unit i water technology - department of chemistry · 2018. 2. 9. · water technology 1.1...

1

UNIT I

WATER TECHNOLOGY 1.1 INTRODUCTION

Water is one of the most abundant commodities in nature. Water is essential for

survival of all living organisms. About 80% of the earth surface is covered by water. The

main sources of water are

i) Rain

ii) River and lakes (surface water)

iii) Wells and springs(ground water)

Among the above (sources of water) rain water is the purest form of water whereas sea

water is the most impure form. For drinking and industrial purposes we need water free

from undesirable impurities. So first we should understand about the nature and types of

impurities present in water so that the water can be treated suitably to remove the

undesirable impurities.

1.2 TYPES OF IMPURITIES IN WATER

The impurities present in the water may be broadly classified into three types

(i) Physical impurities

(a) Suspended impurities

(b) Colloidal impurities

(ii) Chemical impurities

(a) Dissolved salts

(b) Dissolved gases

(iii) Bacteriological impurities due to pathogenic bacteria which spread diseases.

2

1.3 HARD AND SOFT WATER

Water which does not produce lather with soap solution but forms a white precipitate

is called hard water.

Water which lathers easily with soap solution is called soft water.

1.3.1 Hardness of water

Hardness is the characteristic property of water which does not lather with soap

solution due to the presence of Ca2+ and Mg2+ ions. .

Hardness can be detected by treating water with soap.

2C17H35COONa + CaCl2 C17H35COO)2 Ca + 2NaCl

Depending upon the types of dissolved salts present in water, hardness of water can be

classified into two types.

i. Temporary hardness or Carbonate hardness or Alkaline hardness: This is due to the

presence of bicarbonates of calcium and magnesium. It can be removed by boiling the

water or adding lime to the water.

Ca (HCO3)2 Δ CaCO3 + H2O + CO2

Mg (HCO3)2 + 2Ca (OH)2 Mg (OH)2 + 2CaCO3 + 2H2O

ii. Permanent hardness or Non-Carbonate hardness or Non - alkaline hardness: This is

due to the presence of chlorides and sulphates of calcium and magnesium. Permanent

hardness is removed by lime-soda process or zeolite process.

CaCl2 + Na2CO3 CaCO3 + 2NaCl

(soda)

CaSO4 + Na2Ze CaZe + Na2SO4

(soda. zeolite)

3

1.3.2 Expression of hardness as equivalents of CaCO3 (or) CaCO3 standard

The concentration of hardness causing ions are usually expressed in terms of an

equivalent amount of CaCO3. The choice of CaCO3 is standard because its molecular

weight is 100 (equivalent weight =50) and also it is the most insoluble salt that can be

precipitated in water treatment.

If the concentration of hardness producing constituent is x mg/lit., then

Equivalent amount of CaCO3 = x × 100 Molecular weight of hardness producing substance i.e., Amount Equivalent to CaCO3 = Amount of hardness producing salt

× Molecular weight of CaCO3 Molecular weight of hardness producing salt i.e., Amount Equivalent to CaCO3 = Amount of hardness producing salt

× Equivalent weight of CaCO3 Equivalent weight of hardness producing salt

1.4 UNITS OF HARDNESS

i) Parts per million (ppm)

It is defined as the number of parts of CaCO3 equivalent hardness per 106 parts of water.

ii) Milligrams per litre (mg / lit)

It is defined as the number of milligrams of CaCO3 equivalent hardness per 1 litre of

water.

Relationship between ppm and mg/litre

Since weight of1 litre of water = 1 kg

= 1000gms = 1000 × 1000 = 106 mgs

1 mg/lit = 1 mg of CaCO3 equivalent hardness in

106 mgs of water

4

= 1 part of CaCO3 equivalent hardness

in 106 parts of water

= ppm

Thus mathematically ppm and mg/lit unites are equal.

iii) Clarke’s degree (0Cl)

It is defined as the number of parts of CaCO3 equivalent hardness per 70,000 parts of

water.

iv) French degree (0Fr)

It is defined as the number of parts of CaCO3 equivalent hardness per 105 parts of water.

Problems based on Hardness

1. A sample of water contains 100 mgs of MgSO4 per litre. Calculate the hardness in

terms of CaCO3 equivalents.

Solution

Given: The amount of MgSO4 = 100 mgs/lit

The amount of hardness producing salt Amount equivalent to CaCO3 = ---------------------------------------------------------- X 100 Molecular weight of hardness producing salt We know that, the molecular weight of MgSO4 = 120

Amount equivalent to CaCO3 =100 x 100/120 = 83.3 mgs/lit

2. A water sample contains 200 mgs of CaSO4 and 75 mgs of Mg (HCO3)2 per litre.

What is the total hardness interms of CaCO3 equivalent?

Solution:

Name of the hardness producing salts

Amount in mgs/lit

Molecular weight

Amount equivalent to CaCO3

CaSO4 200 136 200 X 100 / 136 = 147 mgs/lit

Mg(HCO3)2 75 146 75 x 100 / 146 = 51.4 mgs/lit

5

Temporary Hardness = Mg (HCO3)2 = 51.4 mgs/lit

Permanent hardness = CaSO4 =147 mgs/lit.

Total hardness = Mg (HCO3)2 + CaSO4

= 51.4 + 147

= 198.4mgs/lit

3. A sample of water contains 25mgs of Ca2+ ions per litre, Calculate its hardness

interms of CaCO3 equivalent?

Solution

Given:

The amount of Ca2+ ions = 25 mgs/lit

We know that, the atomic weight of calcium = 40

Amount of CaCO3 = 25 X 100 / 40

= 62.5 mgs/lit

4. Calculate the carbonate and non-carbonate hardness of a sample water containing

the dissolved salts as given below in mgs/lit Mg(HCO3)2 = 10;Ca(HCO3)2 = 40; MgCl2 =

21.; and NaCl = 20

Solution

Name of the hardness producing

salts

Amount in mgs/lit

Molecular weight

Amounts equivalent to

CaCO3 Mg(HCO3)2 10 146 10 X 100 / 146 =

6.8 mgs/lit Ca(HCO3)2 40 162 40X 100 / 162 =

24.7 mgs/lit MgCl2 21 95 21 X 100 / 95 = 22.1

mgs/lit NaCl 20 NaCl does not contribute any

hardness to water, hence it is ignored.

6

Carbonate hardness = Mg(HCO3)2 + Ca(HCO3)2

= 6.8 + 24.7 = 31.5 mgs/lit

Non-carbonate hardness = MgCl2

= 22.1 mgs/lit

Total hardness = carbonate hardness + Non-carbonate hardness

= 31.5 + 22.1

= 53.6 mgs/lit

Exercises

5. A sample of water contains the following dissolved salts in mgs/lit. Mg (HCO3)2 =

73; CaCl2=111; Ca (HCO3)2=81 and MgSO4=40. Calculate the temporary and permanent

hardness of the water (Atomic weights of Ca, Mg, O, C, Cl, S, H are 40, 24, 16, 12, 35.5, 32

and respectively.

6. A sample of water is found to contain 16.8 mg/lit Mg(HCO3)2, 12 mg/lit MgCl2, 29.6

mg/lit MgSO4, and 5.0 mg/lit NaCl. Calculate the permanent and temporary hardness of

water and express it in ppm. (Atomic weight of Mg=24; H=1; C=12; O=16; Cl=35.5;

Na=23; S=32)

1.5ALKALINITY

Alkalinity in water is due to the presence of soluble (i) hydroxide (OH-) ions, (ii)

carbonate (CO32-) ions and (iii) bicarbonate (HCO3-) ions. These can be determined by

titrimetry using standard acid and phenolphthalein and methyl orange as indicators. The

determination is based on the following reactions.

(i) [OH-] + [H+] H2O

(ii) [CO3]2- + [H+] [HCO3-] P

(iii) [HCO3-] + [H+] H2O + CO2 M

7

Depending upon the anion that is present in water alkalinity is classified into three types.

1. Hydroxide alkalinity - due to (OH-)

2. Carbonate alkalinity - due to (CO3 2-)

3. Bicarbonate alkalinity - due to (HCO3-)

1.5.1 Detection of various alkalinity:

Titration 1:

Known amount of water sample is titrated against a standard acid using

phenolphthalein indicator, the end point indicates the completion of the reaction (i) and

(ii) i.e. neutralization of OH- and CO32- ions. The amount of acid used corresponds to all

the hydroxide ions + one half of the carbonate ions present.

Titration 2:

Similarly the same amount of water sample is titrated against a standard acid using

methyl orange indicator, the end point indicates the completion of reaction (i), (ii) and (iii).

The amount of acid used after the phenolphthalein end point corresponds to one-half of

normal carbonate {+} all the bicarbonates.

i.e., neutralization of OH-, CO32-, and HCO3- ions.

The total amount of acid used (amount of acid upto phenolphthalein end point {+}

amount of acid upto methyl orange end point) represents the total alkalinity (due to

hydroxide, carbonate and bicarbonates).

