Water is the driving force in nature, we never know the worth of water till the river is dry.
Water is a natural wonder and is the most common, important, useful thing for surviving of all the living beings. Without food, living beings can survive for some days but without water nobody can survive. Seventy percent of our body contains water, which regulates life processes such as digestion of food, transportation of nutrients, and excretion of body wastes. It regulates the body temperature by the process of sweating and evaporation. Water acts as a universal solvent; due to this reason, water is widely used in laboratories, irrigation, steam generator, industrial purpose, fire fighting, etc. Besides it is used for bathing, drinking, sanitary purposes, etc.
From an engineer‚s point of view too, very important, without water nothing will happen. It is required in boilers for production of steam, which acts as a source of energy and a coolant in many power and chemical plants and many other industries.
Water present on earth passes through a remarkable cycle of changes (as shown in Figure 1.1):
Figure 1.1 Flow diagram of sources of water
Water may contain various impurities due to
The common impurities present in natural water can be classified into four groups that are as follows and shown in Figure 1.2:
Figure 1.2 Types of impurities
Dissolved gases – NO2, CO2, SO2, etc., which are soluble in water and make it impure.
Dissolved inorganic salts or ions
Cations: Ca2+, Mg2+, Na+, K+, Fe2+, Al3+, Zn2+, etc.
Anions:, , , Cl–, etc.
Inorganic – sand, clay, lime, etc.
Organic – Plant and animal materials like discarded vegetables, dry leaves, dead materials, etc.
Finely divided silica, clay, organic products, colouring matter, etc.
Various pathogenic microorganisms such as bacteria, fungi, virus, etc.
The various types of impurities present in the water impact certain properties in water.
Presence of different chemicals impart colour, odour and taste to the water.
Presence of dissolved salt makes the water hard.
Excess quantities of metals and dissolved gases make the water corrosive in nature.
Presence of pathogenic bacteria in water makes it unfit for drinking or domestic purposes.
Suspended matter create turbidity to the water.
Depending on salts presents in water and reaction with soap, water is categorized into hard water and soft water. Hardness is the characteristic of water by which water does not produce lather with soap. It is due to presence of chlorides, sulphates and bicarbonates salts of magnesium, calcium and other heavy metals [CaCl2, CaSO4, MgCl2, MgSO4, Ca(HCO3)2, Mg(HCO3)2, etc]. When hard water is treated with soap, it does not produce lather, rather it forms a white scum. Soap is the sodium or potassium salt of higher fatty acids like stearic acid [C17H35 COONa – sodium stearate].
Water which can produces lather with soap easily is called soft water.
Depending on salts present in the water, hardness is of two types, i.e., temporary hardness and permanent hardness.
We know that hardness of water is due to the presence of number of dissolved salts in water but for comparing the hardness of different samples of water of varying composition, it is necessary to choose a reference standard. For this purpose, hardness of water is expressed in terms of equivalents of calcium carbonate only. The following are the reasons for choosing CaCO3 as a standard for expressing the hardness:
Hence whatever amount of dissolved salts is present in water, it is first converted into calcium carbonate equivalents by using the formula:
W = Mass of hardness-producing substance in mg/L
E = Equivalent weight of hardness-producing substance
or
Salts responsible for hardness is given in Table 1.1 and method to calculate the hardness.
Various units used for expressing hardness of water are given below:
1 ppm = 1 part of CaCO3 equivalent hardness in 106 parts of H2O
1 mg/L = 1 mg of CaCO3 equivalent per 106 mL of water
= 1 part of CaCO3 equivalent per 106 parts of water
= 1 ppm
1 °Fr = 1 part of CaCO3 equivalent hardness per 105 parts of water.
1 °Cl = 1 part of CaCO3 equivalent hardness per 70,000 parts of water
Relationship between various units of hardness is shown in Table 1.2.
Solution
Ca(HCO3)2 = 16.2 mg/L, Mg(HCO3)2 = 7.3 mg/L
MgCl2 = 9.5 mg/L, CaSO4 = 13.6 mg/L
Solution
CaCO3 equivalent =
For (i) Ca(HCO3)2, CaCO3 equivalent == 10 mg/L
(ii) Mg(HCO3)2, CaCO3 equivalent == 5 mg/L
(iii) MgCl2, CaCO3 equivalent = = 10 mg/L
(iv) CaSO4, CaCO3 equivalent == 10 mg/L
Now
MgCl2 = 0.143°Fr, MgSO4 = 0.572°Fr, CaSO4 = 0.286°Fr, and Ca(HCO3)2 = 2.316°Fr. Calculate the hardness in ppm.
Solution Since 1°Fr = 10 ppm
CaCO3 equivalent =
0.143°Fr of MgCl2 = 1.43 ppm = ppm CaCO3 equivalent = 1.51 ppm
0.572°Fr of MgSO4 = 5.72 ppm = ppm CaCO3 equivalent = 4.77 ppm
0.286°Fr of CaSO4 = 2.86 ppm = ppm CaCO3 equivalent = 2.11 ppm
2.316°Fr of Ca(HCO3)2 = 23.16 ppm = ppm CaCO3 equivalent = 14.29 ppm
= 22.68 ppm
Solution
Strength of hardness in terms of a CaCO3 equivalent
=
Hardness = 720 ppm
Solution
Carbonate hardness = Due to bicarbonate of Ca and Mg
= 250 ppm
Non-carbonate hardness = (Hardness due to permanent Ca2+ and Mg2+) –
(Hardness due to ion)
= (200 + 200) – (250)
Non-carbonate hardness = 150 ppm
We know that there are two types of hardness of water, i.e., temporary and permanent hardness. Temporary hardness is due to bicarbonate of calcium and magnesium, and permanent hardness is due to chlorides and sulphates of calcium and magnesium.
The hardness of water can be determined by complexometric titration by using ethylenediamine tetra acetic acid [EDTA] commonly known as EDTA method.
EDTA Method: It is the most important and more accurate method to determine the hardness of water. EDTA has limited solubility in water, Hence, disodium salt of EDTA is used which is soluble in water.
Principle: EDTA can from complex with salts (Ca2+ and Mg2+) which are present in hard water. Hence, it is known as complexometric titration. Calcium or magnesium ions present in the water sample with ammonical buffer solution form an unstable wine red colour complex with Eriochrome Black T (EBT) indicator. When it is titrated with EDTA solution the metal ions present in water give a stable deep blue colour (M-EDTA) complex and releases the free indicator.
The formula of EDTA is written as
Ethylene diamine tetra acetic acid
Disodium salt of EDTA is
It is represented as Na2H2Y. It ionizes in aqueous solution to give 2Na+ ion and a strong chelating ion represented as H2Y2–.
It is a hexadentate ligand, and it forms complexes with bivalent cations (Mg2+, Ca2+, etc.), and these complexes are stable in alkaline medium (pH 8–10).
EBT may be represented as:
{sodium 4-(1-hydroxy-2-napthylazo)-3-hydroxy-7-nitronapthalene-1-sulphonate}
The EBT has two ionisable phenolic hydrogen atoms, and it is represented as; indicator EBT gives different colours at different pH values.
