Applied Chemistry - I

CHAPTER NO. Water

1

Q. 1. Distinguish between BOD and COD. [Dec’12(3m), May’13(3m)]
Ans.
COD
Sr. BOD
no. The amount of oxygen required by
organic matter in a sample of water for its
1 BOD of water is a measure of amount of oxidation by strong oxidizing agent is
known as chemical oxygen demand.
oxygen
COD determination is faster as compared
Required for the biological oxidation of to BOD as it requires only 3 hours.
Determination of COD is done by
organic refluxing the sample with a known excess
Matter under aerobic conditions at 200C of K2Cr2O7 and 50% H2SO4 and then
titrating the unreacted K2Cr2O7 solution
and for against FAS solution.
COD =(V2−V1)×VN×8×1000ppm
A period of 5 days. Where,
V1 = Volume of FAS for sample titration
2 BOD determination requires 5 days for V2 = Volume of FAS for blank titration
V = Volume of sample taken for the test
completion of experiment. N = Normality of FAS solution
COD is more than BOD as it is the
3 Determination of BOD is based on the measure of oxygen required for the
oxidation of biologically oxidisable and
determination of dissolved oxygen prior to biologically inert organic matter.

and following a 5 days period at 200C.

4 BOD =(DO)1−(DO)2 ppm

x

Where ,

(DO)1 = DO in blank titration

(DO)2 = DO of the sample after

X = total volume of sample diluted
volume to which it was

5 BOD is normally less than COD.

Q. 2. Differentiate between temporary and permanent hardness.
Ans.

Sr. Temporary hardness Permanent hardness
no.
Permanent Hardness Permanent hardness
1 Temporary hardness is hardness that can is hardness (mineral content) that cannot
be removed by boiling.
be removed by boiling or by the addition When this is the case, it is usually caused
by the presence of calcium and
of lime (calcium hydroxide). magnesium sulphates and/or chlorides in
the water, which become more soluble as
2 It is caused by a combination of calcium the temperature rises.

ions and bicarbonate ions in the water.

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2 Applied Chemistry - I

3 Heating water will remove hardness as Hardness of the water can be easily

long as the lime scale that precipitates out removed using a water softener, or ion

is removed. exchange column.

4 Heat destroys the hardness Heat has no effect on permanent hardness

Q. 3. Discuss zeolite process with the help of diagram, chemical reactions and advantages.
Ans. [Dec’12(5m), Dec’14(5m), May 15(5m)]
i.
ii. Zeolite permutit process operates alternatively as the softening run and the regeneration.
iii. During softening process the hard water from the top enter at a specified rate and passes over a bed
iv. of of sodium zeolite kept in a cylinder.
Softened water containing sodium salts is
v. collected at the bottom of the cylinder and is
taken out from time to time.
vi. The cations Ca+2 and Mg+2 are retained in
zeolite bed and soft water rich in Na+ salts is
etc collected. After some time the zeolite bed gets
vii. exhausted.
viii. When zeolite bed gets exhausted, the softening
run is discontinued and regeneration is started.
During regeneration process, the following
three operations are carried out.
(a) Back washing
(b) Salting (or brining) and
(c) Rinsing to get regenerated bed for reuse.
Reactions taking place during softening
process are:

CaCl2 + Na2Ze → CaZe + 2NaCl
CaSO4 + Na2Ze → CaZe + Na2SO4

Reactions of Regeneration of zeolite-permutit bed are:
CaZe + 2NaCl → Na2Ze

MgZe + 2NaCl → Na2Ze
Advantages of zeolite process
(a) Water of about less than 15 ppm of hardness is produced.
(b) The process automatically adjusts itself for different hardness of incoming water.
(c) Water obtained is very clear.
(d) The equipment is compact.
(e) Less skilled workers are required.