Note:

OH- and HCO3- ions cannot exist together in water, because they combine

instantaneously to form CO32- ions.

i.e., OH- + HCO3- CO32- + H2O

e.g. NaOH + NaHCO3 Na2CO3 + H2O

Thus in a water all the three ions (OH-, CO32-, HCO3-) cannot exist together.

8

1.5.2 Determination of various types of alkalinity

Various types involved in the determination of various types of alkalinity are as

follows.

Step I : Experimental procedure

Titration I : Determination of Phenolphthalein alkalinity

Pipette out 100 ml of the given water sample into a clean conical flask. Add 2 to 3

drops of a phenolphthalein indicator and titrate it against N/50 H2SO4 taken in the

burette. The end point is the disappearance of pink colour.

Let the volume of acid consumed be V1 ml.

Titration II: Determination of methyl orange alkalinity

After the completion of titration I, to the same solution, add 2 to 3 drops of methyl

orange indicator, and continue the titration against the same N/50 H2SO4. The end point is

the reappearance of pink color.

Let the volume be V2 ml.

Step II: Calculation

(i) Calculation of Phenolphthalein Alkalinity

The volume of acid used to phenolphthalein indicator V1 = ml

Normality of the acid N1 = 50N

Volume of the water sample V2 = 100 ml

Normality of water sample N2 = ?

According to volumetric law V1N1 = V2N2

2N = 2

11

VNV

9

2N = 501

100V1

Phenolphthalein alkalinity (P) (in terms of CaCO3 equivalent)

= N2 Eq.wt of CaCO3 1000 ppm

= 5010010050V1

ppm

(ii) Calculation of Methyl Orange Alkalinity

Extra volume of acid used to methyl orange end point = V2 ml

Total volume of acid used to methyl orange end point = (V1+V2) ml

The volume of acid used to methyl orange end point} V1 = (V1+V2) ml

Normality of the acid N1 = 50N

Volume of water sample V2 = 100 ml

Normality of alkaline water sample N2 = ?

According to Volumetric law N2 =

501

100VV 21

Methyl orange alkalinity (M)

(interms of CaCO3 equivalent) =

50100100050VV 21

ppm

Conclusions

(i) When P = 0, both OH- and CO32-are absent and alkalinity is only due to HCO3-.

(ii) When P = 1/2 M, only CO32- is present, half of the carbonate neutralization reaction

takes place (i.e CO32- + H+ HCO3-) with phenolphthalein indicator complete

carbonate neutralization reaction (i.e CO32- + H+ HCO3-. HCO3-+ H+ H2O + CO2)

occurs when methyl orange indicator is used. Thus alkalinity due to CO32- = 2P.

P = 10V1

M = 10(V1 + V2)

10

(iii) When P = M, only OH— is present, alkalinity due to OH-= P = M

(iv) When P > 1/2M, besides CO32-, OH- ions are also present. Now half of CO32-

(i.e,HCO3- + H+ CO2 + H2O) equal to (M-P) so, alkalinity is due to complete CO32- =

2(M-P) alkalinity due to OH- = M-2(M-P) = 2P-M

(v) When P < 1/2 M, Besides CO32-, HCO3- ions are also present.

Now alkalinity due to CO32- = 2P

Alkalinity due to HCO3- = (M- 2P)

The data of the above conclusions are tabulated as follows.

Table 1.2 Calculation of alkalinity of water

Result of P end point and P M end point OH

- (ppm) CO32- (ppm) HCO3-(ppm) Nature of alkalinity

P=0 0 0 M Only bicarbonate

P=1/2 M 0 2P 0 Only carbonate

P1/2 M (2P-M) 2(M-P) 0 Hydroxide and

carbonate

P=M P=M 0 0 Only hydroxide

11

Problems based on Alkalinity

1. 100 ml of a raw water sample on titration with N/50 H2SO4 required 10.0 ml of the

acid to phenolphthalein end-point and 14.0 ml of the acid to methyl orange end-

point. Determine the type and extent of alkalinity present in the water sample.

Solution:

Strength of HCl = 0.02 N

phenolphthalein end-point = P = 10.0 ml

methyl orange end-point = M = 14.0 ml

Since P > M,

the water sample must contain only OH- and CO32- alkalinities and there cannot be any

HCO3- alkalinity.

i) Volume of std .HCl required for OH- alkalinity = 2P - M

= (2 x 10.0) ml - 14 ml

= 20 ml- 14.0 ml

= 6.0 ml

Volume of acid consumed to OH- alkalinity V1 = 6.0 ml

ii) Volume of std .HCl required for CO32- alkalinity = 2M – 2P

= 2 x 14.0 ml – 2 x 10.0 ml

= 24 – 20

Volume of acid consumed to CO32- alkalinity V1 = 4.0 ml

1. Calculate the OH- alkalinity

Volume of HCl V1 = 6.0 ml

Strength of HCl N1 = 0.02 N


12

Strength of water sample due to OH- alkalinity N2 = ?

V1 N1 = V2 N2

2N = 2

11

VNV

2N = 10002.06

Strength of water sample due to OH- alkalinity = 0.0012 N

Amount of OH- alkalinity present in

1 litre in terms Of CaCO3 equivalent = Strength of OH- alkalinity

Eq. wt of CaCO3

= 0.0012 N 50

= 0.06 gm x 1000

Amount of OH- alkalinity = 60 ppm

2. Calculate the CO32- alkalinity




Strength of water sample due to CO32- alkalinity N2 = ?

V1 N1 = V2 N2

2N = 2

11

VNV

2N = 10002.04

Strength of water sample due to CO32- alkalinity = 0.0008 N

Amount of CO32- alkalinity present in

1 litre in terms Of CaCO3 equivalent = Strength of of CO32- alkalinity

Eq. wt of CaCO3

13

= 0.0008 N 50

= 0.04 gm x1000

= 40 ppm

Amount of CO32- alkalinity = 40 ppm

Total Alkalinity

Total alkalinity = Alkalinity due to OH— +

Alkalinity due to CO32-

= 60 ppm + 40 ppm

= 100 ppm

2. A water sample is not alkaline to phenolphthalein but, 100 ml of the sample on

titration with N/10 HCl, required 15 ml to methyl orange end point. What are the types

and amounts of alkalinity present in the sample.

Solution:


phenolphthalein end point p = 0

methyl orange end point M = 15 ml

Since P = 0

The water sample contain only HCO3- alkalinity ,

Volume of HCl required to HCO3- alkalinity = M

M = 15 ml

1) Calculate the HCO3- alkalinity




14

Strength of water sample due to HCO3- alkalinity N2 = ?

V1 N1= V2 N2

2N = 2

11

VNV

2N = 1001.015

= 0.015 N

Strength of water sample due to HCO3- alkalinity = 0.015 N

Amount of HCO3- alkalinity present in

1 litre in terms Of CaCO3 equivalent = Strength of HCO3- alkalinity

Eq. wt of CaCO3

= 0.015 N 50

= 0.75 gm x1000

= 750 ppm

Amount of HCO3- alkalinity = 750 ppm

3. 100 ml of a water sample on titration with 0.02 N H2SO4 gave a titre value of 7.8 ml

to phenolphthalein end-point and 15.6m l to methyl orange end-point. Calculate the

alkalinity of the water sample interms of CaCO3 equivalent and comment the type of

alkalinity present.

Solution:

Given:


Volume of the water sample = 100 ml

phenolphthalein end point P = 7.8 ml

methyl orange end point M = 15.6 ml

Given data satisfy the condition P = 1/2 M, Therefore water sample contains only 2

3CO alkalinity not OH - and HCO3- alkalinity,

15

Volume of HCl required to CO32- alkalinity = 2P

= 2 7.8

= 15.6 ml

Calculation for CO32- Alkalinity.




Strength of water sample due to CO32- alkalinity N2 = ?

V1 N1 = V2 N2

2N = 2

11

VNV

2N = 10002.06.15

Strength of water sample due to CO32- alkalinity = 0.00312 N


1 litre in terms Of CaCO3 equivalent = Strength of CO32- alkalinity

Eq. wt of CaCO3

= 0.00312 N 50

= 0.156 gm x1000

= 156 ppm

Amount of CO32- alkalinity = 156 ppm

16

4. 100 ml of a raw water sample on titration with N/50 H2SO4 required 7.5 ml of the

acid to phenolphthalein end-point and 18.0 ml of the acid to methyl orange end-point.

Determine the type and extent of alkalinity present in the water sample.

Solution.


phenolphthalein end-point P = 7.5 ml

methyl orange end-point M = 18.0 ml

If the data satisfy the condition, P < M,

the water sample must contain both CO32- and HCO3- alkalinities and there cannot be any

OH- alkalinity.

i) Volume of std .HCl required for CO32- alkalinity = 2P

= 2 x 7.5 ml

= 15.0 ml

Volume of acid consumed to CO32- alkalinity V1 = 15.0 ml

ii) Volume of std .HCl required for HCO3- alkalinity = M – 2P

= 18.0 ml – 2 x 7.5.0 ml

= 3.0 ml

Volume of acid consumed to HCO3- alkalinity V1 = 3.0 ml

Calculate the CO32- alkalinity




Strength of water sample due to

CO32- alkalinity N2 = ?

