1. water

INFOMATICA ACADEMY CONTACT: 9821131002/9029004242
Degree Sem - I 1 Water
Water
 Syllabus:
 Water
 Impurities in water, Hardness of water, Determination of Hardness of water by
EDTA method and problems. Softening of water by Hot cold lime soda method
and problems. Zeolite process and problems. Ion Exchange process and
problems.
 Drinking water or Municipal water, Treatments removal of microorganisms, by
adding Bleaching powder, Chlorination (no breakpoint chlorination), Disinfection
by Ozone, Electro dialysis and Reverse osmosis, ultra filtration.
 BOD, COD (Definition & significance), sewage treatments activated sludge
process, numerical problems related to COD.
 Introduction:
 Water is found abundantly in nature. It is essential for the survival of plants,
animals & human beings and is equally important for use in industries.
 Earth’s 71% surfaces are covered by water and the remaining 29% by land. About
97% of the earth’s water supply is in the ocean, which is unfit for human
consumption and other uses because of its high salt content.
 Of the remaining 3%, 2% is locked in the polar ice caps and only 1% is available as
fresh water in rivers, lakes, streams, reservoirs and ground water which is suitable
for human consumption.
 Source of water:
The sources of water are broadly classified as:
Sources of Water
1) Rain water
2) Surface water
i. River water
ii. Sea water
iii. Lake water
iv. Underground water

 Impurities in water:
 Suspended impurities:
Suspended impurities are dispersion of solid particles that are large enough to be
removed by filtration or on settling.
 Dissolved inorganic impurities:
Ca & Mg
Ng
Metals &
Oxides
&
Gases
1. Bicarbonates – cause alkalinity & softening
2. Carbonates – cause alkalinity & softening
3. Chlorides – cause taste
4. Sulphates- cause mottled enamel of teeth.
1. Mn – impart black or brown colour
2. Iron oxide – corrosive & impart red colour
3. Lead – lead poisoning
4. As – Arsenic poisoning
1. - corrodes the metal
2. - gives rotten egg smell
1. Bicarbonates – cause alkalinity & hardness
2. Carbonates – cause alkalinity & hardness
3. Chlorides – cause hardness
4. Sulphates- cause hardness
Suspended Impurities
Organic Inorganic
Vegetable and animal matter,
microorganisms, (bacteria
caused disease), algae
protozoa impart color and
odor.
For e.g., clay, sand.
It causes turbidity
– i.e., reduce the clarity
of water.

 Organic impurities:
It may be suspended or dissolved vegetable and animal matter which impart colour,
taste, acidity and produce harmful disease causing germs.
 Hardness:
 Soft and hard water:
 A sample of water which on shaking with soap solution produces lather instantly
is called Soft-water. It does not contain any of the dissolved inorganic salts of
Ca/Mg. The lathering action by soap is due to the reduction of surface tension of
water by addition of soap.
 A sample of water which on shaking with soap solution does not give lather
instantly but produces a white curd – like ppt. is called Hard-water. Such a
sample contains dissolved bicarbonates sulphates, chlorides of Ca/Mg.
 Hardness in water is that characteristic which prevents the lathering of soap.
Reaction of soap (Na / K salt of higher fatty acid like Oleic, Stearic) with 2CaCI &
4MgSO is as follows.
17 35 22C H COONa CaCI   17 35 2
2C H COO Ca NaCI 
Sodium Stearate Calcium stearate
(Insoluble)
17 35 42C H COONa MgSO   17 35 2 42
C H COO Mg Na SO 
Magnesium Stearate
(Insoluble)
 Classification:
Hardness is of two types:
1)Temporary or Carbonate or Alkaline Hardness
2)Permanent or non-carbonate or non-alkaline Hardness
1) Temporary or Carbonate or Alkaline Hardness:
a. Temporary hardness is caused by the presence of dissolved bicarbonates of
calcium, magnesium and other heavy metals like 2 3
,Fe Al 
. Hardness is
destroyed by boiling of water, when bicarbonates are decomposed yielding
insoluble carbonates or hydroxide.
 3 3 2 22
On Heating
Ca HCO CaCO H O CO   
(Insoluble)

   3 222
2On Heating
Mg HCO Mg OH CO   
Or
3 2 2MgCO H O CO   
b. Ca/Mg carbonates or hydroxides thus formed being insoluble is deposited as a
crust or scale at the bottom of the vessel while 2CO escapes out. Precipitates
are filtered off to get Soft-water.
2) Permanent or non-carbonate or non-alkaline Hardness:
a. Permanent hardness or non-carbonate hardness is due to the presence of
chlorides, sulphates & nitrates of Ca, Mg, Fe and other heavy metals. This
cannot be destroyed on boiling. Special softening methods are used.
 Comparison:
Sr.
No.
Temporary Hardness Permanent Hardness
1.
This type of hardness can be removed by
easy means such as boiling of water. The
name temporary indicates the property
which is not difficult to be changed.
This type of hardness cannot
be removed by simple
techniques. Hence the term
permanent is used to describe
these characteristics. Special
softening techniques are used.
2.
It is due to dissolved bicarbonates and
Carbonates of 2 2 2
, , ,Ca Mg Fe 
etc.
It is due to dissolved chlorides,
sulphates & nitrates of
2 2 2
, ,Ca Mg Fe 
etc.
3.
It is caused due to carbonates hence known
as carbonate hardness.
It is caused due to salts other
than carbonates hence known
as non-carbonates hardness.
4.
Temporary hard water if used in steam
production deposits ppts of Carbonates &
hydroxides of 2 2
,Ca Mg 
at the bottom of the
container. These deposits harden to form
scales. The reaction taking place is
 3 3 2 22
Ca HCO CaCO H O CO   
   3 222
2Mg HCO Mg OH CO  
Permanent hard water if used
in steam production forms
scales in boiler.

 Measurements of Hardness:
 The extent of hardness is measured in terms concentration of ions contributing to
hardness.
 It is usually expressed in terms of equivalent amount of 3CaCO .
 The equivalents of 3CaCO is convenient precisely because the molecular weight of
3CaCO is 100 (equivalent weight is 50). The other reasons this compound is being
formed which gets precipitated during water treatment.
 Equivalents of Calcium Carbonate:
 Concentrations of hardness as well as non-hardness constituting ions are
expressed in terms of equivalent amount of 3CaCO . The choice of 3CaCO is
accepted because.
1) Its M. wt. is 100 (Eq. Wt. = 50).
2) It is the most insoluble salt that is precipitated in water treatment.
 Equivalent of 3CaCO for a hardness producing substance is
3Mass of hps Chemical equivalent of
Chemical equivalent of hps.
CaCO

3Chemical equivalent of
Chemical equivalent of hps.
CaCO
Or 3Molecular weight of
Molecular weight of hps.
CaCO
= Multiplication Factor
*hps = Hardness producing substance
 If  3 2
Ca HCO is the hps – 162 parts by mass of  3 2
Ca HCO would react with the
same amount of soap as 100 parts by mass of 3CaCO . Therefore the mass of
 3 2
Ca HCO is multiplied by a factor 100/162 to give mass in terms of 3CaCO .
Salt Molecular Mass Chemical equivalent Multiplication factor
 3 2
Ca HCO 162 81 100/162
 3 2
Mg HCO 146 73 100/146
4CaSO 136 68 100/136
2CaCl 111 55.5 100/111
4MgSO 120 60 100/120
2MgCl 95 47.5 100/95

3CaCO 100 50 100/100
3MgCO 84 42 100/84
2CO 44 22 100/44
 3 2
Mg NO 148 74 100/148
 Units of Hardness:
Hardness of water is expressed in terms of following units.
 Parts per million (ppm)
It is the number of parts of 3CaCO equivalent in 6
10 parts of water.
 Milligram per litre (mg/L)
It is the number of milligram of 3CaCO equivalent hardness present per litre of
water.
 1 litre of water weighs 1 Kg = 1000 gm = 6
10 mg
 1 mg/L = 1 mg of 3CaCO equivalent per 6
10 mg of water.
 1 mg/L = 1 ppm.
 Clarkes Degree 0
Cl
 It is the number of grains (1 / 7000lb) of 3CaCO equivalent hardness per gallon
(10 lb) of water. Or it is the parts of 3CaCO equivalent hardness per 70,000 parts
of water.
 Degree French 0
Fr
 It is the number of parts of 3CaCO equivalent hardness per 5
10 parts of water.
1 ppm = 1 mg / L = 0.1 0 0
0.07Fr Cl
1 mg/L = 1 ppm = 0.1 0 0
0.07Fr Cl
1 0 0
1.43 14.3Cl Fr  ppm = 14.3 mg/L
0
1 10Fr ppm = 10 mg/ L 0
0.7 Cl

 Determination of Hardness of water:
EDTA Titration method or Complex metric Titration Method:
 Theory:
1) The principle of this method is that hardness causing ions like 2 2
,Ca Mg 
forms
a weak unstable complex of wine red colour with indicator EB-T (Eriochrome
Black – T).
2) When added in small amount to hard water buffered to a pH value of 10.0.
3) When such a complex is treated with EDTA, It extracts the metal ions from the
metal ion dye complex to form a stable M-EDTA complex because of great affinity
of EDTA to form stable complexes with metal ions releasing Eriochrome black – T
free.
4) Thus, wine red colour changes to distinct blue marking the end point. End point
indicates that there is complete extraction of metal ions by EDTA. This can be
represented as :
2
2
Ca
Mg


 
 
  
+ Eriochrome Black – T 
Ca EB T
Mg
 
 
 
complex
Unstable complex (wine red)
Unstable complex is replaced by
Eriochrome Black
Ca
T
Mg
 
  
  
Complex + EDTA
EDTA complex
Ca
Mg
 
 
 
  
+ Eriochrome Black-T
(Blue)
EDTA is Ethylene Di-amine tetra acetic acid.
EDTA in the form of its di-sodium salt

2 2
,M Ca Mg 

 Procedure:
Various steps involved in this method are
1) Preparation of Hard water:
Dissolve 1 gm of pure dry 3CaCO in minimum quantity of dilute HCI (1:1) and
then evaporate the solution to dryness on a water bath. Dissolve the residue in
distilled water to make 1 litre solution. Each ml of this solution thus contains 1
mg of 3CaCO equivalent hardness.
2) Preparation of EDTA solution:
Dissolve 4.0 gm of EDTA crystals + 0.1 gm 2MgCI in 1 litre of distilled water.
3) Preparation of Indicator:
Dissolve 0.5 gm of Eriochrome Black – T in 100 ml of alcohol.
4) Buffer solution (pH=10):
Add 67.5 gm of 4NH Cl to 570 ml of conc. 3NH soln. and dilute with distilled
water to 1 litre.
M
O=C C=O
O
N
CH2
N
CH2
O
Structure of EDTA – M Complex

 Titration Procedure:
1) Standardization of EDTA Soln.:
Rinse and fill the burette with EDTA soln. Pipette out 50 ml of standard hard
water in a conical flask. Add 10-15 ml of buffer soln. & 4-5 drops of indicator.
Titrate with EDTA soln. till wine red colour changes to blue. Let volume of EDTA
consumed is 1V ml.
2) Titration of unknown Hard water:
50 ml of unknown hard water mixed with buffer solution & indicator as in step1
& titrated against EDTA solution till wine red colour changes to blue. Let the
volume of EDTA consumed be 2V ml.
3) Titration of Boiled Hard water:
Take 250 ml of tap water in a beaker. Boil it till the volume is reduced to about
50 ml [when all bicarbonates are decomposed to insoluble  3 2
/CaCO Mg OH ].
Filter & wash the ppt. with distilled water, collect filtrate & washings in a 250 ml
measuring flask. Finally make up the volume to 250 ml with distilled water.
Then titrate 50 ml of boiled water against EDTA solution. Let the volume
consumed be 3V ml.
 Calculations:
50 ml of standard hard water = 1V ml of EDTA solution
50 ml of unknown hard water = 2V ml of EDTA solution
50 ml of boiled hard water = 3V ml of EDTA solution
1)50 ml of SHW= 50 mg of 3CaCO equivalent hardness.
50 ml of SHW = 1V ml of EDTA soln.
1V ml of EDTA soln. = 50 mg of 3CaCO equivalent hardness.
1 ml of EDTA soln. = 50/ 1V mg of 3CaCO equivalent
hardness.
2)50 ml of UHW= 2V ml of EDTA soln.
2V ml of EDTA soln. = 2 150 /V V mg of 3CaCO equivalent hardness
50 ml of UHW = 2 150 /V V mg of 3CaCO equivalent hardness
1000 ml of UHW = 2 11000 /V V mg of 3CaCO equivalent hardness.
Total Hardness = 2 11000 / /V V mg L