End point: During titration, the colour of the solution changes from wine red to pure blue.
Reactions involved during titration:
The metal-EDTA complex may be represented as:
[M-EDTA Complex]
Step I: Standardization of EDTA solution: Rinse and fill the burette with EDTA solution. Pipette out 50 ml of standard hard water (S.H.W)/Standard MgSO4 solution in a conical flask. Add 10–15 ml of buffer solution and two drops of EBT indicator. Titrate the flask solution against the EDTA solution from the burette until the colour changes from wine red to pure blue, it is end point. Repeat the procedure to get two concordant readings. Let the volume of EDTA be consumed as V1 ml.
Step II: Determination of total hardness: Titrate 50 ml of unknown water sample with EDTA solution by addition of 10–15 ml of buffer solution and two drops of EBT indicator till the wine red colour changes to pure blue. Let the volume of EDTA be consumed as V2 ml.
Step III: Determination of permanent hardness: Take 250 ml of water sample in a 500 ml beaker and boil gently for half an hour. Cool, filter, and wash the precipitate with distilled water, collecting filtrate and washing in a 250 ml measuring flask, and make the volume up to the mark. Now titrate 50 ml of boiled water sample same as in step I. Let the volume of EDTA be consumed as V3 ml.
1 ml of standard hard water = 1 mg of CaCO3
Step I: Standardization of EDTA solution:
Volume of S.H.W taken for titration = 50 ml
Concordant volume of EDTA used = V1 ml
Now,
V1 ml of EDTA = 50 ml of S.H.W = 50 × 1 mg of CaCO3
1 ml of EDTA =mg of CaCO3
Step II: Determination of total hardness:
Volume of unknown water sample taken for titration = 50 ml
Volume of EDTA used = V2 ml
Now,
50 ml of unknown water sample = V2 ml of EDTA = mg of CaCO3
1 ml of unknown water sample = mg of CaCO3
1 L (1000 ml) of unknown water sample = mg/L or ppm
i.e., total hardness = ppm
Step III: Determination of permanent hardness:
Volume of hard water sample taken after boiling and filtering = 50 ml
Let concordant volume of EDTA used = V3 ml
50 ml of boiled water = V3 ml of EDTA = mg of CaCO3
1 ml of boiled water = mg of CaCO3
1 L (1000 ml) of boiled water = mg/L or ppm
Permanent hardness = mg/L
Hence,
Total hardness = ppm
Permanent hardness = ppm
Temporary hardness = Total hardness – permanent hardness
=
= ppm
Solution
Step-I Standardization of EDTA solution
1L (1000 mL) of SHW = 1 gm (1000 mg) of CaCO3
= 1 mg of CaCO3
46 ml of EDTA solution required = 50 ml of SHW = 50 × 1 mg of CaCO3
1 ml of EDTA solution = mg of CaCO3
Step-II Determination of total hardness
50 ml of the given hard water sample required = 20 ml EDTA solution
= 20 × mg of CaCO3
1 ml of the given hard water sample required = × mg of CaCO3
1L (1000 ml) of the given hard water sample required = ×1000 mg/L
= 434.78 mg/L
Step-III Determination of permanent hardness
50 ml of boiled water sample required = 10 ml of EDTA solution
= 10 × mg of CaCO3 eq.
1 ml of boiled water sample required = mg of CaCO3 eq.
1L (1000 ml) of boiled water sample required = ×1000 mg/L
= 217.39 mg/L
Total hardness = 434.78 ppm
Permanent hardness = 217.39 ppm
Temporary hardness = Total hardness – Permanent hardness
= 434.78 – 217.39
= 217.39 ppm
Solution 1000 ml H2O contains 1.5 g of CaCO3
1 ml of SHW = 1.5 mg of CaCO3
Step-I Standardization of EDTA solution
25 ml EDTA required = 20 ml of SHW = 20×1.5 mg of CaCO3
1 ml EDTA required = mg of CaCO3
Step-II Determination of total hardness
100 ml of hard water sample required = 18 ml EDTA solution
= mg of CaCO3 eq.
1 ml of hard water sample required = mg of CaCO3 eq.
1L (1000 ml) of hard water sample required = mg/L
Total hardness = 216 mg/L
Step-III Determination of permanent (non-carbonate) hardness
100 ml of boiled water sample required = 12 ml of EDTA
100 ml of boiled water sample required = mg of CaCO3 eq.
1 ml of boiled water sample required = mg of CaCO3 eq.
1L (1000 mL) of boiled water sample required = mg/L
= 144 mg/L
Total hardness = 216 ppm
Non-carbonate hardness = 144 ppm
Carbonate hardness = Total hardness – Non-carbonate hardness
= 216 – 144
Carbonate hardness = 72 ppm
Temporary hardness = 25 ppm
Permanent hardness = 20 ppm
Permanent Mg hardness = 15 ppm
Solution Lime requirement = (Temp. hardness + perm. Mg hardness) × Volume of water
= = 444000 mg = 444 g
Soda requirement = (Perm. hardness) × volume of water
= = 318 g
= 444 g
Soda requirement = 318 g
CaCO3 = 10.0 mg/L MgCO3 = 8.4 mg/L
CaCl2 = 11.1 mg/L MgSO4 = 6.0 mg/L
SiO2 = 1.2 mg/L
assuming the purity of lime as 90% and soda as 95%.
Solution
Lime requirement = (Temp. Ca2+ + 2 × Temp. Mg2+ + Perm. Mg2+) × Volume of water × purity factor
=
Lime requirement = 0.5755 kg
Soda requirement = [Perm.(Ca2+ + Mg2+)] × Volume of water × purity factor
=
= 0.3347 kg
Solution
Lime requirement = (H2SO4 + MgSO4 as CaCO3 equivalent) × volume of water
=
=
= 814 kg
For April 2008, total lime requirement = 814 × 30
= 24420 kg
Given cost of lime = Rs. 5/kg
Total cost of lime = 24420 kg ×
= 1,22,100 Rs.
Similarly, Soda requirement per day
= 2226 kg
For April 2008 (30 days), total soda requirement = 2226 × 30 = 66780 kg
Given cost of soda = Rs. 8.00/kg
= 66780 ×
= Rs. 5,34,240
Amount of oxygen dissolved in water (mg/L) is known as dissolved oxygen. At ambient conditions of temperature and pressure, the solubility of oxygen is about 8 mg/L. The amount of dissolved oxygen measures the biological activity of the water bodies, and this is most essential for sustaining aquatic life. Estimation of DO content in a particular water body is of important significance of environmental as well as the industrial point of view. This serves as an indicator of the extent of pollution of water by oxidizable and organic impurities. Further, DO is also responsible for corrosion of boilers and requires to be eliminated.
The Winkler test is used to determine the concentration of DO in a water sample. Here the water sample is treated with a mixture of manganese sulphate and alkaline potassium iodide. Initially formed manganous hydroxide precipitate traps the dissolved oxygen and oxidizes manganous ion (Mn+2) to a brown-coloured precipitate of manganic oxide (MnO(OH)2).
Brown ppt.
