:

(a) Washing and Bathing : Hard water does not form lather easily with
soap is wasted

(b) Drinking : Hard water causes bad effects on our digestive system.
Sometimes, stone formation takes place in kidneys

(c) Cooking : The boiling point of water is increased due to the

presence of salts. Hence, more fuel and time are required for
cooking.

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 (ii) INDUSTRIAL USE:

(a) Textile Industry : Hard water causes wastage of soap. Precipitates

of calcium and magnesium soap adhere to the fabrics and cause
problem

(b) Paper Industry : Calcium and Magnesium salts in water may effect
the quality of paper.

(c) Sugar Industry : Water containing sulphates, carbonates, nitrates

affects the crystallisation of sugar.

(d) Pharmaceutical Industry : Hard water may form some undesirable

products while preparation of pharmaceutical products.

Unit-6 Water MRJ

 (iii) STEAM GENERATION IN BOILERS:
For steam generation, boilers are employed.
If hard water is used in boilers, It may lead to
the following troubles

(a)Boiler Corrosion

(b)Scale and Sludge formation.

(c)Priming and Foaming

(d)Caustic embrittlement Pharmaceutical

industry

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 HARDNESS OF WATER (OR) HARDWATER AND SOFT
WATER:

Hard Water : Those water which does not produce

lather (or) very little lather with soap is called Hard
Water

Soft Water : Soft water readily produce a lot of lather

when mixed with little soap.

The Hardness of water is caused by the presence of

dissolved salts such as Bicarbonates, Sulphates,
Chlorides and Nitrates of bivalent metal ions like
Ca+2 & Mg+2

Unit-6 Water MRJ

 Soap is sodium/potassium salt of higher fatty acids
like stearic, oleic and
palmetic acids.

When soap is mixed with soft water lather is

produced due to stearic acid and sodium
stearate

→ Na – Stearate + H2O → NaOH + Stearic Acid

[C17H35COOH]

→ Stearic Acid + Na-Stearate → Formatioin of lather.

Unit-6 Water MRJ

 When soap comes in contact with HARD WATER,
Sodium stearate will react with dissolved calcium and
magnesium salts and produce calcium stearate or
magnesium stearate which is white precipitate

→ 2Na – Stearate + Ca+2 → Ca – Stearate ↓ + 2Na+

→[2C17H35COONa] + Ca+2 → [(C17H35COO)2 Ca] ↓ + 2Na+

(Soap) (Soluble) (Insoluble) (Soluble)

→ [2C17H35COONa] + Mg+2 → [(C17H35COO)2 Mg] ↓ + 2Na+

(Soluble) (Soluble) (Insoluble) (Soluble)

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 Different Type Of Water have different degree of
hardness. The different types of water are
commercially classified on the basis of degree of
hardness as follows:
Hardness Name Of Water

0-70 mg/litre Soft Water

70-150 mg/litre Moderate Hard Water

150-300 mg/litre Hard Water

> 300 mg/litre Very Hard Water

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 TYPES OF HARDNESS

The hardness of water is of two types

(1) Temporary hardness (or) Carbonate hardness

(2) Permanent hardness (or) Non-Carbonate hardness
(3) Temporary Hardness: Temporary hardness is caused by two
dissolved bicarbonate salts Ca(HCO3) and Mg(HCO3). The
hardness is called “Temporary Hardness”

Because it can be removed easily by means of boiling.

Ca(HCO3)2 Heating→ CaCo3 ↓ + H2O + CO2 ↑

Mg(HCO3)2 Heating→ Mg(OH)2 ↓ + 2CO2 ↑

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 (2) PERMANENT HARDNESS:

Permanent hardness of water is due to the

dissolved chlorides, sulphates and nitrates of
calcium and magnesium.

These salts are CaCl2, CaSo4, Ca(NO3)2, MgCl2,

MgSo4, Mg(No3)2
These Hardness cannot be removed easily by boiling.
Hence it is called “Permanent Hardness”. Only
chemical treatment can remove this hardness.
Total Hardness Of Water = Temporary Hardness +
Permanent Hardness

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 DEGREE OF HARDNESS:

• The Concentration of hardness as well as non-

hardness constituting ions are, usually
expressed in the term of “Equivalent amount of
CaCo3”

• Since this mode permits the multiplication and

division concentration, when required. The
choice of CaCo3 in particular is due to its
molecular weight (m.wt) is “100” (Equivalent
wt = 50), and moreover, It is unsoluble salt that
can be precipitated in water treatment.

