1 Chapter 7

Hardness of Water

Hardness of water is due to the presence of dissolved minerals of Ca2+, Mg2+,

Al3+, Iron and other heavy elements. Specifically due to Calcum Sulphate, Calcium

Bicarbonate, Calcium chloride, Magnesium Sulphate, Magnesium Bicarbonate and

Magnesium chloride.

It can also be defined as soap consuming capacity of water.

Soft water . Hard water

1. It readily lather with soap. It does not readily lather with soap.

2. It will not form insoluble scum It forms insoluble scum with soap.

with soap.

3. It does not affect the cleaning It affects the cleaning action of soap.

action of soap.

4. It does not contain dissolved It does contain dissolved salts of

salts of Ca and Mg. Ca and Mg.

Reaction with Soap

2 R-COONa + Ca2+ (RCOO)2 Ca + 2 Na+

(White insoluble scum)

2 R-COONa + Mg2+ (RCOO)2 Mg + 2 Na+

Scummy Precipitate

Temporary hardness (Carbonate hardness) is caused by the presence of

dissolved bicarbonates of Calcium, Magnesium, Iron and other heavy elements.

It can be eliminated by mere boiling.

Ca(HCO3)2 → CaCO3↓ + H2O + CO2 ↑

Mg(HCO3)2 → MgCO3 + H2O + CO2 ↑

MgCO3

→ Mg(OH)2↓ + CO2 ↑

Permanent Hardness (non-carbonate hardness) is due to the presence of

dissolved chlorides and sulphates of Calcium, Magnesium, Iron and other heavy

elements.

It cannot be eliminated by boiling.

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Degree of Hardness

Degree of hardness is extent of hardness.

It is expressed as weight in milligrams of CaCO3 equivalent to all hardness

causing substance in one million milligram of water. That is parts per million (ppm).

Or it can be expressed as weight in milligrams of CaCO3 equivalent to all hardness

causing substance in one litre of water.

Unit of hardness = ppm or mg/L

Different units used to express degree of hardness

ppm (parts per million) = This is weight in milligrams of CaCO3 equivalent to

all hardness causing substance in one million (106) milligram of water.

mg/L (milligram per litre) = weight in milligrams of CaCO3 equivalents in one

litre of water.

⁰Cl (degree Clark) = weight in grams of CaCO3 equivalents in 70,000 gram

of water.

⁰Fr (degree French) = weight in grams of CaCO3 equivalents in 105 gram of

water.

1 ppm = 1 mg/L = 0.1⁰Fr = 0.07 ⁰Cl

CaCO3 as reference for Hardness

a. CaCO3 is stable, non-hygroscopic and is obtained in pure form. Therefore a

standard hard water solution can be prepared by dissolving accurately

weighed CaCO3 in dilute HCl and make up to a known volume.

b. CaCO3 is insoluble in water. Therefore it can be easily precipitated in water

treatments.

c. Molecular weight of CaCO3 is 100, so mathematical calculations are easy

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Ion Exchange Water softening process (Demineralization process ) ( Explain the ion exchange method for water softening? Or Give a detailed account of the ion exchange method for softening of water? Or Describe how water is purified by ion exchange method for industrial purpose?)

Ion exchange synthetic resin can be used for water purification. They are cross linked organic

polymers with micro-porous structure. They have large number of functional groups that have ion

exchange properties.

Two types of ion exchange resins:

a. Cation exchange resin (Res-H)

They have acidic functional groups like –COOH or –SO3H. Hypothetical example,

These resins can exchange their H+ ions with

cations in water.

M

2 Res-H + Ca2+ → Res2Ca + 2H+

2 Res-H + Mg2+ → Res2Mg + 2H+

b. Anion Exchange resin (Res-OH)

They have basic functional groups like –CH2-NMe3+OH-

These resins can exchange their OH- ions

with anions in water.