The formed manganic oxide precipitate is allowed to settle down for a few minutes and then 2 to 3 ml of concentrated H2SO4 is added to dissolved the precipitate. The liberated iodine is proportional to the dissolved oxygen content of water sample. This is estimated by titrating a standard sodium thiosulphate solution and using a starch solution as an indicator.
From the above equation, we can find that
Therefore, after determining the number of moles of iodine produced, we can calculate the number of moles of oxygen molecule present in the water sample. The oxygen content is usually presented as mg dm–3. The solubility of oxygen in water at ambient conditions of temperature, and pressure is about 8 mg/L.
Chlorides are present in water as salts of calcium, magnesium, sodium and potassium [NaCl, CaCl2, KCl, MgCl2]. The salty taste of water is due to NaCl present in it. Chlorides are not considered harmful if their concentration is less than of 250 mg/L. Other salts such as MgCl2 in water undergo hydrolysis and cause problem in boilers.
When potassium chromate is added as an indicator to the water sample, it dissolves in water and the chromate ions give yellow colour to the sample. Sodium chloride is present in the dissolved state in the given sample of water. When this is titrated against silver nitrate, the silver ions react first with the chloride ions present in the sample and form silver chloride precipitate and sodium nitrate.
AgNO3 + NaCl → AgCl↓ + NaNO3
When all the chloride ions in the sample are precipitated, the excess silver nitrate present reacts with potassium chromate and forms a pale red precipitate of silver chromate.
2AgNO3 + K2CrO4 → Ag2CrO4 + 2KNO3
The appearance of the pale red colour indicates that all chloride ions have been precipitated and indicates the end point of titration. From the titre values, the amount of chloride and salt present in the sample is calculated.
Indicator: Potassium chromate
End point: Yellow to brick red
Take a 50 ml burette and wash it with tap water and distilled water and then rinse it with 0.005 N silver nitrate solution. Fill the burette with the 0.005 N silver nitrate solution and note down the initial reading. Pipette out 10 ml of the given water sample with a clean 10 ml pipette into a clean and dry conical flask. Add two to three drops of potassium chromate as the indicator. The solution in the conical flask turns to yellow colour. Titrate this solution against the 0.005 N silver nitrate solution taken in the burette. The appearance of a brick red colour is the end point of titration. Note down the final burette reading. Repeat the titration until consecutive concordant values are obtained. From the titre values, calculate the amount of chloride and salt present in the given water sample using the given formulae.
Pipette solution (water sample) Burette solution (AgNO3)
Volume of given water sample (V1) = 10 ml Volume of silver nitrate solution (V2) = _____ ml
Normality of given water sample (N1) = ? Normality of silver nitrate solution (N2) = 0.005 N
N1V1 = N2V2
Normality of given water sample
Amount of chloride present in given sample A
= Normality of sample × equivalent weight of chloride = N1 × 35.45 = _____ g/L
Amount of salt present in given sample A
= Normality of the sample × equivalent weight of chloride salt = N1 × 58.45 = _____ g/L.
Burette solution (AgNO3)
Volume of silver nitrate solution (V2) = _____ ml
Normality of silver nitrate solution (N2) = 0.005 N
Dissolved carbon dioxide (CO2) in water contributes to the acidity of water by formation of carbonic acid. Water used for drinking purpose should not contain mineral acidity. Highly acidic water, i.e., having low pH affects aquatic life.
Acidity of water depends on the end point of indicator used. Hydrolysis or dissociation of acids release H+ ions which react with standard alkali (NaOH) during titrations.
The colour change of phenolphthalein indicator indicates neutralization of carbonic acid present in water sample.
In case of methyl orange: Orange to yellow
In case of phenolphthalein: Appearance of pink colour
Take a 50 ml burette and wash it with tap water and distilled water and then rinse it with sodium hydroxide solution. Fill the burette with 0.02 N sodium hydroxide solution and note down the initial reading. Take 100 ml of water sample into a conical flask. Add 4 to 5 drops of methyl orange indicator and colour changes to orange. Titrate the water sample against sodium hydroxide solution until the colour changes to yellow and note down the volume consumed as A ml. To the same solution, add 3 to 4 drops of phenolphthalein indicator and continue the titration until the appearance of pink colour. Note down the volume of sodium hydroxide consumed as B ml. Repeat the titration to get consecutive concordant readings.
Water sample NaOH solution
Normality of water (N1) = ? Normality of NaOH solution (N2) = 0.02 N
Volume of water (V1) = 100 ml Volume of NaOH used for methyl orange = A ml
Volume of NaOH used for phenolphthalein = B ml
N1V1 = N2V2 Volume of NaOH (V2) = A + B
Acidity of water = Normality × Eq. wt. of CaCO3
The capacity of water for neutralizing an acid solution is known as alkalinity of water (or) the capacity of a water to accept protons is known as alkalinity of water.
The alkalinity of water is mainly dependent on the following factors:
Because they consume or have a tendency to take up N+ ions, hence concentration of OH– ions in water increases.
Depending on the ions present, alkalinity of water is broadly classified as
The alkalinity of a water sample may be due to
But there is no possibility with because they combine with each other to form carbonate.
Alkalinity and hardness are expressed in terms of CaCO3 equivalents, ppm, mg/L, etc.
Carbonate hardness in ppm = Alkalinity in ppm
Carbonate hardness in ppm = Total hardness in ppm
Non-carbonate hardness = Total hardness – Carbonate hardness
The type and extent of alkalinity of a water sample can be easily determined by volumetric method. A known volume of water sample is titrated against standard sulphuric acid by using phenolphthalein indicator. The end point is disappearance of pink colour. Further the titrated water sample is titrated against the same standard sulphuric acid by using methyl range indicator. The end point is appearance of red colour and the volume of H2SO4 consumed is noted.
The volume of the standard acid used up to phenolphthalein end point P marks the completion of reactions (i) and (ii), whereas the total volume of the standard acid used from the beginning up to methyl orange end point M corresponds to the completion of reactions (i), (ii) and (iii).
Solution Volume of water sample for titration = 100 mL
Volume used to phenolphthalein end point (A) = 4 mL
Volume used to methyl orange end point (B) = 16 mL
Total volume used to methyl orange end point (A + B) = 20 mL
Phenolphthalein alkalinity (in terms of CaCO3 equivalent)
Similarly, for methyl orange alkalinity,
N3V3 = N4V4
Water Acid
N3 × 100 =
N3 =
Strength = N3 × Eq. wt. of CaCO3
=
Methyl orange alkalinity (M) = ppm
M = 200 ppm
Hence, P < M
P(40) < M(100)
Soand ions are present.
Now, alkalinity due to ions = 2P = 2 × 40 ppm = 80 ppm
alkalinity due to ions = M – 2P = 200 – 80 = 120 ppm
Solution Volume of water sample = 100 mL
For Phenolphthalein alkalinity
For Methyl orange alkalinity,
N3V3 = N4V4
(Water) (Acid)
N3 × 100 = 0.03 × 15
N3 =
Strength = N3 × Eq. wt of CaCO3
=
Methyl orange alkalinity (M) =
M = 225 ppm
Hence
P(122.5) = M
So alkalinity due to ions,
Alkalinity due to ions = 2P or M = 225 ppm
Solution For Phenolphthalein alkalinity,
N1V1 = N2V2
(Water) (Acid)
N1 × 50 = × 20
N1 =
Strength = N1 × Eq. wt. of CaCO3
=
= ppm
= 400 ppm
For Methyl orange alkalinity,
N3V3 = N4V4
(Water) (Acid)
N3 × 50 = × 25
N3 =
Strength = N3 × Eq. wt. of CaCO3
=
=
= 500 ppm
Hence, P > M
P(400) > M
So OH– and ions are present.