Unit-6 Water MRJ

 The method of calculating degree of hardness will be clear from the
following formula

•Hardness causing in salt in terms of CaCo3

Amount of the hardness causing salt x 100

Molecular weight of hardness causing salt
UNITS OF HARDNESS:

These are 4 different units in which the hardness of water is expressed

as given below

(1) Parts per million (PPM): PPM is the number of parts of CaCo3
equivalent hardness per 106 parts ofwater.

i.e., 1 PPM = 1 part of CaCo3 equivalent hardness in 106 parts of

water.

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 (2) Milli grams Per Litre (mg/litre): mg/L is the number of milligrams of
CaCo3 equivalent hardness present per litre ofwater.

i.e., 1 mg/L = 1 mg of CaCo3 equivalent hardness of 1 L of water.

But 1 L water weights = 1 kg of water

1 kg = 1000 gms
= 1000 x 1000 mg
= 106 mg

∴1 mg/L = 1 mg of CaCo3 equivalent per 106 mg ofwater

= 1 part of CaCo3 equivalent per 106 parts of water

∴1 mg/L = 1 ppm

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 Degree Of Clark (ocl) :

ocl is number of grains (1/7000 lb) of CaCo3 equivalent hardness per

gallon (10 lb) of water.

(or)

It is defined as the number of parts of CaCo3 equivalent hardness per

70,000 parts of water.

∴1ocl = 1 grain of CaCo3 eq. hardness per gallon of water.

(or)

1ocl = 1 part of CaCo3 eq. hardness per 70,000 parts of water

∴1 ppm = 0.07ocl

Unit-6 Water MRJ

 Degree Of French (oFr) :

oFr is the number of parts of CaCo3 equivalent hardness per 105 parts
of water.

1oFr = 1 part of CaCo3 equivalent hardness per 105 parts of water

∴0.1o Fr = 1 ppm

Note: The hardness of water can be converted into all the four unitsby
making use of the following interconversion formula

1 ppm = 1mg/L = 0.07ocl = 0.1oFr

1ocl = 1.43oFr = 14.3 ppm = 14.3 mg/L

Unit-6 Water MRJ

 PROBLEMS:

(1) A sample of water gives an analysis 13.6 mg/L of CaSO4. 7.3 mg/L of
Mg(HCO3)2. Calculate the total hardness and permanent hardness.

Sol:
Salt Quantity Present (mg/L) M.Wt Eq. of CaCo3
CaSO4 13.6 136 13.6x100 = 10
136
Mg(HCo3)2 7.3 146 7.3 x 100 = 5
146

The Total hardness of H2O = Temporary hardness + Permanent Hardness

= 5 + 10 = 15 mg/L

Permanent hardness = 10 ppm (or) 10 mg/L

Unit-6 Water MRJ

 PROBLEM

(2) Calculate the total hardness of 1000 litre of a sample of water

containing the following impurities 16.2 mg/L of Ca(HCO3), 11.1mg/L
of CaCl2, 60 mg/L of MgSo4 and Ca(HCO3)2, 11.1 mg/L of CaCl2, 60
mg/L of MgSo4 and 19 mg/L if MgCl2.
Salt Quantity Present (mg/L) M.Wt Eq. of CaCo3
Sol:
Ca(HCO3)2 16.2 162 16.2x100 = 10
162
CaCl2 11.1 111 11.1x100 = 10
111
MgSO4 60 120 60x100 = 50
120
MgCl 2 19 95 19x100 = 20
95

Total hardness of H2O = Temporary hardness + Permanent Hardness

= 10 + 10 + 50 + 20 = 90 mg/L

Total hardness for 1000 litres

Unit-6 Water MRJ
= 90 x 1000 = 90,000 mg/L
 PROBLEM

(3) A Sample of hard water contains the following dissolved salts per litre.
CaCl2 = 111 mgs, CaSO4 = 1.36 mgs, Ca(HCO3)2 = 16.2 mgs,
Mg(HCO3)2 = 14.6 mgs, Silica = 40 gms, Turbidity = 10 mgs.
Calculate the temporary, permanent and total hardness of water in ppm,
Ocl & OFr

