WATER and ITS TREATMENT.pdf

water and its treatment

2. Water Technology:Hardness of water.
Determination of hardness of water by EDTA
titration. Numericalsbased on hardness of water
and EDTA method.
Softening Methods:Hot and Cold Lime-Soda
Method, Zeolite Method and Ion-Exchange
Method. Numerical based on lime-soda and
zeolite method.
Drinking water purification:Removalof
microorganisms, by adding bleaching powder,
chlorination (no breakpoint chlorination),
Ozonization, Desalination by Reverse Osmosis,
Ultrafiltration.
2

3. 3

4. 4
Sources
Surface Water Under ground water
Rain water
River water
Lake water
Sea water
Spring & Well water

5. River water –dissolved minerals
Cl
-
, SO
4
2-
, HCO
3
-
of
Na+, Mg
2+
, Ca
2+
and Fe
2+
suspended impurities-Organic matter,
sand, rock
composition is NOT constant –depend on contact
with soil.

Lake water: High in organic and less in minerals.
composition is constant.
5

6. Rain water–pure form
dissolved organic and inorganic particles
and dissolved industrial gases CO
2, NO
2,SO
2
etc
Underground water-free from organic
impurities due to filtering action of the soil
Sea water–very impure; too saline for
industrial use except cooling
6

7. Suspended impurities
inorganic (clay, sand) organic (oil, plant,
and animal matter)
Colloidal impurities-finely divided silica and
clay
Dissolved impurities –salts and gases
Microorganisms –bacteria, fungi and algae
7

8. Hard water: do not produce lather with soap
solution readily. It may be temporary or permanent
and due to the presence of Ca, Mg, Fe & Al ions
Soft water: produces lather readily with soap
solution. It contains little amount of Ca, Mg, Fe and
Al ions or no such ions
8

9. Hardness prevents the lathering of soap.
due to the presence of salts of Ca, Mg, Al, Fe
and Mndissolved in it.
Soap–Na or K salts of long chain fatty acids
C
17H
35COOH
2C
17H
35COONa + CaCl
2→ (C
17H
35COO)
2Ca↓ + 2NaCl
2C
17H
35COONa + MgSO
4→ (C
17H
35COO)
2Mg↓+ Na
2SO
4
Hardness of water
9

10. Temporary Hardness-caused by
dissolved bicarbonates of Ca and Mg
Also known as ‘alkaline or carbonate
hardness’
Permanent Hardness–dissolved Cl-,
NO3-and SO
4
2-
of Ca, Mg, Fe and Al etc
10

11. Due to dissolved bicarbonates of Ca & Mg in water
It is so called since it can be easily removed simply by
boiling
Heat decomposes bicarbonates of Ca & Mg as follows
Also known as ‘alkaline or carbonate hardness’
Determined by titration with H
2SO
4/HClusing
Phenolphthalein & methyl orange as indicator
11

12. CaCl
2, MgCl
2, CaSO
4, MgSO
4, FeSO
4, Al
2(SO
4)
3

Cannot be destroyed on boiling the water

Also known as non-carbonateor non alkaline
hardness

non alkalinehardness = Total hardness –
alkaline hardness
12

13. Temporary Hardness Permanent Hardness
Itiscausedduetothepresenceof
bicarbonatesandcarbonatesof
Ca
2+
,Mg
2+
.SaltslikeCa(HCO
3)
2,
CaCO
3,Mg(HCO
3)
2,MgCO
3,etc.
Itiscausedduetothepresenceof
Cl
-
,SO
4
2-
,nitratesofCa
2+
,Mg
2+
otherthancarbonatesand
bicarbonates.SaltslikeCaCl
2,
CaSO
4,Ca(NO
3)
2,MgCl
2,MgSO
4,
Mg(NO
3)
2.
Itcanberemovedbyboiling. Itcannotberemovedbyboiling,but
needschemicaltreatment.
Itisalsoknownascarbonateor
alkalinehardness.
Itisalsoknownasnon-carbonateor
non-alkalinehardness.
Temporaryhardnessleadsto
formationofloosedeposits
(sludge)ofcarbonatesand
hydroxidesofCa
2+
,Mg
2+
,ifused
inboilers
Permanenthardnessleadsto
formationofharddeposits(scales)
ifusedintheboilers
13

14. Domestic use:
Washing : no lather formation ,wastage of soap
Bathing: no lather formation. Also the resulting pptsticks
on body
Cooking: due to dissolved salts boiling point of water is
elevated, causing wastage of time & fuel.
Drinking: bad effect on metabolic system. Calcium
oxalate stones may develop in urinary tracts, if used
regularly.
14

15. Industrial diasadvantages:
Textiles: loss of soap during washing of yarn, & fabrics.
Pptof Ca& Mg sticks on fabric. Fe, Mnsalts leave
colored spots on fabrics
Sugar: crystallization of sugar is affected due to
presence of sulfate ions.
Dyeing: Dissolved Ca, Mg, Fe salts react with dye
forming undesirable pptsgiving poor shades of color on
the fabric.
Paper:Ca and Mg salts react with chemicals to provide
smooth finish to paper. Iron salts affect colourof the
paper.
Pharmaceuticals: If used in drug, injections, ointments–
results in undesirable products in them causing ill health
15

16. Hardness is expressed as equivalent amount
(equivalents) of CaCO
3
Reason: Molar mass is exactly 100, and is the most insoluble salt that can
be precipitated in water treatment.
Equivalents of
CaCO
3 =
( mass of hardness producing substance in mg/L) x100 / (eq.wtof
substancex2)
units –mg/L = ppm
parts of CaCO3 equivalents in hardness in 10
6
parts of water
16