17

V1 N1 = V2 N2

2N = 2

11

VNV

2N = 10002.015

= 0.003 N

Strength of water sample due to CO32- alkalinity = 0.003N


1 litre in terms Of CaCO3 equivalent = Strength of CO32- alkalinity

Eq. wt of CaCO3

= 0.003 N 50

= 0.15 gm x 1000

Amount of CO32- alkalinity = 150ppm

Calculate the HCO3- alkalinity




Strength of water sample due to HCO3- alkalinity N2 = ?

V1 N1 = V2 N2

2N = 2

11

VNV

2N = 10002.03

N2 = 0.0006 N

Strength of water sample due to HCO3- alkalinity = 0.0006 N

Amount of HCO3- alkalinity present in

1 litre in terms Of CaCO3 equivalent = Strength of of HCO3- alkalinity

Eq. wt of CaCO3

18

= 0.0006 N x 50

= 0.03 gm x1000

= 30 ppm

Amount of HCO3- alkalinity = 30 ppm

Total Alkalinity

Total alkalinity = Alkalinity due to CO32- +

Alkalinity due to HCO3-

= 160 ppm + 30 ppm

= 190 ppm

1.6 ESTIMATION OF HARDNESS BY EDTA METHOD

EDTA is Ethylene Diamine Tetra Acetic acid. The structure of EDTA is

HOOCH2C CH2COOH

N-CH2-CH2-N

HOOCH2C CH2COOH

Since, EDTA is insoluble in water, its disodium salt is used as a sequestering or

complexing agent.

Principle:

The amount of hardness causing ions (Ca2+and Mg2+) can be estimated by titrating the

water sample against EDTA using Eriochrome-Black-T indicator (EBT) at a pH of 8-10. In

order to maintain the pH, buffer solution (NH4Cl-NH4OH mixture) is added. Only at this

pH such a complexation is possible.

When the EBT indicator is added to the water sample, it forms wine red coloured

weak complex with Ca2+and Mg2+ ions.

19

[ Ca2+ & Mg2+] + EBT pH = 8-10 Ca EBT complex Mg

wine red coloured weak complex

When this solution is titrated against EDTA, it replaces the indicator and form stable

EDTA complex. When all the hardness causing ions are complexed by EDTA, the

indicator is set free. The colour of the free indicator is steel blue. Thus the end point is

change of colour from wine red to steel blue.

Ca EBT complex + EDTA pH = 8-10 Ca Mg EDTA + EBT Mg

wine red coloured weak complex Stable complex Steel blue

1.6.1 Preparation of solutions:

a) EDTA Solution

It is prepared by dissolving 4 gms of EDTA in 1000 ml of distilled water.

b) Standard hard water

1 gm of pure CaCO3 is dissolved in minimum quantity of HCl and then made up to

1000 ml using distilled water.

1 ml of standard hard water 1 mg of CaCO3

c) EBT Indicator

0.5 gms of EBT is dissolved in 100 ml of alcohol.

d) Buffer solution

67.5 gms of NH4Cl and 570 ml of NH3 are dissolved and the solution is made upto

1000 ml using distilled water.

20

1.6.2 Experimental procedure:

(i) Standardisation of EDTA

Pipette out 50 ml of standard hard water into a clean conical flask. Add 10 ml of buffer

solution and 4-5 drops of EBT indicator and titrate it against EDTA solution taken in the

burette. The end point is the change of colour from wine red to steel blue

Let the volume of EDTA consumed be V1 ml

(ii) Estimation of total hardness of water sample

Pipette out 50 ml of the given hard water sample into a clean conical flask. Add 10 ml

of buffer solution and 4-5 drops of EBT indicator and titrate it against EDTA as before.

Let the volume of EDTA consumed be V2 ml

(iii) Estimation of permanent hardness of water sample

Take a 100 ml of the same hard water sample in a 250 ml beaker. Boil it for 15 minutes.

During boiling temporary hardness gets removed. Cool and filter the solution and make

upto 100 ml in a standard flask by adding distilled water. Pipette out 50 ml of the made up

solution into a clean conical flask and titrate against EDTA as before.

Let the volume of EDTA consumed be V3 ml.

Calculation:

(i) Standardization of EDTA using standard hardwater

1 ml of std. hard water = 1 mg of CaCO3

50 ml of std. hard water = 50 mgs of CaCO3

50 ml of std. hard water consumes = V1 ml of EDTA

V1 ml of EDTA 50 mgs of CaCO3 equivalent of hardness

(Or)

1 ml of EDTA 1V

50 mgs of CaCo3 equivalent of hardness

21

(ii) Estimation of total hardness of water sample

50 ml of the given hard water sample consumes} = V2 ml of EDTA

= 1

2 V50V mgs of CaCO3 equivalent of hardness

[1 ml of EDTA = 1V

50 mgs of CaCO3]

1000 ml of the given sample hard water sample

= 50

1000V50V1

2

= 1

2

VV

1000 {mgs of CaCO3 equivalent of hardness}

Total Hardness = 1

2

VV

1000 ppm

(iii) Estimation of permanent hardness of water sample

50 ml of same hard water samples after boiling, filtering, etc, consumes}

= V3 ml of EDTA

= 1

3 V50V mgs of CaCO3 equivalent of hardness

1000 ml of the given hard water sample}

= 50

1000V50V1

2

= 1

3

VV

1000 {mgs of CaCO3 equivalent of hardness}

Permanent Hardness = 1

3

VV

1000 ppm

22

(iii) Temporary hardness

Temporary hardness = Total hardness - Permanent hardness

= 1

3

1

2

VV

1000VV

1000

Temporary hardness = 121

VVV

1000 ppm

Problems based on Hardness

1. 100 ml of a water sample requires 20 ml of EDTA solution for titration. 1 ml of EDTA

solution is equivalent to 1.1 mgs of CaCO3.Calculate the hardness of the water in ppm

unit.

Solution

Given:

1 ml of EDTA solution = 1.1 mgs of CaCO3

20 ml of EDTA solution = 20 x 1.1 mgs of CaCO3

= 22 mgs of CaCO3

100 ml of water sample requires = 20 ml of EDTA

= 22 mgs of CaCO3

1000 ml of water sample, = 100100022 mgs of CaCO3

= 220 mgs/lit or 200 ppm

2. In an estimation of hardness of water by EDTA titration, 50 ml of a sample of water

required 15 ml of 0.02M EDTA solution to reach the end point. Calculate the hardness

of water.

Estimation of hardness

Volume of water sample V1 = 50ml

Strength of water sample M1 = ?

Volume of std EDTA V2 = 15 ml

23

Strength of b std EDTA M2 = 0.02 M

V1M1 = V2M2

1M = 1

22

VMV

1M = 5002.015

2M = 0.006 M

Strength of hard water sample = 0.006 M

Amount of hardness present in per liter in

terms of CaCO3 equivalent = Strength of hard water x M .W of CaCO3

= 0.006 M X 100

= 0.6 gm x1000

Amount of hardness = 600 ppm

3. 100 ml of a water sample of consumed 19.5 ml of 0.01 M EDTA for titration using

Erio-chrome Black-T indicator. In another experiment, 100 ml of the same water sample

was boiled to remove the carbonate hardness, the precipitate was discarded and then it

is allowed to reach the room temperature. The filtrate CaCO3 solution was then titrated

against 0.01 M EDTA which consumed 10 ml of EDTA, in the presence of EBT

indicator. Calculate (i) the total hardness (ii) permanent hardness of non carbonate

hardness (iii) carbonate hardness, in terms of mg/lit of CaCO3.

Titration: I

Strength of EDTA = 0.01M

Estimation of total hardness.



Volume of std EDTA V2 = 19.5 ml


24

V1M1 = V2M2

1M = 1

22

VMV

1M = 10001.05.19

1M = 0.00195 M


Amount of total hardness present in per

liter in terms of CaCO3 equivalent = Strength of hard water x M .W of CaCO3

= 0.00195 M X 100

= 0.195 gm x1000

Amount of total hardness = 195 ppm

Titration: II

Estimation of permanent hardness.

Volume of boiled water V1 = 100 ml

Strength of boiled water M1 = ?



V1M1 = V2M2

1M = 1

22

VMV

1M = 10001.010

1M = 0.001 M

Strength of boiled water = 0.001 M

Amount of permanent hardness present in per liter in

25

Terms of CaCO3 equivalent = Strength of boiled water x M .W of CaCO3

= 0.001M X 100

= 0.1 gm x1000

Amount of permanent hardness = 100 ppm


Temporary hardness = 195 ppm - 100 ppm

Temporary hardness = 95 ppm

4. Calculate permanent and temporary hardness from the following. 250 ml of a water

sample is boiled for 1 hr. It is then cooled nd filtered. The filtrate is made upto 250 ml

again with the addition of distilled water. 20 ml of this solution requires 10 ml of 1/100

EDTA with EBT-indicator and NH4Cl - NH4OH buffer.