3)50 ml of BHW= 3V ml of EDTA soln.
3V ml of EDTA soln. = 3 150 /V V mg of 3CaCO equivalent hardness
50 ml of BHW = 3 150 /V V mg of 3CaCO equivalent hardness
1000 ml of BHW = 3 11000 / /V V mg L
Permanent Hardness =
3 11000 / /V V mg L
Temporary hardness = Total hardness – Permanent hardness
= 2 1 3 11000 / 1000 / /V V V V mg L  
= 1 2 31000 / /V V V mg L  
Temporary Hardness =
1 2 31000 / /V V V mg L  
EDTA titration method is preferred because of greater accuracy, convenience and
more rapid procedure.
 Theory questions:
1) Define temporary hardness and permanent hardness.
2) How is temporary hardness in water eliminated?
3) What is hardness? Name the substances that cause permanent hardness?
4) Differentiate between hard water & soft water.
5) Why is hard water unsuitable for boilers?
6) Write in brief “Disadvantages of hard water in different industries”.
7) Differentiate between temporary hardness and permanent hardness.
8) What are the impurities associated with water? Give their effects if it is used in
various industries.
9) What is hardness? Explain how you would determine hardness of water using
EDTA method.
 Type - 1 Problems on Carbonate & Non – Carbonate:
 Note the following three points
 1 ppm = 1 mg/l
 Multiplying factor =
 3Mol. wt. 100
Mol. wt.of the impurity
of CaCO 

 3CaCO equivalent in ppm = impurity quantity in ppm  (M.F.)
(M.F.  Multiplying Factor)
1) A water sample contains  3 2
32.4 /Ca HCO mg lit ,  3 2
29.2 /Mg HCO mg lit &
4 13.5 /CaSO mg lit . Calculate all the three hardness.
a. Calculations of 3CaCO equivalents:
Impurity Qty.(ppm) Mol. wt. M. F. 3CaCO Equivalent (ppm)
 3 2
Ca HCO 32.4 162 100/162  32.4 100 /162 20 
 3 2
Mg HCO 29.2 146 100/146  29.2 100 /146 20 
4CaSO 13.5 136 100/136  13.5 100 /136 10 
b. Temporary hardness =    3 32 2
20 20 40Mg HCO Ca HCO ppm   
c. Permanent hardens = 4CaSO = 10 ppm
d. Total hardness = Tempo. + Perm. = 40 + 10 = 50 ppm.
2) Calculate the carbonate & non-carbonate hardness of a water sample containing
2 9.5MgCl ppm , 4 48MgSO ppm ,  3 2
16.2Ca HCO ppm , 12KCl ppm ,
 3 2
14.6Mg HCO ppm .
a. Calculations of 3CaCO equivalents :
Impurity Qty.(ppm) Mol. wt. M.F. 3CaCO Equivalent (ppm)
2MgCl 9.5 95 100/95  9.5 100 / 95 10 
4MgSO 48 120 100/120  4.8 100 /120 40 
 3 2
Ca HCO 16.2 162 100/162  16.2 100 /162 10 
KCl 12 Does not cause hardness no calculations
 3 2
Mg HCO 14.6 146 100/146  14.6 100 /146 10 
b. Carbonate hardness =    3 32 2
10 10 20Ca HCO Mg HCO ppm   
c. Non carbonate hardness = 2 4 10 40 50MgCl MgSO ppm   

3) Calculate tempo, perm & total hardness of water sample containing
 3 2
7.3Mg HCO ppm ,  3 2
16.2Ca HCO ppm , 2 9.5MgCl ppm , 4 13.6CaSO ppm ,
 3 2
Mg HCO 7.3 146 100/146  7.3 100 /146 5 
 3 2
Ca HCO 16.2 162 100/162  16.2 100 /162 10 
2MgCl 9.5 95 100/95  9.5 100 / 95 10 
4CaSO 13.6 136 100/136  13.6 100 /136 10 
b. Tempo. hardness due to    3 32 2
5 10 15Mg HCO Ca HCO ppm   
c. Perm. Hardness due to 2 4 10 10 20MgCl CaSO ppm   
d.Total hardness = Tempo + Perm. = 15+20 = 35 ppm
4) Calculate tempo. & perm. hardness of water sample containing
 3 2
16.8 /Mg HCO mg l , 2 19 /MgCl mg l , 4 24 /MgSO mg l ,
 3 2
29.6 /Mg NO mg l , 3 4 /CaCO mg L , 3 10 /MgCO mg l .
Impurity Qty. (ppm) Mol. wt. M.F. 3CaCO Equivalent (ppm)
 3 2
Mg HCO 16.8 146 100/146  16.8 100 /146 11.5 
2Mgcl 19 95 100/95  19 100 / 95 20 
4MgSO 24 120 100/120  24 100 /120 20 
 3 2
Mg NO 29.6 148 100/148  29.6 100 /148 20 
3CaCO 4 100 100/100  4 100 /100 4 
3MgCO 10 84 100/84  10 100 / 84 11.9 
b. Tempo. hardness due to  3 3 32
11.5 4 11.9 27.4Mg HCO CaCO MgCO ppm     
c. Perm. Hardness due to  2 4 3 2
20 20 20 60Mgcl MgSO Mg NO ppm     

5) A water sample contains 2 19MgCl ppm , 3 5CaCO ppm ,  3 2
29.5Ca HCO ppm ,
4 13CaSO  . Find all the three harnesses.
a. 3CaCO equivalents
Impurity Qty. (ppm) Mol. wt. M.F. 3CaCO Equivalent (ppm)
2MgCl 19 95 100/95  19 100 / 95 20 
3CaCO 5 100 100/100  5 100 /100 5 
 3 2
Ca HCO 29.5 162 100/162  29.5 100 /162 18.21 
4CaSO 13 136 100/136  13 100 /136 9.56 
b. Tempo. hardness due to  3 3 2
5 18.21 23.21CaCO Ca HCO ppm   
c. Perm. Hardness due to 2 4 20 9.56 29.56MgCl CaSO ppm   
d.Total hardness = Tempo. + Perm. = 23.21 + 29.56=52.77 ppm
6) What is the total hardness of a water sample having the following impurities in
mg/l?  3 2
162Ca HCO  , 2 22.2CaCl  , 2 95MgCl  , 20NaCl 
a. 3CaCO equivalents:
 3 2
Ca HCO 162 162 100/162  162 100 /162 100 
2CaCl 22.2 111 100/111  22.2 100 /111 20 
2MgCl 95 95 100/95  95 100 / 95 100 
NaCl 20 - - -
b. Total hardness = Hardness due to
 3 2 22
100 20 100 220Ca HCO CaCl MgCl ppm     
7) Classify the following impurities into temporal, permanent & non-hardness causing
impurities:
 3 2
Ca HCO , 4MgSO , 2CaCl , 2CO , HCl ,  3 2
Mg HCO , 4CaSO & NaCl .

How many grams of 2CaCl dissolved per litre gives 150 ppm of hardness?
a. Tempo. hardness causing impurities:  3 2
Ca HCO ,  3 2
Mg HCO
b. Perm. Hardness causing impurities: 4CaSO , 4MgSO , 2CaCl
c. Non-hardness causing impurities: 2CO , HCl , NaCl
d. 3 2 2 22CaCO HCl CaCl H O CO    
MW = 100 MW = 111
(1 Mole) (1 Mole)
1 ppm 3CaCO hardness  1.11 gm of 2CaCl
 150 ppm 3CaCO hardness  2
150
1.11 166.5
1
gm CaCl 
e. Thus, 166.5 gm 2CaCl dissolved per litre will give 150 ppm of hardness.
8) Classify the following salts into temporal & permanent hardness causing & also
calculate their 3CaCO equivalents.
 3 2
16.2 /Ca HCO mg L , 4 1.2 /MgSO mg L , 2 12.7 /FeCl mg L , 94 /NaCl mg L
a. Tempo. hardness causing salts:  3 2
Ca HCO
b. Perm. Hardness causing salts: 4MgSO , 2FeCl
c. Non-hardness causing salts: NaCl
d. 3CaCO equivalents:
Salt Qty. (ppm) Mol. wt. M.F. 3CaCO Equivalent (ppm)
 3 2
Ca HCO 16.2 162 100/162  16.2 100 /162 10 
4MgSO 1.2 120 100/120  1.2 100 /120 1 
2FeCl 12.7 127 100/127  12.7 100 /127 10 
NaCl 94 - - -
9) Two water samples A & B were analysed for their salt content.
a. Sample A was found to contain 168 mg 3 / .MgCO lit
b. Sample B was found to contain 820 mg  3 2
/Ca NO lit & 2 mg 2 /SiO lit .
Calculate the total hardness of each sample & state which sample is more-hard.
i. Sample A:

 It contains 168 mg 3 /MgCO lit
 3CaCO e.g.
100
168 200
84
ppm  = tempo. Hardness = total hardness.
ii. Sample B:
 It contains 820 mg  3 2
/Ca NO lit
 3CaCO e.g.
100
820 500
164
ppm  = perm. Hardness = total hardness
2SiO does not cause any hardness
Sample B is more hard.
 Type – 2 Problems on EDTA method:
1) Calculate the total hardness for given water sample:
a. 50 ml SHW containing 1 mg pure 3 /CaCO ml consumed 20 ml. EDTA solutions.
b. 50 ml water sample consumed 30 ml EDTA solution using EBT.
a. Standardization of EDTA:
20 ml EDTA solution  50 ml SHW
 1 ml EDTA solution
1
50
20
ml  SHW
50
20
 3CaCO equivalent
b. Calculation of total hardness :
50 ml HW sample 30 ml EDTA solution
 1 ml HW
1
30
50
ml  EDTA solution
30 50
50 20
mg  3CaCO eq
1000 ml HW
30
50

50
 1000
20
mg 3CaCO eq
 Total hardness = 1500 ppm
2) A SHW contains 15 g/L 3CaCO . 20 ml of this water required 25 ml EDTA solution.
100ml of sample water required 18 ml EDTA solution. The same sample after
boiling required 12 ml of EDTA solution. Calculate temporal hardness of the water
sample.
a. Concentration of SHW = 15 g/l = 15 mg/ml 3CaCO .
b. 20 ml of SHW = 20 15 300 mg  of 3CaCO
20 ml of SHW = 25 ml of EDTA solution
 1 ml of EDTA solution
20
15 12
25
mg   of 3CaCO

c. 100 ml of water sample = 18 ml of EDTA
 1000 ml of water sample
1000
18 180
100
   ml of EDTA
1 ml of EDTA = 12 mg of 3CaCO
 180 ml of EDTA = 180 12 2160 mg  of 3 /CaCO L
 Total hardness = 2160 mg /L
d.100 ml of boiled water = 12 ml of EDTA of 12 12 144 mg   of 3CaCO
 1000 ml of boiled water
1000
144 1440 /
100
mg l   3CaCO
Perm. Hardness = 1440 mg/ L.
e. Tempo. Hardness = total hardness – perm. hardness
= 2160-1440= 720 mg/L or ppm
3) 0.5 gm. of 3CaCO was dissolved in HCI & the solution mode up-to 500 ml. with
distilled water. 50ml of the solution required 45 ml of EDTA solution for titration.
50 ml of hard water sample required 15 ml of EDTA & after boiling & filtering
required 10 ml of EDTA solution. Calculate total & tempo. hardness of the hard
water sample.
a. Concentration of SHW = 0.5 g 3CaCO / 500 ml of distilled water
= 500 mg in 500 ml = 1 mg / ml 3CaCO
b. 50 ml of SHW 4.5 .ml EDTA
i.e. 45 ml EDTA  50 mg 3CaCO eq. hardness
Now, 50 ml water sample  15 ml EDTA
Hardness of sample
50
15
45
mg
 
  
 
3CaCO eq. for 50 ml hardness
Hardness/L of sample
50 1000
15 333.33
45 50
ppm
 
    
 
 Total hardness = 333.33 ppm
c. 50 ml water sample after boiling  10 ml EDTA
Perm. Hardness of sample 3
50
10
45
mg CaCO
 
  
 
eq. for 50 ml
Perm hardness of 1 litre sample
50 1000
10 222.22 /
45 50
mg L
 
    
 
 Perm. Hardness = 222.22 ppm
d.Tempo. Hardness = total hardness – perm. hardness
= 333.33 – 222.22 = 111.11 ppm

4) 50 ml of SHW containing 1mg of pure 3 /CaCO ml consumed 20 ml of EDTA. 50 ml of
the water sample consumed 30 ml. of the same EDTA solution using EBT. After
boiling & filtering, 50 ml of the water sample required 10 ml of the same EDTA for
titration. Calculate the total & perm. Hardness of the water sample.
a. Concentration of SHW = 1mg/ml 3CaCO
b. 50 ml of SHW = 50 mg of 3CaCO
50 ml of SHW = 20 ml of EDTA
1 ml of EDTA =
1 50
50
20 20
  of 3CaCO eq. = 2.5 mg of 3CaCO eq.
c. 50 ml water sample = 30 ml of EDTA
1000 ml of water sample =
1000
30 600
50
ml  of EDTA
1000 ml of EDTA = 2.5 mg of 3CaCO eq.
600 ml of EDTA = 600 2.5 1500 mg  of 3CaCO eq.
Total hardness = 1500 ppm
d.50 ml of boiled water = 10 ml of EDTA = 10 2.5 mg 3CaCO
1000 ml of boiled water =
1000 10 2.5
500
50
mg
 
 3CaCO
 Perm hardness = 500 ppm
5) 1 gm. of 3CaCO was dissolved in 1 lit. of distilled water. 50 ml of solution required
45 ml EDTA for titration. 50 ml of hard water required 25 ml of EDTA for titration.
The same sample of water after boiling consumed 15 ml of EDTA for titration.
Calculate the total, perm. & tempo. hardness of water.
a. Concentration of distilled i.e SHW water = 1 gm / lit = 1 mg/ml.
50 ml of SHW = 50 mg of 3CaCO
50 ml of SHW = 45 ml EDTA
1 ml of EDTA =
50
45
mg 3CaCO eq.
b. 50 ml of hard water sample – 25 ml EDTA
1000 ml of hard water sample
1000
25
50
ml  EDTA = 500 ml EDTA
=
50
500 555.55
45
  of 3CaCO eq.
Total hardness = 555.55 ppm

c. 50 ml of boiled water = 15 ml EDTA
1000 ml of boiled water
1000
15
50
ml 
=
1000 50
15 333.33
50 45
  
Perm hardness = 333. 33 ppm
d.Tempo hardness = Total hardness – Perm. Hardness
= 555.55-333.33 = 222.22 ppm
6) A standard hardware contains 15 gm/l calcium carbonate. 20 ml of this water
required 25 ml of EDTA solution. 100 ml of sample water required 18 ml of EDTA
solution. The same sample after boiling required 12 ml of EDTA solution. Calculate
temporary hardness of water.
Solution:
Given: Concentration of SHW = 15 g/lit
= 15000 mg/1000 ml
= 15 mg/ml
Volume of std. water = 20 ml
Volume of EDTA for SHW = 25 ml = 1V
Volume of water sample = 100 ml
Volume of EDTA for sample = 18 ml = 2V
Volume of water sample (after boiling) = 100 ml
Volume of EDTA for sample (after boiling) = 12 ml = 3V
20 ml SHW  25 ml EDTA Soln
1 ml EDTA 
20
15
25
 
  
mgs of 3CaCO Equivalent hardness
 300
25
Mgs of 3CaCO Equivalent hardness
 12 mgs of 3CaCO equivalent hardness
Now,
100 ml water sample  18 ml EDTA solution.
  18 21 Mgs of 3CaCO Equivalent hardness
 216 mgs of 3CaCO Equivalent hardness
Per litre  2160 mgs of 3CaCO Equivalent hardness
Total hardness  2160 ppm.