17. Eq. wt= Molar mass/ no of charge on ion
CaCO
3 MM/2
NaCl MM/1
AlCl
3 MM/3
Al
2(SO
4)
3 MM/6
17

18. A water sample contains 408 mg of CaSO
4
per liter. Calculate the hardness in terms of
CaCO
3equivalents
Hardness = mg/L of CaSO
4x 100/MM(CaSO
4)
= 408 mg/L x 100/136
= 300 mg/L = 300 ppm
18

19. How many milligrams of MgCO
3dissolved per
liter gives 84 ppm of hardness?
Hardness = mg/L of MgCO
3x
100/MM(MgCO
3)
84 ppm = ppm of MgCO
3x 100/84
ppm of MgCO
3= 84 ppm x (84/100)
= 70.56 ppm
= 71 mg/L
19

20. Calculation of hardness caused by each ion.
Na+ -20 mg/L Ca
2+
-15 mg/L
Mg
2+
-10 mg/L Sr
2+
-2 mg/L
Al
3+
-0.3 mg/L
Equvalentsof
CaCO
3 =
( mass of hardness producing substance in mg/L) x100 / (eq.wtof
substancex2)
Cation Eq.wt Hardness
Ca
2+
40.0/2 37.5
Mg
2+
24.4/2 41.0
Sr
2+
87.6/2 2.3
Al
3+
27.0/3 1.7
Total hardness = 82.5 ppm
Example 3
20

21. A water containing the following salts per litre: Ca(HCO
3)
2= 8.1 mg; Mg(HCO
3)
2=
7.5 mg;CaSO
4= 13.6 mg; MgSO
4= 12 mg; MgCl
2 = 2 mg; NaCl= 4.7 mg.
Calculate temporary, permanent and total hardness of water sample.
(At. wt. Ca =40, Mg =24, H =1, C = 12, O = 16, S = 32)
Ans:Calculation of CaCO
3equivalents:
Constituent Multiplication factor CaCO
3equivalent(ppm) Types of Hardness
Ca(HCO
3)
2= 8.1 mg/L 100/162 8.1 x 100/162 = 5 mg/L Temp
Mg(HCO
3)
2= 7.5 mg/L 100/146 7.5 x 100/146 = 5.14 mg/L Temp
CaSO
4= 13.6 mg/L 100/136 13.6 x 100/136 = 10 mg/L Perma
MgSO
4= 12 mg/L 100/120 12 x 100/120 = 10 mg/L Perma
MgCl
2= 2 mg/L 100/95 2 x 100/95 = 2.11 mg/L Perma
Temporary hardness =Mg(HCO3)2+Ca(HCO3)2
=5+5.14 =10.14 ppm
Permanent hardness= CaSO4 + MgSO4 + MgCl2
10+ 10+ 2.11=22.11 ppm
Total hardness= permanent hardness+ temporary hardness
=10.14 +22.11 =32.25 ppm
21

22. 22
Classificationhardness in
mg/L
hardness in
ppm
soft 0-60 <60
Moderately hard 61-120 60-120
Hard 121-180 120-180
Very hard ≥ 181 > 181
Hardness of water classifications

23. Step1: EDTA solution is standardized with standard hard
water
Step2: EDTA is used to titrate unknown hard water sample
23
Ethylenediaminetetraaceticacid (EDTA) is a reagent
that forms EDTA-metal complexes with many metal
ions (but not with alkali metal ions such as Na
+
and
K
+
). In alkaline conditions (pH›9) it forms stable
complexeswith the alkaline earth metal ions Ca
2+
and Mg
2+
. The EDTA reagent can be used to
measure the total quantity of dissolved Ca
2+
and
Mg
2+
ions in a water sample. Thus the total
hardness of a water sample can be estimated by
titration with a standard solution of EDTA.

24. Determining total amount of Ca
2+
& Mg
2+
ions in water
Titrating sample with standard EDTA solution using an
organic dye indicator EBT (EriochromeBlack T)
EDTA is a weak acid with the structure given below
(a) Structure of ethylene
diaminetetraaceticacid, EDTA
(b) Structure of tetracarboxylate
ion, [EDTA]
4–
, formed by
dissociation of EDTA
24

25. EriochromeBlack-T[EBT]istheindicatorusedin
thedeterminationofhardnessbycomplexometric
titrationwithEDTA.
andEDTA[disodiumsaltofethylenediamine
tetraaceticacid]formahexadentateligand
Initsprotonatedform,EriochromeBlackTis
blue.Itturnsredwhenitformsacomplexwith
calcium,magnesium,orothermetalions.
25

26. EBT + Mg
2+
/Ca
2+
(aq) pH 10 [Ca-EBT] / [Mg-EBT]
wine red or purple complex
[Ca-EBT] / [Mg-EBT] + EDTA[Ca-EDTA] / [Mg-EDTA]
+ + EBT
blue or bluish green
EDTA
EDTA-metal complex
Eriochrome
Black T
26

27. Standardization of EDTA solution
Pipette 50 ml of std hard water(known concentration) in conical
flask.
Add 5 ml buffer soln(pH=10), 3-4 drops of EBT indicator.
Solution is titrated against EDTA till colourchanges from wine red
to deep blue (V1 ml)
For total hardness : 50 ml hard water sample titrated
against EDTA (V2 ml)
Permanent hardness : 50 ml hard water sample is boiled
for 10-15 min, filtered, diluted with D/W to make up to
50 ml & titrated against EDTA (V3 ml)
Temporary Hardness= Total hardness –Permanent
hardness
27