Q. 4. Explain activated sludge process with the help of flow sheet diagram. [Dec’12(6m)]
Ans.
i. It is based on biological oxidation of soluble
organic material by bacteria.
ii.
(CH2O) + O2 → Biomass + CO2
iii. Organic N → NH4+ + NO3
Organic P → PO43-
Activated sludge process involves extensive
aeration of the sewage water and the process of
aerobic oxidation being enhanced by the
addition of part of sludge from previous
oxidation process into sewage water.
This added sludge is known as activated sludge

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as it contains large no. of aerobic bacteria and other micro-organisms.
iv. Mixture of sewage water and activated sludge is sent to aeration tank where it is aerated and agitated

for several hours.
v. During this process organic matters are oxidized.
vi. After completing the process, the effluent is sent to sedimentation tank where sludge is deposited and

water free from organic matter is drained off.
vii. A part of settled sludge is sent back for settling fresh batch of sewage. This water after chlorination

is discharged into lakes, streams, rivers and seas
viii. About 50% of biodegradable carbon present in waste water is converted into biomass and remaining

50% into CO2. Sludge can be used as a fertilizer.

Q. 5. Explain demineralization of water by ion exchange method. [May 13(5m), May 14(5m)]
Ans.

i. Ion exchange resins are used for softening of water. They are organic polymers with long chains with
cross links and having functional groups through which various ions are exchanged. The resins are
porous and insoluble in water.

ii. The principle of ion exchange method is based on ability of ion exchange resins to exchange their
functional groups like H+ with cations like Ca++, Mg++, Na+ and (OH)-with all anions present. The
process of softening in the ion exchange involves passing a raw water through cationic exchange
resin and followed by passing it through the anion exchange resin

iii. As the raw water passes through the cation exchange resin we get
R-H3 + MgCl2 → R-Mg + 2HCl

Thus sulphates, chlorides, bicarbonates get converted into suphuric, hydrochloric and carbonic
Acids. The acidic water emerging from the cation exchange bed is passed through the anion
exchange bed where the anions are exchanged for the OH ions of resin.

R-(OH)2 + H2SO4 → R-SO4 + 2H2O
R-(OH)2 + 2HCl → R-Cl2 + 2H2O
iv. The water emerging from the anion exchange bed is free from both cations and anions and hence is
completely demineralized. However water may contain some dissolved gases. Therefore it is passed
through degasifiers where it is heated and the escaping gases are removed by applying vacuum.
v. The cation and anion exchange resin are regenerated when they get saturated and can be used over
and again. The softened water has very low hardness of less than 2ppm and thus can be safely used
for higher pressure boilers.

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4 Applied Chemistry - I

Q. 6. With the help of chemical equations explain the lime soda process. [May 13(6m),Dec’13(5m)]

Ans. In this method hardwater is treated with calculated amount of slaked lime and soda ash in reaction

tanks, so as to convert hardness producing chemicals into insoluble compounds which are then removed by

settling and filtration. If this process is carried out at room temperature it is called Cold lime soda process
and if carried out at higher temperature(50-600C) is known as hot lime soda process.

Treatment with lime:
1. Removal of temporary hardness of Ca+2
Ca(HCO3)2 + Ca(OH)2 → 2CaCO3 + 2H2O
2. Removal of Temporary hardness of Mg+2
Mg(HCO3)2 + 2Ca(OH)2 → Mg(OH)2 + 2CaCO3 + 2H2O
3. Removal of Permanent hardness of Mg+2
MgCl2 + Ca(OH)2 → Mg(OH)2 + CaCl2
MgSO4 + Ca(OH)2 → Mg(OH)2 + CaSO4
Mg(NO3)2 + Ca(OH)2 → Mg(OH)2 + Ca(NO3)2

4. Removal of CO2
CO2 + Ca(OH)2 → CaCO3 + H2O

5. Removal of acids
2HCl + Ca(OH)2 → CaCl2 + 2H2O
H2SO4 + Ca(OH)2 → CaSO4 + 2H2O

6. Removal of bicarbonates of Na+ and K+
2NaHCO3 + Ca(OH)2 → CaCO3 + Na2CO3 + 2H2O
2KHCO3 + Ca(OH)2 → 2CaCO3 + K2CO3 + 2H2O