Now,
100 ml water sample (after boiling)  12 ml EDTA solution
  12 12 mgs of 3CaCO equation
 144 mgs of 3CaCO Equation
Per litre  1440 ppm
Permanent Hardness  1440 ppm
Temporary Hardness = 2160 – 1440  720 ppm
Ans.: Total hardness = 2160 ppm
Permanent Hardness = 720 ppm
Temporary Hardness = 1440 ppm
7) 0.5 gm of 3CaCO was dissolved in HCI and the solution made up-to 500 ml with
distilled water. 50 ml of the solution required 48 ml of EDTA solution for titration.
50 ml of water sample required 15 ml of EDTA and after boiling and filtering
required 10 ml of EDTA solution. Calculate temporary hardness of water.
Solution:
Concentration of S.H.W. = 0.5 gm 3CaCO / 500 ml D.W.
= 500 mgs in 500 ml water
= 1 mg/ml
Now,
50 ml SHW required  48 ml EDTA solution.
i.e. 48 ml EDTA solution  50 mgs 3CaCO equivalent hardness
1 ml EDTA solution 
50
48
 
 
 
mgs 3CaCO equivalent hardness
Now,
50 ml water sample  15 ml EDTA solution
Hardness of sample 
50
15
48
 
 
 
mgs 3CaCO equivalent for 50 ml Sample
Hardness per litre of sample 
50 1000
15 /
48 50
mgs lit
 
  
 
750
20 /
48
mgs lit
 
  
 
Total hardness  312.50 ppm
Now,
50 ml water sample after boiling  10 ml EDTA solution

Permanent hardness of sample
50
10
48
 
  
 
mgs 3CaCO equivalent for 50 ml
Permanent hardness of one litre sample =
50 1000
10 /
48 50
mgs lit
 
  
 
=
500
20 /
48
mgs lit
 
 
 
Permanent hardness of sample = 208.33 ppm
Temporary hardness = Total hardness – perm. Hardness
= 312.50-208.33
Temporary hardness of sample = 104.17 ppm
Ans.: Total hardness = 312.50 ppm
Permanent hardness = 208.33 ppm
And Temporary hardness = 104.17 ppm
8) 1 gm of 3CaCO was dissolved in 1 litre of distilled water, 50 ml of this solution
required 45 ml of EDTA solution for titration. 50 ml of hard water required 25 ml of
EDTA for titration. The same sample of water after boiling consumed 15 ml of EDTA
for titration. Calculate the hardness of water.
Solution:
Given: Concentration of SHW = 1g/lit
Quantity of SHW (1g/lit) = 50 ml
Quantity of EDTA consumed by 50 ml SHW = 45 ml
Quantity of hard water sample = 50 ml
Quantity of EDTA consumed = 25 ml
Quantity of EDTA consumed after boiling = 15 ml
Hardness =?
Standardization of EDTA:
Standard hardware has 1 g i.e. 1 1000 100  mg of 3CaCO equivalent hardness
per lit.
100 /mgs lit
=
100
0.1 /
1000
mg ml 3CaCO equivalent hardness
50 ml SHW = 350 1 mg CaCO 350 mgs CaCO
45 ml EDTA = 50 ml SHW 350 mgs CaCO
1 ml EDTA =
50
1.11
45
 mgs of 3CaCO equivalent hardness

Calculation of total hardness:
50 ml of H.W. sample = 25 ml EDTA
= 25 1.11 mgs of HW per 100 ml
= 0.2775 mgs of 3CaCO
Total H per litre = 277.5 mgs of 3CaCO
Calculation of permanent hardness:
50 ml of boiled water = 15 ml of EDTA
=15 1.11 mgs of 3CaCO in 100 ml
Hence per litre = 166.5 mgs of 3CaCO
temporary total permanentH H H 
= 277.5 −166.5
= 111 mgs of 3CaCO
Total hardness = 277.5 ppm
Permanent hardness = 166.5 (mgs of 3CaCO ) ppm
Temporary hardness = 111 ppm
9) 50 ml sample of water required 7.2 ml of N/20 disodium EDTA for titration. After
boiling and filtration the same volume required 4 ml of EDTA. Calculated each type
of hardness.
Solution:
As
1
20
EDTA solution =
20
N
3CaCO And 1000 ml of 1 normal, 3CaCO corresponding to 50 gm 3CaCO
1 ml of
20
N
3CaCO correspond to
0.005
20
of 3CaCO
Now, 50 ml of hard water correspond to 7.2 ml of
20
N
EDTA
i.e. correspond to (0.005/ 20 1/ 20 7.2  (outside gms. of 3CaCO ))
1000 ml of hard water correspond to 0.005/ 20 1/ 20 7.2 1000 gm   3CaCO
Hardness is equal to,
0.005/ 20 1/ 20 7.2 0.036
1000
1000 1000
 
 
= 360 PPM or m gm/lit

10) 20 ml of standard hard water containing 1.2 g 3CaCO per litre required 35 ml of
EDTA. 50 ml of hard water sample required 30 ml of the same EDTA. 100 ml of
hard water sample after boiling required 25 ml of the same EDTA. Calculate the
various hardness.
Solution:
Given: Weight of 3CaCO = 1.2 gm / litre
Voltage of SHW = 20 ml
Voltage of EDTA solution for SHW = 35 ml
Voltage of EDTA  2V = 30 ml
Voltage of EDTA  3V = 25 ml (for 100 ml sample)
To calculate the all types of hardness,
20 ml SHW = 35 ml. EDTA
Concentration of SHW = 1.2 g/lit
= 1200 mg/100 ml.
= 1.2 mg/ml.
20 ml SHW 20 1.2 mgs  hardness
 24.0 mgs
Thus, 35 ml EDTA  24 mgs equivalent 3CaCO eq. hardness
1 ml SHW
24
35
 mgs equivalent 3CaCO eq. hardness
Total hardness is per 50 ml = 1
24
35
V
 
 
 
mgs of equivalent 3CaCO eq. hardness
=
24
30
35
 
 
 
mgs of equivalent 3CaCO eq. hardness
= 20.57
Per litre =
1000
20.57 411.4
50
ppm
 
  
 
Permanent hardness per 100 ml = 2
24
35
V
 
 
 
mgs of equivalent 3CaCO eq.
hardness
=
24
25
35
 
 
 
mgs of equivalent 3CaCO eq.
hardness
= 17.14
Per litre =
1000
17.14 171.4
1000
ppm
 
  
 

Temporary hardness = Total H – Perm. Hardness
= 441.4-171.4
= 240.0 ppm
Ans.: Total Hardness = 411.4 ppm
Permanent hardness = 171.4 ppm
Temporary hardness = 240.0 ppm
 Unsolved Problems:
1) What is the carbonate and non-carbonate hardness of a sample of water which has
the following impurities per litre?
 3 2
50.25Ca HCO mg  3 2
8.2Ca NO mg
2 22.2CaCl mg 3 20.2KNO mg
3 16.8MgCO mg 4 3.8FeSO mg
4 6.0MgSO mg
Express the results as ppm, Clarke degree and French degree.
Ans.: [Carbonate hardness = 51 ppm or 0
5.1 3.56Fr or C
Non-carbonate hardness = 27.5 ppm or 0
2.75 or1.92Fr CI ]
2) Find carbonate and non-carbonate hardness of a sample of water which has the
following impurities per lit?
3 12.5 /CaCO mg lit 2 9.5 /MgCl mg lit
3 16.8 /MgCO mg lit 2 22 /CO mg lit
2 11.1 /CaCl mg lit 3 13.6 /NaHCO mg lit
Ans.: [Carbonate hardness = 32.5 ppm, Non-carbonate hardness = 20 ppm]
3) 50 ml of standard hard water containing 1 mg of pure 3CaCO ; per ml consumed 30
ml of EDTA. 50 ml of water sample consumed 35 ml of EDTA solution using
Eriochrome Black T indicator, 50 ml of water sample required 28 ml of EDTA using
same indicator. Calculate temporary and permanent hardness.
Ans.: [Temporary hardness = 234 ppm, Permanent hardness = 935.2 ppm &
Total hardness = 1169.2 ppm]

4) 0.25 gm of 3CaCO is dissolved in dilute HCI and diluted to 250 ml. 25 ml of this
solution required 24.0 ml of EDTA using Eriochrome Black –T indicator. 50 ml of a
hard water sample required 22.55 ml of the same EDTA. 100 ml of the water after
boiling and filtering required 120 ml of the said EDTA. Calculate the hardness in
the sample (permanent and temporary).
Ans.: [Total Hardness = 468 ppm, Permanent hardness = 333 ppm &
Temporary hardness = 135 ppm]
5) What is carbonate and non-carbonate hardness of a sample of water which has the
following impurities per litre :
 3 2
146 /Mg HCO mg lit 2 95 /MgCl mg lit
50 /NaCl mg lit  3 2
81 /Ca HCO mg lit
2 111 /CaCl mg lit 4 68 /CaSO mg lit
Express the results as ppm,
Ans.: [Carbonate hardness = 150 ppm =15 10.5Fr Cl
Non carbonate hardness = 250 ppm = 25 17.5Fr Cl ]
 3 2
81 /Ca HCO mg lit 2 95 /MgCl mg lit
38.5 /NaCl mg lit 2 22.2 /CaCl mg lit
 3 2
58.2 /Mg HCO mg lit 2 4.4CO lit
Express the results as ppm,
Ans.: [Carbonate hardness = 89.86 ppm = 6.29 8.986Cl Fr
Non carbonate hardness = 250 ppm = 8.4 12Cl Fr ]
7) 0.1 g of 3CaCO was dissolved in dil. HCI and diluted to 100 ml. 50 ml of this
solution required 40.0 ml of EDTA solution for titration. 50 ml of hard water sample
required 25 ml of EDTA solution, 50 ml of same water sample on boiling, filtering
required 20 ml of EDTA. Calculate hardness in ppm.
Ans.: [Temporary hardness = 125 ppm, Permanent hardness = 500 ppm
Total hardness = 625 ppm]

8) Calculate the hardness of water sample, whose 20 ml required 30 ml of EDTA. 10
ml of calcium solution, whose strength is equivalent to 300 mg of calcium
carbonate per 200 ml required 20 ml of EDTA solution.
 Softening of Hard Water:
It means the process of removing or reducing concentration of hardness causing salts
from water.
There are mainly 3 methods as follows:
 Lime Soda Process:
 Principle
In this method hard water is treated with calculated amount of slaked lime.
 2
Ca OH   and Soda ash  2 3Na CO in reaction tanks. So as to convert hardness
producing chemicals into insoluble compounds which are then removed by setting
and filtration.
Lime requirement for softening =
74
100
[Temp. 2
2Ca 
  Temp. 2
Mg 
+ Perm.  2 3 2
2 2 4/Mg Al Fe CO H HCI H SO   
   
3 2HCO NaAIO  ] all in terms of eq. 6
Volume of water 100
kg
10 % purity
  .
Water Softening Methods
1) Lime Soda 2) Zeolite
(or Permutit)
3) Ion Exchange
(or Demineralisation)
Cold Hot
ContinuousBatch wise