Now, alkalinity due to OH– = 2P – M = 2(400) – 500
= 300 ppm
alkalinity due to = 2(M – P) = 2(500 – 400)
= 200 ppm
Hard water contains large amounts of bicarbonates, sulphates, and chlorides of calcium and magnesium salts. It causes number of problems in domestic use, industrial use, and in boilers.
Tea, coffee and other drinks prepared with hard water gives an unpleasant taste.
Due to formation of calcium oxalates, stones are formed in kidneys.
Washing: Hard water does not give much lather with soap, as most of the soap is consumed for removing calcium and magnesium salts present in water.
Bathing: It produces sticky scum on both tub and body.
After prolongedusage, kettle also gets damaged due to scale formation.
For the generation of steam a huge quantity of water is used in boilers and is known as boiler feed water. If water used for boilers is hard, it may create number of problems like caustic embrittlement, corrosion, scale and sludge formation, priming and forming, etc. This is very dangerous because at high pressure the same causes explosions. Hence water which is used in boilers should be softened and should be pure before feeding into the boilers.
Boiler-feed water should satisfy the following requirements:
The potable water or drinking water should satisfy the following essential requirements:
Purification of water for potable use involves mainly the following steps:
The commonly used coagulants are the salts of iron and aluminium, e.g., alum (K2SO4 Al2(SO4)3 . 24H2O), ferrous sulphate (FeSO4 . 7H2O), sodium aluminate (NaAlO2), etc. These coagulants react with alkaline salts and form a thick gelatinous precipitate known as Flock. Flock has the property to attract finely suspended particles and form big flock, which settles down rapidly. This process is called flocculation.
A few commonly used coagulants and their reactions are as follows:
The NaOH produced precipitate of Mg salts as Mg(OH)2
The precipitates obtained by using suitable coagulants in water get settled down during sedimentation.
Figure 1.3 Sand filter
It consists of three layers. The upper layer consists of fine sand (about 50 cm thick) and is a thick layer. The middle layer consists of coarse sand (about 20 cm thick), and the bottom layer consists of gravels (about 30 cm thick). It is provided with an inlet for sedimented water and an under drain channel at the bottom for the exit of filtered water. Sedimented water enters the sand filter from the top and is uniformly distributed over the fine sand layer. As the water percolates through the sand bed, finely suspended particles and most of the germs and bacteria are retained by the top layer. Clear, filtered water is collected in the under drain channel, from where it is drawn out.
The rate of filteration becomes slow after some time due to clogging of pores of the top sand layer by the impurities retained in the pores. Therefore, the portion of the top fine sand layer is scrapped off and replaced by a new sand layer. The filter is put to use again.
Figure 1.4 Vertical pressure filter
Impure, sedimented water is mixed with a small amount of alum solution, and then water is forced through filter bed under pressure. Alum forms the slimy layer on the filter bed, and this helps in the removal of colloidal and bacteriological impurities. The function of deflector plate, which is provided at the top, is to distribute the slimy layer of alum uniformly over the top of the filter bed. Filtered water, as it comes out from the bottom of filter, is under pressure and can thus be pumped directly. These filters are widely used for industrial purposes.
Several methods have been adopted for sterilization of water. Some of them are given below:
Chlorine is a powerful germicide and most commonly used disinfectant. Chlorine used for this purpose can be used directly as a gas or as chlorine water.
It reacts with water to form hypochlorous acid and nascent oxygen, both of which are powerful germicides.
Apparatus: The apparatus used for disinfection by chlorine is known as chlorinator (Figure 1.5). It is a large tower containing number of baffle plates. From the top of the tower, proper dose of chlorine and water is introduced. They get thoroughly mixed during their passage through the tower, and treated water is taken out from the bottom.
Figure 1.5 Chlorinator
Bleaching powder is a strong oxidizing agent and is having 30 per cent available chlorine. When water is treated with bleaching powder, hypochlorous acid is formed. It releases nascent oxygen and the nascent oxygen thus released deactivates the enzymes of microorganisms; due to this, metabolic activities will stop and the microorganisms get killed.
About 1 kg of bleaching powder is sufficient for 1000 kilolitres of water, but allow the water to stand for several hours.
By mixing of chlorine and ammonia in 2:1 ratio, chloramine is formed.
Whenever water is treated with chloramine, hypochlorous acid is formed and with release of hypochlorous acid it provides greater safeguard from recontamination.
So, HOCl + germs → germs are killed.
The nascent oxygen is very powerful oxidizing agent, which kills all the bacteria and germs present in water.
Apparatus: The reaction of ozone and water is carried out in ozone sterilizer (Figure 1.6). During the treatment of water, water is allowed to enter from top to bottom, and ozone is allowed to enter from bottom to top, which kills the germs when they come in contact with each other. Sterilized water is collected at the bottom of the tank. The contact time for ozone and water is about 10–15 minutes.
Figure 1.6 Ozone sterilizer
In all these methods, water is forced through membranes made of synthetic polymers, cellulose acetate, or even ceramics by the application of high pressure in the range of 10 to 50 atm. pressure. Microfilteration and ultrafilteraton membranes with pores of 0.002 to 10 μm in diameter can filter off most bacteria and colloidal particles but not viruses and ions. Nanofilteration soften water by removing hardness causing metal ions, and reverse osmosis is used for desalination of sea water.
Chlorination of water is done carefully in a controlled manner with the dip or break is called break-point chlorination. Added chlorine consumed for different reactions such as
With this method not only living organisms but also organic impurities and free NH3 present in water are destroyed. The point at which free residual chlorine begins to appear is called break-point chlorination. It is also known as free residual chlorination.
Break-point chlorination shows four stages of sterilization as shown in Figure 1.7:
Figure 1.7 Break-point chlorination curve
Stage I: Initially, with the lower dosage of Cl2, there is no free residual chlorine since all the added Cl2 gets consumed in complete oxidation of reducing substances present in water.
Stage II: As the amount of Cl2 dose is increased, the amount of residual Cl2 also shows steady increase. This stage corresponds to the formation of chloro-organic and chloramines compounds without undergoing oxidation.
Stage III: As more amount of Cl2 is applied, the amount of free residual chlorine also decreases, due to oxidation of chloro-organic and chloramines. When the oxidation destruction is complete, it reaches a minima.
Stage IV: After minima, the added Cl2 is not used in any reaction. Thus, the residual Cl2 keeps on increasing in direct proportion to added Cl2.
The point ‘C‚ is called break point. It is a point at which free residual chlorine begins to appear. Thus, break-point chlorination helps in eliminating disagreeable odour and bad taste in water.