M

2 Res-OH + SO42- → Res2SO4 + 2OH-

Res-OH + Cl- → Res-Cl + OH-

The H+ ions obtained from cation exchanger react with OH- obtained from anion exchanges

to form water.

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Process

Service Cycle:

Step1: The hard water is passed through a cylindrical tank packed with cation exchange resin

bed. The cation exchange removes cations like Ca2+ and Mg2+ etc from water and equivalent

amount of H+ ions are released from resin to water.

Step2: The hard water is then passed through another column packed with anion exchange

resin. It removes all hardness causing anions like SO42-, Cl- and HCO3

- from water and

equivalent amount of OH- ions are released from resin to water.

Regeneration Cycle:

After some time the resin get exhausted.

The exhausted cation exchange resin is regenerated by passing a solution of dil HCl.

Res2Ca + 2HCl 2 Res-H + CaCl2

Res2Mg + 2HCl 2 Res-H + MgCl2

The exhausted anion exchange resin is regenerated by passing a solution of dil NaOH.

Res2SO4 + 2NaOH 2 Res-OH + Na2SO4

ResHCO3 + NaOH Res-OH + NaHCO3

ResCl + NaOH Res-OH + NaCl

Advantages:

It produces water of very low hardness (2 ppm)

It can be used for highly acidic and alkaline water.

Disadvantages:

Costly.

It cannot remove dissolved non-ionic salts and dissolved organic substances

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EDTA Method For Total, Permanent and Temporary Hardness Estimation

Complexometric titration: A type of volumetric analysis based on complex formation.

Complexing agent is EDTA (ethylene-diamine-tetracetic-acid) -- 4 acidic proton (YH4)

Titrant used is disodium salt of EDTA (Na2H2Y or H2Y2-). Disodium salt is obtained in high purity.

EDTA can combine with metal ion in 1:1 ratio to form a chelate structure.

Indicator used is Eriochrome-Black-T (EBT)

EBT is added to hard water sample to be analyzed. At pH 10, EBT complexes with metal ions to form

wine red coloured M-EBT complex.

The hard water is titrated against EDTA. Free metal ions (Ca2+, Mg2+) complexes with EDTA. Towards

the endpoint, no free metal ions available (only M-EBT and M-EDTA complexes). M-EBT complex is

less stable than M-EDTA complex. Therefore M-EBT complex decomposes to EBT and free the metal

ion. The metal ion complexes with EDTA. At pH 10, free EBT is blue in colour. So at the endpoint

colour change is from wine red to blue.

To maintain the pH at 10 we use basic buffer solution (NH4OH - NH4Cl).

Procedure:

Step1: Preparation of Standard Hard water solution: Dissolve accurately weighed X g of CaCO3 in

dilute HCl and make upto 1000 ml in a standard flask.

Molarity (M1) = wt per litre / Mol wt = X/100 M = .0X M

Normality = wt per litre / equivalent wt = X/50 M

Role of Buffer NH4Cl + NH4OH to maintain pH 10. M-EDTA complex is more stable at pH 10. M-

EBT complex formation is more favourable at pH10.

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Step2: Standardize the EDTA solution (Finding out the molarity or normality of EDTA M2)

V1 ml V2 ml Standard Hard water (at pH=10) Vs EDTA (EBT indicator wine red colour to blue colour) M1 M2

Step3: Estimate the total hardness of unknown water sample.

V3 ml V4 ml Standard Hard water (at pH=10) Vs EDTA (EBT indicator wine red colour to blue colour) M3 M4=M2

Hardness = M3 x 100000 ppm

Step5: Estimate the permanent hardness of unknown water sample: Pipette out a known volume of

hard water sample (V5 ml). Boil it to precipitate Ca- and Mg-bicarbonates as calcium carbonate and

magnesium hydroxide. Filter it and quantitatively remove the precipitates by washing with distilled

water. To the washing add 10 ml of buffer and few drops of EBT indicator. Titrate against EDTA

solution. Volume of EDTA consumed (V6 ml). From this we can calculate the normality of temporary

hardness removed sample water M5.