7. Removal of alums
FeSO4 + Ca(OH)2 → CaSO4 + Fe(OH)2
Al2(SO4)3 + 3Ca(OH)2 → 3CaSO4 + 2Al(OH)3
NaAlO2 + 2H2O → Al(OH)3 + NaOH
2NaOH + CaCl2 → Ca(OH)2 + 2NaCl

Treatment with soda:
1. Removal of Permanent hardness of Ca+2
CaCl2 + Na2CO3 → CaCO3 + 2NaCl
CaSO4 + Na2CO3 → CaCO3 + Na2SO4
Ca(NO3)2 + Na2CO3 → CaCO3 + 2NaNO3

Q. 7. What happens when temporary hard water is boiled? Give equation to explain. [Dec’13(3m)]
Ans.
i. Temporary hardness is defined as the hardness due to carbonates, hydroxides and bicarbonates of
calcium, magnesium and other metals.
ii. When hard water is boiled, temporary hardness is mostly removed by mere boiling of water, where
bicarbonates are decomposed producing insoluble carbonates or hydroxides.
iii.
For eg. Ca(HCO3)2 → CaCO3↓ + H2O + CO2 ↑
The Ca/Mg carbonates or hydroxides thus formed being almost insoluble are deposited as a crust or
scale at the bottom of the vessel, while CO2 escapes out in air.

Q. 8. Write short notes on : [Dec’13(6m), May 14(3m)]

Ans.

A. Reverse osmosis

i. The reversal of solvent flow, from higher concentration solution to lower concentration solution

through a semi permeable membrane, by applying an external pressure slightly higher than the

osmotic pressure of higher concentration solution, is known as reverse osmosis.

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ii. In osmosis, the solvent flows from low
concentration solution to higher concentration
solution, through the semi permeable membrane,
until difference in water levels creates a sufficient
pressure to counteract the original flow. The
differences in levels represent osmotic pressure
solution.

iii. In the reverse osmosis, we apply external pressure
on the higher concentration solution slightly higher
than its osmotic pressure.

iv. The flow of solvent is from higher concentration
solution to lower concentration solution in reverse
osmosis. Thus in reverse osmosis we separate water
from its contaminants rather than contaminants from
water.

B. Electro dialysis
i. It is based on the principle that the ions
present in saline water migrate towards
respective electrodes through ion
selective membranes. These membranes
are thin, rigid and permeable to either
cation or anion. The anode is placed near
anion selective membrane while the
cathode is placed near the cation
selective membrane.
ii. Under the pressure of about5-6 kg/m3
saline water is applied perpendicular to the direction of the water flow. Fixed positive charges
inside the membrane repel positively charged ions (Na+) and permit negatively charged ions
(Cl-) to pass through.

iii. Similarly the fixed negative charges inside the membrane repel chloride ions but permit
sodium ions. Thus the water in one compartment is deprived of salts while the salt
concentration in the adjacent compartment is increased. Thus alternate streams of pure water
and brackish water are obtained.

C. Ultrafiltration
i. Some of the toxic chlorinated organisms are removed by filtering industrial waste activated
charcoal as follows: Alderine, Diedrin, Endrin, DDT etc are removed nearly 99%.
ii. Synthetic organic ion exchange resins are very useful for removal of industrial waste
chemicals; styrene-divinyl-benzene copolymer can remove chlorinated pesticides by
absorption at the surface. Ionic dies from textile mill wastewater can be eliminated by using
cation and anionic exchange resin

iii. In ultrafiltration, the solution is pushed through a membrane which contains pores of size 2-
10000 nm whereby big molecules are retained and the effluent that passes are free of the big
molecules. In reverse osmosis the membrane pores are smaller 0.04-600nm. Both of the
techniques are extensively used for purification of industrial waste water in metal, textile,
paper and food industries.

Q. 9. Give the principle of estimation of hardness of water using EDTA method.
[May 14(3m), Dec’14]
Ans.
i. The di-sodium salt of Ethylene diamine tetra acetic acid (EDTA) forms complexes with Ca+2 and
Mg+2, as well as with many other metal cations in aqueous solution.