Soda requirement for softening =
106
100
[Perm.  2 2 3 2
2 4 3 2/Ca Mg Al Fe H HCI H SO HCO NaAlO    
     ] all terms in
3CaCO of eq. 6
Volume of water 100
kg
10 % purity
  .
 Reactions with lime:
1) It neutralize any free acid present
  2 22
2 2HCI Ca OH CaCl H O  
 2 4 4 22
2H SO Ca OH CaSO H O  
2) To precipitate iron and aluminium salts as hydroxide
     2 4 42 33
3 2 3Al SO Ca OH Al OH CaSO   
   4 42 2
FeSO Ca OH Fe OH CaSO   
   2 22 3
1
2 2
2
Fe OH H O O Fe OH   
3) To precipitate dissolved 2CO as 3CaCO
 2 3 22
CO Ca OH CaCO H O  
4) To precipitate calcium bi-carbonic as 3CaCO
   3 3 222
2 2Ca OH Ca OH CaCO H O   
5) To precipitates magnesium salts as hydroxides
     3 3 22 22
2 2Mg HCO Ca OH CaCO Mg OH H O   
   2 22 2
MgCl Ca OH Mg OH CaCl   
   4 42 2
MgSO Ca OH Mg OH CaSO   
6) convert bicarbonate ions (like ) into carbonate
 3 3 2 2 32
2 NaHCO Ca OH CaCO H O Na CO    
 Reactions with Soda:
Soda removes all soluble permanent hardness due to calcium salts as:
2 2 3 3 2CaCl Na CO CaCO NaCl   
4 2 3 3 2 4CaSO Na CO CaCO Na SO   

 Cold-Lime Soda Process:
 It is carried out in steel plated large rectangular tank provided with mechanical
stirrer.
 A calculated quantity of lime and soda added with water at room temp. Mixed
thoroughly by using mechanical stirrer.
 Lime and soda convert soluble salts of Ca and Mg into insoluble salts. Small
quantity of alum is added it coagulates insoluble impurities.
     2 3 4 24 32
3 2 3 6Al SO Ca HCO Al OH CaSO CO   
 Impurities are settle down to the bottom i.e. sludge. Sludge is removed through
sludge and wash with water.
 The clear softened water is collected through a float pipe and sent to the filtering
unit.
 As the process completed after 2 to 3 hours, the same procedure is repeated again
to obtain soft water. Hence, process is known as batch process.
 Continuous Process:
 This process is continuous and time saving process.
 In this process cylindrical container is used for the process with mechanical stirrer.
 Hard water, lime, soda and coagulant are passed together from the top of the inner
chamber.
 Stirring is done continuously, mixture flows down and soluble salts present in the
water are converted into insoluble salts.
 Coagulates collects these impurities together. These impurities are settle down at
the bottom in the form of sludge.
 Salt water moves upward, it is filtered by wood fibre filter, so all the solid impurities
are removed.
 Soft water is obtained from the outlet present at the top of the chamber.

 Continuous Hot Lime Soda Process:
 Plant contain reaction tank in which raw water, chemicals and steam are
thoroughly mixed.
 Impurities are settle down to the conical sedimentation vessel.
 It contain sand filter which ensures complete removal of sludge from the softened
water.
 Chemicals are added at higher temp.  80 150C to C so the process is called as Hot
lime soda process. Due to higher temperature

1) The reaction are fast.
2) Precipitation is more complete
3) Setting rate and filtration rates increased.
4) Co-agulants and chemicals needed are in smaller quantities.
5) Dissolved gases are eliminated from water to a certain extent.
 Advantages of Lime-soda process:
1) It is a very economical process.
2) pH of water increased so corrosion of pipes is reduced.
3) To certain extent, iron and manganese are also reduced.
4) Certain quantity of minerals are reduced from water.
5) Reduces pathogenic bacteria from water to certain extent.
6) Less amount of co-agulants are required.
 Disadvantages of Lime-soda process :
1) In cold lime soda process softened water is about 50-60 ppm is not good for use
in high pressure boilers. In hot process it is 15-30 ppm which is also quit high
for pressure boilers.
2) Careful operation and skilled supervision are required.
3) Disposal of large quantity of sludge formed during process is a problem.
 Zeolite Softener:

 Principle of Zeolite – Permutit Process
When hard water is passed over a bed of sodium Zeolite 2
Ca 
and 2
Mg
ions are
taken up by Zeolite and release equivalent ions in exchange of them.
 During softening [Ze : Zeolite]
2 2 2CaCl Na Ze CaZe NaCl  
2 2 2MgCl Na Ze MgZe NaCl  
4 2 2 4CaSO Na Ze CaZe Na SO  
4 2 2 4MgSO Na Ze MgZe Na SO  
 3 2 32
2Ca HCO Na Ze CaZe NaHCO  
 3 2 32
2Mg HCO Na Ze MgZe NaHCO  
 Regeneration of Zeolite:
2 22CaZe NaCl Na Ze CaCl  
2 22MgZe NaCl Na Ze MgCl  
Used Brine Regenerated Washing
Zeolite Zeolite
 Process of softening:
1) During process, hard water form top enter at a specified rate and over a bed of
sodium Zeolite.
2) Soften water containing sodium salts is collected at the bottom of the cylinder
and is taken out from time to time.
3) The Cations 2
Ca 
and 2
Mg 
are retained in Zeolite bed and soft water rich in
Na 
is collected.
4) Exhausted Zeolite is treated with 10% brine solution to get 2Na Ze (Regenerated
Zeolite). It contains three processes (a) Back washing (b) salting (c) Rinsing.
 Advantages:
1) It removes the hardness almost completely water about 10 ppm hardness is
produced.
2) The equipment used is compact, occupying a small space.
3) There is no sludge formation as no impurities are precipitated.
4) It is quite clean.
5) It requires less time for softening.

 Disadvantages:
1) The treated water contains more sodium salts.
2) The method only replaces 2
Ca 
and 2
Mg 
but leaves other acidic ions in soft
water e.g. 2
3 3,HCO CO 
.
 Limitations of Zeolite Process:
1) The presence of minerals in hard water, destroy the Zeolite bed.
2) Zeolite is process which used to replace only contains like ,Ca Mg 
ions but
onions like 3 3,CO HCO 
remain in 2H O.
3) Flow of water through Zeolite, if water contains turbidity may clog the pores of
Zeolite.
 Ion Exchange Process:
 The process of removal of ions (both cations and anions) from water sample is
known as ion-exchange process.
 Ion-exchange resins are insoluble, cross linked, long chain. High molecular weight
organic polymers which are permeable due to their micro-porous structure and the
functional groups attached to the chains are involved in the ion-exchanging
properties.
 Water which is free from both cations and onions is known as deionised or
demineralised water.
 Cations- Exchange Resin:
 Resin containing acidic functional group 3,COOH SO H  exchange their H
ions
with other cations is known as cation exchange resin.
2
2 2RH Ca RCa H 
  
2
2RH Mg RMg H
  
 E.g. Divinyl benzene copolymer, dowex 50, Duolite etc.
 Anion exchange Resin:
 Resins containing basic functional groups  2 ,NH OH  exchange their anions with
other is known as anion exchange resin.
 E.g. Amberlite 400, Zeolite – FF
  2 4 4 22
' ' 2R OH H SO R SO H O  
  2 22
' 2 ' 2R OH CHI R Cl H O  
  2 3 3 22
' ' 2R OH H CO R CO H O  

 When capacities of cation and anion exchangers to exchange H
and OH
ions
respectively are lost, they are then said to be exhausted.
 Demineralization of water:
 Process:
1)The hard water is passed through cation exchange column, which removes all
cation like 2
Ca 
, 2
Mg 
etc. from it and equivalent amount of H 
ions are
released from this column to water.
2
22 2RH Ca R Ca H  
  
2
22 2RH Mg R Mg H  
  
2)After cation exchange column, hard water passed through anion exchange
resin, which removes all anions like 2
4 ,SO Cl 
present in the water, and
equivalent amount of released in water.
ROH Cl RCl OH  
  
2
4 2 42 2ROH SO R SO OH  
  
2
3 2 32 2ROH CO R CO OH  
  
H 
And OH
released from both the columns get combined to produce water
molecule.
2H OH H O 
 
Water coming out from the exchange is free from cations as well as anions. It
is called as demineralised or deionised water.

 Regeneration:
When capacities of cation and anion exchanger to exchange H 
and OH
ions
are lost, then they are said to be exhausted.
1) The exhausted cation exchange column is regenerated by passing a solution of
dil. HCI or dil. 2 4H SO .
 2
2 2 2 washingR Ca H RH Ca  
  
2) The exhausted anion exchange is regenerated by passing solution of dil.
NaOH.
 2
2 4 42 2 washingR SO OH ROH SO  
  
Regenerated ion exchange resins are then used again.
 Advantages:
1) Process can be used to soften highly acidic or alkaline waters.
2) Produces water with very low hardness (abt. 2ppm)
 Disadvantages:
1) The equipment’s are costly and more expensive chemicals are needed.
2) If water contains turbidity, then the output of the process is reduced.
 Mixed Bed Deioniser:
Mixed bed deioniser consist essentially a single cylinder containing an intimate
mixture of hydrogen exchanger and strongly basic anion exchange.
 Regeneration of mixed ion Exchange:

 Comparison between Ion-Exchange and Zeolite Process:
Sr. No. Ion – Exchange Process Zeolite Process
1.
This process can produce
softened water with residual
hardness ranging between 0 to 2
ppm.
This process produces water with
residual hardness in the range 5-
10ppm.
2.
Water obtained is suitable for all
types of boilers especially high
pressure boilers.
Water obtained is not suitable for use
in high pressure boilers but can be
used in low and medium pressure
boilers.
3.
The cat ion and anion exchange
beds used are more expensive.
Hence, capital cost is high.
Zeolite is comparatively cheap, hence
capital cost is lower.
4.
The softening plant is not
compact, hence occupies more
space.
The softening plant is compact, hence
occupies less space.
5.
The process effectively removes
all the hardness and non-
hardness producing salts.
This process can remove only
2 2 2 2
, , ,Ca Mg Fe Mn   
ions. Hence
softened water contains salts like
3 2 4, ,NaCl NaHCO Na SO in dissolved
form.
6.
This process is useful for acidic
as well as alkaline water.
This process is not useful for highly
acidic water as it destroy the Zeolite
bed.
7.
De-ionized water does not cause
any of the boiler problems.
Soft water contains 3 2 3,NaHCO Na CO
that lead to caustic embrittlement in
boilers.
 Comparison between Zeolite process Lime-Soda process:
Sr. No. Zeolite process Lime soda process
1.
Residual hardness of water
obtained is in the range 5-10
ppm
Water of residual hardness 15-60
ppm is obtained.

2. Capital cost is high.
The process is relatively cheap as
the cost of chemicals & that of the
plant is low.
3.
Operating cost is low, as
exhausted Zeolite bed can be
regenerated by passing brine
solution through it.
Operating cost is higher because all
the chemicals used. i.e. lime, soda
and coagulants are consumed in
the process and cannot be
regenerated.
4.
The softener is compact &
occupies less space.
The size of softening plant depends
upon the quantity of hard water to
be softened. Hence. Size is not
compact, but increase with
quantity of water & thus occupies
more space.
5.
This process cannot be used for
hot water, acidic & turbid
water.
This process does not have any
such limitations.
 Type- 3 Problem on Lime Soda Process:
Note the following points:
1) Convert all the line (L) and soda (S) consuming impurities in terms of 3mgCaCO
equivalent/lit.
2) Impurities like NaCl , KCl , 2SiO , 2 3Fe O , 2 4Na SO should be ignored as they do not
impart hardness & do not consume L or S.
3) 3CaCO , 3MgCO ,  3 2
Fe HCO should be considered as temporary hardness & to be
taken for L calculations only.
4)  3 2
Mg HCO Quantity should be doubled.
5) For 3NaHCO & 3KHCO , their 3CaCO eq. must be added for L calculations &
subtracted for S calculations.
6) For 2NaAlO : Subtract from L
7) For 4FeSO ,  2 4 3
Al SO , 3AlCl : Add in L and S both
8) Given below is the table for calculations for L-S requirements.
The reactions are avoided for convenience only.