V5 ml V4 ml Standard Hard water (boiled and removed temporary hardness) + pH 10 buffer Vs EDTA (EBT indicator wine red colour to blue colour) M5 M6=M2

Hardness = M5 x 100000 ppm

Temporary Hardness = Total hardness – Permanent Hardness.

MUNCIPAL WATER TREATMENT (for drinking) [Potable water]

Water from most of the natural sources is not fit for drinking. They contain various types of

impurities. So the water should be purified before use. The water which is safe for drinking purpose

is called POTABLE water.

Characteristics of potable water: a. It should be clear and odourless. b. It should be free from disease producing micro organisms and germs. c. It should be free from objectionable dissolved gases like H2S, NH3 etc. d. It should be free from dissolved minerals like nitrates, nitrites etc. e. It should be reasonably soft. f. Total dissolved solids should be less than 500 ppm.

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Steps in municipal water treatment

Screening: Water is passed through bar screens or mesh screens to remove floating

materials like wood pieces, leaves etc.

Sedimentation: Water is allowed to stand undisturbed in big tanks for 6 to 12 hrs. During

this most of the suspended insoluble particles settle under the force of gravity.

Coagulation: Finely divided particle exist as colloidal suspension which will not settle

under the influence of gravity. They can be made to settle down by adding certain chemicals called

coagulants. Common coagulants are alum, sodium aluminates etc.

Filtration: It removes colloidal impurities, remaining insoluble impurities and bacterial

impurities. Sand filters are used for this purpose.

Removal of Micro Organisms: Disinfection or Sterilization: By disinfection disease producing

bacteria and germs are destroyed. Chemicals used for disinfection or sterilization are called

disinfectants. Chlorine, chloramines, bleaching powder, ozone etc are important disinfectants.

Disinfection can be achieved by passing UV radiation.

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DISINFECTION (Sterilization) of water

Disinfection or Sterilization: By disinfection disease producing bacteria and germs are destroyed.

Chemicals used for disinfection or sterilization are called disinfectants. Chlorine, chloramines,

bleaching powder, ozone etc are important disinfectants. Disinfection can be achieved by passing

UV radiation.

Chlorination: Widely used sterilization process. Here chlorine gas or chlorine

water is used for disinfection. The apparatus used is chlorinator.

The chlorine reacts with water to form hyprochlorous acid (HOCl). This HOCl decomposes to

HCl and nascent oxygen [O]. The chlorine, HOCl and [O] are powerful germicides. The nascent

oxygen is a strong oxidizing agent which oxidizes all other organics mater present in water.

Advantages Effective and economic It leaves no salt impurities Chlorine is available in pure form It requires little space for storage.

Disadvantages Excess chlorine if present causes unpleasant taste and odour. It causes irritation to mucous membrane. Requires facility to store Cl2 disinfectant Excess chlorine is removed by passing the chlorinated water through activated charcoal.

Break Point Chlorination: It is also called free residual chlorination. It is a controlled

chlorination process. Only the required amount of chlorine dose is added. Break point chlorination

curve has four stages.

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Stage I: No residual chlorine in the sample. All chlorine added is consumed for oxidation and destruction.

Stage II. The amount of residual chlorine increases with increasing dose of chlorine. Some organic compounds defy oxidation at lower chlorine concentrations. chloro-organics and chloramines are formed.

Stage III. Amount of residual chlorine decreases with increase in applied chlorine dose. Here the chlorine added is consumed for oxidation of chloro-organics and chloramines etc.

Complete oxidation and disinfection occur at break point. Stage IV. The applied chlorine dose is not consumed so left us residual chlorine. Advantages: It destroys all pathogens. It oxidizes all organic compounds, ammonia and other reducing substances. Remove colour, bad taste and odour from water. Prevents the growth of algae, weed etc.

UV rays disinfection: UV disinfection means exposing water to UV radiations.