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6 Applied Chemistry - I

ii. Thus, in a hard water sample, the total hardness can be determined by titrating Ca+2 and Mg+2 present
in an aliquot of the sample with Na EDTA solution, using NH4Cl. NH4OH buffer solution of pH 10
and Erichrome Black T as the metal indicator.

iii. At pH 10, EBT indicator forms wine red coloured unstable complex with Ca+2/Mg+2 ions in hard
water .

iv. This complex is broken by EDTA solution during titration, giving stable complex with ions; and
releasing EBT indicator solution which is blue in colour. Hence the colour change is from wine red
to blue. Thus noting the colour change, the point of equivalence can be trapped and hardness of
water can be determined by using this method.

Q. 10. Define and explain significance of BOD and COD. [May 14(3m)]

Ans.

i. BOD of water is a measure of amount of oxygen Required for the biological oxidation of organic

Matter under aerobic conditions at 200C and for a period of five days.

Significance of BOD:

i. The higher the BOD of a sample the higher the amount of decomposable organic matter in the

sample and higher the pollution of the sample.

ii. Thus BOD gives an idea about the extent of pollution at any time in the sewage sample and helps

in pollution control.

iii. The amount of oxygen required by organic matter in a sample of water for its oxidation by strong

oxidizing agent is known as chemical oxygen demand

Significance of COD:

i. It helps in designing the water treatment plant.

ii. It helps in deciding the disposal of domestic effluents in various types of water streams.

Q. 11. Discuss the following treatment methods for municipal water.
Ans.
i. Bleaching powder
Bleaching powder reacts with water as,
CaOCl2 + H2O → Ca(OH)2 +Cl2
Cl2 + H2O → HCl + HOCl
(Hypochlorus acid)
HOCl → HCl + [O]
Nascent oxygen
In this reaction above, the HOCl deactivates the enzymes present in the cells of micro organisms.
Thus the metabolic activity of micro-organisms gets affected, thereby making the micro organisms
inactive. Finally it dies
ii. *Chlorination
The reaction of chlorine is as,
Cl2 + H2O → HCl + HOCl
With very small concentration of chlorine. HOCl thus formed destroys the bacteria, as it is a
powerful germicide.
The apparatus used for chlorination is called as chlorinator.
Application of Cl2 can be as gas or concentrated solution of water.
The raw water and the concentrated chlorine solution (0.3-0.5ppm
Cl2) is passed through the chlorinator. The water and Cl2 solution get
mixed thoroughly in the chlorinator due to baffle plates, and sterilized
water is collected using outlet.
The disinfection is governed by nature of substances used, and
sterilization by chlorination is governed by temperature of (i) water
(ii) time of contact (iii) pH of water.
iii. *By ozone
One common method for disinfecting wastewater is ozonation ( also

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known as ozone disinfection). Ozone is an unstable gas that can destroy bacteria and viruses. It is
formed when oxygen molecules (O2) collide with oxygen atoms to produce ozone (O3).

3O2 → 2O3 (on silent electric discharge)
Ozone is generated by an electrical discharge through dry air or pure oxygen and is generated onsite
because it decomposes to elemental oxygen in a short amount of time. After generation, ozone is fed
into a down-flow contact chamber containing the wastewater to be disinfected. Form the bottom of
the contact chamber, ozone is diffused into fine bubbles that mix with the downward flowing
wastewater.

Q. 1. Calculate temporary, permanent and total hardness of water sample containing

Mg(HCO3)2 =7.3 ppm, Ca(HCO3)2 = 16.2 ppm. MgCl2 = 9.5 ppm , CaSO4 = 13.6 ppm.
[Dec’12(3m)]

Ans.