Sol: Salt Quantity Present (mg/L) M.Wt Eq. of CaCo3

CaCl2 111 mg/ L 111 111x100 = 1 0 0
111
CaSO4 1.36 mg/ L 136 1.36x100 = 1
136
Ca(HCO3)2 16.2 mg/ L 162 16.2x100 = 10
162
Mg(HCO3)2 14.6 mg/ L 146 14.6x100 = 10
146

Note: Si & Turbidity must not be considered because they do not cause
hardness to water.

Unit-6 Water MRJ

 Temp. hardness of H2O = Hardness of Ca(HCO3)2+ Mg(HCO3)2 interms
of CaCO3 equivalents

= 10 + 10 = 20 mg/L

Permanent hardness = Hardness of CaCl2+ CaSO4

interms of CaCO3 equivalents

= 100 + 1 = 101 mg/L

Conversion of hardness:

1 ppm = 1mg/L = 0.07ocl = 0.1o Fr

Total hardness of the sample of water = 121 ppm = 121 mg/L
= 121 x 0.07 = 8.47ocl and
= 121 x 0.1 = 12.1o F

Permanent hardness = 101 mg/L, 101 ppm, 7.07ocl, 10.1o Fr

Total hardness = 20 mg/L, 20 ppm, 1.4ocl and 2o Fr

Unit-6 Water MRJ

 PROBLEM

(04) 1 litre of water from an underground reservoir in Tirupati town in

A.P showed the following analysis for contents:
Mg(HCO3)2 = 42 mg; Ca(HCO3)2 = 146 mg; CaCl2 = 71 mg;
NaOH = 40 mg; MgSO4 = 48 mg; Organic impurities = 100 mg;

Calculate temporary and permanent and total hardness of the sample of

10,000 lit. of water

Note: NaOH & Organic impurities donot causes any hardness.

The Bicarbonate salt cause temp hardness,

While other are responsible for Permanent hardness.

Mg(HCO3)2 = 42 mg in 10,000 (or) 42/10,000ppm

Its CaCO3 eq’s are = [42/10000] x [100/146]

Unit-6 Water MRJ

 Similarly for Ca(HCO3)2 it is [146/10000] x [100/162]
For CaCl2 it is [71/10000] x [100/111]
For MgSO4 it is [48/10000] x [100/120]

Temporary Hardness = [42/10000] x [100/146] +

[146/10000] x [100x162]

Permanent Hardness = [71/10000] x [100/111] + [48/10000] x [100x120]

Total Hardness = Temporary Hardness + Permanent Hardness

= ............ + ………
= ……... ppm

Unit-6 Water MRJ

 PROBLEM

(5) Calculate the temporary & permanent hardness of water in ocl,

containing the following dissolved salts. CaCO3=50 mg/L,
MgCl2=9.5 Mg/L, CaCl2=2.2 mg/L and MgSO4=12 mg/L

Note: CaCO3 is an insoluble salt. It does not cause hardness. If CaCO3 is

given as H.C.S, It must be considered as Ca(HCO3)2 whose hardness
is expressed in the term of CaCO3 equivalent
Salt Quantity Present (mg/L) M.Wt Eq. of CaCo3
Sol:
CaCO3 50 mg/ L 100 50 m g/ L
MgCl 2 9.5 mg/ L 95 9.5x100 = 10
95
MgSO4 12 mg/ L 120 12x100 = 10
12
CaCl2 22.2 mg/ L 111 22.2x100 = 20
111

Unit-6 Water MRJ

 Temporary Hardness Of Water = [Hardness Of CaCO3]
= 50 ppm/50 mg/L
= 50 x 0.7 = 3.5 ocl

Permanent Hardness Of Water = Hardness of MgCl2 + MgSO4 + CaCl2

= 10 + 10 + 20
= 40 mg/L
= 40 x 0.7
= 2.8o cl

Unit-6 Water MRJ

 DETERMINATION OF HARDNESS OF WATER BY EDTAMETHOD:-

1. This is a Complexometric method where

Ethlene Diamine Tetra Acetic Acid (EDTA) is thereagent

2. EDTA forms complexes with different metal ions at differentpH.

3. Calcium & Magnesium ions form complexes with EDTA at pH 9-10.

To maintain the pH 9-10 NH4Cl, NH4OH buffer solution is used.