Short wave length UV radiations have germicidal acitivity. The germicidal activity is proportional to UV dose = UV light intensity x time. Source for short wavelength UV radiation is a mercury lamp. Irradiation of UV light result in a photochemical damage to DNA/RNA.

It destroys all pathogens. Simple to install and requires little supervision no disinfection by products. Absence of chemical smell and taste in purified water. No disinfection byproducts. Does not require any facilities for the storage and transport of chemicals. Disadvantages Expensive method. It does not improve the taste, odour and colour of water.

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Disadvantages of Hard water in industries

Scale and Sludge formation in boilers

Boilers convert water into steam. The dissolved solids remain in boilers. The concentration

of dissolved salts increases progressively. They start precipitating out.

Sludge: If the precipitate is loose and non-sticky then it is called sludge. They occur in the

form of suspension. Common sludge deposits are MgCO3, MgCl2, CaCl2, MgSO4 etc.

Disadvantages of sludge formation:

Wastage of fuel - Sludge deposits are poor conductors of heat. Sludge also elevate

the boiling point.

Lowers the safety of boilers: Overheating may result in a crack.

Choking of pipes. The sludge precipitates in colder regions, bents, joints etc and will

result in choked flow.

Scale: If the precipitate is hard and sticky, then it is called scale. They stick very firmly to the

inner walls of the boiler, hence very difficult to remove. Examples of common scale deposits are

Mg(OH)2, CaSO4, calcium and magnesium silicates etc.

Disadvantages of scale formation: Wastage of fuel: Scale deposits are poor conductors of heat. Lowers the safety of boilers: Overheating may result in a crack. Choking of pipes: The scale deposit will result in choked flow. Explosion of boilers: Uneven expansion of boiler metal and scale result in the

development of crack on the scales. Sudden entry of the water through these cracks to very hot metal result in the conversion of water into steam. These leads to the development of high pressure inside the boiler and the boiler may explode.

Priming and Foaming:

Priming: Boiler convert water to steam. The steam generated may carry small

droplets of water. This wet stem formation is called priming.

Foaming: Formation of small bubbles at the surface of water that persist for longer

time. These bubbles are carried along with steam.

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Disadvantage of priming and foaming: The liquid droplet and bubbles carried by steam may

carry dissolved solids and suspended impurities and will get deposited in engine valves, turbine

blades and decreases their life span. It also leads to corrosion.

Boiler Corrosion.

Disadvantages of Hard water in home

“Hard water isn’t a health hazard but can be a nuisance within the home.”

Hard water with a hardness of 60 to 100 ppm are recommended for drinking. It can act as

calcium/magnesium supplement.

It decreases the cleansing action of soap. Wastage of soap and detergents.

It produces stains in kitchen utensils and bathroom fittings.

REVERSE OSMOSIS (RO) Desalination Water Softening Method

Brackish water is the water which contains high concentrations of dissolved

salts, example sea water. The process of removal common salt from water is called

desalination.

Osmosis: In a system where two solutions which differ in their concentration

are separated by a semi-permeable membrane, solvent molecules pass through the

membrane from the side having lower solute concentration to the side having higher

solute concentrations. This spontaneous flow depends on osmotic pressure and

concentration gradient.

Reverse Osmosis: If we apply an external pressure in excess of the osmotic

pressure on the concentrated side, the solvent flow is reversed. That is solvent is

forced to move from the side having higher solute concentration to the side having

lower solute concentrations.

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This principle can be used for desalination of sea water.

Pure water is separated from dissolved minerals. This membrane filtration is called

super filtration.

Advantage

It removes ionic as well as non-ionic substances (even organic matter).

But ion exchange process removes only ionic compounds.

It can remove colloidal silica which is not removed by ion-exchange

demineralization.

High maintenance cost for Ion-echange process , comparatively RO

process is cheap.

Sewage treatment (sewerage)

Sewage is the liquid waste. It contains Human waste, household waste, industrial waste, fertilizers

etc. Major portion of sewage is water and rest is constituted by organic and inorganic matter. They

can be either biodegradable or non-biodegradable.