Titration: I





Strength of std EDTA M2 = 0.01 M

V1M1 = V2M2

1M = 1

22

VMV

1M = 2001.010

1M = 0.005 M Strength of boiled water = 0.005 M


Terms of CaCO3 equivalent = Strength of boiled water x M .W of CaCO3

= 0.005 M X 100

= 0.5 gm x1000


26

5. 100 ml of a water sample consumed 25.0 ml of 0.01 M EDTA for the titration using

EriochromeBlack-T indicator. Calculate the total hardness.

Estimation of hardness.




Strength of std EDTA M2 = 0.01 M

V1M1 = V2M2

1M = 1

22

VMV

1M = 10001.025

1M = 0.0025 M



Terms of CaCO3 equivalent = Strength of hard water x M .W of CaCO3

= 0.0025 M X 100

= 0.25 gm x1000


27

6. 0.25 gm of CaCO3 was dissolved in HCl and the solution made upto one litre with

distilled water. 100 ml of the above solution consumed 25 ml of EDTA solution on

titration.100 ml of hard water sample required 30 ml of same EDTA solution on

titration. 100 ml of this water, after boiling cooling and filtering required 11 ml of

EDTA solution on titration. Calculate the temporary permanent and total hardness of

water.

Calculate the strength of given std water,

Amount / Lit Strength of Std Water = _________________________ Molecular Weight

Strength of std Water = 100

gm25.0

Strength of Std water = 0.0025 M

Titration: I

Standardization of EDTA :

Volume of std water V1 = 100 ml

Strength of std water M1 = 0.0025M

Volume of EDTA V2 = 25 ml

Strength of EDTA M2 = ?

V1M1 = V2M2

1M = 1

22

VMV

1M = 250025.0100

M2 = 0.01 M Strength of EDTA = 0.01 M

28

Titration: II

Estimation of total hardness.





V1M1 = V2M2

1M = 1

22

VMV

1M = 10001.030

M2 = 0.003 M Strength of hard water sample = 0.003 M

Amount of total hardness present in per liter in

terms of CaCO3 equivalent = Strength of hard water x M .W OF CaCO3

= 0.003M X 100

= 0.3 gm x1000

Amount total hardness = 300 ppm

Titration: III






V1M1 = V2M2

1M = 1

22

VMV

1M = 10001.011

M2 = 0.0011 M Strength of boiled water = 0.0011 M

29


terms of CaCO3 equivalent = Strength of boiled water x M .W of CaCO3

= 0.0011M X 100

= 0.11 gm x1000



Temporary hardness = 300 ppm - 110 ppm

Temporary hardness = 190 ppm

7. 100 ml of a molecular water sample of requires 18.5 ml of EDTA solution for

titration. 20 ml of the same EDTA solution was required for the titration of 100 ml of

standard hard water containing 1 gm of CaCO3 per litre. Calculate the hardness of water

in ppm unit.

Calculate the strength of given std hard water.

Amount / Lit Strength of Std Water = _________________________ Molecular Weight

Strength of std Water = 100gm1

Strength of Std water = 0.01

Titration: I

Standardization of EDTA :

Volume of std water V1 = 100 ml

Strength of std water M1 = 0.01M

Volume of EDTA V2 = 20 ml

Strength of EDTA M2 = ?

V1M1 = V2M2

2M = 2

11

VMV

30

2M = 2001.0100

M2 = 0.05 M Strength of EDTA = 0.05 M

Titration : II

Estimation of hardness.



Volume of std EDTA V2 = 18.5 ml


V1 M1 = V2 M2

V1M1 = V2M2

1M = 1

22

VMV

2M = 10005.05.18

2M = 0.00925 M Strength of hard water sample = 0.00925 M


terms of CaCO3 equivalent = Strength of hard water x M .W of CaCO3

= 0.00925 M X 100

= 0.925 gm x1000


31

1.7 BOILER FEED WATER

The water fed into the boiler for the production of steam is called boiler feed water. Water

used in steam making should be free from dissolved salts and gases, suspended

impurities, silica and oil. If it is used in boilers, these impurities lead to following

problems.

(i) Scale and sludge formation

(ii) Boiler Corrosion

(iii) Priming and foaming

(iv) Caustic embrittlement

1.7.1 Scale and sludge formation

When water is converted into steam in boilers, the concentration of dissolved salts in

water increases progressively. When the salts concentration reach their saturation point,

they are thrown out in the form precipitate, i.e., the least soluble one gets precipitated first.

If the precipitated matter is soft and slimy, it is called sludge and when the precipitate

forms an adherent coating on the inner walls of the boiler, it is called scale.

Figure 1.1 a) sludge b) scale

32

Table 1.1. Difference between Sludge and Scale

1.7.2 Boiler corrosion

Corrosion in boilers is due to the presence of

a) dissolved oxygen,

S.NO Sludge Scale

1 Sludge is a loose, slimy and non-adherent

precipitate. Scale is a hard, adherent coating.

2 The main sludge forming substances are

MgCO3, MgCl2, MgSO4 and CaCl2 etc.

The main scale forming substances

are Ca (HCO3)2, CaSO4, Mg (OH)2.

3

Disadvantages: Sludges are poor

conductors of heat. Excess of sludge

formation decreases the efficiency of boiler.

Disadvantages: Scales act as

thermal insulators. It decreases the

efficiency of boiler. Any crack

developed on the scale, leads to

explosion.

4 Prevention: Sludge formation can be

prevented by using softened water.

Prevention: Scale formation can be

prevented by dissolving using

acids like HCl, H2SO4.

5 (ii) Sludges can also be removed by “blow –

down operation”.

(ii) Scale formation can be

removed by

(a)External treatment

(b) Internal treatment

6

(iii) Blow down operation is a process of

removing a portion of concentrated water

by fresh water frequently from the boiler

during steam production.

(iii) They can also be removed by

applying thermal shocks, scrapers,

wire brush, etc.,

33

b) dissolved carbon dioxide

c) dissolved salts like MgCl2

a) Dissolved oxygen: The concentration of oxygen in water used in all types of boilers

should be minimum (0.01 ppm 0.05 ppm). Dissolved oxygen in water attacks the boiler

material at high temperature.

4Fe + 2H2O + O2 4Fe (OH)3

b) Dissolved carbon dioxide: Dissolved carbon dioxide in water produces carbonic

acidic, which is acidic and corrosive in nature.

Ca (HCO3)2 CaCO3 + H2O + CO2

CO2 + H2O H2CO3

c) Dissolved MgCl2: When water containing dissolved MgCl2 is used in the boiler, HCl is

produced which attacks boiler in a chain like reaction producing HCl again and again

which corrodes boiler severely.

MgCl2 + H2O Mg (OH)2 + 2HCl

Fe + 2HCl FeCl2 + H2

FeCl2 + 2H2O Fe (OH)2 + 2HCl

1.7.2.1 Removal of dissolved oxygen and carbon dioxide

i) Chemical method

Sodium sulphite, hydrazine are some of the chemicals used for removing oxygen.

2Na2SO3 + O2 2Na2SO4

N2H4 + O2 N2 + 2H2O

Hydrazine is found to be an ideal compound for removing dissolved oxygen in the

water, since the products are water and inert N2 gas.

Carbon dioxide can be removed from water by adding of calculated amount of

NH4OH in to water

2NH4OH + CO2 (NH4)2CO3 + H2O

Corrosion by acids can be avoided by the addition of alkali to the boiler water.

34

ii) Mechanical de-aeration method

Dissolved oxygen along with CO2 can be removed by mechanical deaeration.

Deaeration: It is a mechanical method of removal of dissolved gases such as O2 and CO2.

The solubility of a gas is directly proportional to pressure and inversely proportional to

temperature (Daltons law and henrys law). These two principles are made use in the

design of mechanical deaerator.

In this process, water is allowed to fall slowly on the perforated plates fitted inside the

tower. The sides of the tower are heated, and vacuum pump is also attached to it. The high

temperature and low pressure produced inside the tower reduce the dissolved oxygen

content of the water.

Figure 1.2 Mechanical deaeration of water

1.7.3. Priming and foaming (Carry over)

During the production of steam in the boiler, due to the rapid boiling, some droplets of

liquid water are carried along with steam. Steam containing droplets of liquid water is

35

called wet steam. These droplets of liuid water carry with them some dissolved salts and

suspended impurities. The phenomenon is called carry over. It occurs due to priming and

foaming.

Priming:

Priming is the process of formation of wet steam. It is caused by

(i) very high water level

(ii) high steam velocity

(iii) sudden steam demands leading to sudden boiling.