4. An Alcoholic solution of Eriochrome Black-T (EBT) is used asan

indicator

5. The disodium salt of EDTA under the trade name Triplex-III isused
for complexation

Unit-6 Water MRJ

 EDTA

Disodium Salt Of EDTA

Ca++

M-EDTACOMPLEX

Unit-6 Water MRJ

 BASIC PRINCIPLE:

• When hardwater comes in contact with EDTA, at pH 9-10, the Ca+2 &
Mg+2 forms stable, colourless complex withEDTA.

Ca+2
+ + EBT pH 9-10 → Ca-EBT
Mg+2 Mg-EBT (Complex)

(from hard water) (unstable, wine red colour)

• To the hard water sample the blue coloured indicator EBT is added
along with the NH4Cl, NH4OH buffer solution. EBT forms an
unstable, winered complex with Ca+2 & Mg+2

Ca+2
pH 9-10→ Ca-EDTA
+ + EDTA
Mg+2 Mg-EDTA (Complex)

(from hard water) (stable, colourless)

Unit-6 Water MRJ
 • The winered coloured [Ca-EBT, Mg-EBT] complex is titrated with
EDTA replaces EBT from Ca-EBT complex and form stable colourless
Mg-EBT

[Ca-EDTA] [Mg-EDTA] complex releasing the blue coloured

indicator EBT into H2O

• Hence the colour change at the end point is winered to blue

colour

• The titration is carried out in the following steps

1. PREPARATION OF STANDARD HARD WATER:

Dissolve 1gm of pure, dry CaCO3 in minimum quantity of dilute

HCl and evaporate the solution to dryness on a waterbath.

Unit-6 Water MRJ

 • Dissolve the residue in distilled water to make 1 litre in a standard flask
and shake well.

Strength of Std. hard water 1 lit = 1gm CaCO3 =1000mg CaCO3

1ml =1mgs of CaCO3

(2) PREPARATION OF EDTASOLUTION:

Dissolve 4 gms of pure EDTA crystals along with 0.1 gm of MgCl2 in one
litre of distilled water.

Unit-6 Water MRJ

 (3) PREPARATION OF INDICATOR (EBT):

Dissolve 0.5 gms of Erichome Black-T in 100 ml of alcohol.

(4) PREPARATION OF BUFFER SOLUTION:

Add 67.5 gm of NH4Cl to 570 ml of concentrated ammonia solution and

dilute with distilled water to one litre

(5) STANDARDISATION OF EDTASOLUTION:

Pipette out 50 ml of standard hard water solution into a conical flask.Add

2-3 ml of buffer solution and 2-3 drops of EBT indicator.

Titrate the wine red coloured complex with EDTA taken in a burette after
rinsing it with EDTA solution till the wine red colour changes to clear
blue.

Unit-6 Water MRJ

 Not the burette reading and let the volume be “V1”-ml. Repeat the titration
to
get concurrent values.

(6) STANDARDISATION OF HARD WATER SAMPLE:

Pipette out 50 ml of the water sample into a 250ml conical flask, add 2-3 ml
of buffer solution and 2-3 drops of EBT indicator.

• Titrate the wine red coloured solution with EDTA taken in the burette
till a clear blue coloured endpoint is obtained

• Let the volume of EDTA be “V2” ml. Repeat the titration to

get concurrent values

Unit-6 Water MRJ

 (7) STANDARDISATION FOR PERMANENT HARDNESS:

Pipette out 50 ml of hard water sample in a beaker and boil till thevolume
reduces to 20 ml. All the bicarbonates of Ca++ and Mg++ decomposes to
CaCO3 and Mg(OH)2

• Cool the solution and filter the water into a flask, wash the beaker
and precipitate with distilled water and add the washing to conical
flask.

• Add 2-3 ml of buffer solution and 2-3 drops of EBT indicator andtitrate
with EDTA solution taken in the burette till a clear blue colour end
point is obtained.