Sr. No. Constituent Need Sr. No. Constituent Need
1.
 3 2
Ca HCO
 Tempo. Ca
L 6. 2CO L
2.  3 2
Mg HCO  Tempo. Mg 2L 7.
(free acids likeH HCl
2 4 etc.)H SO
L + S
3.
2
Mg 
(Perm. Mg
2 4from orMgCl MgSO
or  3 2
)Mg NO
L + S 8. 4FeSO L + S
4. 3HCO
 3. .e g NHCO L - S 9.  2 4 3
Al SO , 3AlCl (L + S)
5.
2
Ca 
( . fromPerm Ca
2CaCl or 4CaSO  3 2
or )Ca NO
S 10. 2NaAlO -L
9) 100 parts by mass of  3 2
74 parts ofCaCO Ca OH
  2 2 2 2 3
2 3 2
74
Temp. 2 Temp. Perm.
100
L Ca Mg Mg Fe Al CO H HCO NaAlO      
           
10) 100 parts by mass of 2 4 17.9Na SO 
  2 2 2 3
3
106
S Perm.
100
Ca Mg Fe Al H HCO     
       
 Problems:
1) Calculate the quantity of lime & soda required for softening of 100000 lit. of water
containing following impurities in ppm:
 3 2
30.2Ca HCO  ,  3 2
20.8Mg HCO  , 2 28.1CaCl  , 2 8.78MgCl  , 4 35CaSO  , 4 6.7MgSO  ,
2 4 17.9Na SO  . The purity of lime is 70% & soda 85%
Solution:
a. 3CaCO equivalent:
Impurity
Qty.
(ppm)
Mol. wt. M.F. 3CaCO equivalent (ppm) Need
 3 2
Ca HCO 30.2 162 100/162  30.2 100/162 18.64  L
 3 2
Mg HCO 20.8 146 100/146  20.8 100/146 14.25  2 L

2CaCl 28.1 111 100/111  28.1 100/111 25.31  S
2MgCl 8.78 95 100/95  8.78 100/ 95 9.16  L + S
4CaSO 35 136 100/136  35 100/136 25.73  S
4MgSO 6.7 120 100/120  6.7 100/120 5.58  L + S
2 4Na SO 17.9 - - -
b.  2 2 2
6
74 Vol. of water 100
Lime Tempo. 2 tempo. Perm. kg
100 10 Purity of lime
Ca Mg Mg   
      
 
  6
74 100000 100
18.64 2 14.25 9.16 5.58 kg
100 10 70
      
 6.541 kg Ans.
c.  2 2 3 2
2 4 3
106
Soda [ Perm. or
100
Ca Mg Al Fe H HCl H SO HCO     
     
6
Vol. of water 100
] kg
10 purity of soda
 
   6
106 100000 100
25.31 9.16 25.73 5.58 8.228 kg Ans.
100 10 85
      
2) Calculate amount of lime (90% pure) & soda (98 % pure) for the treatment of 1
million lit. of water containing:
 3 2
8.1 ppmCa HCO  , 2 33.3 ppmCaCl  , 3 91.5 ppmHCO  , 2 38 ppmMgCl  ,
 3 2
14.6 ppmMg HCO  . The coagulant  2 4 3
Al SO was added at the rate of 17.1 ppm.
Solution:
Impurity
Qty.
(ppm)
Mol.
wt.
M.F. 3CaCO equivalent (ppm) Need
 3 2
Ca HCO 8.1 162 100/162  8.1 100/162 5  L
 3 2
Mg HCO 14.6 146 100/146  14.6 100/146 10  2 L
2CaCl 33.3 111 100/111  33.3 100/111 30  S
2MgCl 38 95 100/95  38 100/ 95 40  L + S

3HCO
91.5 122 100/122  91.5 100/122 75  L - S
 2 4 3
Al SO 17.1 114 100/114  17.1 100/114 15  L + S
b.      3 3 2 3 3 2 42 2 3
74
L [ eq. of 2
100
CaCO Ca HCO MgCl Mg HCO HCO Al SO     
6
Vol. of water 100
] kg
10 % of Purity
 
   6
74 1000000 100
5 40 20 15 75 127.44 kg Ans
100 10 90
       
c.  3 2 2 2 4 3 63
eq. of
100 10 % of purity
S CaCO MgCl CaCl Al SO HCO       
   6
106 1000000 100
40 30 15 75 10.82 kg Ans.
100 10 98
      
3) Calculate the qty. of lime & soda required for softening 50000 lit. of water containing
the following salts in mg/lit.
 3 2
8.1Ca HCO  ,  3 2
7.3Mg HCO  , 4 13.6CaSO  , 4 12MgSO  , 2 23.75MgCl  , 4.7.NaCl 
Solution:
Salt
Qty.
(ppm)
 3 2
Ca HCO 8.1 162 100/162  8.1 100/162 5  L
 3 2
Mg HCO 7.3 146 100/146  7.3 100/146 5  2 L
4CaSO 13.6 136 100/136  13.6 100/136 10  S
4MgSO 12 120 100/120  12 100/120 10  L + S
2MgCl 23.75 95 100/95  23.75 100/ 95 25  L + S
NaCl 4.7 - - - -
b.      3 3 2 42 2
74
[ Tempo. 2 Tempo. Perm
100
L Ca HCO Mg HCO MgCl MgSO    
6
Vol. of water 100
] kg
10 % of purity
 
   6
74 50000 100
5 2 5 23.75 10 1.803 kg Ans.
100 10 100
         

c.  2 4 4 6
Perm.
100 10 % of purity
S MgCl MgSO CaSO      
   6
106 50000 100
25 10 10 2.385 kg Ans.
100 10 100
     
(Note:- We have assumed that L & S are 100% pure)
4) Calculate the amount of lime (90% pure) & soda (95% pure) required softening of
50000 lit. with the following impurities in ppm:
 3 2
156Mg HCO  , 2 4 4.9H SO  , 2 23.75MgCl  , 5.6NaCl  , 2 111CaCl  , 2 16.2SiO  .
Solution:
Impurity
Qty.
(ppm)
Mol.
wt.
M.F. 3CaCO equivalent (ppm) Need
 3 2
Mg HCO 156 146 100/146  156 100/146 106.85  2 L
2 4H SO 4.9 98 100/98  4.9 100/ 98 5  L + S
2MgCl 23.75 95 100/95  23.75 100/ 95 25  L + S
NaCl 5.6 - - - -
2CaCl 111 111 100/111  111 100/111 100  S
2SiO 16.2 - - - -
b.   3 3 2 4 2 62
74 Volume of water 100
eq. of 2
100 10 % of purity
L CaCO Mg HCO H SO MgCl        Kg
   6
74 50000 100
2 106.85 5 25 10.018 kg Ans.
100 10 90
      
c.  3 2 4 2 2 6
eq. of
100 10 % of purity
S CaCO H SO MgCl CaCl       Kg
   6
106 50000 100
5 25 100 7.252 kg. Ans.
100 10 95
     
5) Calculate the amount of lime and soda required for softening 50000 lit. of hard water
containing following impurities in ppm.
3 144MgCO  , 2 95MgCl  , 3 25CaCO  , 2 111CaCl  , 2 3 25Fe O  , 2 4 15Na SO  .
Assume that both; the lime and soda are 100% pure.

Solution:
Impurity
Qty.
(ppm)
3MgCO 144 84 100/84  144 100 / 84 171.42 2 L
2MgCl 95 95 100/95  95 100 / 95 100 L + S
3CaCO 25 100 100/100  25 100 /100 25  L
2CaCl 111 111 100/111  111 100/111 100  S
2 3Fe O - - - - -
2 4Na SO - - - - -
b.  3 3 2 3 6
74 Vol. of water
eq. of 2
100 10
L CaCO MgCO MgCl CaCO      
  6
74 50000
2 171.42 100 25
100 10
     Kg
 17.31 kg. Ans.
c.  3 2 2 6
106 Vol. of water
eq. of
100 10
S CaCO MgCl CaCl    
   6
106 50000
100 100 10.6 kg. Ans.
100 10
   
6) Calculate the quantity of lime and soda required for softening of 1,00,000 litres of
water containing following impurities in ppm.
The purity of lime is 70% and soda 85%.
 3 2
30.2Ca HCO  ,  3 2
20.8Mg HCO  , 2 28.1CaCl  , 2 8.78MgCl  , 4 35.0CaSO  , 4 6.7MgSO  ,
2 4 17.9Na SO  . (Dec. 2008, 8 Marks)
Solution:
Conversion in 3 eq.CaCO
Salt/Impurity
Qty.
(ppm)
M.F.
3CaCO
equivalent (ppm)
Requirement
Lime/Soda/Both
 3 2
Ca HCO 30.2 100/162 18.64 Lime

 3 2
Mg HCO 20.8 100/146 14.25 2 Lime
2CaCl 28.1 100/111 25.31 Soda
2MgCl 8.7 100/95 9.16 Lime + Soda
4CaSO 35.0 100/136 25.73 Soda
4MgSO 6.7 100/120 5.58 Lime + Soda
2 4Na SO 17.9 - - Does not consume L or Soda
2 4Na SO has to be ignored as it does not consume lime or soda.
2 2
6
Lime = temporary 2 temporary permanent kg
100 10 Purity of lime
Ca Mg 
      
  6
74 100000 100
18.64 2 14.25 9.16 5.58
100 10 70
      
 
5
6
74 10 100
61.88 kg
100 10 70
  
6.5416 kg
  2 2 3 2
2 4 3
106
Soda = [ permanent or ]
100
    
6
Vol. of water 100
kg
10 Purity of soda
 
 
5
6
106 10 100
25.31 9.16 25.73 5.58 kg
100 10 85
     
 
5
6
106 10 100
65.98 kg
100 10 85
  
8.228 kg
7) Calculate the amount of Lime (85% pure) and Soda (85 % of pure) required for
softening 6
10 litres of water containing the following constituents:
 3 22
162 ppm, 9.5 ppmCa HCO MgCl 
 3 2
58.5 ppm, 7.3 ppmNaCl Mg HCO 
236.5 ppm, 44.0 ppmHCl CO 
2 4111 ppm, 60 ppmCaCl MgSO 
 At. wt. 40, 24, 32, 16, 35.5, 12, H 1, 28Ca Mg S O Cl C Si       

Solution:
Conversion in 3CaCO equivalents:
Salt/Impurit
y
Qty.
(ppm)
M.F.
3CaCO equivalent
(ppm)
Requirement
Lime/Soda/Both
 3 2
Ca HCO 162 100/162 100 Lime
 3 2
Mg HCO 7.3 100/146 05 2 Lime
2MgCl 9.5 100/95 10 L + S
NaCl 58.5 100/58.5 100 -
HCl 36.5 100/36.5 50 L + S
2CO 44 100/44 100 L
2CaCl 111 100/111 100 Soda
4MgSO 60 100/60 50 L + S
Here, NaCl does not consumer lime or soda
Now, 2 274
[temporary 2 temporary permanent
100
L Ca Mg 
   
  2 2 3
2 4 3 2 6
Vol. of water 100
or ] ]
10 Purity of lime
Mg Fe Al H HCl H SO HCO NaAlO    
      
 
6
6
74 10 100
100 2 0.05 10 50 100 50
100 10 80
L           
 
6
6
74 10 100
320 296 kgs
100 10 80
   
  2 2 3 2 1
2 4 3
106
Soda [ or ]
100
     
6
Vol. of water 100
10 Purity of soda
 
 
6
6
106 10 100
10 50 100 50
100 10 85
     
 
6
6
106 10 100
210
100 10 85
  
261.9 kgs

8) Calculate lime (90%) pure and soda (95%) pure required for softening of 20000 liters
of water containing following impurities:
 3 2
81 mg per literCa HCO  , 3 42 mg per literMgCO  , 2 4.1 mg per literNaAlO  ,
3.65 mg per literHCl  ,  3 2
82 mg per literCa NO  , 4.5 mg per literNaCl 
(At wt. 40, 1, 12, 16, 24, 23,Ca H C O Mg Na      27, 35.5, 14)Al Cl N   (Dec. 2007)
Solution:
Calculations of 3CaCO equivalent:
Salt/Impurity
Qty.
(ppm)
M.F.
3CaCO equivalent
(ppm)
Requirement
Lime/Soda/Both
 3 2
Ca HCO 81 81 100/162 50 L
3MgCO 42 42100/84 50 L
2NaAlO 4.1 4.1100/82 5 L – S
HCl 3.65 3.65100/36.5 10 L + S
 3 2
Ca NO 82 82 100/164 50 S
NaCl 4.5 - - -
  2 2 2 2 3
2 4
74
Lime [Temporary 2 Temporary Permanent or
100
Ca Mg Mg Fe Al H HCl H SO     
      
3 2 6
Vol. of water 100
] kg
10 % of purity
HCO NaAlO
  
  6
74 100 1
50 2 50 10 5 20000
100 90 10
         
  6
74 100 1
155 20000 kg lime
100 90 10
   
2.549 kg
  2 2 3 2
2 4
106
Soda [Permanent or
100
Ca Mg Al Fe H HCl H SO    
    
3 6
Vol. of water 100
] kg
10 % of purity
HCO
  
  6
106 00 1
50 10 5 20000
100 95 10

     
  6
106 00 1
55 20000
100 95 10

   
0.61 kg

9) Calculate the amount of lime and soda needed to soften 50,000 litres of water
containing the following impurities per litre of water:
   2 3 3 2 42 2
222 mg, 296 mg, 324 mg, 196 mgCaCl Mg NO Ca HCO H SO    and
organic matter = 130 mg. (Dec. 2009)
Solution:
Calculations of 3CaCO equivalents for impurities:
Salt/Impurity
Qty. in
(mg/L)
M.F.
3CaCO equivalent
(ppm)
Requirement
Lime/Soda/Both
2CaCl 222 222 100/111 200 S
 3 2
Mg NO 296 296100/148 200 L + S
 3 2
Ca HCO 324 324100/162 200 L
2 4H SO 196 196100/98 200 L + S
Organic
matter
130 Does not consume Lime or Soda
Calculations of quantity of Lime required for softening 50000 litres of water.
  2 2 2 2 3
2 4
74
[Temporary 2 Temporary Permanent or
100
L Ca Mg Mg Fe Al H HCl H SO     
      
3 2 6
Vol. of water 100
] kg
10 % of purity
HCO NaAlO
   
All in term of their 3CaCO equivalents,
  6
74 50000 100
200 200 200 kg
100 10 100
L     
 
74 5
600 kg
100 100
  22.2 kg
  2 2 3 2
2 4
106
[Permanent or
100
S Ca Mg Al Fe H HCl H SO    
    