(iv) improper boiler design

Priming can be controlled by

(i) keeping the water level lower

(ii) good boiler design providing mechanical steam purifier

(iii) avoiding rapid changes in the steaming rate, caused by sudden steam

demands.

Foaming:

The formation of stable bubbles above the surface of water is called foaming. These

bubbles are carried along with steam leading to excessive priming.

It may be caused by presence of oil, grease in water and finely divided sludge

particles.

Foaming can be prevented by

(i) adding antifoaming agents like synthetic polyamides

(ii) finely divided sludge particles, oil and grease can be removed by the addition

of coagulants such as sodium aluminate, ferrous sulphate etc.

36

1.7.4 Caustic embrittlement (Inter-crystalline cracking)

As water evaporates in the boiler, the concentration of sodium carbonate increases in

the boiler. Sodium carbonate is used in softening of water by lime soda process, due to

this some sodium carbonate maybe left behind in the water. As the concentration of

sodium carbonate increases, it undergoes hydrolysis to form sodium hydroxide.

Na2CO3 + H2O → 2NaOH + CO2

This NaOH flows into the minute hair cracks and crevices usually present on the boiler

material by capillary action and dissolves the surrounding area of iron as sodium ferroate.

Fe + 2NaOH Na2FeO2 + H2

This causes brittlement of boiler parts, Particularly stressed parts like bends, joints,

rivets etc, causing even failure of the boiler.

Prevention :

Caustic embrittlement can be prevented by using sodium phosphate as softening

agent instead of sodium carbonate

By adding tanning lignin to the boiler water, which blocks the minute hair cracks

1.8 Softening (Or) Conditioning Methods

Water used for industrial purposes should be free from hardness producing

substances, suspended impurities and dissolved gases etc. The process of removing

hardness producing salts from water is known as softening (or) conditioning of water.

Softening of water can be done in two methods

i. External treatment

ii. Internal treatment

1.8.1 External treatment

It involves the removal of hardness producing salts from the water before feeding into

the boiler. The external treatment can be done by the following method demineralization

or ion-exchange process.

37

1.8.1.1 Ion exchange (or) demineralisation process:

This process removes almost all the ions (both anions and cations) present in the hard

water.

The Soft water, produced by lime-soda and zeolite processes, does not contain

hardness producing Ca2+ and Mg2+ ions, but it will contain other ions like Na+,K+,SO42-,Cl-

etc. On the other hand D.M.(Demineralised) water does not contain both anions and

cations.

Thus a soft water means it is not demineralized water whereas a demineralized water

means it is a soft water.

This process is carried out by using ions exchange resins, which are long chain, cross

linked, insoluble organic polymers with a micro porous structure. The functional groups

attached to the chains are responsible for the ion exchanging properties.

Figure 1.3 Demineralisation Process

38

Cation exchanger:

Resins containing acidic functional groups (-COOH,-SO3H) are capable of exchanging

their H+ ions with other cations of hard water. Cation exchange resin is represented as

RH2.

Examples

i. Sulphonated coals.

ii. Sulphonate polystyrene.

R-SO3H: R-COOH = RH2

Anion exchanger:

Resin containing basic functional groups (-NH2, -OH) are capable of exchanging their

anions with other anions of hard water. Anion exchange resin is represented as R (OH)2.

Examples

i) Cross-linked quaternary ammonium salts.

ii) Urea-formaldehyde resin

R-NR3OH; R-OH; R-NH2 = R (OH)2

Process

The hard water first passed through a cation exchange column,

Which absorbs all the cations like Ca2+,Mg2+,Na+,K+,etc.,present in the hard water.

RH2 + CaCl2 RCa + 2HCl

RH2 + MgSO4 RMg + H2SO4

RH + NaCl RNa + HCl

The cation free water is then passed through a anion exchange column, which absorbs

all the anions like Cl-, SO42-, HCO3-, etc., Present in the water.

R’ (OH)2 + 2HCl R’Cl2 + 2H2O

R’ (OH)2 + H2SO4 R’SO4 + 2H2O

39

The water coming out of the anion exchanger is completely free from cations and

anions. This water is known as determineralised or deionized water.

Regeneration

When the cation exchange resin is exhausted, it can be regenerated by passing a

solution of dil. HCl or H2SO4

RCa + 2HCl RH2 + CaCl2

RNa + HCl RH + NaCl

Similarly, when the anion exchange resin exhausted, it can be regenerated by passing a

solution of dil.NaOH.

R’Cl2 + 2NaOH R’(OH)2 + 2NaCl.

Advantage of Ion-Exchange Process

i. Highly acidic or alkaline water can be treated by this process

ii. The water obtained by this process will have very low hardness

(nearly 2 ppm).

Disadvantage of ion exchange process

i. Water containing turbidity, Fe and Mn cannot be treated, because turbidity

reduces the output and Fe, Mn form stable compound with resin.

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

1.8.2 Internal conditioning

It involves the removal of scale forming substance, which were not completely

removed in the external treatment, by adding chemicals directly into the boiler. These

chemicals are also called boiler compounds.

40

1.8.2.1 Carbonate Conditioning

Scale formation can be avoided by adding Na2CO3 to the boiler water. It is used only

in low pressure boilers. The scale forming salt like CaSO4 is converted into CaCO3, which

can be removed easily.

CaSO4 + Na2CO3 CaCO3 + Na2SO4

1.8.2.2 Phosphate Conditioning

Scale formation can be avoided by adding sodium phosphate. It is used in high

pressure boilers. The phosphate reacts with Ca2+and Mg2+salts to give soft sludge’s of

calcium & magnesium phosphates.

3CaSO4 + 2Na3PO4 Ca3 (PO4)2 + 3Na2SO4

Generally 3 types of phosphates are employed.

(a) Trisodium phosphate – Na3PO4 (Too alkaline) – Used for too acidic water.

(b) Disodium hydrogen phosphate – Na2HPO4 (Weakly alkaline) – Used for

weakly acidic water.

(c) Sodium dihydrogen phosphate – NaH2PO4 (Acidic) – Used for alkaline water.

1.8.2.3 Calgon Conditioning

Calgon is sodium hexameta phosphate Na2 [Na4(PO3)6]. This substance interacts with

calcium ions forming a highly soluble complex and thus prevents the precipitation of scale

forming salt.

2CaSO4 + Na2[Na4(PO3)6] Na2[Ca2(PO3)6] + 2Na2SO4

The complex Na2 [Ca2(PO3)6] is soluble in water and there is no problem of sludge

disposal.

41

1.9 TREATMENT OF WATER FOR DOMESTIC SUPPLY

Rivers and lakes are the most common source of water used by municipalities. These

water should be free from colloidal impurities, domestic sewages, industrial effluents and

disease producing bacteria. Hence domestic supply of water involves the following stages

in the purification process.

i. Screening

It is a process of removing the floating materials like leaves, wood pieces, etc. from

water. The raw water is allowed to pass through a screen, having large no of holes, which

retains the floating materials and allow the water to pass.

ii. Aeration

The process of mixing water with air is known as aeration. The main purpose of

aeration is to remove gasses like CO2, H2S, and other volatile impurities causing bad taste

and odour to water and remove ferrous and manganous salts as insoluble ferric and

manganic salts.

Sources of water Screening Aeration

Filteration Coagulation Sedimentation

Sterilisation (or)

Disinfection

42

iii. Sedimentation

It is a process of removing suspended impurities by allowing the water to stand

undisturbed for 2-6 hours in a big tank. Most of the suspended particles settle down at the

bottom, due to forces of gravity, and they are removed. Sedimentation removes only 75%

of the suspended impurities.

iv. Coagulation

Finely divided clay, silica, etc do not settle down easily and hence cannot be removed

by sedimentation. Such impurities are removed by coagulation method.

In this method certain chemicals, called coagulants like alum, Al2 (SO4)3 etc., are added

to water. When the Al2 (SO4)3 is added to water, it gets hydrolyzed to form a gelatinous

precipitate of Al (OH)3. The gelatinous precipitate of Al (OH)3 entraps the finely divided

clay and colloidal impurities, settles to the bottom and can be removed easily.

v. Filtration

It is the process of removing bacteria, colour, taste, odour and suspend particles by

passing the water through filter beds containing fine sand, coarse sand and gravel.

The sand filter consists of a tank containing a thick top layer of fine sand followed by

coarse sand, fine gravel and coarse gravel. When the water passes through the filtering

medium, it flows through the various beds slowly. The rate of filtration decreases slowly

due to clogging of filtration becomes very slow, the filtration is stopped and the thick top

layer of fine sand is scrapped off and replaced with clean sand. Bacteria are also partly

removed by this process.