• Note the burette reading. Let the volume be “V3” ml

Unit-6 Water MRJ

 CALCULATIONS:

50ml of std. hard water

= V1 ml of EDTA
50 x 1mg of CaCO3 = V1 ml of EDTA
1 ml of EDTA = 50/V1 mgs of CaCO3 eq.
Now 50ml of given hard water = V2 ml of EDTA
= V2 x 50/V1 mgs of CaCO3 eq.
1L(1000ml) of given hard water = 1000 V2/V1 mgs of CaCO3 eq.

Total Hardness of water = 1000 V2/V1 mg/l or

= 1000 V2/V1 ppm

Unit-6 Water MRJ

 Now 50 ml of Boiled water = V3 ml of EDTA
= V3 x 50/V1 mgs of CaCO3 eq.

1000ml(1lit.) boiled water = 1000 V3/V1 mgs of CaCO3 eq.

Permanent Hardness = 1000 V3/V1 mgs/l or ppm

and Temporary Hardness = Toal – Permnent Hardness

= 1000[ V2/V1 –V3/V1] ppm

= 1000[ V2-V3/V1] ppm or mg/l

Unit-6 Water MRJ

 Numerical: 1
• 1 gm of CaCO3 was dissolved in HCl and the solution was made upto
one litre with distilled water. 50 ml of the above solution required 30 ml
of EDTA solution on titration. 50 ml of hard water sample required 40
ml of the same solution of EDTA for titration. 50 ml of the hard water
after boiling, filtering etc. required 30 ml of the same EDTA solutionfor
titration.
Calculate the temporary hardness of the water

Soln: Molarity of CaCO3 solution (M3) = 1/100 = 0.01 M

Molarity of EDTA solution (M2) = V1 M1
V2

V1= volume of CaCO3 solution = 50 ml

M1= Molarity of CaCO3 solution = 0.01 M
V2= volume of EDTA = 30 ml
M2= V4 M4 = 50 x 0.01 = 0.016 M
n2 30

Unit-6 Water MRJ

 • Molarity of Hard Water Solution (M3) = V2 M2
V3

V2= volume of EDTA = 40 ml

M2= Molarity of EDTA = 0.016 M
V3= volume of hard water = 50 ml
M3= 40 x 0.016 = 0.0128 M
50

Total Hardness of water = 0.0128 x 100 x 1000

= 1280 ppm

Permanent hardness of water: = V4 M4 = V2 M2

n4 n2

M4 = V 2 M2
V4

Unit-6 Water MRJ

 n4=1; V2= volume of EDTA = 30 ml
n2=1; M2= molarity of EDTA = 0.016
V4= volume of permanent hardness containing water = 50

M4 = 30 x 0.016 = 0.0096 M
50

Permanent hardness of water = 0.0096 x 100 x 1000

= 930 ppm

Temporary hardness = Total hardness – Permanent hardness

= 1280 – 960
= 320 ppm

Unit-6 Water MRJ

 Numerical: 2
▪ 1 g of CaCO3 was dissolved in dil. HCl & diluted to 1000 ml. 50 ml of
this solution required 48 ml of EDTA solution for titration. 50 ml of
hard water sample required 15 ml of EDTA solution for titration. 50 ml
of same water sample on boiling, filtering etc required 10 ml of EDTA
solution.

Calculate the different kind of hardness in ppm

Soln:

Volume of EDTA consumed for 50 ml of standard hardwater (V1) = 48 ml,

Volume of EDTA consumed for 50 ml of given sample (V2) = 15 ml
Volume of EDTA consumed for 50 ml of boiled water (V3) = 10ml

Unit-6 Water MRJ

 • Total Hardness = V2 x 1000 mg/L
V1
= 15 x 1000 = 312.5 mg/L
48

• Permanent Hardness = V3 x 1000 ml

V1
= 10 x 1000
48
= 208.3 mg/L

• Temporary Hardness = Total Hardness – Permanent Hardness

= 312.5 – 208.3
= 104.2 mg/L

Unit-6 Water MRJ

 Softening of water
ION EXCHANGE PROCESS/DETERMINERALISATION PROCESS:

Ion exchange resins are insoluble, cross-linked, long chain organic polymers.
The functional groups attached to the chains can exchange hardness
producing cations and anions present in the water

PROCESS: The process involves the following steps:

1) The first chamber is packed with cation exchange resin (RH+). When
the hard water is passed through a bed of cation exchange resin it
exchanges H+ with Ca+, Mg+2, K+, Na+ etc of hard water.