Salt Quantity in Multiplication CaCO3 Types of hardness
Mg(HCO3)2 ppm or factor equivalent
mg/L Carbonate or temporary
7.3 × 100 05 hardness
7.3 146

Ca(HCO3)2 16.2 16.2 × 100 10 Carbonate or temporary
MgCl2 9.5 162 10 hardness
CaSO4 13.6 10 Non-Carbonate or
9.5 × 100 temporary hardness
95 Non-carbonate or temporary
hardness
13.6 × 100
136

∴ Temporary hardness = Mg (HCO3)2 + Ca(HCO3)2 = 05 + 10 = 15 ppm

Also, Permanent hardness = MgCl2 + CaSO4 = 10 + 10 = 20ppm

Total hardness = temporary + permanent = 15 + 20 = 35 ppm

Q. 2. Calculate amount of lime (90%) pure and soda (98% pure) for the treatment of 1 million litres
Ans. of Water containing Ca(HCO3)2 =8.1ppm, CaCl2 =33.3ppm, HCO3- = 91.5ppm, MgCl2 =

38ppm, Mg(HCO3)2 =14.6ppm. The coagulants Al2(SO4)2 was added at the rate of 17.1ppm.
[Dec’12(6m)]

Salt/ impurity Quantity (ppm) Multiplication CaCO3 Requirement of
Factor equivalents lime(L) or
soda(S)
ppm

Ca(HCO3)2 8.1 100/162 5 L

CaCl2 33.3 100/111 30 S
HCO3- 100 75 +L -S

91.5
61 × 2

MgCl2 38 100/95 40 L+S

Mg(HCO3)2 14.6 100/146 10 2L

Al2(SO4)3 17.1 100/114 15 +L+S

L = 74 [temp Ca+2 + 2 × temp Mg+2 +permanent ] vol.of water × 100
106 of purity
100 %

= 74 [5 + 2×10 + 40 + 30 + 15 + 75] ×110066 × 100 = 74 [185] × 100 = 152 kg
90 100 90
100

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8 Applied Chemistry - I

S = 106 [permanent (Ca+2 + Mg+2 + Al+3 + Fe+2 + H+ - HCO3-] × vol.of water × % 100 kg in terms of
100 106 of purity

CaCO3

= 106 [30 + 40 + 15 – 75] ×110066 × 100 kg = 10.8 kg
100 98

Q. 3. The hardness of 1,00,000 litres of water completely removed by passing through zeolite
Ans.
softener, the softener than requires 400 litres of NaCl solution containing 100g/litre NaCl for

regeneration. Calculate the hardness of the water sample. [Dec’12(4m)]

Let the hardness of the water sample be x mg/L

Now 1 litre of NaCl solution contains 100 g. NaCl.

∴ 400 litre of NaCl solution contains = 400 ×100 = 40000 gms NaCl = 40000×103 mgs NaCl

Since 58.5 mgs of NaCl = 50 mgs of CaCO3 equivalents.

∴ 40000 × 103 mg NaCl = (40000 × 103 × 50 ) mgs of CaCO3 hardness
58.5

= 34188034 mgs eq. CaCO3 hardness

But the total quantity of water sample = 100000 litres

∴ 100000 litre of water = 34188034 mgs CaCO3

∴ 1 litre of water = 341088034 mgs CaCO3 = 341.88 mg/L CaCO3 eq. hardness
100000

Hardness of water sample = 342 ppm

Q. 4. What is the total hardness of sample of water which has the following impurities in mg/l.

Ca(HCO3)2 =162 CaCl2= 22.2 MgCl2 =95 NaCl =20 [May 13(3m)]

Ans.

Impurity Quantity of Multiplication CaCO3 eq. ppm. Types of
mg/L factor hardness

Ca(HCO3)2 162 100/162 100 Temporary

CaCl2 22.2 100/111 20 Permanent

MgCl2 95 100/95 100 permanent

NaCl Does not contribute to hardness.