Dechlorination: Excess of chlorine after the break-point chlorination gives unpleasant taste and odour in water. The excess of Cl2 may be removed by filtering the treated water over activated carbon. Over chlorination may also be removed by treating the water with SO2, Na2SO3, and Na2S2O3.
Superchlorination: Superchlorination is the addition of excess amount of chlorine for disinfection of water. It destroys the pathogenic microorganisms as well as organic impurities by oxidation.
Prechlorination: Prechlorination is the treatment of water with chlorine before filtration. In this process high chlorine is required to satisfy the chlorine demand of filterable matter. With prechlorination the quality of water is superior because unpleasant tastes and odours due to chlorinated products may be absorbed during filtration. This process is highly expensive.
Post-chlorination: Post-chlorination is the treatment of chlorine with water after filtration. In this method treated water may have unpleasant taste and odour, but it is cheaper than prechlorination due to lower chlorine demand.
In all the industries, boilers are used for generating steam. Boiler-feed water is the water required for generation of steam and with the safety, economy and efficiency concerns it should be of very good quality.
Depending upon the operating pressures, boilers are classified into low-pressure (10–15 kg/cm2), medium-pressure (15–35 kg/cm2), high-pressure (50–140 kg/cm2), very high-pressure (150–225 kg/cm2) and supercritical boilers (>225 kg/cm2).
Depends upon the quality of the feed water, so many problems may arise in the boilers. Some of them are listed hereunder.
In the boilers, when water is vaporized to steam gradually the concentration of dissolved salts increases. When the concentration of salts reaches their saturation, they are thrown out in the form of precipitates. Sludge is the soft, slimy and non-adherent layer of precipitate inside the boiler and also called mud. Hard adhering coating of precipitate inside the boiler walls is called scale. Scale and sludge are shown in Figure 1.8.
Figure 1.8 Sludge and scale
Composition: The main composition of sludge includes MgCO3, MgCl2, CaCl2, MgSO4, etc.
Formation of Scales: Due to the following reactions, scales are formed.
In low pressure boilers, calcium bicarbonate decomposes and gives calcium carbonate.
At high pressure boiler formed CaCO3 is soluble and gives calcium hydroxide, whose solubility decreases with the temperature and deposit as scale.
With increase of temperature the solubility of calcium sulphate decreases, and consequently gets precipitated as hard scale. This scale is quite adherent and difficult to remove.
At high temperatures, the magnesium salts undergo hydrolysis and give magnesium hydroxide.
Minute amounts of silica present in water form and deposit as calcium or magnesium silicates and stick very firmly to the inner side of the boiler surface.
Example: (i) CaCO3 scale can be dissolved by using 5%–10% HCl.
(ii) CaSO4 scale can be dissolved by adding EDTA since Ca-EDTA complex is highly soluble in water.
Carryover: Carrying of suspended and dissolved solids along with wet steam is called carryover.
The following are the causes of foaming:
It is the disintegration or decay of boiler material either due to chemical or electrochemical reaction with its environment.
Carbon dioxide (CO2) dissolves in water to form a weak carbonic acid.
The liberated HCl reacts with boiler material in chain-like reaction.
Hydrazine is found to be an ideal compound for removing DO because the products are water and nitrogen gas, which do not form hard products, while due to sodium sulphite (Na2S) and sodium sulphide (Na2SO3), there is a formation of sodium sulphate (Na2SO4), which decomposes and gives SO2, and it forms sulphurous acid (H2SO3) in steam condensate.
Figure 1.9 Mechanical deaeration of water
Caustic embrittlement is the special type of boiler corrosion caused by the use of highly alkaline water. With this phenomena boiler material becomes brittle with the accumulation of caustic substances.
During the softening of water by lime soda process, usually small amount of free Na2CO3 is present. In high-pressure boilers, sodium carbonate decomposes and gives sodium hydroxide and this makes the boiler water ‘caustic.‚
The concentration of NaOH is increased by evaporation of water, and attacks the boiler material by giving sodium ferroate (Na2FeO2), which decomposes and forms rust.
This is an electrochemical phenomenon and can be explained on the basis that a concentration cell is formed due to concentration difference of sodium hydroxide in the boilers particularly at highly stressed parts like joints, rivets, etc. The dilute NaOH region in the boiler acts as a cathode and the concentrated NaOH region acts as an anode and undergoes corrosion.
In water, there is a formation of scale-like impurities in the boiler. This scale formation may be minimized by the following treatments:
Some important internal treatment methods are:
When calcium sulphate is converted into calcium carbonate by the addition of sodium carbonate, CaCO3 acts as a loose sludge, which can be removed by blow-down operation.
Carbonate conditioning is not used in high-pressure boilers because excess of Na2CO3 might be converted into NaOH due to hydrolysis, which causes caustic embrittlement.
The three sodium orthophosphates may be used depending upon the alkalinity of the boiler-feed water.
The sodium hydroxide reacts with magnesium salt and converts it into magnesium hydroxide Mg(OH)2 precipitates.
The gelatinous precipitate of Al(OH)3 and Mg(OH)2 entraps colloidal and finely suspended impurities along with oil drops and silica. The loose slimy precipitate can be easily removed by blow-down operation.
The most common methods for softening of water are given below:
It is a very important and popular process for softening of water.
Principle: This method involves the treatment of water sample with calculated quantities of lime [Ca(OH)2] and soda (Na2CO3), which react with calcium and magnesium salts to form insoluble precipitates as calcium carbonate (CaCO3) and magnesium hydroxide (Mg(OH)2). To accelerate the precipitation of CaCO3 and Mg(OH)2, certain substances are added, known as “coagulants” or “flocculants.”
Functions of lime: For removing temporary hardness, permanent magnesium hardness, free mineral acids, iron and aluminium salts, dissolved CO2 and H2S in water, lime acts as a good agent.
In this case, permanent magnesium hardness is converted to permanent calcium hardness.
Functions of soda: When lime is used to remove the hardness or mineral acids, it has been found that permanent calcium hardness (CaCl2) and (CaSO4) is introduced in water. Soda is very effective to remove permanent calcium hardness as follows:
Important Points about Calculation of Lime and Soda
Requirement of lime and soda for the constituents responsible for hardness is given in Table 1.4.
Formula for Lime and Soda Requirement
100 parts by mass of CaCO3 are equivalent to
Lime Requirement
Soda Requirement
Process: Lime soda process is of two types:
Figure 1.10 Cold lime soda softener
Hard water (containing Ca2+, Mg2+, or other heavy metals) + lime + soda
(i) Addition of coagulants or flocculent
(ii) Proper setting time
↓
ppts of CaCO3 + Mg(OH)2 settle out.
Figure 1.11 Hot lime soda process
In the hot process, sodium carbonate (Na2CO3) is used for softening because it decomposes into sodium hydroxide under high pressure and temperature.
Difference between cold and hot lime soda process is given in Table 1.5.
Zeolite is known as permutit, i.e., boiling stone. Zeolite process is widely used to soften water. Zeolites are hydrated alumino silicate minerals. or
Sodium aluminium orthosilicate, and it is represented as Na2O·Al2O3·xSiO2·yH2O (x = 2 – 10, y = 2 – 6) represented as Na2Z.