2RH+ + Mg2+ Cl2→ R2Mg2+ + 2H+ Cl-

2RH+ + Ca2+ Cl2 → R2Ca2+ + 2H+ Cl-

2RH+ + CaSO42+ → R2Ca+2 + 2H+ + SO4-

Unit-6 Water MRJ

Unit-6 Water MRJ
 Thus, the hardness producing cations (Ca2+, Mg2+ etc) areremoved
1) The second chamber is packed with anion exchange resin. The water
coming out of the first chamber contains H+, Cl-, SO42- and HCO3- ions.
It is now passed through anion exchange resin bed which can exchange
OH- ions with anions like Cl-, SO42- and HCO3-

RlOH- + Cl- → RlCl- + OH-

2RlOH- + SO42- → Rl2SO42- +2OH-

RlOH- + HCO3- → RlHCO3- +OH-

Thus, hardness producing anions like Cl-, SO42- and HCO3- are removed.

Unit-6 Water MRJ

 3) Thus, H+ ions produced from first chamber combine with OH- ions
produced from second chamber to form water.

H+ + OH- → H2O

Hence, the water produced from ion-exchange process is completely free

from all cations and anions of salts.

Unit-6 Water MRJ

Unit-6 Water MRJ
 REGENERATION OF RESINS:

Regeneration of Resin the resin bed gets exhausted, when used for a long
period and can be regenerated:

(a) The exhausted cation exchange resin can be regenerated by passing dil.
HCl (H+)

R2Mg2+ + 2H+ → 2RH+ + Mg2+

R2Ca2+ + 2H+ → 2RH+ + Ca2+
(b) The exhausted an ion exchange resin can be regenerated by passing dil.
NaOH (OH-)

RlCl + OH- → RlOH- + Cl-

R2SO42- + 2OH- → 2RlOH- + SO42-
R2HCO3- + OH- → RlOH- + HCO -3

Unit-6 Water MRJ

 ADVANTAGES:
1) The softened water by this method is completely free from all salts and
fit for use in boilers

2) It produces very low hardness nearly 2 ppm

3) Highly acidic or alkaline water can be treated by this process

DISADVANTAGES:

1) The equipment is costly

2) More expensive chemicals are required for regeneration

3) Turbid water cannot be treated by this method

Unit-6 Water MRJ

 Numericals:
1) After treating 10000 liters of water by ion exchanger, the cationic resins
required 200 L of 0.1N HCl and anionic resin required 200 L of NaOH
solution for regeneration. Find the hardness of above samples of water.

Solution: In an ion exchanger all hardness causing cations are removed by

cation exchanger while anion exchanger removes anions of the constituents
present in water so that amount of acid used for regeneration refers as
hardness of water.

Hardness in 10000 L of water = 200 L of 0.1N HCL = 200 L of 0.1 N CaCO3

= 200 L of 0.1 N CaCO3 =20 L of 1 N CaCO3
Total Hardness in 10000 L = 20 x 50g of CaCO3 eq. = 1000 gms of CaCO3
1 L water contains =1000/10000 = 0.1 gms of CaCO3 eq.
Hardness in ppm = 0.1 x 1000mg = 100 mgs of CaCO3
Hardness of water = 100 ppm

Unit-6 Water MRJ

 DESALINATION OF BRACKISH WATER:

Water containing high concentrations of dissolved solids

with a peculiar salty or brackish taste is called brackish
water

Sea water is an example of brackish water containing about

3.5% of dissolved salts. This water cannot be used for
domestic and industrial applications unless the dissolved
salts are removed by desalination.

Commonly used methods are:

1) Electrodialysis

2) Reverse Osmosis

Unit-6 Water MRJ

 1) Electrodialysis: Electrodialysis is based on the principle that the
ions present in saline water migrate towards their respective
electrodes through ion selective membranes. Under the influence of
applied e.m.f.

2) The unit consists of a chamber containing two electrodes, the cathode

and anode

3) The chamber is divided to 3-compartments with the help of thin, rigid,

ion-selective membranes which are permeable to either cation or anion.