3 6
Vol. of water 100
] kg
10 % of purity
HCO
  
All in term of their 3CaCO equivalents
  6
106 50000 100
200 200 200 kg
100 10 100
S     
  6
106 50000 100
600 kg
100 10 100
S   
31.8 kgS 

10) Calculate the amount of lime (85% pure) and soda (95% pure) required to soften one
million litre of water which contains
3 3 2 2 212.5 ppm, 8.4 ppm, 22.2 ppm, 9.5 ppm, 33.0 ppm,CaCO MgCO CaCl MgCl CO    
7.3 ppmHCl  , 3 16.8 ppmNaHCO  . (May 2010, 5 Marks)
Solution: Calculations of 3CaCO equivalents for impurities:
Salt/Impurity
Qty. in
(mg/L)
M.F.
3CaCO
equivalent
(ppm)
Requirement
Lime/Soda/Both
3CaCO 12.5 12.5 100/100 12.5 L
3MgCO 8.4 8.4100/84 10 L
2CaCl 22.2 22.2100/111 20 S
2MgCl 9.5 9.5100/95 10 L + S
2CO 33 33100/44 75 L
HCl 7.3 7.3100/236.5 10 L + S
3NaHCO 16.8 16.8100/284 10
Add in Lime
Subtract from
Soda
Calculation of Lime (85% pure) required for one million litres of water.
  2 2 2 2 3
2 4
74
100
      
3 2 6
Vol. of water 100
] kg
10 % of purity
HCO NaAlO
   
All in terms of their 3CaCO equivalents
 
6
6
74 10 100
12.5 10 10 75 10 10 kg
100 10 85
       
 
74 100
127.5 5 kg
100 85
  
110.95 kg
Quantity of Soda (95%) required for one million litres of water.
  2 2 3 2
2 4 3
106
[Permanent or ]
100
S Ca Mg Al Fe H HCl H SO HCO     
     
6
Vol. of water 100
kg
10 % of purity
 

All in term of their 3CaCO equivalents
 
6
6
106 10 100
20 10 10 10 kg
100 10 95
S      
 
6
6
106 10 100 106 30
30 kg = kg
100 10 95 95
 
     
33.474 kg
11) Calculate the quantity of lime and soda required for softening 50,000 liters of water
containing the following salts per litre:
   3 3 4 4 22 2
8.1 mg, 7.5, 13.6 mg, 12 mg, 2 mg,Ca HCO Mg HCO CaSO MgSO MgCl    
4.7 mgNaCl  . (Dec. 2010, 5 Marks)
Solution:
Calculations of 3CaCO equivalents for impurities
Salt/Impurity
Qty. in
(mg/L)
M.F.
3CaCO equivalent
(ppm)
Requirement
Lime/Soda/Both
 3 2
Ca HCO 8.1 8.1 100/162 5 L (Temporary)
 3 2
Mg HCO 7.5 7.5100/146 5.13 2L (Temporary Mg )
4CaSO 13.6 13.6100/136 10 S(Permanent)
4MgSO 12 12100/120 10 L + S(Permanent)
2MgCl 2 8.1100/95 2.1 L + S(Permanent)
NaCl 4.7 4.7100/58.5 8.0
Does not react with
lime and soda
Lime required for softening:
   3 3 2 42 2
74
Temp. 2 temperature Perm. Perm.
100
Ca HCO Mg HCO MgCl MgSO      
3 6
Vol. of water 100
all in terms of equivalent
10 Purity of lime
CaCO  
  6
74 50000 100
5 2 5.13 2.1 10
100 10 100
        
0.74 27.36 0.05  
1.012 kg
Soda required for softening:

 2 4 4 3
106
Perm. Perm. Perm. all in terms of equivalents
100
MgCl MgSO CaSO CaCO  
6
Vol. of water 100
10 Purity of soda
 
  6
106 50000 100
2.1 10 10
100 10 100
    
1.06 22.1 0.05  
1.1713 kg
12) Calculate the amount of lime (90% pure) and (95% pure) required to soften 50000
litres of same water containing the following impurities in ppm:
 3 22
2 4
2 2 4 2
155 23
6.9 5
18.4 111
Mg HCO MgCl
NaCl H SO
Na S O CaCl
 
 
 
(May 2011, 5 Marks)
Solution:
Salt/Impurity
Qty. in
(mg/L)
M.F.
3CaCO equivalent
(ppm)
Requirement
Lime/Soda/Both
2CaCl 111 111 100/162 100 S
2MgCl 23 23100/95 24.2 L + S
NaCl 6.9 6.9100/58.5 Does not react With Lime and Soda
 3 2
Mg HCO 155 155100/146 106.2 2L
2 2 4Na S O 18.4 18.4100/142 Does not react With lime and Soda
2 4H SO 5 5100/98 5.1 L + S
   2 2 2 2 3
2 4
74
[Temp. 2 Temp. Perm. /
100
      
3 2 36
Vol. of water 100
] kg all in terms of their equivalents.
10 % of Purity
HCO NaAlO CaCO
   
 2 3 2
6
2 4 3
274 50000 100
L kg
100 10 90in terms of equivalent
MgCl Mg HCO
H SO CaCO
    
    
 
 
74
24.2 2 106.2 5.1 0.05 kg
100
     

8.9429 kg of lime
Quantity of Soda:
   2 2 3
2 4 3
106
Perm. , , ,etc. , ,etc.
100
S Ca Mg Al H HCl H SO NaHCO   
     
3 6
Vol. of water 100
all in terms of their equivalents
10 % of Purity
CaCO  
2 2
6
2 4 3
106 50000 100
in terms of equivalent100 10 95
CaCl MgCl
H SO CaCO
  
   
 
 
106
100 24.2 5.1 0.050
100
   
106
129.3 0.050
100
  
6.8529 kg
13) Calculate the quantity of lime (90% pure) and soda (95% pure) required for softening
50000 litres of water containing the following impurities.
 3 2
81 mg /Ca HCO l , 2 95 mg /MgCl l ,
4 68 mg /CaSO l , 2 50 mg /SiO l ,
 3 2
146 mg /Mg HCO l , 2 4 49 mg/H SO l (Dec. 2011, 5 Marks)
Solution:
Salt/Impurity
Qty. in
(mg/L)
M.F.
3CaCO equivalent
(ppm)
Requirement
Lime/Soda/Both
 3 2
Ca HCO 81 100/162 50 L
2MgCl 95 100/95 100 L + S
4CaSO 68 100/136 50 S
2SiO - - - -
 3 2
Mg HCO 146 100/146 100 2L
2 4H SO 49 100/98 50 L + S
Calculation of required of Lime in kg for 50000 litres of water
Lime required

  2 2 2 2 3
2 4
74
100
      
3 2 36
Vol. of water 100
] kg all in term of their equivalents.
10 % of purity
HCO NaAlO CaCO
   
  6
74 50000 100
50 2 100 50 100 kg
100 10 90
       
 
74 5
400 kg 164.4 kg
100 9
  
Quantity of Soda:-
  2 2 3 2
2 4 3
106
Permanent or
100
       
36
Vol. of water 100
kg all in term of their equivalents
10 % of purity
CaCO 
  6
106 50000 100
100 50 50 kg
100 10 95
    
 
106 50
200 kg
100 95
 
111.5 kg
14) Calculate lime(95% pure) and soda (90% pure) required for softening one million
litres of water containing the following constituents:
 3 2
81 mg / ,Ca HCO l  3 2
73 mg /Mg HCO l ,
4 68CaSO  /mg l , 2 95MgCl  /mg l ,
 3 2
4.8Mg NO  /mg l , 2 4 14.7 /H SO mg l
Solution:
Salt/Impurity
Qty. in
(mg/L)
M.F.
3CaCO equivalent
(ppm)
Requirement
Lime/Soda/Both
 3 2
Ca HCO 81 100/162 50 L
 3 2
Mg HCO 73 100/146 50 2L
4CaSO 68 100/136 50 S
2MgCl 95 100/95 100 L + S

 3 2
Mg NO 14.8 100/148 10 L + S
2 4H SO 14.7 100/98 15 L + S
Calculations of requirement of Lime in kg for 50000 litres of water:
Lime required
  2 2 2 2 3
2 4
74
100
      
3 2 36
Vol. of water 100
] kg all in term of their equivalents
10 % of purity
HCO NaAlO CaCO
   
 
6
6
74 10 100
50 2 50 100 10 15 kg
100 10 95
        
 
74 100
275 kg = 214.2 kg
100 95
 
Quantity of Soda:
  2 2 3 2
2 4 3
106
[Permanent or ]
100
     
36
Vol. of water 100
kg all in term of their equivalents.
10 % of purity
CaCO 
 
6
6
106 10 100
50 100 10 15 kg
100 10 90
     
 
106 100
175 kg
100 90
 
206.1 kg
 Problems Based on Zeolite Process:
1) 800 litres of raw water were softened by zeolite. After it got exhausted required 40 lit.
of NaCl solution containing 110 g/lit for its regeneration. Calculate the hardness of
water.
Solution:
a. Concentration of NaCl Solution = 110 g/l
b. 40 lit. of NaCl consumed = 40 × 110 = 4400 gm of NaCl
= 4400 ×
50
58.5
eq. of CaCO3
= 3760.7 gm eq. of CaCO3
c. 800 lit of water = 3760.7 gm of CaCO3 eq.
3 3
3760.7
1 lit of Water 4.7 gm of eq. 4700.85 mg eq.
800
CaCO CaCO   
Hardness of water sample = 4700.85 ppm (Ans.)

2) One lit. of H.W. containing 4.5 gm of CaCl2 was passed through a permuted softener.
Calculate the quantity of NaCl produced in the soft water.
(Dec. 08)
Solution:
a. 2 2Na Ze NaCl CaZe 
111 2 58.5
b. 2NaCl quantity =4.5gm
NaCl Produced =
4.5 2 58.5
4.74
111
 
 gm
 Quantity of NaCl in soft water =4.74gms
3) 15000 lit of H.W was passed through zeolite softener. The exhausted zeolite required
120lit of NaCl having 30 gm/lit. of NaCl. Calculate the hardness of water.
(Dec. 11)
Solution:
a. 1 lit of NaCl solution=30gms of NaCl.
120 lit of NaCl solution= 3
120 30 10  mg of NaCl
 3
3600 10 mg of NaCl 3 50
3600 10
58.5
  
5
30.762 10  mg of 3CaCO eq.
b. 15000 lit of water 5
30.762 10  mg of 3CaCO eq.
1 lit of water
5
30.762 10
205.1
15000
ppm

 
4) An exhausted zeolite softener was regenerated by passing 300 lit. of NaCl solution
having strength of 150 gm/lit of NaCl. If the hardness of the water sample was 480
ppm calculate the total volume of water softened by the softer.(May 12)
Solution:
a. Amount of NaCl present in 300 lit of NaCl= 300 150
45000gm of NaCl
b. Quantity of NaCl in terms of 3CaCO eq. hardness
100
45000
117
 
3
38.46 10  g of 3CaCO eq.
6
38.46 10  mg of 3CaCO eq.
c. It is given that 480 ppm hardness is present in 1 lit of water.

6
38.46 10  ppm hardness is present in
6
38.46 10
480

4
8.01 10  lit of water.
5) The hardness of 100000 lit was removed by passing through zeolite softener. The
softener then required 400 lit NaCl solutions containing 100g/lit NaCl for regeration.
Calculate the hardness of the water sample.(Dec 12)
Solution:
a. 1 lit of NaCl solution=100gms of NaCl.
 400 lit of NaCl solution 3
400 100 10   mg of NaCl
 3
40000 10 mg of NaCl 3 50
40000 10
58.5
  
5
341.88 10  mg of 3CaCO eq.
b. 100000 lit of water 5
341.88 10  mg of 3CaCO eq.
1 lit of water
5
341.88 10
341.88
100000
ppm

 
6) A zeolite softener was completely exhausted & was regenerated by passing 150 lit of
NaCl solution containing 50g/lit of hardness 450 ppm can be softened by this
zeolite?(June 13)
Solution:
7500g of NaCl
b. Quantity of NaCl in terms of 3CaCO eq. hardness
100
7500
117
 
2
64.10 10  g of 3CaCO eq.
6
6.4 10  mg of 3CaCO eq.
c. It is given that 450 ppm hardness is present in 1 lit of water.
6
6
6.4 10
450

4
Thus, the zeolite softener can soften 4
1.424 10 lit of water.
7) An exhausted zeolite softener was regenerated by passing 150g/l of NaCl. The
hardness of water is 600 ppm. Calculate the total volume of water that is softened by
this softener. (Dec. 13)

22500gm
b. Quantity in terms of 3CaCO eq.
100
22500
117
  6
19.23 10  g of 3CaCO mg
c. It is given that 600 ppm is present in 1 lit of water.
6
6
19.23 10
600

3
Thus, the zeolite softener can soften 3
32.05 10 lit of water.
8) The hardness of 50000 lit of sample of water was removed by passing it through a
zeolite softener. The softener required 200 lit NaCl solutions containing 12.5g/lit of
NaCl for regeneration. Calculate the hardness of the sample of water.
Solution:
a. 1 lit of NaCl solution=12.5 gms of NaCl.
200 lit of NaCl solution= 3
200 12.5 10  mg of NaCl
3 50
200 12.5 10
58.5
   