The artificial process of sewage treatment is sewerage. Harmful compounds are converted into

harmless compounds before discharge.

The various steps involved are shown in the scheme below.

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Preliminary process

a. Screening: In this process large floating matter like wood piece, leaves, plastic etc and

large solids are removed. Screening is done using bar screens or mesh screens.

b. Sedimentation or settling process: Suspended solids and colloidal impurities are

removed from sewage. Sometime coagulants like alum are added to precipitate colloidal

impurities. Sewage is kept undisturbed in big tanks for several hours, particles settle

down due to the force of gravity.

Secondary or biological treatment:

It is biodegradation process under aerobic or anaerobic condition.

Aerobic degradation : Trickling filter method: During aerobic degradation the aerobic bacteria convert Carbon content in organic matter is

carbon dioxide, Nitrogen to ammonia then to nitrate or nitrite. One of the commonly

employed aerobic digestion method is Trickling filter method

It consists of a rectangular or circular tank of about 2 m depth.

The filtering media is a bed of gravel, broken bricks or coal.

The top layer of bed contains large number of aerobic bacteria.

The sewage is allowed to trickle down by means of a rotating arm distributor.

The aerobic bacteria grow on the top surface. Biodegradable matters are aerobically

decomposed. Carbon content in organic matter is carbon dioxide, Nitrogen to ammonia then

to nitrate or nitrite etc.

About 90 % BOD can be removed.

Anaerobic degradation: UASB process (Upflow Anaerobic Sludge

Blanket process): During anaerobic degradation process large portion of carbon is converted into methane

and remaining carbon is converted into carbon dioxide. Commonly employed anaerobic

degradation process is UASB process (Upflow Anaerobic Sludge Bed or Blanket process).

Bottom of UASB reactor contains sludge blanket.

Sludge blanket is biologically formed granules.

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These granules are colonies of anaerobic bacteria (due to lack of air).

The sewage for treatment is fed from the bottom.

In the anaerobic decomposition biogas is formed.

The biogas is collected from the gas collecting dome at the top of the reactor.

Tertiary or Advanced treatment: The main function of this treatment is to decrease the amount of nitrogen and phosphorus compounds and disinfection. Precipitation: The biologically treated sewage is mixed with lime (CaO). The phosphorous contents will precipitate as Calcium Phosphate. Nitrogen Stripping (Degasification): Ammonia gas is removed by passing the effluent through a series of heated baffle plates. Disinfection: After the removal of P and N, the effluent is chlorinated to kill all pathogens. DISSOLVED OXYGEN and OXYGEN DEMANDING WASTE

Dissolved oxygen is the oxygen that is dissolved in water. Dissolved oxygen is essential for

aquatic life.

It enters water through two natural process – a. Diffusion from atmosphere b. Photosynthesis by

aquatic plants.

DO of 5-6 ppm is required to support a healthy aquatic ecosystem.

Factors influence the level of DO.

1. Temperature: Cold water holds more DO than warm water.

2. Decomposable waste. Most of human related waste or natural wastes are biodegradable.

During the decomposition of such waste in water the aerobic bacteria consumes the DO and

level of DO considerably fall down. These waste are called oxygen demanding waste.

Biological Oxygen Demand (BOD):

BOD is the amount of oxygen required by microbes for biodegradation of organic waste under

aerobic condition.

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BOD is related to the amount of biologically oxidisable impurity present. Ie it is an indication of

quality of water.

BOD is Milligrams of oxygen consumed per litre of sewage water during 5 days of incubation at 20oC.

Procedure:

One litre of sewage water is taken. It is mixed with fresh water. Dissolved oxygen (DO) content is

measured using a DO meter or titration.

Close the bottle. Incubate it at 20oC for 5 days. Find out the DO content.

The difference in Dissolved Oxygen in mg/L is BOD.