4) The anode is placed near anion selective membrane while the cathode
placed near cation selective membrane

5) The anion selective membrane is containing positively charged

functional groups such as R4N+ and is permeable to anions only

6) The cation selective membrane consists of negatively charged functional

groups such as RSO3- and is permeable to cationsonly.

Unit-6 Water MRJ

Unit-6 Water MRJ
 ▪ Under the influence of applied e.m.f. across the electrodes the cations
move towards cathode through the membrane and the anions move
towards anode through the membrane.

▪ The net result is depletion of ions in the central compartment, while it

increases in the cathodic and anodic compartments.

▪ Desalinated water is periodically drawn from the central compartment

while concentrated brackish water is replaced with fresh sample.

ADVANTAGES OF ELECTRODIALYSIS:

▪ The unit is compact.

▪ The process is economical as for as capital cost and operational

expenses are concerned.

Unit-6 Water MRJ

 REVERSE OSMOSIS:

▪ When two solutions of unequal concentration are separated by a

semi-permeable membrane which does not permit the passage of
dissolved solute particles, i.e., molecules and ions.

▪ Flow of solvent takes place from the dilute solution to concentrated

solution this is called as “OSMOSIS”.

▪ If a hydrostatic pressure in excess of osmotic pressure is applied on the

concentrated side the solvent is forced to move from higher
concentration to lower concentrated side across. Thus, solvent flow is
reversed hence this method is called “Reverse Osmosis”

▪ Thus, in reverse osmosis pure water is separated from the contaminated

water. This membrane filtration is also called “Super Filtration” or
“Hyper-Filtration”.

Unit-6 Water MRJ

 METHOD OF PURIFICATION:

The reverse osmosis cell consists of a chamber fitted with a

semi-permeable membrane, above which sea water/impure
water is taken and a pressure of 15 to 40 kg/cm2 is applied
on the sea water/impure water. The pure water is forced
through the semi permeable membrane which is made of
very thin films of cellulose acelate. However superior
membrane made of Polymethacrylate and Polyamide
polymers have come to use

Unit-6 Water MRJ

 ADVANTAGES:

▪ Both ionic and non-ionic colloidal and high molecule weight organic
matter is removed from the water sample

▪ Cost of purification of water is less and maintenance cost is less

▪ This water can be used for high pressure boilers.

Unit-6 Water MRJ

 WATER FOR DOMESTIC USE &
TREATMENT OF WATER FOR MUNICIPALSUPPLY:

The following are the specification of water for drinking purpose:

▪ This water should be clear, colourless and odourless.

▪ The water must be free from pathogenic bacteria and dissolved gases
like H2S.

▪ The optimum hardness of water must be 125 ppm and

pH must be 7.0 to 8.5

▪ The turbidity in drinking water should not exceed 25ppm

▪ The recommended maximum concentration of total dissolved solids in

potable water must not exceed 500ppm

Unit-6 Water MRJ

 Ultrafiltration
Ultrafiltration is separation process using membranes with pore size in the
range 0.1 to 0.001 micron. It is membrane filtration in which hydrostatic
pressure forces a liquid against a semipermeable membrane. Suspended solids
and high molecular weight retained, while water and low molecular weight
solutes pass through the membrane.
Ultrafiltration does this by pressuring the solution flow, which is tangential to
the surface of supported membrane. The solvent and other dissolved
component that pass through membrane are known as permeate and the
component that donot pass through membrane are known as retanate.
Generally low applied pressure are sufficient high rate filtration.

Unit-6 Water MRJ

 Ultrafiltration

Suspended Proteins Sugar H2O

Imp.
Semipermeable

membrane

Applications
1) It is used in industry and research for purifying and concentrating
macromolecular solutions (e.g. protein solution)
2) To separate suspended solids from solution.
3) It used in paint recovery in automotive industry.
4) It is also used in to remove milk products from milk.

Unit-6 Water MRJ

 BOD: Biological Oxygen demand:

Biochemical oxygen demand (BOD) represents the amount

of oxygen consumed by bacteria and other microorganisms
while they decompose organic matter under aerobic (oxygen is
present) conditions at a specified temperature. ... The decay of
organic matter in water is measured as biochemical or
chemical oxygen demand.
A known volume of sample has its initial DO content recorded
and after a five day incubation period at 20°C, the sample is
removed from the incubator and the final DO content is taken.