28. 0.5gCaCO
3wasdissolvedinminimumamountof10%HCland
thesolutionwasdilutedupto500mLwithdistilledwater.50mLof
thissolutionrequired48mLofEDTAsolutionforreachingtheend
pointofthetitration.50mLofanunknownwatersamplerequired
15mLofthesameEDTAsolutionforthetitrationandafterboiling
andfilteringthesame50mLofunknownwatersamplerequired
10mLofEDTAsolutionfortitration.Calculatethetotal,
permanent,andtemporaryhardnessoftheunknownwatersample
28
Ans. Total hardness = 312.5 ppm; Temporary hardness = 104.2
ppm; Permanent hardness = 208.3 ppm

29. •Determining the strength of 1 ml of standard hard water (SHW):
0.5 g CaCO
3= 500 mg CaCO
3= 1 mg CaCO
3/ml
500 ml 500 ml
•Standardization of EDTA:
48 ml EDTA = 50 ml SHW
= 50 mg CaCO
3
1 ml EDTA = (50/48) mg CaCO
3= 1.042 mg CaCO
3
•Determining total hardness of water sample:
50 ml unknown = 15 ml EDTA
= (15 x 1.042) = 15.63 mg CaCO
3
1000 ml unknown = (15.63 x 1000)/ 50 = 312.6 mg CaCO
3/ L = 312. 6 ppm
•Determining permanent hardness of water sample:
50 ml boiled unknown = 10 ml EDTA
= (10 x 1.042) = 10.42 mg CaCO
3
1000 ml boiled unknown = (10.42 x 1000)/50 = 208.4 mg CaCO
3/ L = 208.4 ppm
•Determining temporary hardness of water sample:
Total hardness = Temporary hardness + Permanent hardness
Temporary hardness = Total hardness -Permanent hardness
= 312.6 -208.4 = 104.2 ppm
29

30. 0.25gCaCO
3wasdissolvedinminimumamountof10%HCland
thesolutionwasdilutedupto200mLwithdistilledwater.25mLof
thissolutionrequired35mLofEDTAsolutionforreachingtheend
pointofthetitration.25mLofanunknownhardwatersample
required30mLofthesameEDTAsolution,toreachtheendpoint
ofthetitration.Calculatethetotalhardnessoftheunknownwater
sample
30

31. •Determining the strength of 1 ml of standard hard water (SHW):
0.25 g CaCO
3= 250 mg CaCO
3= 1.25 mg CaCO
3/ml
200 ml 200 ml
•Standardization of EDTA:
35 ml EDTA = 25 ml SHW
= (25 x 1.25) mg CaCO
3 = 31.25 mg CaCO
3
1 ml EDTA = (31.25/35) mg CaCO
3= 0.89 mg CaCO
3
•Determining total hardness of water sample:
25 ml unknown = 30 ml EDTA
= (30 x 0.89) = 26.78 mg CaCO
3
1000 ml unknown = (26.7 x 1000)/ 25 = 1068 mg CaCO
3/ L = 1068 ppm
31

32. 0.6gCaCO
3wasdissolvedinminimumamountof10%HCland
thesolutionwasdilutedupto500mLwithdistilledwater.50mLof
thissolutionrequired32mLofEDTAsolutionforreachingtheend
pointofthetitration.100mLofanunknownwatersamplerequired
14mLofthesameEDTAsolutionforthetitration,andafterboiling
andfilteringthesame100mLofunknownwatersamplerequired
8.5mLofEDTAsolutionfortitration.Calculatethetotal,
permanent,andtemporaryhardnessoftheunknownwatersample
32

33. •Determining the strength of 1 ml of standard hard water (SHW):
0.6 g CaCO
3= 600 mg CaCO
3= 1.2 mg CaCO
3/ml
500 ml 500 ml
•Standardization of EDTA:
32 ml EDTA = 50 ml SHW
= (50 x 1.2) mg CaCO
3 = 60 mg CaCO
3
1 ml EDTA = (60/32) mg CaCO
3= 1.875 mg CaCO
3
•Determining total hardness of water sample:
100 ml unknown = 14 ml EDTA
= (14 x 1.875) = 26.25 mg CaCO
3
1000 ml unknown = (26.25 x 1000)/ 100 = 262.5 mg CaCO
3/ L = 262.5 ppm
•Determining permanent hardness of water sample:
100 ml boiled unknown = 8.5 ml EDTA
= (8.5 x 1.875) = 15.9375 mg CaCO
3
1000 ml boiled unknown = (15.9375 x 1000)/100 = 159.37 mg CaCO
3/ L
= 159.37 ppm
•Determining temporary hardness of water sample:
Total hardness = Temporary hardness + Permanent hardness
Temporary hardness = Total hardness -Permanent hardness
= 262.5 -159.37 = 103.13 ppm
33

34. 50 ml of a standard hard water containing 1.2 mg of pure CaCO
3per mL
consumed 40 mL of EDTA. 100 ml of water sample consumed 40 ml of
same EDTA solution using EBT indicator. Calculate the total hardness of water
sample in ppm or mg/L.
Ans:
1 ml of standard hard water contain 1.2 mg of CaCO
3
50 mL of standard hard water consumed = 40 ml of EDTA
50x1.2 mg of CaCO3= 40 mL of EDTA
1 mL of EDTA= 50x1.2/40 mg of CaCO3
60/40= 1. 5 mg of CaCO3
Now 100 mL of hard water sample= 40 mL of EDTA
100 mL hard water= 40x 1.5 mg of CaCO3
1000 ml of given hard water= 40x 1.5x1000/100 mg of CaCO3/L
Total hardness of water=600 ppm or mg/L
34