Zeolites are of two types:
Example: Natrolite (Na2OAl2O3·3SiO2·2H2O)
In the zeolite process for softening hard water, the raw water is percolate through a bed of zeolite (Na2Z), which is packed in a vertical cylindrical tank as shown in the Figure 1.12. The zeolite bed retains the Ca2+ and Mg2+ ions from hard water by exchanging with Na+ ions thereby the out-flowing water contains sodium salts.
Figure 1.12 Zeolite softener
This process removes both temporary and permanent hardness. After long use, the zeolite bed gets exhausted. It can be regenerated by using chemicals.
Regeneration: When the zeolite bed is completely converted into calcium and magnesium zeolite, it no longer works as softener. It gets exhausted. At this stage, supply of feed water is stopped, and the exhausted zeolite is regenerated by treating with a concentrated (10%) brine (NaCl) solution.
The regenerated zeolite bed thus obtained is used again for softening operation. Zeolite process reduces hardness to 0–15 ppm.
Difference between zeolite and lime soda process is given in Table 1.6.
In this process the cations and anions present in water and which can produce hardness are removed by ion-exchange resins. Resins are long, cross-linked organic polymers with a porous structure. Ion-exchange resins are mainly (1) cation-exchange resins and (2) anion-exchange resins.
Sulphonation form of cation-exchange resin
Process: Both cation exchanger and anion exchanger are inter-connected with a pipe as shown in the Figure 1.13. The hard water is first passed through cation-exchange resin chamber, which removes all the cations (e.g., Ca2+ and Mg2+) from it, and equal amount of H+ ions are released from its column to water.
Figure 1.13 Demineralization by ion exchangers
Cation-exchange reaction
After passing through cation-exchange chamber, the hard water is now pumped to ‘anion-exchange resin‚ chamber where all anions like Cl–, , etc., are removed, and equal amount of OH– ions are released from this resin bed to water.
H+ and OH– ions released from reactions in equivalent amount get combined to produce water molecules.
Thus, the treated water is completely free from cations as well as anions, so it is known as demineralized or deionized water.
After some time of usage (depending on water) of cation/anion exchange resins will exhaust, and it is most important to regenerate.
Regeneration of cation-exchange resins: The cation-exchange resins are regenerated by addition of dil. HCl or H2SO4:
Regenerationof anion-exchange resin: The anion-exchange resins are regenerated by addition of dil. NaOH:
After regeneration of both resins, columns are washed with deionized water, and the washed product is passed to sink.
Mixed-bed deionizer: As shown in Figure 1.14, mixed-bed deionizer contains a single cylindrical vessel with a mixture of a strong cation exchanger and a strong anion exchanger, and is the most efficient process than separate column exchanger process. Hard water which pass through the mixed bed contacts number of times with both exchangers and purifies the water. Purified water is having less than 1 ppm hardness and also this is a most widely used convenient method.
Figure 1.14 Mixed-bed deionizer
Regeneration of resins: When the resins get exhausted, the mixed bed is backwashed with water. The lighter anion exchangers get displaced to form an upper layer above the heavier cation exchangers. Then the anion exchangers are regenerated by passing NaOH solution from the top and then rinsed with deionized water. The lower layer of cation exchangers is generated by passing H2SO4 solution and is finally rinsed with deionized water. The two beds are then mixed again by forcing compressed air through it. Now the regenerated bed is ready for use again.
Desalination or desalting involves the removal of dissolved salts (e.g., NaCl) from water. The salinity of water is due to dissolved NaCl and to a smaller extent of other inorganic salts.
Natural saline water such as sea water contains more than 35,000 ppm while brackish water contains dissolved salts in the range 1000–3500 ppm.
Desalination of saline water may be achieved by any of the two approaches:
As in
As in
Reverse osmosis and electrodialysis are more important in large-scale operations, and operation and principle involved in reverse osmosis are discussed.
This technique worksbased on the principle of osmosis. Reverse osmosis is a process by which a solvent such as water is purified of solutes by being forced through a semipermeable membrane through which the solvent, but not the solute may pass. It is exactly opposite of osmosis and hence it is known as reverse osmosis.
Generally the tendency of a fluid, i.e., water, to pass through a semipermeable membrane into a solution where the solvent concentration is higher, thus equalizing the concentrations of materials on either side of the membrane is known as osmosis. But when pressure is applied on the concentrated side, the solvent will flow in the reverse direction. Reverse osmosis uses 100–150 micron thick membrane made from cellulose acetate or polymeric membranes having pores in the range of 0.0001–0.001 μm in diameter; it allows only water to pass through it and not to the salt. The water molecules diffuse through the membrane while the contaminants get concentrated in the effluent stream and are discharged.
Process: In this process, a high pressure (≈15–40 kg cm–2) is applied to the sea water or brackish water, which is to be treated (as shown in Figure 1.15). The semi-permeable membrane allows only the solvent molecule (pure water) to pass through it. Thus dissolved ionic and non-ionic solvents are left behind, and water get purified from salt. Generally, we use membrane made up of polymethacrylate and polyamide polymers for this process.
Figure 1.15 Reverse osmosis unit
Electrodialysis is another efficient technique used for the desalination of saline water and is a membrane process.
Principle: Under the influence of an electric potential across a salt water solution, charged ions move towards respective electrodes through ions and selective membrane.
The membranes are cation or anion selective, which basically means that either positive ions or negative ions will flow through cation-selective membrane consisting of sulphated polystyrene, which allows only cations to flow through and rejects anions. However, anion-selective membrane consists of polystyrene with quaternary ammonia, which allows only anions and rejects cations. Multiple membranes alternatively allow cation or anions to flow through. Hence, with this method we can get fresh water from saline water.
Process: The process is carried out in a special type of the cell called electrodialysis cell (as shown in Figure 1.16). It consists of two electrodes and ion selective membranes which are permeable to either cation or anion. The anode is placed near anion-selective membrane while the cathode is placed near the cation-selective membrane. The anion selective membrane has positively charged functional groups such as R4N+ and therefore allows negatively charged ions only to pass through them. Similarly, a cation-selective membrane has negatively charged functional groups such as RCOO– and allows only positively charged ions to pass through it. Saline water under a pressure of around 5–6 kg/m2 is passed from the top of the cell and it passes between membrane pairs.
Figure 1.16 Electrodialysis of sea water
When an emf is applied across two electrodes the cations (Na+) present in salt water move towards cathode through cation selective membrane and anions (Cl–) move towards the anode through anion selective membrane. As a result, the concentration of ions in alternate compartments 2, 4, 6 etc. decreases, while it increases in the alternate compartment 1, 3, 5 etc. Thus water in the even number compartments becomes pure and is collected from the bottom of the cell. Similarly, water in the odd number compartment becomes rich in the saline water i.e. it becomes concentrated saline water. It is collected from a separate outlet at the bottom of the cell.