3
2.136 10  mg of 3CaCO solution
b. 50000 lit of sample water 3
2.136 10  mg of 3CaCO solution.
1 lit of sample water
3
2.136 10
42.375
50000
ppm

 
9) One lit of hard water containing 4.5 gm of CaCl2 was passed through a permutit
softener. Calculate the quantity of NaCl produced in the soft water. (Dec 08)
Solution:
a. 2 2Na Ze NaCl CaZe 
111 2 58.5
b. 2CaCl quantity =4.5gm
NaCl Produced =
4.5 2 58.5
4.74
111
 
 gm
 quantity of NaCl in soft water =4.74gm
10) 800 litres of raw water was softening by zeolite softener. After it got exhausted
required 40 liters of NaCl solution. Containing 110 gm per litre of NaCl for its
regeneration. Calculate the hardness of water. (Dec. 2007)

Solution:
Concentration of NaCl solution:
=100 gm/liter NaCl
= 110/1000 ml.
Volume of Raw water = 800liter
To calculate hardness of water
 40 liter of NaCl consumed= 40 110 gm NaCl
= 4400 gm of NaCl consumed for regeneration.
=
4400 50
58.5

equivalent of 3CaCO
= 3760.7 gms equivalent of 3CaCO
Thus,
800 litres of water=3760 gms 3CaCO equivalents
 1 litre of water =
3760.7
800
gms 3CaCO equivalents
= 4.7 gms 3CaCO equivalents
= 4700.85 gms 3CaCO equivalents
Thus, hardness of water sample = 4700.85 ppm.
Hardness of water sample = 4700.85 ppm.
11) One litre of hard water containing 4.5 gm of 2CaCl was passed through a permutit’s
softener. Calculate quantity of NaCl produced in soft water.
Solution:
Volume of hard water = 1 litre
Quantity of 2CaCl = 4.5 gm
To calculate = Quantity of NaCl produced in soft water
2 2Na Ze NaCl CaZe 
111 2 58.5
Now, since 2CaCl quantity = 4.5 gm
NaCl produced =
4.5 2 58.5
111
 
=4.74 gm
Quantity of NaCl is for water = 4.74 gm.

12) A Zeolite softener was completely exhausted and was regenerated by passing 100
litres of sodium chloride solution containing 120 gm per litre of NaCl. How many
litres of sample of water of hardness 500 ppm can be softened by this softener?
Given:
Volume of NaCl solution= 100 litres
Quantity of NaCl =120 gms/litres
Hardness of water= 500 ppm
To calculate = Volume of water softened.
Solution:
100 litres of NaCl used, in regeneration and concentration of NaCl is 120
gm/litres.
Quantity of NaCl consumed = 100 120 = 12000 gms
Reaction:
2 22CaZe NaCl CaCl Na Ze  
Exhausted  2 58.5
Zeolite Brine soln.
3CaCO equivalents = 6100
12000 10
2 58.5
 
  
 
mgs
6
10.26 10  mgs 3CaCO eq.
Now, Let V litres of 500 ppm (i.e. 500mg/lit.) of water consume 6
10.26 10 mgs 3CaCO
in NaCl.
6
500 10.26 10V   
6
10.26 10
500
V
 
   
 
litres
 4
2.052 10  Litres
20520 Litres
20520 litres of water was softened by zeolite bed.
13) The hardness of 3500 litres of water was completely removed by zeolite softener. The
zeolite had required 25 litres of 100 gm/lit of NaCl to regenerate. Calculate the
hardness of the water.
Solution:
Using regeneration reaction
2 22CaZe NaCl Na Ze CaCl  
i.e 2 32NaCl CaCl CaCO 

  32 58.5 gms 111gm 100 gm of CaCO  
Now,
Quantity of NaCl in regeneration,
25 100 
= 2500 gms NaCl
Thus,
 2 58.5 gmsNaCl  100 gm 3CaCO
2500 100
2500 gm
2 58.5
NaCl

 

= 21.36.75 gms of 3CaCO
=213675 mgs of 3CaCO
3500 litres of hard water = 213675 mgs 3CaCO
 1 litres of hard water =
213675
3500
= 61.05 ppm
 Hardness of water sample =61.05 ppm
14) 15000 litres of hard water was passed through a zeolite softener. The exhausted
zeolite required 120 litres of NaCl having 30 g / litre of NaCl. Calculate the hardness
of water.
Solution:
Let the hardness of the water sample be x mg/l
Now, 1 litre of NaCl contains 30g of NaCl =30000 mg NaCl.
 120 litres of NaCl contains 120 30000 3600000  mg of NaCl.
58.5 mg of NaCl =50 mgs of 3CaCO equivalent hardness
50
3600000
58.5
 
= 30, 46,923 mgs of 3CaCO equivalents hardness
But the total quantity of the water sample = 75,000 litre
 15000 lit of water = 3046923 mgs of 3CaCO
 1 lit of water =
3046923
15000
mgs of 3CaCO
= 205 mg/l of 3CaCO
= 205 ppm 3CaCO
Hardness of water sample = 205 ppm

15) An exhausted Zeolite softener was regenerated by passing 300 litres of NaCl solution
having strength of 150 g per litre of NaCl. If the hardness of water sample was 480
ppm, calculate the total volume of water softened by this softener. (May 2012)
Solution:
Here, let the quantity of the water sample be x litre.
Now, the hardness of the water sample is 480 ppm
Hardness = concentration of NaCl  litres of NaCl
= 150000 300
5
450 10  mg of NaCl
5 50
450 10
58.3
   mg of 3CaCO
5
384.6 10  mg
If 1 litre of water = 48 mgs of 3CaCO equivalent hardness
x litres of water 5
384.6 10  of 3CaCO equivalent hardness
5
384.6 10
80128
480
x

   litre
80128 litres of water sample can be softened.
 Type - 5 Problems on Ion Exchange Process:
Note: Only one problem is solved below as the numerical problems on this process are
very much rare.
1) After treating 104 lit. of water by ion exchanger, the cationic resin required 200 lt. of
0.1 N HCl & anionic resin required 200 lit of 0.1 N NaOH solutions find the
hardness of the water sample.
2) Hardness in 104 lit of water = 200 lit. of 0.1 N HCl
= 200 lit of 0.1 N CaCo3 eq.
= 200 0.1 lit. of 1 N CaCo3 eq.
= 20 lit. of 1 N CaCo3 eq.
= 20 50 gm of CaCo3 eq
= 1000 gm of CaCo3 eq.
3) Hardness in 1 lit. of water = 4
1000
10
gm of CaCo3
= 100 mg of CaCo3 eq.
 Hardness of the water sample = 100 mg/lit.

 Unsolved Problems:
1) Calculate the quantity of lime and soda required for softening 5,00,000 litre of
water containing following impurities.
Ca (HCO3)2 = 32.4 ppm; CaSO4 = 40.8 ppm;
Mg (HCO3)2 = 43.8ppm; MgCl = 5.7 ppm;
CaCl2 = 22.2 ppm; MgSO4 = 3.6 ppm;
[Ans.: Lime 32.9 kg., Soda = 31.27 kg]
2) A sample of water has the following impurities per litre :
Mg(HCO3)2 = 58.4 mg; SiO2 = 20 mg;
Ca(HCO3)2 = 32.4 mg; Na2SO4 = 435 mg
Calculate the amount of soda required for treatment of 20,000 litres of water.
(Hint: SiO2 and Na2SO4 do not contribute to hardness)
[Ans.: Lime = 1.48 kg., Soda not required]
3) A sample of water contains the following impurities per litre :
Mg(HCO3)2 = 146 mg; MgSO4 = 12 mg;
CaCl2 = 111 mg; Ca(NO3)2 = 16.4 mg
Lime used is 75% pure while soda used is 85% pure.
Calculate the amount of lime and soda required for softening 10,000 litters of water.
20% excess chemicals are to be used for treatment.
[Ans.: Lime = 2.072 kg., Sods = 1.371 kg.]
4) Calculate the amount of lime (81 % pure) required for treatment of 30,000 litres of
water whose analysis is as follows :
Ca (HCO3)2 = 40.5 mg/lit.; NaCl = 58.0 mg./lit.;
SiO2 = 15 mg/lit.; FeSO4 = 75.9 mg./lit.;
MgCl2 = 19 mg/lit.; CaCl2 = 55.5 mg/lit.;
MgCO3 = 84 mg/lit.;
[Ans. Lime = 8.085 kg., Soda not required]
5) Calculate the amount of lime (80 % pure) and soda (95 % pure) required to soften
and soda (95 % pure) required to soften one million litre of water which contains.
CaCO3 = 125 mg/lit.; MgCl2 = 95 ppm;
O2 = 66 ppm; MgCO3 = 84 ppm;
CaCl2 = 222 pm; HCl = 14.6 ppm;
NaHCO3 = 33.6 ppm
(Hint: Both MgCO3, CaCO3 are to be considered as temporary harness.)
[Ans.: Lime = 606.8 kg., Soda = 335.85 kg.]

6) Calculate the amount of lime and soda required for softening 10,000 litres of water
containing the following dissolved salts.
Ca(HCO3)2 = 162 mg; Mg(HCO3)2 = 73 mg;
CaSO4 = 136 mg; MgCl2 = 95 mg;
NaCl = 585 mg
7) A water sample was found to contain the following dissolved salts;
 3 2
Ca HCO =48.6 ppm; 4MgSO =4.8 ppm;
 3 2
Mg HcO =29.2 ppm ; 2CaCI =33.3ppm ;
Calculated the cost of chemicals required for softening 1 million litres of water by
the lime soda process. Cost of 90 % pure lime is Rs. 62 per kg and that of soda is
Rs. 29 per kg.
[Ans.: Lime = 2.72 kg; Cost = Rs. 4038.3., Soda = 2.35 kg; Cost = Rs. 68.15.]
8) Calculated the quantity of lime and soda for softening 70,000 litres of water
containing the following dissolved salts.
 3 2
Ca HCO =8.1 mg ; 2CaCl =11.1 mg ;
NaCl=66 mg; 4MgSO =18.0 mg;
4MgSO =18.0 mg;  3 2
Mg HCO 7.3 mg ;
[Ans.: Lime = 1.55., Soda = 2.59 kg.]
9) Calculate the amount of lime and soda needed for softening water containing the
following dissolved salts per litre:
 3 2
Ca HCO =16.2 mg ; 2MgCl =9.5 ppm;
 3 2
Mg HCO =73 mg; 4CaSO =136 mg ;
NaCl=58.5 mg [Ans.: Lime = 88.8 kg., Soda = 116.6 kg.]
10) Calculate the amount of lime and soda required to soften 3,00,000 litres of water
which contained the following dissolved salts :
2CO =300 ppm ;  3 2
Mg NO =86.7 ppm ;
2MgCl =95 ppm;  3 2
Mg NO =26.9 ppm ;
4MgSO =8.4 ppm; 2 4H SO =16.9 ppm

11) A sample of hard water was found to contain:
3CaCO =120 mg/lit ; 3MgCO =8.4 mg/lit ;
2SiO =50 mg/lit .; 4CaSO =8.6 mg/lit.;
4MgSO =60 mg/lit.; 2MgCl =19mg/lit.
Calculate the amount of lime and soda required for the softening of 5 million litres
of water, if the purity of lime of 90 % and that of soda is 88 %.
[Ans.: Lime = 803 kg., Soda = 483.75 kg.]
12) Calculate the amount of lime and soda required to soften 3 million litres of water
having following composition.
Free 2CO =25 ppm ; +2
4MgSO as Mg =30 ppm;
 3 32
Mg HCO as MgCO =16 ppm; 2+
4CaSO as Ca =15 ppm;
 3 32
Ca HCO as CaCO =120 ppm .
[Ans.: Lime = 530 kg., Soda = 198.76 kg.]
13) A sample of water contains the following salts expressed in gm/lit.
 3 2
Ca HCO =0.343;  3 2
Mg HCO =0.175;
4CaSO =0.235; 2MgCl =0.147.
Calculate the amount of pure lime and soda for softening 1000 litres of water.
14) A sample of hard water contains following dissolved salts.
 3 2
Ca HCO =48.5 ppm;  3 2
Mg HCO =36.5 ppm;
4MgSO =35 ppm; 4CaSO =43 ppm;
2CaCl =27.75 ppm; NaCl=50 ppm.
15) A water sample on analysis gave the following compositions:
Calcium bicarbonate: 4.86 mg/lit.;
Magnesium chloride: 38 mg/lit.;
Calcium chloride: 333 mg/lit.;
Magnesium sulphate: 54.4 mg/lit.;
Calcium sulphate: 54.4 mg/lit.;
Magnesium bicarbonate: 29.2 mg/lit.
Calculate the carbonate and non-carbonate hardness of the sample and the amount
of lime and soda required to soften 5000 litres, of hard water. The purity of lime is
80 % and that of soda is 75 %. [Ans.: Lime = 0.4 kg., Soda = 2.71 kg.]