35. External:using soft water
Internal: is a corrective treatment for the removal
of some impurities left out in the external
treatment
Internal treatment involves addition of certain
chemicals to the boiler water. The added
chemicals either precipitate the scale forming
impurities as a sludge (which can be removed by
a blow-down operation) OR the added chemical
convert the scale forming impurities into
compounds that stay in the dissolved form in the
boiler thereby causing no harm
35

36. 36
Boilers are used in several industries and powerhouses for
steam generation
If the boiler feed water contain the excess of impurities then it
causes the following defects:
(a) Scale & sludge formation
(b) Corrosion
(c) Caustic Embrittlement
(d) Priming & foaming
Continuous evaporation of water in boilers results in
concentration of the dissolved salts

37. 37
Eventuallythesaltsprecipitate
Thesoft,loose&slimypptissludge
Collected in colder partsof the boiler, where the flow
of water is slow & in area of bends.
Substances which have greater solubility in hot water
than in cold water form sludge e.g. MgCO
3, MgCl
2,
CaCl
2, MgSO
4

38. 38
(a) Disadvantages:
i) it is a bad conductor of heatso a lot of energy is
wasted.
ii) chocking & poor water circulation
iii) Excessive sludge obstruct pipeline flow, plug
opening, gauge-glass connection, thereby disturbing
the normal functioning of the boiler

39. 3
9
(b)RemovalofSludge:
Sludgebeinglooseandslimy,canbeeasilyremoved
bymeansofawirebrush
(c) Prevention:
i) using soft water
ii) blow down operation i.e. drawing off a portion of
the concentrated water and replacing with fresh water

40. 4
0
When the precipitated salts deposit as a hard layer on the
inner walls of the boiler, they are called scales
It is hard & adhering coating,which deposits & sticks
firmly
It is formed as follows-
(i) Decomposition of bicarbonate
Ca(HCO)3 → CaCO3 ↓ + H2O + CO2
Though scale due to CaCO3 is soft, it is the main cause of
scale formation in low pressure boilers. But in high
pressure boilers CaCO3 reacts with water under hot
conditions to give Ca(OH)
2 (sludge).

41. 4
1
(ii) Deposition of CaSO4:
Scales could also be formed by the deposition of CaSO
4
that precipitates from the hot water in the boiler, and gets
deposited on the inner walls of the boiler
It is soluble in cold water but less soluble in hot water &
almost insoluble in super heated water.
(iii) Magnesium salts: dissolved Mg salts undergo
hydrolysis (at prevailing high temp. inside the boiler)
forming Mg(OH)
2ppt ;a soft type of scale
(iv) Silica: deposits as CaSiO
3&MgSiO
3& deposits very
firmly on inner side & difficult to remove.

42. External Treatment ( Softening methods): The
process of removing hardness causing salts from
water is called as softening of water. Mainly the
following three methods are used for softening:
• Lime soda process
• Zeolite or permutitprocess
• Ion exchange or deionization or
demineralization process
42

43. Principle:In this method hard water is treated
with calculated amounts of slaked lime
[Ca(OH)
2]. Soda ash [Na
2CO
3] in reaction tanks,
so as to convert hardness producing chemicals
into insoluble compounds which are then
removed by settling & filtration.
• Usually 10% extra chemical is added for
better results.
43

44. Limeremoves hardness due to temporary
Calcium & all types of magnesium hardness
Lime react with free acids, Fe, Al salts, CO
2
gas & produce extra calcium hardness.
Sodaremoves all the soluble permanent
hardness due to calcium salts (i.e. that which
is present originally as well as that which is
introduced during removal of Mg2+ , Fe2+ ,
Al3+ , HCl, H
2SO
4, CO
2etc. by lime )
44

45. Lime removes free acids :
2HCl + Ca(OH)
2 CaCl
2+ 2H
2O
H
2SO
4+ Ca(OH)
2 CaSO
4+ 2H
2O
Lime removes temporary hardness:
Ca(HCO3)
2+ Ca(OH)
2 2CaCO
3+ 2H
2O
Mg(HCO3)
2+ 2Ca(OH)
2 2CaCO
3+Mg(OH)
2+ 2H
2O
Lime removes permanent hardness;
MgCl
2/MgSO
4+ Ca(OH)
2 Mg(OH)
2+ CaCl
2/CaSO
4
Lime removes dissolved iron and aluminiumsalts:
Al
2(SO
4)
3+ 3Ca(OH)
2 2Al(OH)
3+ 3CaSO
4
FeSO
4+ Ca(OH)
2 Fe(OH)
2+ CaSO
4
45

46. Removes all soluble permanent hardness due
to Ca salts
CaCl
2+ Na
2CO
3→ CaCO
3+ 2NaCl
CaSO
4+ Na
2CO
3→ CaCO
3+ Na
2SO
4
The method discussed is called Cold Lime
Soda Process
46

47. Amount of Lime required = 74/100 [temp Ca
hardness + (2 ×temp. Mg hardness) + perm
Mg hardness + Salts like Fe, Al + CO2 +
Acids + HCO3
-
-NaAlO2 ] ×volume of water
×100/ % purity
Amount of Soda = 106/100 [perm Ca
hardness + perm Mg hardness + Salts like Fe,
Al + Acids -HCO3
-
] ×volume of water ×
100/ % purity
47