Chemical Oxygen Demand (COD):

It is the amount of oxygen required for the chemical oxidation of all oxidisable impurities present in

sewage. It is related to the amount of biologically oxidisable and other oxidisable impurities present

in sewage.

COD is milligrams of oxygen equivalent to all oxidisable impurities present in one litre of sewage. The

oxidation is carried out using strong oxidizing agents like potassium dichromate.

Procedure:

Oxidizable impurities in sewage can be determined by oxidation with potassium dichromate. The

amount of oxygen that is equivalent to dichromate consumed is COD.

Step1: A known volume of sewage is treated with known excess of potassium dichromate + dil

sulphuric acid + a catalyst and refluxed for two hours. Then find out the amount of un-reacted

potassium dichromate by back titrating against Mohrs Salt.

Step2: A blank experiment is conducted with fresh water with the same amount of potassium

dichromate used for step 1.

From the difference of values (Step 2-Step 1) COD can be calculated.

BOD COD

1. Amount of oxygen required for the biological Amount of oxygen equivalent for the

oxidation of biodgreadable matter . chemical oxidation of all oxidizable impurities.

2. Aerobic bacteria oxidizes the impurities. Oxidizing agents are used.

3. BOD determination takes five days. COD determination takes 2-3 hours.

4. it is equivalent to biologically oxidizable It is equivalent to all oxidizable impurities

impurities.

5. BOD value < COD value. COD value > BOD value.

6. Maintaining 20oC for analysis is must. High temperature is usually opted.

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Aerobic process Anaerobic process

1. Natural Biological degradation by microbes Natural biological degradation by microbes

in presence of oxygen. in the absence of oxygen.

2. During oxidation pollutants are broken down broken down to methane gas, CO2, Acetic

CO2, H2O, Nitrates, Phosphates and biomass acid and relatively little amount biomass.

3. Products are non-combustible. Biogas is formed.

4. Products does not have offensive smell. Most of the products have offensive smell.

4. Is a fast process. Is relatively slow.

Numerical problems

Q. In a hardness estimation by EDTA method 50 ml of a sample water required 30

ml of std EDTA solution. Calculate hardness of sample water? Given that 1ml of

EDTA solution is equivalent to 1.8 mg of CaCO3.

Here you are provided with a standardized EDTA

Soln: 1 ml of EDTA solution is equivalent to 1.8 mg of CaCO3

Hardness estimation 50 ml sample water consumed 30 ml of EDTA Ie 50 ml sample water Ξ 30 x 1.8 mg of CaCO3.

Then, 1 ml of sample water Ξ

mg of CaCO3.

Therefore 1000 ml sample water Ξ

mg of CaCO3 in

Then hardness = 1080 ppm. .....................................................................................................................................

Q. 50 ml of standard hard water containing 1mg of CaCO3/L consumed 20 ml of

EDTA. 50 ml of sample water consumed 25 ml of EDTA. Calculate the total

hardness. Here you are provided with standard hard water solution, u have to

standardize EDTA then find out the total hardness

Soln: Standard Hard water 1mg of CaCO3 per litre That means 1000 ml Std Hard water contains 1mg of CaCO3 Then 1 ml Std Hard water contains 1/1000 mg = 0.001 mg of CaCO3

Standardization of EDTA. 20 ml of EDTA for 50 ml standard hard water Ie 20 ml of EDTA Ξ 50 ml X 0.001 mg/ml of CaCO3

Then, 1 ml EDTA Ξ

= 0.0025 mg of CaCO3.

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Estimation of total hardness 50 ml sample water consumed 25 ml of EDTA Ie 50 ml sample water Ξ 25 x 0.0025 mg of CaCO3.

Then, 1 ml of sample water Ξ

mg of CaCO3.

Therefore 1000 ml sample water Ξ

mg of CaCO3 in

Then hardness = 1.25 ppm. ..................................................................................................................................