[Ans.: calcium, magnesium]
[Ans.: Lake]
[Ans.: CaCO3]
[Ans.: 1, 0.1, 0.07]
[Ans.: EBT]
[Ans.: OH–, ]
[Ans.: CaSO4]
[Ans.: carry-over]
[Ans.: Hydrazine]
[Ans.: caustic, NaOH]
[Ans.: sodium sulphate]
[Ans.: coagulant]
[Ans.: CaCO3, Mg(OH)2]
[Ans.: natrolite, Na2O . Al2O3 . 4SiO2 . 2H2O]
[Ans.: sodium]
[Ans.: dilute HCl or dilute H2SO4]
[Ans.: hydrogen, H+]
[Ans.: hydroxide, OH–]
[Ans.: Ion-exchange]
[Ans.: Tannin, agar-agar]
[Ans.: demineralized or deionized]
[Ans.: sodium hexameta phosphate, Na2[Na4(PO3)6]]
[Ans.: 7–8]
[Ans.: Al(OH)3]
[Ans.: K2SO4 . Al2(SO4)3 . 24H2O]
[Ans.: chloramines, ClNH2]
[Ans.: NaCl]
[Ans.: ions]
[Ans.: cations, anions]
[Ans.: osmosis]
[Ans.: d]
[Ans.: d]
[Ans.: b]
[Ans.: a]
[Ans.: b]
[Ans.: b]
[Ans.: b]
[Ans.: c]
[Ans.: a]
[Ans.: b]
[Ans.: c]
[Ans.: b]
[Ans.: c]
[Ans.: b]
[Ans.: b]
[Ans.: b]
[Ans.: c]
[Ans.: d]
[Ans.: c]
[Ans.: b]
[Ans.: c]
[Ans.: a]
[Ans.: c]
[Ans.: c]
[Ans.: a]
[Ans.: a]
[Ans.: b]
[Ans.: a]
[Ans.: c]
[Ans.: a]
[Ans.: b]
[Ans.: b]
[Ans.: d]
[Ans.: c]
[Ans.: a]
[Ans.: b]
[Ans.: c]
[Ans.: d]
[Ans.: d]
[Ans.: c]
Ans.: Sea water, rain water, ground water, and surface water.
Ans.: Due to the presence of dissolved bicarbonates of Ca and Mg in water.
Ans.: Hardness is the characteristic property, which produces white scum on treating with soap solution.
Ans.: Because hard water produces insoluble white precipitate on treating with soap.
Ans.: The concentration of hardness-producing salts is expressed in terms of calcium carbonate (CaCO3) equivalent.
Ans.: Hardness (CaCO3 equivalent) =
W = Weight of hardness-producing substance in ppm
E = Equivalent weight of hardness-producing substance
Ans.: ppm: 1 part of CaCO3 equivalent hardness present in 106 parts of water.
mg/L: Number of mg of CaCO3 equivalent hardness present in 1L of water.
Clarke's degree: Number of parts of CaCO3 equivalent hardness present in 70,000 parts of water.
French degree: Number of parts of CaCO3 equivalent hardness present in 105 parts of water.
Ans.: 1 ppm = 1 mg/L = 0.07°Cl = 0.1°Fr
Ans.: The Ca2+ ions give precipitates with soaps. On heating ions, they are converted to ions, which precipitate in kettles/boilers with Ca2+ ions.
Ans.: Because addition and subtraction of concentration of hardness-causing constituents are easy. Its molecular mass is 100.
Ans.: Temporary hardness: Mg(HCO3)2 and Ca(HCO3)2
Permanent hardness: CaCl2, MgCl2, CaSO4, MgSO4, FeSO4, etc.
Ans.: Oxygen, carbon dioxide, and sulphur dioxide.
Ans.: On boiling, temporary hardness is removed by precipitating as
Ca(HCO3)2 CaCO3↓ + H2O + CO2↑
Mg(HCO3)2 Mg(OH)2↓ + H2O + CO2↑
Ans.: (i) Liquid chlorine (ii) Bleaching powder (iii) Chloramine
Ans.: Because chloramine
(i) is quite stable
(ii) does not impart bad taste to treated water
(iii) imparts good taste to treated water
Ans.: It involves addition of sufficient amount of chlorine to water in order to
(i) oxidize organic matter
(ii) reduce substance and
(iii) free ammonia
and leaves behind mainly free chlorine for disinfecting disease-producing bacteria.
Ans.: (i) It oxidizes organic matter, NH3, and reducing substances completely.
(ii) It removes colour in water.
(iii) It destroys all the disease-producing bacteria completely.
(iv) It removes odour from water.
(v) It prevents any growth of weeds in water.
Ans.: (i) Suspended impurities
(ii) Colloidal impurities
(iii) Dissolved impurities
Ans.: Usually it is a solution containing 1 g of CaCO3 equivalent hardness in 1 liter, i.e., 1000 ppm of hardness water.
Ans.: The process of removing of finely suspended impurities as well as colloidal impurities by adding requisite amount of coagulant to water before sedimentation.
Ans.: Scale formation can be avoided in low-pressure boilers by adding substances like kerosene, tannin, agar-agar, etc., which get adsorbed over the scale-forming precipitates, thereby yielding non-sticky and loose deposits, which can be removed by blow-down operation.
Ans.: Indicator: EBT
End point: Wine red to pure blue
Ans.: The indicator used in this titration (EBT) shows colour change at a pH value of about 10. So alkaline buffer (NH4Cl) is used.
Ans.: Soft water may contain Na+, Cl–, and ions, so it is not demineralized, whereas demineralized water does not contain any cation and anion.
Ans.: Because zeolite-softened water contains large quantities of sodium salts like NaCl, Na2SO4, etc., which avoids caustic embrittlement.
Ans.: Because CO2 forms carbonic acid (H2CO3) on reacting with water. So boiler’s wall material can be attacked slowly by carbonic acid and becomes weaker and weaker progressively.
Ans.: Softening of water means removing hardness-producing salts from water.
Ans.: Water should be softened before using in boilers otherwise it may cause various boiler problems like
(i) scale and sludge formation
(ii) priming and foaming
(iii) boiler corrosion.
Ans.: When water is irradiated by UV radiations, microorganisms and bacteria are killed. This so-called disinfection of water by UV radiation.
Ans.: It is hydrated sodium alumino-silicate having formula Na2O . Al2O3 . xSiO2 . yH2O,
where x = 2–10
y = 2–6
It is represented as Na2Z, and Na+ ions are capable of exchanging by M2+ (Ca2+ or Mg2+) present in water sample.
Ans.: If silica is present in water, it causes formation of very firmly sticking deposits of calcium silicate (CaSiO3) and magnesium silicate (MgSiO3) scales in the boilers, which are very difficult to remove.
Ans.: NaAlO2 + H2O→Al(OH)3 + NaOH
1 eq. 1 eq. of Ca(OH)2
Ans.: Because, OH– and HCO3– react to form CO32–
+ → + H2O.
Ans.: Mg(HCO3)2 + 2Ca(OH)2 → 2CaCO3↓ + Mg(OH)2↓ + 2H2O
Thus from the above equation, mole of Mg(HCO3)2 ≡ 2 mol of Ca(OH)2
Ans.: No, because reaction proceeds faster in hot lime soda process, and the precipitate and sludge formed settle down rapidly. Thus, no coagulants are required in hot lime soda process.