Total hardness = Temporary Hardness + permanent Hardness

= Ca(HCO3)2 + CaCl2 + MgCl2 = 100 + 20 + 100 = 220 mg/L or ppm

Q. 5. 0.5 g of CaCO3 was dissolved in dilute HCl and diluted to 500 ml, 50 ml of this solution

required 45 ml of EDTA solution for titration. 50 ml of hard water sample required 15 ml of

EDTA solution for titration. 50 ml of same water sample on boiling, filtering requires 10 ml of

EDTA solution. Calculate the temporary permanent and total hardness in ppm . [May 13(6m)]

Ans. Concentration of standard hard water = 0.5 gm CaCO3 /500ml = 500 mgs CaCO3 /500ml

= 1 mg/ml

Now 50 ml SHW required = 45 ml EDTA = 50 mgs CaCO3

i.e. 45 ml EDTA solution = 50 mg CaCO3 eq. hardness

∴ 1ml EDTA solution = 50 mgs CaCO3 Eq. Hardness
45

Now 50ml water sample = 15 ml EDTA solution

∴ Hardness of water sample = 50 × 15 mgs CaCO3 eq.H
45

∴ Total hardness = 333.33 ppm

Now 50 ml sample after boiling = 10ml EDTA solution

∴ Hardness of sample = (10 ×5450 ) mgs CaCO3 eq.H
×5405 1000
∴ Per litre = (10 × 50 ) mg CaCO3 eq.H

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Hence, Permanent hardness = 222.22 ppm
Temporary hardness = total Hardness – Permanent hardness = 333.33 – 222.22 = 111.11ppm

Q. 6. A zeolite softener was completely exhausted and was regenerated by passing 150 litres of NaCl
Ans.
solution, containing 50 g/l of NaCl. How many litres of water sample of hardness 450ppm can

be softened by this zeolite container? [May 13(4m)]

Let the quantity of water be x litres.

Hardness of the water given is 450ppm

Hardness = concentration of NaCl × litres of NaCl

= 150 × 50 × 1000 = 7500000 mgs of NaCl

= 7500000 × 50 mgs CaCO3 eq.
58.5
= 6.410 × 106 mgs CaCO3 eq.H

If 1 litre water = 450 mgs CaCO3 eq.H

Then x litre of water = 6.41 × 106 CaCO3 eq.H
∴ x = 6.410×106 litres = 0.01424501 × 106 litres = 14245 litres

450

Q. 7. Two samples of water A and B analysed for their salt content:

(i) Sample A was found to contain 168mg MgCO3 per litre.

(ii) Sample B was found to contain 820mg Ca(NO3)2 per litre amd 2mg SiO2 per litre.

Calculate the total hardness of each sample and state which sample is harder. [Dec’13(3m)]

Ans. Sample A = MgCO3 = 168 mg/L

CaCO3 equivalent of MgCO3 = 168 × 100 = 200 ppm
84
Sample B = Ca (NO3)2 = 820 mg/L ….. SiO2 = 2 mg/L

CaCO3 equivalent of Ca (NO3)2 = 820 × 100 = 500 ppm
164

SiO2 does not contribute to hardness.

Thus hardness of sample B is 500 ppm.

Q. 8. 50ml of standard hard water containing 1mg of pure CaCO3 per ml consumed 20ml of EDTA.
Ans.
50 ml of the water sample consumed 30ml of same EDTA solution using Erichrome Black T

indicator. After boiling and filtering, 50ml of the water sample required 10ml of the same

EDTA for titration. Calculate the total and permanent hardness of water sample. [Dec 13(6m)]

Standard hard water concentration = 1mg/ml

Vol. of EDTA for 50ml of SHW (V1) = 20 ml

Vol. of EDTA for 50 ml sample (V2) = 30 ml

Vol. of EDTA for 50ml sample after boiling(V3) = 10ml

To calculate all types of hardness in sample.