16) Sample of water was found to contain the following impurities :
3CaCO =140 mg/lit ; 4CaSO =136 mg/lit ;
3MgCO =8.4 mg/lit ; 4MgSO =60 mg/lit ;
2SiO =20 mg/lit ; 2MgCl =38 mg/lit .
Calculate the quantity of lime (90 % pure) and soda (95 % pure) required for
softening 5 million litres of water. [Ans.: Lime = 904.5 kg., Soda = 892.6 kg.]
17) A water sample on analysis, gave the following data.
2MgCl =95 ppm; 4MgSO =120 ppm;
2 4H SO =49 ppm; 4CaSO =272 ppm ;
2SiO =4 ppm.
Calculate the amount of lime (95 % pure) and Soda (97 %) needed for treating 1
million litres of water. [Ans.: Lime = 194 gm., Soda = 491.75 gm.]
18) A water sample containing following salts,
2MgCl =9.5 ppm; 4MgSO =60.0 ppm;
4CaSO =27.2 ppm; 2 4H SO =4.9 ppm;
2SiO =0.4 ppm.
Was softened by lime soda process. Calculate the quantity of lime and soda (90 %
pure each) required to soften 1 million litres of water.
[Ans.: Lime = 4.1 kg., Soda = 100 kg.]
19) Calculate the amount of lime (90 % pure) and soda (95 % pure) required for
softening of 5
10 litres of hard water containing the following constituent.
 3 2
Ca HCO =162 mg/lit ; HCI=73 mg/lit ;
2MgCl =10 mg/lit ; 2CO =22.0 mg/lit ;
NaCl=117 mg/lit ; 2CaCl =111 mg/lit ;
 3 2
Mg HCO =3 mg/lit ; 4MgSO =30.0 mg/lit .
[Ans.: Lime = 580.99 kg., Soda = 1012.35kg.]
20) A water sample on analysis gave the following data :
2+
Ca =40 ppm; 2-
3CO =15 ppm ;
2+
Mg =72 ppm; +
K =20 ppm ;
2CO =20 ppm .
Calculate the lime (88 % pure) and soda (89 % pure) required to soften 2 million
litres of the water sample. [Ans.: Lime = 580.99 kg., soda = 1012.35 kg.]

 3 2
Ca HCO =50.25 mg ;  3 2
Ca NO =8.2 mg;
2CaCl =22.2 mg ; 3KNO =20.2 mg ;
3MgCO =16.8 mg ; 4FeSO =3.8 mg ;
4MgSO =6.0 mg.
Express the results as ppm, Clarke degree and French degree.
[Ans.: Carbonate hardness = 51 ppm or o o
51 Fr or 3.56 C]
Non-carbonate hardness = 27.5 ppm or o o
27.5 Fr or 1.92 Cl ]
22) Find carbonate and non-carbonate hardness of a sample of water which has the
following impurities per lit.
3CaCO =12.5 mg/lit ; 2MgCl =9.5 mg/lit ;
3MgCO =16.8 mg/lit ; 2CO =22 mg/lit ;
2CaCl =11.1 mg/lit ; 3NaHCO =13.6 mg/lit .
[Ans.: Carbonate hardness = 32.5 ppm, Non-carbonate hardness = 20 ppm]
23) 50 ml of standard hard water containing 1 mg of pure 3CaCO ; per m/consumed 30
ml if EDTA. 50 ml of water sample consumed 35 ml of EDTA solution using
Enochrome Black T indicator, 50 ml of water sample required 28 ml if EDTA using
same indicator. Calculate temporary and permanent hardness.
[Ans.: Temporary hardness = 234 ppm, Permanent hardness = 935.2 ppm.
Total hardness = 1169.2 ppm]
24) 0.25 gm of 3CaCO is dissolved in dilute HCI and diluted to 250 ml. 25 ml of this
solution required 24.0 ml of EDTA using Eriochrome Black T indicator. 50 ml of a
hard water sample required 22.5 ml of the same EDTA. 100 ml of the water after
boiling and filtering required 120 ml of the said EDTA. Calculate the hardness in
the sample (permanent and temporary.]
[Ans.: Total Hardness = 468 ppm, Permanent Hardness = 333
ppm, Temporary Hardness = 135 ppm]
25) What is carbonate and non-carbonate hardness of a sample of water which has the
following impurities per litre
 3 2
Mg HCO =146 mg/lit ; 2MgCl =95 mg/lit ;
NaCl=50 mg/lit ;  3 2
Ca HCO =81 mg/lit ;

2CaCl =111 mg/lit ; 4CaSO =68 mg/lit .
Express the results as ppm, o o
Cl and Fr .
[Ans.: Carbonate hardness o o
= 150 ppm = 15 Fr = 10.5 Cl ,
Non-carbonate hardness o o
= 250 ppm = 25 Fr = 17.5 Cl ]
 3 2
Ca HCO =81 mg/lit ; 2MgCl =95 mg/lit ;
NaCl=38.5 mg/lit ; 2CaCl =22.2 mg/lit ;
 3 2
Mg HCO =58.2 mg/lit ; 2CO =4.4 lit
Express the results as ppm. Clarke’s degree and French degree.
[Ans.: Carbonate hardness o o
= 89.86 ppm = 6.29 Cl = 8.986 Fr,
Non-carbonate hardness o o
= 120 ppm = 8.4 Cl = 12 Fr]
27) 0.1 g of 3CaCO was dissolved in dil. HCI and diluted to 100 ml. 50 ml of this
solution required 40.0 ml of EDTA solution for titration. 50 ml of hard water
sample required 25 ml EDTA solution, 50 ml of same water sample on boiling,
filtering required 20 ml of EDTA. Calculate hardness in ppm.
[Ans.: Temporary hardness = 125 ppm, Permanent hardness = 500 ppm, Total
hardness = 625 ppm]
28) The hardness of 3500 lit. of water was completely removed by zeolite softener. The
zeolite had required 25 lit. of 100 g/lit of 2NaCl to regenerate. Calculate hardness of
water. (May 2011, 5 M)
29) Zeolite softener was completely exhausted & was regenerated by passing 100 lit of
2NaCl solution containing 120 gm/lit. of NaCl. How may lit. of sample of water of
hardness 500 ppm cab be softened by this softer? (May 2010, 5 Marks)
[Ans. 20520 lit.]
 Theory Question:
1) Difference between temporary hardness and permanent hardness.
2) Discuss the process of softening of water by using zeolites. What are the limitations
of this process?
3) Describe the demineralization (Ion-exchange/de-ionisation) process for softening of
hard water. Are there any drawbacks of this method?

 Drinking Water/Municipal Water:
 Drinking water is fit for human consumption. Common specification for drinking
water are:
1) It should be colourless, odourless and good in taste.
2) It should be free from objectionable dissolved gases, other impurities and
turbidity.
3) Total hardness should be less than 500 ppm.
4) It should be free from disease producing micro-organisms.
5) Alkalinity should not be high. pH should be 8.
 The process of destroying the pathogenic bacteria, micro-organisms etc. from water
and making it safe for use is known as disinfection.
 The chemicals added to water for killing the bacteria or other micro-organisms is
known as Disinfectants.
 Disinfection:
 Municipal water treatment does not include removing the dissolved salts present in
the water.
 After using various techniques such as sedimentation, coagulation & flirtation,
suspend impurities are removed from water. But it still contains very small
percentage of pathogenic bacteria and therefore it is unfit for drinking.
 The process of destroying bacteria, micro-organisms etc. from the water is known as
Disinfection. The chemicals added to water for killing the bacteria are known as
Disinfectants. They mainly include Bleaching Powder, Chlorine and Ozone.
 These are briefly explained below.
 Common Methods of Water Disinfection:
1)By adding bleaching powder:
Bleaching powder 2CaOCl (calcium hypochlorite) is widely used as bleaching agent.
About 10 kg of bleaching powder per 1000 kilometres of water is mixed and water
allowed to stank undisturbed for several hours. Chemical action produces
hypochlorous acid and nascent oxygen, both are powerful germicide.
 2 2 22
CaOCl +H O Ca OH +Cl
2 2Cl +H O HOCl + HCL
Hypochlorous acid
2HOCl+H O HOCl+HCl Nascent oxygen

Drawbacks:
i. IntroduceCa in water, thereby making it harder.
ii. Chlorine content of water increases.
iii. It gives bad taste and smell to treated water.
2)By Chlorination:
a. Chlorine (gas or conc. Solution form) produces hypochlorous acid which is
powerful germicide
2 2Cl + H O HOCl + HCl
Hypochlorous acid
Bacteria + HOCl Bacteria are killed
b. The death of microorganisms results from chemical reaction of HOCl with the
enzymes in the cells of the organism. Since enzyme is essential for the metabolic
process, death of micro-organism results due to inactivation of enzyme.
c. Chlorinator is a large lower which contain number of baffle plates. Water and
proper quality of concentrated chlorine solution are introduced at its top. During
their passage through the lower, they get thoroughly mixed. About 0.3 to 0.5 ppm
of chlorine is sufficient for filtered water. Disinfected water is taken out from the
outlet at the bottom of the chlorinator.
 Factors affecting on Efficiency of Chlorine:
i. Temperature of water:
Death rate of micro-organisms is directly proportional to temp. Thus
efficiency increases with rise in temp.
ii. Time of Contact:
The death rate is maximum in the beginning and decreases with time.

iii. pH if water:
At lower pH values (between 5 to 6.5) a small contact period is required to kill
same percentage of Organisms.
 Advantages:
i. Effective, economics, and most ideal disinfectant.
ii. It requires very little space for storage.
iii. It can be used at low as well as high temp.
iv. It introduces no salt impurities in the treated water.
 Disadvantages:
i. Excess of chlorine imparts unpleasant taste and odour to water.
ii. It is more effective below 6.5 pH and less effective at higher pH values.
3)Disinfection by Ozone:
a. Ozone is an excellent disinfectant, which produces by passing silent elective
discharge through cold and dry discharge.
silent
2 3electric discharge
3 2O O
 3 2
Nascent Oxygen
O O O 
b. Ozone is highly unstable, hence breaks down and liberate nascent oxygen.
c. Nascent oxygen is very powerful oxidising agent and kills all the bacterias as well
as oxidises the organic matter present in water.
d.Ozone is injected into the water and then this mixture is allowed to come in
contact in sterilizing tank.
e. The disinfected water is removed from the top. The contact period is about 10.15
minutes, and Ozone dose strength is 2-3 ppm.
 Advantages:
i. During disinfection it also removes colour, odour and taste without giving any
residue.

ii. Its excess is not harmful, as 3O is unstable and decompose into Oxygen.
 Disadvantages:
This method is expensive hence, not employed for disinfection of municipal water
supply.
4)Electrodialysis:
The process of removing common salts (NaCl) from the water is known as
Desalination.
In Electrodialysis, ions present in the saline water migrate towards respective
electrode through ion selective membrane by passing direct current.
a. In this method desalination is carried out by electro-dialysis method.
b. When direct current is passed through saline water the sodium ions ( Na
) start
moving towards negative pole (cathode); while chloride ions (Cl) start moving
towards positive pole (anode) through membranes.
c. As a result, the concentration of brine decreases in the central compartment,
while increases in two side compartments.
d. Desalinated water (pure water) is removed from the central compartment from
time to time, while conc. brine is replaced by sea water.
e. For more efficient separation, ion-selective membranes are employed.
Cation Selective membrane:
Cation selective membrane is permeable to cations only. They possess functional
group like 3RSO
, RCOO
, etc….

Anion Selective membrane:
Anion selective membrane is permeable to anions only. They possess functional
group like 4R N
.
 Advantages:
i. It is most compact unit.
ii. Installation and operation of plant is economical.
iii. If electricity is easily available, it is best suited.
 Reverse Osmosis and Ultrafiltration:
 Reverse Osmosis:
 Normal Osmosis:
In Normal Osmosis Process, solvent flows from low concentration solution to
high concentration solution through the semipermeable membrane, until
difference in water level creates a sufficient pressure to counteract the original
flow. The difference in levels represents osmotic pressure of the solution.
In reverse osmosis, external pressure is applied on higher concentration solution
slightly higher than its osmotic pressure. Hydrostatic Pressure in excess of
osmotic pressure is applied on concentration side, the solvent flows reverses
 In R.O. solvent is forced to move from conc. Side to dil. Side.
 In R.O. Method, pure solvent (water) is separated from its contaminates, rather
than removing contaminants from the water. This membrane filtration sometime
also called Super-Filtration or Hyper-Filtration.

 Method:
1)Pressure is applied to the sea water (impure water) to force its pure water out
through the out through the semi-permeable membrane, leaving behind the
dissolved solids.
2)The membrane consists of very thin films of cellulose acetate, affixed to either
side of a polymers are used.
 Advantages:
1)This method removes ionic as well as non-ionic, colloidal and high molecular
weight impurities.
2)It removes colloidal silica, which is not removed by demineralization
3)The life time of membrane is quite high about 2 years.
4)Low capital cost simplicity, low operating cost and high reliability.
 Applications of Reverse Osmosis:
1)Desalination of sea water
2)For drinking water purification, as it provides pathogen free water
3)Waste-water recovery in the food & beverages processing industries.
4)Production of ultra-pure water for use in semiconductor industry
5)Milk can be concentrated prior to cheese making at the farm level.
 Ultrafiltration:
 Ultrafiltration is a variety of membrane filtration in which hydrostatic pressure
forces a liquid against a semipermeable membrane.
 Suspended solids and solutes of high molecular weight are retained, while
water and low molecular weight solutes pass through the membrane.

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