Ans.: Water should not be soft for drinking purposes because soft water is plumbosolvent, i.e., it attacks lead used in plumbing.
Ans.: Reverse osmosis removes all ionic, non-ionic, colloidal, and high molecular weight organic matter.
Ans.: The treated water still contains some residual hardness.
Ans.: When Na2SO4 is added to boiler-feed water, it blocks hair cracks, thereby preventing infilteration of caustic soda solution in these areas. So by this way, caustic embrittlement is prevented by using Na2SO4 in boiler-feed water.
Ans.: In calgon conditioning, the added calgon forms soluble complex compound with CaSO4, thereby it prevents the scale and sludge formation in water.
Na2[Na4(PO3) 6] 2Na+ + [Na4(PO3)6]2–
calgon
2CaSO4 + [Na4(PO3)6]2– → [Ca2(PO3)6]2– + 2Na2SO4
soluble complex
This soluble complex does not cause any problem in the boilers.
On the other hand, in phosphate conditioning, sodium phosphate is added to boiler water so that precipitate of calcium phosphate is formed. Although this precipitate is non-adherent and soft, it has to be removed by frequent blow-down operation.
2Na3PO4 + 3CaSO4 → Ca3(PO4)2↓ + 3Na2SO4
Hence, calgon conditioning is better than phosphate conditioning.
Q.1 Complete the following equations:
a. 1 ppm = ______mg/L = ______°French = ______°Clark
b. Ca(HCO3)2
c. Mg(HCO3)2+ 2Ca(OH)2
d. Na2[Na4(PO3)6] + 2CaSO4
Q.2 Explain the following:
a. Scale and sludge formation and their disadvantages
b. Caustic embrittlement
c. Boiler corrosion
Q.3 What is hardness of water? How is it determined by EDTA method?
Q.4 Describe the continuous lime soda process of softening hard water. Compare continuous cold lime soda process with hot lime soda process.
Q.5 How is true exhausted zeolite bed regenerated? Give the merits and demerits of zeolite process.
Q.6 hat are the requirements of water for domestic use?
Q.7 A water sample contains Ca(HCO3)2 = 32.4 mg/L, Mg(HCO3)2 = 29.2 mg/L, and CaSO4 = 13.5 mg/L. Calculate the temporary and permanent hardness.
[Ans.: 40 ppm,10 ppm]
Q.8 Calculate the hardness of water containing the following salts:
CaSO4 = 28 mg/LMg(HCO3)2 = 22 mg/L
MgCl2 = 30 mg/LCaCl2 = 85 mg/L
[Ans.: Temporary hardness = 15.07 ppm
Permanent hardness = 128.7 ppm]
Q.9 1 g of CaCO3 was dissolved in dil. HCl, and the solution was diluted to 1 liter. 50 ml of this solution required 45 ml of EDTA solution. 50 ml of hard water required 18 ml of EDTA solution during titration in ammonia buffer using EBT indicator. On the other hand, 50 ml of boiled water sample required 9 ml of EDTA solution under the same condition. Calculate each type of hardness in ppm.
[Ans.: Total hardness = 400 ppm
Permanent hardness = 200 ppm
Temporary hardness = 200 ppm]
Q.10 0.28 g of CaCO3 was dissolved in HCl and the solution made up to 1 liter with distilled water. 100 ml of the above solution required 28 ml of EDTA solution on titration. 100 ml of a hard water sample required 33 ml of same required solution on titration. After 100 ml of this water, cooling and filtering and then titrated 10 ml of EDTA solution. Calculate the temporary and permanent hardness.
[Ans.: 230 mg/L, 100 mg/L]
Q.11 Explain the ion-exchange method of purifying the water. Discuss their use and regeneration, giving the reaction involved.
Q.12 Write a short note on break-point chlorination.
Q.13 Pure soft water is not fit for drinking purpose. Why?
Q.14 Write the principle of lime-soda process? Why should we use coagulants along with lime and soda? Why is water softened by zeolite process that is unfit for use in boilers?
Q.15 Explain reverse osmosis process for desalination of sea water.
Q.16 A water sample contains the following impurities: Ca2+ = 20 ppm, Mg2+ = 18 ppm, = 183 ppm, and = 24 ppm. Calculate the amount of lime and soda needed for softening.
[Ans.: Lime = 185 mg/L
Soda = Zero mg/L]
Q.17 Water sample on analysis gave the following results: Mg(HCO3)2= 70 mg/L, CaCl2 = 220 mg/L, MgSO4 = 120 mg/L Ca(NO3)2 = 164 mg/L. Calculate the quantity of lime (80% pure) and soda (90% pure) needed for softening the 10,000 liters of water.
[Ans.: Lime = 1.81 kg
Soda = 4.68 kg]
Q.18 A water sample contains the following constituents in ppm: Mg(HCO3)2 = 73, MgCl2 = 95, MgSO4 = 12, CaSO4 = 68, Ca(HCO3)2 = 81, and NaCl = 4.8. Calculate the cost of chemicals required for softening 20,000 liters of water, if purity factor for lime is 95% and soda is 90%. The costs per 100 kg each of lime and soda are Rs. 75 and Rs. 2480, respectively.
[Ans.: Lime cost = Rs. 3.03;
Soda cost = Rs. 93.44]
Q.19 What do you mean by pre-chlorination, post-chlorination, and superchlorination. Write the significance of break-point chlorination.
Q.20 What do you mean by screening, sedimentation, and coagulant sedimentations? How are colloidal impurities removed from water?
Q.21 What are the factors that cause alkalinity in water? How is alkalinity of water determined by titrimetric method ?
Q.22 Write a short note on the followings:
a. Phosphate conditioning
b. Calgon conditioning
c. Colloidal conditioning
d. Carbonate conditioning
e. EDTA conditioning
1. 200 ml of water sample require 25 ml of N/50 H2SO4 for neutralization to phenolphthalein end point. After that methyl orange was added to this, and further acid required was 35 ml H2SO4. Calculate the type and amount of alkalinity of water as CaCO3 in ppm.
[Ans.: = 250 ppm
= 50 ppm]
2. Calculate the amount of lime (88.3 % pure) and soda (99% pure) required to soften 24,000 liters of water per day, which contains the following:
CaCO3 = 1.85 ppm CaSO4 = 0.34 ppm MgCO3 = 0.42 ppm
MgCl2 = 0.76 ppm MgSO4 = 0.90 ppm NaCl = 2.34 ppm
[Ans.: Lime = 88.49 kg
Soda = 46.25 kg]
3. Calculate the amount of lime and soda needed for softening 106 liters of water sample, which contains Mg2+ = 36 ppm, Ca2+ = 20 ppm, and = 183 ppm.
[Ans.: Lime = 222 kg
Soda = 53 kg]
4. A water sample contains the following: Ca2+ = 120 ppm, Mg2+ = 120 ppm, CO2 = 132 ppm, = 122 ppm, and K+ = 40 ppm. Calculate the amount of lime 80% pure and soda 90% pure for softening 106 liters of water sample.
[Ans.: Lime = 832.5 kg
Soda = 824.4 kg]
3.14.131.212