48. The chemical reactions taking place during
lime soda treatment are slow & pptof CaCO
3
& Mg(OH)
2 are fine & produce super-
saturated solution. As a result after
deposition occurs in pipes, boiler tubes etc
clogging the valves & leading to corrosion.
To avoid this;
Thorough mixing of chemicals & hard water.
Sufficient time allowed to complete reactions
48

49. Accelerators i.e. Substances that bring down
the fine particles of precipitates e.g. Activated
charcoal
Coagulants i.e. Substances which help in the
formation of coarse precipitates are added
e.g. alum
Proper sedimentation chamber for
precipitation to settle, before filtration being
carried out
Process is carried out at room temperature &
at temperature between 50 °C-150 °C
49

50. When the chemicals are added at room
temperature. At this temperature the precipitates
are finely divided & do not settle easily, nor they
can be filtered. It is necessary therefore to add
coagulants like alum, sodium aluminate etc
NaAlO
2+ 2H
2O → NaOH+ Al(OH)
3
This process provides water containing a residual
hardness of 50-to 60 ppm.
50

51. 51

52. When the chemicals are added at higher
temperature 80
0
C to 150
0
C the process is
known as Hot process. At higher temperature
reactions are fast
precipitation is more complete
settling rate & filtration rates are increased
So less amount of chemicals are needed.
Hot-Lime soda process produces water of
comparatively lower residual hardness of 15-
30 ppm
52

53. 53

54. Advantage:
Very economical
pH is increased , there by corrosion is reduced
to a certain extent iron & manganese are also
reduced.
Disadvantages:
Hardness after Cold process is about 50 ppm &
30 ppm by Hot process. These values are high
for pressure boilers.
Careful operation & skilled supervision is
required
54

55. Cold lime soda process Hot lime soda process
1. It is carried out at room temperature
(25-30°C)
1. It is carried out at high temperature (85-
100°C)
2. It is a slow process 2. It is a rapid process
3. Use of coagulant is a necessary3. No coagulant required
4. Filtration is not easy 4. Filtration is easy as viscosity of water is low
5. Residual hardness is 50-60 ppm 5. Residual hardness is 15—30 ppm
6. Dissolved gases are not removed6. Dissolved gases are removed
7. It has low softening capacity7. It has high softening capacity
55

56. Calculate the quantity of lime and soda required for softening 50,000 litres
of water containing the following salts per litre: Ca(HCO
3)
2= 8.1 mg;
Mg(HCO
3)
2= 7.5 mg;CaSO
4= 13.6 mg; MgSO
4= 12 mg; MgCl
2 = 2 mg;
NaCl= 4.7 mg.
Calculation of CaCO
3equivalents:
Constituent Multiplication factor CaCO
3equivalent Requirementlime
L/soda S
Ca(HCO
3)
2= 8.1
mg/L
100/162 8.1 x 100/162 = 5
mg/L
L
Mg(HCO
3)
2= 7.5
mg/L
100/146 7.5 x 100/146 =
5.14 mg/L
L
CaSO
4= 13.6
mg/L
100/136 13.6 x 100/136 = 10
mg/L
S
MgSO
4= 12
mg/L
100/120 12 x 100/120 = 10
mg/L
L/S
MgCl
2= 2 mg/L 100/95 2 x 100/95 = 2.11
mg/L
L/S
56

57. Soda required for softening
= 106/100 [CaSO
4+ MgSO
4+ MgCl
2] x vol. of water
= 106/100[ 10 + 10 + 2.11]mg/L x 50,000 L
= 1171.83 mg = 1.17 Kg
Lime required for 50,000 L water
= 74/100[ Ca(HCO
3)
2+ 2 Mg(HCO
3)
2+ MgSO
4+ MgCl
2] x vol. of water
= 74/100[ 5 + (2 x 5.14) + 10 + 2.11] mg/L x 50,000 L
= 10,13,430 mg = 1013.43 g = 1.0134 Kg
Numericals lime and soda process
57

58. Hydrated sodium aluminosilicate
Na
2O.Al
2O
3.x SiO
2.yH
2O
Zeolites is capable of exchanging reversibly
their sodium ions for hardness producing
ions in water.
Natural Zeolites : are non-porous more
durable & are derived from green sands.
Synthetic Zeolites : porous & possess gel
structure & have higher exchange capacity
per unit weight.
58

59. Theory:When hard water is passed over a
bed of sodium zeolite , Ca2+ , Mg2+ ions are
taken up by the zeolite simultaneously
releasing equivalent Na+ ions in exchange for
them.
CaCl
2+ Na
2Ze → CaZe+ 2NaCl
MgSO4 + Na
2Ze → MgZe+ Na
2SO4
59

60. When Zeolite is completely converted into
calcium & magnesium Zeolites, it ceases to
soften water i.e. it gets exhausted. It is
generated by treating with 10% brine
solution.
CaZe+ 2NaCl → Na
2Ze + CaCl
2
MgZe+ 2NaCl → Na
2Ze + MgCl
2
60

61. Process:Hard water enters from top at a
specified rate & passes over a bed of sodium
zeolite kept in a cylinder. Softened water is
collected at the bottom of cylinder & is taken
out from time to time.
61

62. Turbid water should not be admitted,
otherwise it will block the pores of zeolite &
make them inactive.
Any colored Fe ions must be removed earlier
because, it is difficult to regenerate it from
iron zeolite.
Mineral acid present in water must be
neutralized earlier with soda otherwise that
may destroy zeolite bed.
62

63. Water of about 10 ppm hardness is produced.
Process automatically adjusts itself for
different hardness of incoming water.
Requires less skill in maintenance as well as
operation.
63