Unit-6 Water MRJ

 The BOD value is then calculated from the depletion and the
size of the sample used. The DO readings are usually in parts
per million (ppm). Higher BOD indicates more oxygen is
required, signifying lower water quality. Low BOD means less
oxygen is being removed from water, so the water is usually
more pure.
BOD is often used in waste water treatment plants as an index of the
degree of organic pollution in water.
Industries that discharge waste water into municipal sanitary sewers or
waterways are facing strict regulations on levels of BOD. Solid materials
in wastewater can consist of organic and/or inorganic materials and
organisms and the solids must be significantly reduced by treatment or
they can increase BOD when discharged.
BOD= (Dob -Dos) x dilution factor
Dob = dissolved oxygen in the blank
Dos = dissolved oxygen of sewage after incubation

Unit-6 Water MRJ

 Chemical oxygen demand (COD)

The oxygen required to oxidize the organic matter present in a

given waste water can be theoretically calculated .If the
organic compounds and their concentration are known the
oxygen demand of the sample can be accurately computed but
it is impossible to know the details of organic compounds
Present in raw water or waste water. The COD is therefore
determined by performing a laboratory test On given water
sample using a strong oxidant like dichromate solution.

Unit-6 Water MRJ

 It is expressed in milligrams per liter (mg/L) also referred
to as ppm (parts per million), which indicates the mass of
oxygen consumed per liter of solution.
Definition:
COD is the total amount of oxygen required to chemically
oxidize the bio degradable and non biodegradable
organic matter.
All organic matter are converted in to CO2 & H2O. COD values
are always higher thanBOD. COD can be determined in 1 ½
hrs.

Unit-6 Water MRJ

 Determination of COD:

Water sample is refluxed in strong acidic solution with a

known excess amount of potassium dichromate along
with H2SO4 and HgSO4 as catalyst. After digestion, the
remaining unreacted K2Cr2O7 is titrated with Ferrous
Ammonium Sulfate (FAS) to determine K2Cr2O7
consumed. This gives us the oxidizable organic matter in
terms of oxygen equivalent.

Unit-6 Water MRJ

 (x-y) x 8 x N
COD= x 1000 mg/l
v

x = Volume of FAS required for blank

y= Volume of FAS required for test.
N = Normality of FAS.
v = Volume of the sewage sample taken

Unit-6 Water MRJ

 Problem :1. 6 ml of wastewater is diluted to 300 ml distilled water in
standard BOD bottle. Initial DO in the bottle is determined to be 8.5 mg/l.
DO after 5 days at 20 C is found to be 5 mg/l. Determine BOD of waste
water and compute the ultimate BOD.

BOD = Dob- Dos x dilution factor

𝒎𝒍 𝒐𝒇 𝒔𝒂𝒎𝒑𝒍𝒆 𝒂𝒇𝒕𝒆𝒓 𝒅𝒊𝒍𝒖𝒕𝒊𝒐𝒏

= Dob-Dos x
𝒎𝒍 𝒐𝒇 𝒔𝒂𝒎𝒑𝒍𝒆 𝒃𝒆𝒇𝒐𝒓𝒆 𝒅𝒊𝒍𝒖𝒕𝒊𝒐𝒏

𝟑𝟎𝟎
= 8.5 – 5 x
𝟔
= 175 ppm

Unit-6 Water MRJ

 Problem : 25ml of waste water required 20.3 ml of 0.001M K2Cr2O7
solution for complete oxidation. The blank titration reading was 29.7
ml. Calculate COD of Water sample

(x-y) x 8 x N
COD= x 1000 mg/l
v
x = Volume of K2Cr2O7 required for blank = 29.7 ml
y= Volume of K2Cr2O7 required for test. =20.3ml
N = Normality of K2Cr2O7. = 0.001 M
v = Volume of the sewage sample taken= 25 ml

Unit-6 Water MRJ

 (29.7-20.3) x 8 x .001
COD = x 1000 mg/l
25

COD = 3.008 ppm

Unit-6 Water MRJ