43

vi. Sterilization (Or) Disinfection

The process of destroying the harmful bacterias is known as sterilization or

disinfection. The chemical used for this purpose is called as disinfectants. This process can

be carried over by the following methods

1.By Boiling

When water is boiled for 10-15 minutes all the harmful bacteria are killed and water

becomes safe for use

Disadvantages

1. Boiling alters the taste of drinking water

2. It is impossible to employ it in municipal water works.

Water

Fine sand

Coarse sand

Fine gravel

Coarse gravel

Water outlet

1.4 Sand filter

44

2. By Using Ozone

Ozone is a powerful disinfectant and is readily absorbed by water. Ozone is highly

unstable and break down to give nascent oxygen.

O3 O2 + [O]

The nascent oxygen is a powerful oxidizing agent and kills the bacteria.

Disadvantages

1. This process is costly and cannot be used in large scale

2. Ozone is unstable and cannot be stored for long time

3. By Using Ultra Violet Radiations

UV rays are produced by passing electric current through mercury vapour lamp. This

is particularly useful for sterilizing water in swimming pool.

Disadvantages

1. It is costly

2. Turbid water cannot be treated.

4. By Chlorination

The process of adding chlorine to water is called chlorination. Chlorination can be

done by the following methods.

a) By Adding Chlorine Gas

Chlorine gas can be bubbled in water and it is a very good disinfectant.

45

b) By Adding Chloramine

When chlorine and ammonia are mixed in the ratio 2:1, a compound chloramine is

formed.

Cl2 + NH3 ClNH2 + HCl

When chloramine is added to water it decomposes slowly to give chlorine. It is a better

disinfectant than chlorine.

c) By Adding Bleaching Powder

When bleaching powder is added to water, it produces hypochlorous acid (HOCl).

HOCl is a powerful germicide.

CaOCl2 + H2O Ca (OH)2 + Cl2 Bleaching powder Cl2 + H2O HCl + HOCl Hypochlorous acid HOCl + Bacterias Bacterias are killed 1.9.1Break Point Chlorination

Water contains the following impurities

i) Bacteria’s

ii) Organic impurities

iii) Reduction substances (Fe2+, H2S, etc.)

iv) Free ammonia

Chlorine may be added to water directly as a gas or in the form of bleaching powder.

When chlorine is a applied to water, the result obtained can be depicted graphically as

given below. The graph shows the relationship between the amount of the chlorine added

to water and the residual chlorine.

46

It is seen from the graph that initially the applied chlorine is used to kill the bacteria

and it oxidises all the reducing substances present in the water and there is no free

residual chlorine.

As the amount of applied chlorine increases, the amount of combined residual

chlorine also increases. This is due to the formation of chloramines and other

chlorocompounds.

At one point, on further chlorination the oxidation of chloramines and other impurities

starts and there is a fall in the combined chlorine content. Thus the combined residual

chlorine decreases to a minimum point at which oxidation of chloramines and other

impurities complete and free residual chlorine begins to appear, this minimum point is

known as ”break point chlorination”.

Applied chlorine

Figure 1.5 Break point chlorination

Thus the break point chlorination eliminates the bacterias, reducing substances, organic

substances responsible for the bad taste and odour, from the water.

Formation of chloramines&

chloro compounds

Destruction of

chloramine &chloro

compounds

Kill

the

bact

eria

, and

oxi

datio

n of

re

duci

ng c

ompo

unds

by

chlo

rine

Resi

dual

chl

orin

e

Free residual chlorine

47

1.10. Reverse osmosis

When two solutions of different concentrations are separated by a semi – permeable

membrane, solvent (water) flows from a region of lower concentration to higher

concentration. This process is called Osmosis. The driving force in this phenomenon is

called osmotic pressure.

Figure 1.6 Reverse Osmosis

If a hydrostatic pressure, in excess of osmotic pressure supplied on the higher

concentration side, the solvent flow is reversed i.e., solvent flows from higher

concentration side to lower concentration side. This process is called reverse osmosis.

Thus, in the process of reverse osmosis pure water is separated from salt water. This

process is also known as super – filtration. The membranes used are acetate, cellulose

butyrate, etc.

Advantages:

1. The life time of the membrane is high, and it can be replaced within few minutes.

2. It removes ionic as well as non – ionic, colloidal impurities.

3. Due to low capital cost, simplicity, low operating, this process is used for converting

sea water into drinking water.

48

IMPORTANT QUESTIONS

1. Explain how sterilization of water carried out using chlorine? Give the mechanism.

2. (i) What is meant by carbonate and non-carbonate hardness of water ?Explain with

example

(ii) What is break-point chlorination? State its significance

3. Write a detailed procedure for the determination of various forms of alkalinity.

4. (i) Define the term desalination with a neat diagram, describe desalination by

reserve osmosis’ method.

(ii) What are coagulants? Write the mechanism of coagulation process with suitable

example.

5. What are the boiler troubles? How are they caused? Suggest steps to minimize the

boiler troubles.

6. What is desalination? How is this achieved by reverse osmosis?

Explain break point chlorination, State its significant.

7. Compare the zeolite process with ion-exchange process in water softening. How will

you regenerate the used up reagents.

8. (i) How is the hardness of water determined by EDTA method

(ii) Describe briefly the different steps in the purification of water for drinking

purposes. What is the usage of breakpoint chlorination? (or) Out line the various

stages of domestic water treatment in sequence.

9. What is meant by sterilization of water? What are the chemicals that are normally

used for this purpose? Explain break-point chlorination.

(i) Distinguish between softwater and demineralized water.

(ii) How do you estate the total hardness of water by EDTA method? Explain.

10. Explain how sterilization of water carried out using chlorine? Give the mechanism.

49

11. Explain with a neat sketch the various steps in the treatment of water for municipal

supply.

12. What is the principle of EDTA method? Describe the estimation of hardness by

EDTA method.

(or)

Explain the principle and procedure involved in the determination of permanent and

temporary hardness by EDTA method.

13. Describe de-mineralisation process of water softening. Explain the reaction involved.

14. What is potable water? What are the steps taken to obtain pure drinking?

15. How is water disinfected by chlorine.

16. Describe the principle and method involved in the determination of different types

and amount of alkalinity of water.

17. 0.28 gm of CaCO3 was dissolved in HCl and the solution made upto one litre with

distilled water. 100 ml of the above solution required 28 ml of EDTA solution on

titration.100 ml of hardwater sample required 33 ml of same EDTA solution on

titration. 100 ml of this water, after boiling cooling and filtering required 10 ml of

EDTA solution on titration. Calculate the temporary and permanent hardness of

water.

18. Explain the various steps involved in the domestic water treatment.

19. How is the exhausted resin regenerated in an ion-exchanger? What are the merits

and demerits of ion-exchange method?

20. What are the requisites of water for municipal supply. How is raw water treated for

domestic purpose.

21. Discuss the causes and prevention of priming and foaming.

22. What is the principle of reverse osmosis. How is it used for desalination process.

23. Discuss desalination by reverse osmosis process.

24. Explain

(i) Phosphate conditioning.

50

(ii) Sedimentation with coagulation.

25. What is desalination. Name the different methods of desalination. Explain any one in

detail.

26. What is caustic embrittlement. How can it be prevented.

27. What are scales and sludges. Describe the disadvantages of scale and sludge

formation.

28. What are the problems one would face when hard water is used in boiler industries.

Question Bank

1. Define hard water and soft water.

Water, which does not produce lather with soap solution, but produce white

precipitate (scum), is called hard water.

Water which produce lather readily with soap solution is called soft water.

2. Define hardness of water.

Hardness is the property or characteristics of water which does not produce lather

with soap.

Hardness can be detected by treating water with soap.

2C17H35COONa + CaCl2 (C17H35COO)2 Ca + 2NaCl

Soap Hardness causing substance Hard soap

3. Distinguish between carbonate hardness (CH) and non-carbonate hardness (NCH).

S.NO Carbonate hardness (or)

Temporary hardness

Non-carbonate hardness(or)

permanent hardness

1 It is due to the presence of

bicarbonates of calcium and

It is due to chlorides and

sulphates of calcium and

51

magnesium. magnesium.

2 It can be removed by boiling

the water.

It cannot be removed by

boiling the water

3 It is also called as alkaline

hardness.

It is also called as non-alkaline

hardness.

4. How is the hardness of water expressed? (or) bring out the significance of calcium

carbonate equivalents?

The concentration of hardness producing salts are usually expressed interms of an

equivalent amount of CaCO3.

Significance: Its molecular weight is a whole number and it is the most insoluble

salt.of the concentration of hardness producing salt is X mgs/lit, then

Amount equivalent to

X x molecular weight of CaCO3 CaCO3 = -------------------------------------------------------------- Molecular weight of hardness producing substances 5. Write the units of hardness (or) bring out the relationship between ppm and mg/lit.

(i) parts per million (ppm) : it is defined as the no of parts of CaCO3 equivalent

hardness per 106 parts of water.

(ii) milligrams per litre(mg/lit) : it is defined as the no of milligrams of CaCO3

equivalent hardness per 1 litre of water.