64. Treated water contains more sodium salts
The method only replaces Ca
2+
& Mg
2+
ions
by Na
+
ions, but leaves all (HCO
3
-
& CO
3
--
) in
soft water.
Such soft water containing NaHCO3 ,
Na2CO3 etc, when used in boilers , NaHCO3
decomposes to give CO2 which cause boiler
corrosion & Na2CO3 hydrolyses to NaOH
causing Caustic Embrittlement.
64

65. Zeolite process Lime soda process
1.This process produces water of very low hardness ~10
ppm.
This process produces water of hardness of 15-60 ppm
depending on whether it is a hot or cold process.
2.The cost of the plant and zeolite is higher. Hence, capital
cost is higher.
The capital cost is lower.
3.The exhausted zeolite bed can be regenerated with brine
which is very cheap. Hence, operating cost is less.
The chemicals required are consumed in this process
thus operating cost is higher.
4.The plant is compact and occupies less space. The size of
plant depends on the hardness of water being treated
The plant occupies more space. The size of the plant
depends on the amount of water being handled
5.Cannot be used for hot water, acidic water and water
having turbidity and suspended impurities.
The process is free from such limitations.
6.This process can operate under pressure and can be
designed for fully automatic operations.
This process cannot be operated under pressure.
7.This process does not involve cumbersome operations like
settling, coagulation and filtration.
This process involves all the problems associated with
settling, coagulation and filtration.
8.The water contains larger amounts of sodium salts and
greater percentage of dissolved salts.
Treated water contains lesser percentage of dissolved
solids and lesser quantities of sodium salts.
9.This process adjusts itself to water of different hardness.Reagent doses must be adjusted for waters of different
hardness.
10.Salts causing temporary hardness are converted into
NaHCO
3which will be present in the softened water. The
insoluble CaCO
3and Mg(OH)
2in water creates problems
when used as feed water in boilers.
Temporary hardness is completely removed.
65

66. 66
A zeolite softener was completely exhausted and was regenerated by passing
100 litresof sodium chloride solution containing 120 gram per litreof NaCl.
How many litresof sample of water of hardness 500 ppm can be softened
by this softner?
Volume of NaClsolution = 100 litre
Quantity of NaCl= 120 gms/litre
Hardness of water = 500 ppm
To calculate = volume of water softened
100 litreof NaClused in regeneration and concentration of NaClis 120 gm/litre
Quantity of NaClconsumed = 100x 120
= 12000 gms
Reaction : CaZe+ 2NaCl CaCl
2 + Na
2Ze
i.e. 2NaCl= CaCl2=CaCO3 /2 (58.5) g= 111g=100g of CaCO3
Exhausted zeolite = 2x 58.5 brine solution
Hence CaCO3 equivalents = (12000 x 100/2 x 58.5) =10256.41 g/L of CaCO3 eq.
= 10.26 x 10
6
mgs CaCO3 eq.
Now let V litreof 500 ppm (i.e. 500mgs/litr) of water consume 10.26 x 10
6
mgs of
CaCO
3eq. of NaCl
V x 500 = 10.26 x 10
6
V= (10.26 x10
6
/ 500) litre
=20520 litresof water softened by zeolite bed.

67. Reversible exchange of ions of same charge between mobile
liquid phase & an insoluble solid (stationary phase)
Ion-exchanger is an insoluble material liberating counter ions
by electrolytic dissociation
Cationexchanger: High mol.wt, cross-linked polymer
containing sulfonic (-SO
3H), carboxylic (-COOH), or
phenolic (-OH) as a part of resin and equivalent amount of
cations
H-R (resin) + Na
+
Na-R (resin) + H
+
2NaR (resin) + Ca
2+
CaR
2(resin) + 2Na
+
67

68. Cationexchange Anion exchange
68

69. Anion exchanger : Is a polymer containing quaternary
ammonium groups (-N
+
R
2) containing equivalent amount of
anions, OH
-
, Cl
-
etc
2ROH + SO
4
2-
R
2(SO
4) + 2OH
-
(resin) (solution) (resin) (solution)
Water from which all the cations& anions are removed is called
demineralizedor deionizedwater.
Water is first passed to cationexchanger in acid form. All
cationswill be exchanged for H
+
ions
Next, water coming from cationexchanger is passed into
anion exchanger in basic form and anions are exchanged for
OH
--
69

70. Hard water is first passed through cation
exchange bed which removes all cationslike
Ca2+ , Mg2+ , Na+ & release H+ ions. Thus
Cl
-
, SO
4
--
, OH
-
are converted into
corresponding acids HCl, H
2SO
4, & H
2CO
3
After this the acidic hard water is passed
through an anion exchange bed which
removes all anions present in water &
equivalent amount of OH-ions is released
from it to form water.
70

71. H+ & OH-ions combine to give ion-free
water molecules called De-ionized or De-
mineralized water.
The water is finally freed from dissolved
gases by passing it through a de-gasifier
the water obtained by this process is very
near to distill water.
71
Major Impurity=> CaSO
4
2RSO
3H + Ca
2+
(RSO
3)
2Ca + 2H
+
2RNH
3OH + SO
4
2-
(RNH
3)
2SO
4+ 2OH
-
H
+
+ OH
-
H
2O

72. Example: If water contains 10 grains per gallon
of hardness, how many gallons of water would
the resin remove? The tank holds 500 cubic feet
of resin with capability of removing 45,000
grains per gallon per cubic foot.
Gallons = cubic feet x grains per cubic foot/grains per gallon
 = 500 cubic feet x 45,000 grains/cubic foot /10
grains per gallon
 = 22,500,000 grains/ 10 grains/gallon
 = 2,250,000 gallons before requiring regeneration
72