Since 50ml Standard hard water = 20ml

EDTA = 50 mg CaCO3

Hence 1ml EDTA = 50/20 mg

CaCO3 = 2.5 mg CaCO3 eq. hardness

Total hardness = [ V2 × 2.5 × 105000] =1500ppm
1000
Permanent hardness =[ V3 × 2.5 × 50 ] = 500ppm

Hence temporary hardness = Htotal – Hpermanent = 1500-500 = 1000ppm

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10 Applied Chemistry - I

Q. 9. An exhausted zeolite softener was regenerated by passing 150 litres of NaCl solution having
Ans.
strength of 150g/l of NaCl. If the hardness of the water is 600 ppm. Calculate the total volume

of water that is softened by the softener. [Dec 13(4m)]

Volume of NaCl solution = 150 litres

Weight of NaCl = 150 gm/L

Hardness of water sample = 600 ppm

∴ quantity of NaCl consumed = 150 × 150 = 22500 gm =22.5 × 106 mg

CaCO3 equivalent of NaCl = 22.5 × 106 × 50 = 19.23 × 106 litres
58.5 600

= 32050 litres water.

Volume of water softened = 32050 litres.

Q. 10. Classify the following salts into temporary and permanent hardness causing salts and also
Ans. calculate their calcium carbonate equivalents. May 14 (3m)

i. Ca(HCO3)2 -16.2 mg/L
ii. MgSO4 -1.2 mg/L
iii. FeCl2 -12.7 mg/L
iv. NaCl -94 mg/L

Salt Type of hardness Multiplication factor CaCO3 equivalent
for mg/L
Ca(HCO3)2 Temporary
MgSO4 Permanent CaCO3 equivalent 10
FeCl2 Permanent 16.2 × 100/162 01
NaCl 1.2 × 100/120 10
12.7 × 100/127

Does not contribute

Q. 11. Calculate the quantity of lime and soda required for softening 50,000 litres of water containing
Ans.
The following salts per litre. Ca (HCO3)2 =8.1 mg; Mg (HCO3)2 = 7.3 mg; CaSO4 = 13.6 mg;

MgSO4 = 12 mg: NaCl = 4.7 mg; MgCl2 = 23.75 mg [May 14(6m)]

Salt Quantity mg/L Multiplication CaCO3 Requirement of
factor equivalent ppm lime(L) &
Ca(HCO3)2 8.1 Soda(S)
Mg(HCO3)2 7.3
13.6 8.1 × 100/162 5 L
CaSO4 12
MgSO4 23.75 7.3 × 100/146 5 L
MgCl2 4.7
NaCl 13.6 × 100/136 10 S

12 × 100/120 10 L+S

23.75 × 100/95 25 L+S

Does not contribute to hardness

Lime = L = 17040[temp Ca+2 + 2 × temp Mg+2 +permanent] vol.of water ×% 100
106 of purity

= 74 [05 + 2×5 +10 +25] × 50000 × 1
106
100
= 74 [50] × 50000 × 10-6 = 1.85 kg
vol.of water 100
100 106 of purity
106
Soda = S = 100 [permanent (Ca+2 + Mg+2 + Al+3 + Fe+2 + H+ - HCO3-] × × %

= 106 [10 + 10 + 25] × 50000 × 1
100 106
106
= 100 [45] × 5000 × 10-6

= 2.385 kg

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Q. 12. The hardness of 50,000 litres of a sample of water was removed by passing it through a zeolite
Ans.
softener. The softener required 200 litres of NaCl solution containing 125g/L of NaCl for

regeneration. Calculate the hardness of the sample of water. [May 14(4m)]

Using regeneration reaction
CaZe + 2NaCl → Na2Ze +CaCl2

i.e. 2 NaCl = CaCl2 = CaCO3

∴ 2 × 58.5 gms = 111 gm = 100 gm of CaCO3

Now,

Quantity of NaCl in regeneration = 125 × 200 = 25000 gms of NaCl

Thus,

(2 × 58.5) gms NaCl = 100 gm CaCO3

∴ 25000 gm NaCl = 25000×100 = 21367 gms of CaCO3 = 2136000 mgs of CaCO3
2×58.5
21365000
Since, 50000 litres of hard water = 50000 = 427.3 mgs of CaCO3 = 427.3 ppm

∴ Hardness of water sample = 427.3 ppm

 All the best !!!

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