(iii) clarke’s degree (°Cl) : it is defined as the no of parts of CaCO3 equivalent hardness

per 70,000 parts of water.

(iv) French Degree (°Fr) : It is defined as the number of parts of CaCO3 equivalent

hardness per 105 parts of water.

Relationship between ppm and mg/lit:

1 mg/lit= 1 mg of CaCO3 equivalent in 106 parts of water

= 1 part of CaCO3 equivalent hardness in 106 parts of water.

52

= 1 ppm.

6. A water sample contains 73 mg of Mg (HCO3)2 per litre. Calculate the hardness

interms of CaCO3 equivalents.

Given : Amount of Mg (HCO3)2 = 73 mg/lit

Mol. Wt. of Mg (HCO3)2 = 146

Amount equivalent to CaCO3 = = 50 mgs/lit

7. How does Eriochromo-Black-T indicator function as an indicator in EDTA titration?

Why NH4Cl-NH4OH buffer is used in EDTA titration.

This complexation is possible only at pH = 8-10, that can be maintained by this buffer.

8. What is meant by permanent hardness of water?mention the salts responsible for

the permanent hardness of water.

This is due to the presence of chloride and sulphates of calcium and magnesium.

CaCl2, CaSO4, MgCl2 and MgSO4 are responsible for permanent hardness of water.

9. Draw the structure of EDTA. What happens when EDTA is added to hard water ?

HOOCH2C CH2COOH N - CH2 - CH2 - N HOOCH2C CH2COOH When EDTA is added to hardwater, it forms stable complex with hardness producing ions

like Ca and Mg the pH range = 8 -10

[Ca++, Mg++] + EDTA ------------ [ Ca, Mg EDTA] complex

10. Why is water softened before using in boiler ?

53

If hard water obtained from natural sources is fed directly to the boilers, the following

troubles may arise.

1. Scale and sludge formation

2. Priming and foaming (carry over)

3. Caustic embrittlement

4. Boiler corrosion.

11. What are scales and sludges ?

1. Scales

Scale forms hard and adherent coating on the inner walls of the boiler, it is called scale.

Scales are formed by the substances like Ca (HCO3)2, CaSO4 and MgCl2.

2. Sludge

If the precipitate is loose and slimy it is called sludge. Sludges are formed by

substances like MgCl2, MgCO3, MgSO4 and CaCl2. They have greater solubilities in hot

water than cold water.

12. What is meant by priming and foaming? How can they be prevented?

Priming is the process of production of wet steam. Priming can keep prevented by

controlling the velocity of steam and keeping the water level lower.

Foaming is the formation of stable bubbles above the surface of water. Foaming can be

prevented by adding coagulants like sodium aluminate and antifoaming agents like

synthetic polyamides.

13. What is meant by caustic embrittlement? How is it prevented?

Caustic embrittlement means intercrystalline cracking of boiler metal.

Prevention

Caustic embrittlement can be prevented by

54

(i) Using sodium phosphate as softening agent instead of sodium carbonate

(ii) By adding tannin, lignin to the boiler water, which blocks the hair cracks

14. How does boiler corrosion arise?

Boiler corrosion arise due to the presence of

(i) Dissolved oxygen

(ii) Dissolved carbon dioxide

(iii) Dissolved salts

15. What are the requirements of drinking and boiler feed water?

(i) Boiler feed water Must have zero hardness and free from dissolved gases like O2, CO2.

(ii) Drinking water (i) pH of water should be in the range of 7.0 – 8.5. (ii) Total hardness and dissolved solids of water should be less than 500 ppm

16. What are the advantages of ion-exchange process.

(i) Highly acidic or alkaline water can be treated by this process.

(ii) The water obtained by this process will have very low hardness (nearly 2 ppm).

17. How is exhausted resin regenerated in ion-exchange process?

When the cation exchange resin is exhausted, it can be regenerated by passing a

solution of dil. HCl or H2SO4.

RCa + 2HCl --- RH2 + CaCl2

RNa + HCl --- RH + NaCl

Similarly, when the anion exchange resin is exhausted, it can be regenerated by

passing a solution of dil. NaOH.

R’Cl2 + 2NaOH --- R’(OH)2 + 2NaCl

18. Give some examples for cation exchange resins.

55

(i) Sulphonate coals

(ii) Sulphonate polystyrene

19. Give some example for anion exchange resin.

(i) Cross – linked quaternary ammonium salts

(ii) Urea – formaldehyde resin

20. How is water demineralized in an ion - exchanger?

The water containing ions (both anion and cations) are passed through ion exchange

columns, which absorb all the ions (anions and cations ) as shown below.

Cation exchanger : RH2 + CaCl2 ------ RCa + 2HCl.

Anion exchanger : R(OH)2 + 2HCl ---- R’Cl2 + 2H2O.

21. Write the structure of sand filter.

The sand filter contains a thick top layer of fine sand followed by coarse sand, fine

gravel and coarse gravel.

22. How is boiler corrosion, due to dissolved oxygen, removed? How?

Sodium sulphite, hydrazine are some of the chemicals used for removing

dissolved oxygen from water.

2Na2SO3 + O2 -- 2Na2SO4

N2H4 + O2 -- N2 + 2H2O

23. What is aeration of water? Mention its purpose.

The process of mixing water with air is known as aeration. The main purpose of

aeration is

(i) To remove gases like CO2, H2S and other volatile impurities causing bad taste and

odour to water.

56

(ii) To remove ferrous and manganous salts as insoluble salts as insoluble ferric and

manganic salts.

24. What is carbonate conditioning? For what type of boiler it is used?

Scale formation can be avoided by adding Na2CO3 to the boiler water. It is used only

in low pressure boilers. The scale forming salt like CaSO4 is converted into CaCO3, which

can be removed easily.

CaSO4 + Na2CO3 --- CaCO3 + Na2SO4.

25. Write briefly on disinfection of water by UV treatment.

When the water containing bacteria is irradiated by UV light, all the bacterias are

killed out. This process is known as disinfection. This is useful for sterilizing water in

swimming pool.

26. Write the principle involved in the desalination of water by reverse osmosis.

(Or)

What is meant by ‘reverse osmosis’? How is it applied in the desalination of water?

If the process in excess of osmotic pressure is applied on the higher concentration side,

the solvent flow is reversed i.e., solvent flows from higher concentration to lower

concentration side. This process is called reverse osmosis.

Salt water is taken as higher concentration and water is taken as solvent of pressure is

applied on the salt water, the water flows from salt water to water side.

27. Define the term break point chlorination. (Or)

What is break point chlorination? Explain

Break point chlorination is the point at which all the impurities are removed and free

chlorine begins to appear.

28. What are the disadvantages of using ozone in disinfection of water?

57

(a) This process is costly and cannot be used in large scale.

(b) Ozone is unstable and cannot be stored for long time.

58

UNIT-II

ELECTROCHEMISTRY

Introduction

Electrochemistry deals with chemical changes produced by an electric current and

the production of electricity by chemical reactions. All electrochemical reactions involve

transfer of electrons and are redox reactions. Electrochemistry is a branch of physical

chemistry, which deals with the relationship between chemical energy and electrical

energy. Electrochemistry is the basis for batteries, cells, corrosion and its prevention,

metal coatings, etc. And holds a central position in the study of chemistry for a number of

reasons:

(i) It acts as a bridge between thermodynamics and the rest of the chemistry because

it provides techniques both for measuring the thermo dynamical state functions

and for predicting equilibrium concentrations of dissolving and reacting ions.

(ii) It is of enormous commercial importance because of the costly destruction caused

by corrosion and the exciting possibilities of fuel cells which generate electricity

directly from fuel.

From the above valid reasons, the subject electrochemistry consists of major

technological importance.

59

2.1. Basic Definitions:

2.1.1 Conductors

Conductors are the substances or materials, which allow electric current to pass through

them.

Examples: all metals, graphite, fused salts, aqueous solutions of acids, bases and salts.

2.1.2 Non-conductors or Insulator

Non-conductors or Insulators are substances, which do not conduct electric current

through them.

Examples: Wood, rubber.CCl4

2.1.3 Conductors are classified into two types namely:

(i) Metallic conductors

(ii) Electrolytic conductors.

Metallic conductors are the substances, which conduct electricity, but are not

decomposed by the process. Conduction takes place by the transference of electrons. No

chemical change takes place in the conductor during electrical conduction. As the

temperature increases, the conducting power of a conductor generally decreases.

Examples: All metals, graphite.

Electrolytic conductors are the substances, which conduct electricity, but

decomposition of electrolyte takes place during the process. Conduction takes place by the

movement of ions in solution or in fused electrolyte. Chemical change takes place in the

conductor/electrolytic solution during electrical conduction. As the temperature

increases, the conducting power of a conductor/electrolyte also generally increases.

Examples: Acids, bases and electrovalent salts.

60

Table 2.1 Differences between metallic conduction and electrolytic conduction

S.No Metallic conduction Electrolytic conduction

1 It involves the flow of electrons in

a conductor

It involves the movement of ions in a

solution.