73. 73

74. When ion exchange capacities are lost (
resins are exhausted) the supply of water is
stopped. The exhausted cationexchanger i.e.
regenerated by passing dilute HClor H
2SO
4.
RCaor RMg+ 2HCl → 2RH + CaCl
2 or
MgCl
2
The exchange bed is washed with de-ionized
water & washings( contained Ca
2+
, Mg
2+
, Cl
-
,
SO
4
--
) is passed to sink or drain.
74

75. The exhausted anion exchanger is
regenerated by treating it with a dilute NaOH
solution.
R’Cl
2+ 2NaOH → R’(OH)
2+ 2NaCl
The exchanged bed is washed with de
ionized water & washings( containing NaCl,
Na
2SO
4) is passed to sink.
The regenerated ion-exchange resins are
used again.
75

76. Advantages:
(i) The process can be used to soften highly
acidic or alkaline water,
(ii) Water of low hardness(0-2 ppm) thus very
good for high pressure boilers.
Disadvantages:
(i) The equipment is costly
(ii) Turbidity should be below 10 ppm
76

77. 77
S.
No.
Zeolite process Ion exchange Process
1. This process can produce softened water with
residual hardness 0-15 ppm.
This process can produce softened water with residual
hardness 0-2 ppm.
2. The resultant water is not suitable for high
pressure boilers. Water can be used for low or
medium pressure boilers.
The resultant water is suitable for high pressure boilers.
Water can be used for low or medium pressure boilers
3. Zeolite softer is cheap, hence capital cost is
low.
Cationand anion exchanger beds are expensive, hence
capital cost is more.
4. Softening plant is compact and required less
space.
Softening plant is not compact and required more space.
5 The process is not suitablefor acidic water
processing.
The process is useful for acidic as well for alkaline water
processing.
6. Soft water obtained not suitable for boilers due
to the presence of (NaHCO3)2 and NaOH,
therbycausing caustic embrittlement in boiler.
Soft water does not cause caustic embrittlement in boiler,
hence suitable for boilers as it is free from Na+ oins.

78. Colorless and odorless; good in taste
Turbidity should be less than 10 ppm
No objectionable dissolved gases like H
2S
or minerals such as Pb, As, Cr, Mnsalts.
Alkalinity should not be high; pH 7.0 –8.5
Total ions hardness less than 500 ppm
Free of harmful microorganisms.
Cl
-
, F
-
, and SO
4
2–
less than 250, 15 and
250 ppm, respectively
78

79. Drinking water purification consists of:
A). Removal of suspended impurities: which
involves
1) Screening, 2) Sedimentation, 3) Filtration
B). Removal of micro-organisms: which involves
following method used for destroying the
disease causing micro-organisms
1) Boiling, 2) Addition of bleaching powder,
3) Chlorination 4) Addition of Chloramine
(ClNH
2), 5) Ozonization
79

80. A). Removal of suspended impurities:
1. Screening: The impure water is passed
through screens having a large number of holes
where the bigger sized floating material is
trapped and removed
2. Sedimentation:
Water with sediments is allowed to settle in
large tanks (approx. 5m deep) to allow the
suspended particle to settle under gravity.
Retention period is about 2 –6 hours
80

81. 3. Filtration: In this process the supernatant
water is passed through a bed of fine sand and
other proper sized granular material. During
this process most of the bacteria and micro-
organisms are removed
81

82. B). Removal of micro-organisms:
Water on removal of suspended impurities, has
a small percentage of pathogenic bacteria which
must be removed before the water can be used
for drinking purposes. The process of
destroying or removing of the pathogenic
disinfection, and the chemicals or substances
used for the disinfection are called as
disinfectants
82

83. B.Removal of micro-organisms:
Methods used for disinfection include:
1. Chlorination
2. Ozonization
3. UV treatment
C. Advanced treatments
1. Reverse osmosis
2. Ultrafiltration
83

84. 84
•Chlorine Gas (Cl
2)
•Chlorine Dioxide (ClO
2)
•Bleach (sodium hypochlorite)
•Chloramine T (ammonia +chlorine)
ChlorinationProcess:Methodisusedtokill
certainbacteriaandothermicrobesintap
wateraschlorineishighlytoxic.Inparticular,
chlorinationisusedtopreventthespreadof
waterbornediseasessuchascholera,dysentery,
typhoidetc.

85. Commonly used disinfectant in water
used directly as a gas or conc. solution.
It produces HOCl, a powerful germicide
 0.3-0.5 ppm chlorine is sufficient
85
1. Bleaching powder (CaOCl
2)
CaOCl
2+H
2O → Ca(OH)
2+ HCl+ HOCl
Enzymes of microorganism get deactivated
by HOCl
Excess imparts bad taste and smell
Not stable during storage
Introduces Ca to water and thus increases
hardness
2. Chlorination

86. ClO
2in some respects is chemically similar to
chlorine or hypochlorite, the familiar
household bleach. However, ClO2 reactions
with other organic molecules are relatively
limited as compared to chlorine. When ClO2 is
added to a system –whether a wound or a
water supply –more of the biocide is available
for disinfection and not consumed by other
materials.
86
Chlorine Dioxide (ClO
2)

87. (Ozonizationmethod of disinfection)
Ozone is an excellent disinfectant and is produced
by passing an electric discharge through cold and
dry oxygen
Ozone if highly unstable and breaks down
liberating nascent oxygen
Nascent oxygen is a powerful oxidizing agent and
oxidizes any organic matter, as well as destroys the
bacteria present in H
2O
3O
2 2O
3(Oxygen) (Ozone)
(electric discharge)
O3 O2 + [ O ]
(Nascent oxygen) (Ozone)
87

88. B. Removal of micro-organisms: (Ozonization
method of disinfection)
Ozonizationprocess:
Ozone is injected into water, and the two are
allowed to come into contact for 10 –15
minutes in a sterilizing tank
The amount of ozone injected is 2 –3 ppm
The disinfected water is removed from the
top
88

89. O
3→ O
2+ O
oxygen atom is a
powerful oxidizing
agent.
2 –3 ppm is injected
10 –15 min contact
time
Expensive method
89

90. Osmosis is a naturally
occurring phenomenon
and one of the most
important processes in
nature. It is a process
where a weaker saline
solution will tend to
migrate to a strong
saline solution. Examples
of osmosis are when
plant roots absorb water
from the soil and our
kidneys absorb water
from our blood.
Reverse Osmosis,
commonly referred to as
RO, is a process where
you demineralize or
deionize water by
pushing it under
pressure through a semi-
permeable Reverse
Osmosis Membrane
90

91. 91

92. Desalination of Water (Reverse Osmosis) Advanced
methods
Reverse osmosis is the method used for
desalination (i.e. removal of NaCl) of water
Principle of osmosis: when two solutions of unequal
concentrations are separated by a semi permeable
membrane (that selectively allows the diffusion of
only the solvent molecules, but not the solute
particles viz. molecules, ions, etc.) the flow of
solvent occurs from the dilute solution side to the
concentrated solution side due to osmosis
92

93. If however, hydrostatic pressure in excess of osmotic
pressure is applied on the concentrated solution side,
the solvent flow reverses i.e. the solvent is now
forced to flow from the concentrated solution side to
the dilute solution side across the semi permeable
membrane. This process is termed as reverse
osmosis
Thus, in reverse osmosis, instead of removing
contaminants from water, the pure water is separated
from impure water and the application of hydrostatic
pressure (in excess of osmotic pressure) makes the
water to flow from the impure side (more
concentrated) to the pure side (less concentrated)
93

94. Such type of membrane filtration is often
termed as superfiltrationor hyperfiltration
Process: In this process a pressure of the
order of 15 –20 kg/cm
2
is applied to the
impure water (to be treated) to force pure
water out through the semi permeable
membrane. This leaves behind the dissolved
solute particles (both ionic as well as
molecular)
The semi permeable membrane consists of a
very thin film of cellulose acetate, affixed to
either side of a perforated tube
94

95. The process removes both ionic as well as non-
ionic, colloidal and high molecular weight
organic polymers
It removes colloidal silica which is not removed
by demineralization
The maintenance cost is low and only involves
the replacement of the semi permeable
membrane which takes a few minutes thereby
providing uninterrupted pure water supply.
At the same time the membrane cost is
reasonably low and lasts for approx. 2 years
95

96. ♦Ultrafiltration like reverse osmosis is a
cross flow separation process
♦It is fundamentally similar to reverse
osmosis except for the size of the molecules
it retains
♦Thus, low molecular weight organics, and
ions such as Na+, Ca2+, Mg2+ and Cl─ pass
through the membrane while high molecular
weight species (macromolecules or
supermolecules: 10
3
–10
6
Daltons) are
removed
96

97. ♦The pore size of the membrane used in
ultrafiltration is in the range of 0.1 –0.001
microns
♦The type and the amount of species left over in
the permeate depend on
the characteristic of the membrane,
the operating conditions and the
quality of the feed
97

98. 98

99. 99
UFcanbeusedfortheremovalofparticulatesandmacromoleculesfrom
rawwatertoproducepotablewater.Ithasbeenusedtoeitherreplaceexisting
secondary(coagulation,flocculation,sedimentation)andtertiaryfiltration
(sandfiltrationandchlorination)systemsemployedinwatertreatmentplants
orasstandalonesystemsinisolatedregionswithgrowingpopulations.
Whentreatingwaterwithhighsuspendedsolids,UFisoftenintegratedinto
theprocess,utilizingprimary(screening,flotation,filtration)andsomesecondary
treatmentsaspre-treatmentstages.UFprocessesarecurrentlypreferred
overtraditionaltreatmentmethodsforthefollowingreasons:
Nochemicalsrequired(asidefromcleaning)
Constantproductqualityregardlessoffeedquality
Compactplantsize
Capableofexceedingregulatorystandardsofwaterquality,
achieving90-100%pathogenremoval
In many cases UF is used for pre filtration inreverse
osmosis(RO) plants to protect the RO membranes

100. 1 mg/litre ≡ 1 ppm
1
0
Cl ≡ 14.3 ppm
1
0
Fr ≡ 10 ppm
10
0CaCO3 equivalent=
weight of hardness causing salt
equivalent weight of hardness causing salt
×equivalent weight of CaCO3 Strength of EDTA=
strength of SHW x volume of SHW
volume of EDTA
Soda = 106/100 [perm Ca hardness + perm Mg hardness + Salts like Fe, Al + Acids - HCO3
-
]
× Vol of water x 100/%pure ………. In kg
10
6
Lime = 74/100 [Temp Ca
2+
+ (2 × Temp Mg
2+
) + perm Mg
2+
+ Fe
2+
, Al
3+
+ H
+
(HCl/H2SO4) +
HCO3
-

- NaAlO2] ×Vol of water x 100/%pure ………. In kg
10
6
Hardness=
strength of EDTA x volume of EDTA
volume of water sample
×1000=????????????/??????

101. Thank you
10
1