2 It does not involve any transfer of

matter.

It involves transfer of electrolyte in the

form of ions.

3 Conduction decreases with

increase in temperature.

Conduction increases with increase in

temperature.

4 No change in chemical properties

of the conductor.

Chemical reactions occur at the two

electrodes.

2.1.4. Electrical Conductance

The unit of electrical current is ampere. The unit of quantity of electricity is

coulomb. When one ampere (I)of current is passed for one second(t), then the quantity of

current passed is one coulomb(Q).

(i.e.,) Q = I x t Coulomb

Ohm’s law:

This law can be stated as, at constant temperature, the strength of the current

flowing through a conductor is directly proportional to the potential difference and

inversely proportional to the resistance(R) of the conductor.

RVI , RIV , V = volt ampereI , R = ohm

Specific resistance:

The resistance (R) ohms offered by the material of the conductor to the flow of

current through it is directly proportional to its length (l) and inversely proportional to the

area of cross section (a) of the conductor. Thus,

61

alR

ρ called the specific resistance and it is resistance in ohms which one meter cube of

material offers to the passage of electricity through it, Unit of specific resistance is ohm-

meter.

Specific conductance:

The reciprocal of specific resistance is called as specific conductance (or) specific

conductivity (k) [k is called ‘kappa’].k is defined as the conductance of one meter cube of

an electrolyte Solution

al.

R11k

al.

R1k

Unit of specific conductance is ohm-1 m-1 (or) mho.m-1

Since ohm-1 = mho

k = Ohm1 .

2cmcm

k = Ohm-1.cm-1

Also, 1 siemen = 1 mho.

k is also expressed as S.m-1.

Equivalent conductance:

Equivalent conductance ( C) is defined as the conductance of an electrolyte

solution containing one gram equivalent of the electrolyte. It is equal to the product of

specific conductance (k) of the electrolytic solution and the volume (V) of the solution that

contains one gram equivalent of the electrolyte.

Vkc

In general if an electrolyte solution contains C gram-equivalents in1, 000 cc of the

62

(C is concentration of solution in normality )solution the volume of the solution containing

1 gram equivalent will be

c1000kc mho.m2.(gram.equiv) -1

c values depend on the type of the electrolyte, concentration of the solution and

temperature.

2.2 Kohlraush’s Law

This law states that, ‘‘at infinite dilution wherein the ionisation of all electrolytes is

complete, each ion migrates independently and contributes a definite value to the total

equivalent conductance of the electrolyte’’. Consider an electrolyte AB in aqueous

solution. It dissociates as

Am Bn mAn+ + nBm-

and

are the cationic and anionic equivalent conductance at infinite

dilutions and n+ and m– correspond the valency of cations and anions furnished by each

molecule of the electrolyte

, NaCl = Na+ + Cl–

, BaCl2 = 1/2 Ba2+ + Cl–

, AlCl3 = 1/3 Al3+ + Cl–

For weak electrolytes,

, CH3COOH = H+ + – CH3COO–

, NH4OH = NH4+ + – OH–

63

Application of Kohlraush’s law:

The important use of Kohlraush’s law is to deduce the values of weak

electrolytes correctly by arithmetically combining the values of strong electrolytes in

appropriate manner.

For example

of CH3COOH, which is a weak electrolyte, is deduced from values of NaCl,

HCl, and CH3COONa in such a manner that of CH3COOH is obtained. Sodium acetate

(CH3COONa) is a strong electrolyte and it ionises to acetate (CH3COO–) and sodium (Na+)

ions at all concentrations in water. Applying Kohlraush’s law,

CH3COONa + HCl – NaCl = CH3COOH

This method produces agreeable values of for weak electrolytes. Similarly

NH4OH can be deduced as,

NH4OH = NH4Cl + NaOH – NaCl

Example 1 :The equivalent conductances at infinite dilution of HCl, CH3 COONa and

NaCl are 426.16,91.0 and 126.45 ohm–1 cm2 gm.equiv-1 respectively. Calculate the of

acetic acid.

CH3COOH = CH3COONa + HCl - NaCl

= 426.16 + 91.0 – 126.45

CH3COOH = 391.16 ohm–1 cm2 gm.equiv-1

64

2.3 Conduct metric Titration

This method of estimation is applicable to any titration in which there is a sharp

change in conductance at the end point .The principle involved in these titrations is that

electrical conductance depends upon the number and mobility of the ions. In these

titrations it is a necessary to observe the following.

1. To keep the temperature constant throughout the experiment.

2. The titrant solution should be 10 times stronger than the solution to be titrated so

that the volume change is as little as possible.

Conductance is followed during the course of titration and the values are plotted

against volume of the titrant added. Conductrometric curves should be straight line or

nearly straight lines and only few readings on each sides of the end point are required

.The end point of titration is the point of intersection of the two straight lines.

2.3.1. Titration of a strong acid against a strong base (HCl Vs NaOH)

Consider the titration of a strong acid (say, HCl) with a strong base(NaOH). A

definite volume of the acid is pipetted out in to the conductivity cell or in to a beaker

where the conductivity cell is dipping and the alkali is taken in the burette.

At the beginning of the titration the conductance of the HCl acid solution is due to

the H+ and Cl - .As alkali is added gradually from the burette, fast moving H+ ions are

replaced by slow moving sodium ions and the reaction is.

H+ + Cl – + [Na++OH-] Na+ + Cl- + H2O

Hence on continuous addition of sodium hydroxide, the conductance goes on

decreasing till the neutralization of acid is completed. After the neutralization, further

addition of alkali results in the increase in conductance of the solution in the cell.The

65

points will lie on two straight lines as in figure (A).The point of intersection of these two

straight lines AB and CD gives the volume of alkali required for the neutralization

| Volume of NaOH

2.3.2. Titration of a weak acid against a strong base (CH3COOH Vs NaOH)

Consider the titration of a weak acid acetic acid against a strong alkali, sodium

hydroxide. At the beginning of the titration the conductance of the solution will be very

low because acetic acid is a weak electrolytes and is feebly ionized. When a small quantity

of NaOH is added from the burette to acetic acid, the conductance will increases due to

the formation of highly ionized sodium acetate.

CH3COOH + [Na++OH-] CH3COO-+ Na+ + H2O

After the completion of neutralization of the acid, any further addition of alkali will

show a sharp increase in conductance due to the sodium and the fast moving hydroxide

ions. The plot of conductivity against volume of alkali added is shown in the figure (B).the

point of intersection straight lines gives the end point

Con

duct

ance

(ohm

-1)

A

A D

B C

l

l

66

Volume of NaOH

Fig 2.2

2.3.3 Titration of a mixture of acids (CH3COOH + HCl) against a strong base (NaOH)

(CH3COOH + HCl ) Vs NaOH

When a mixture of acetic acid and hydrochloric acid is titrated against sodium

hydroxide, usually a combination of figures (2.1 &2.2) will be obtained. It is as shown in

the figure 2.3. In these titrations the strong acid (HCl) will get titrated first and the titration

of the weak acid (CH3COOH) will commence only after the complete neutralization of the

strong acid. It is evident from the figure 2.3 , that the first end point corresponds to the

neutralization of HCI, the second end point corresponds to the neutralization of acetic

acid.

Volume of NaOH Fig 2.3

Con

duct

ance

(ohm

-1)

A

D

B

C

l

l

l

Con

duct

ance

(ohm

-1)

A D

B

C l

l

l

E

l

l

67

2.3.4 Precipitation titration (BaCl2 Vs Na2SO4)

Let us now see how the conductance varies during the course of titration of barium

chloride against sodium sulphate. Here BaSO4 gets precipitated; while chloride ions

remain unchanged.

[Ba2++2Cl-]+[2 Na+ + SO42-] BaSO4 +2 Na+ +2Cl-

As a result of addition of Na2SO4 Solution to BaCl2 solution, Ba2+ ions in solution

are replaced by free Na+ ions. Since the mobility of Na+ ions is less than that of Ba2+ ions,

so the conductance of resulting solution falls. However, if the addition of Na2SO4 Solution

is continued beyond the equivalence point, the conductance value starts increasing, due to

the rise in total concentration of free ions in solution,i.e., there will now be increase in free

Na+ and SO42- ions, in addition to the Na+ and Cl-ions already present. This is shown in

fig.2.4

Volume of Na2SO4

Fig 2.4

Advantages of conductometirc titration

1) No special care is necessary near the end point it is ascertained graphically.

2) Colored solution which cannot be titrated by ordinary volumetric methods with the

help of indicators can be titrated.

Con

duct

ance

(mho

-1) A D

B

C

l

l

l

68

3) The titrations of weak bases can be performed conductometrically .These are not

possible in conventional volumetric titrations because they do not produce sharp

change in the colour of indicator

4) Very dilute solutions can be t

unit i water technology - department of chemistry · 2018. 2. 9. · water technology 1.1...

Documents