Q. 50 ml of standard hard water containing 1mg of CaCO3/ml consumed 20 ml of

EDTA. 50 ml of sample water consumed 25 ml of EDTA. Calculate the total

hardness. Here you are provided with standard hard water solution, u have to

standardize EDTA then find out the total hardness

Soln: Standard Hard water 1mg of CaCO3 per ml That means 1ml Std Hard water contains 1mg of CaCO3

Standardization of EDTA. 20 ml of EDTA for 50 ml standard hard water Ie 20 ml of EDTA Ξ 50 ml X 1 mg/ml of CaCO3

Then, 1 ml EDTA Ξ

= 2.5 mg of CaCO3.

Estimation of total hardness 50 ml sample water consumed 25 ml of EDTA Ie 50 ml sample water Ξ 25 x 2.5 mg of CaCO3.

Then, 1 ml of sample water Ξ

mg of CaCO3.

Therefore 1000 ml sample water Ξ

mg of CaCO3 in

Then hardness = 1250 ppm.

Or

Standard hard water = 1 mg of CaCO3 / ml that is in 1000 ml it will contain 1000 mg

of CaCO3. Hardness of standard hard water = 1000 ppm.

Volume factor of water=

Volume reciprocal factor of EDTA=

=

Total Hardness = (Volume factor of water x Volume reciprocal factor of EDTA x

Hardness of standard hard water) = 1.25 x 1 x 1000 = 1250 ppm

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..................................................................................................................................

Q. Calculate the hardness of 0.01 M AlCl3 solution .

Hardness (in ppm) = Molarity x Valency of Hardness causing salt x 50000 ppm

= Normality x 50000 ppm (Al3+)

Soln: Hardness (in ppm) = 0.01 M x 3 x 50000 = 1500 ppm

..................................................................................................................................

Q. Calculate the hardness of 0.01 N AlCl3 solution .

Hardness (in ppm) = Normality x 50000 ppm (Al3+)

Soln: Hardness (in ppm) = 0.01 N x 50000 = 500 ppm

..................................................................................................................................

Q. Calculate the hardness of 0.01 M CaCO3 solution .

Hardness (in ppm) = Molarity x Valency of Hardness causing salt x 50000 ppm

Soln: Hardness (in ppm) = 0.01 M x 2 x 50000 = 1000 ppm

..................................................................................................................................

Q. The hardness of a solution is 1000 ppm. Calculate the molarity solution .

Hardness (in ppm) = Molarity x Valency of Hardness causing salt x 50000 ppm

= Normality x 50000 ppm (Take standard Ca2+CO3

valence)

Soln: Molarity =

=

= 0.01 M

..................................................................................................................................

Q. Calculate the hardness of 0.01 M Al2(SO4)3 solution .

Hardness (in ppm) = Molarity x Valency of Hardness causing salt x 50000 ppm (Al3+)

Soln: Hardness (in ppm) = 0.01 M x 3 x 50000 = 1500 ppm

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Q. Calculate the temporary and permanent hardness of water sample having the

following analysis.

Soln:

( )

∑

Total hardness (due to Mg(HCO3)2 = 73 mg/L, Ca(HCO3)2 = 162 mg/L, CaSO4 = 136 mg/L, MgCl2 = 95

mg/L, CaCl2 = 111 mg/L)

( )

Temporary hardness (due to Mg(HCO3)2 = 73 mg/L, Ca(HCO3)2 = 162 mg/L)

( )

Permanent hardness = Total – temporary hardness = 450 - 150 = 300 ppm.

..................................................................................................................................

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Q. Calculate the temporary, permanent and total hardness of water in ppm having the following

composition. Ca2+ = 200 ppm, Mg2+= 96 ppm, HCO3- = 976 ppm, Cl- = 146 ppm and SO4

2- = 96 ppm,

Na+ = 112 ppm?

Note: Total hardness depends on the amount of calcium and magnesium ions (no need to

consider the anions). Temporary hardness is due to bicarbonate anion. For permanent hardness take

the difference.

Ans.

( )

∑

Total hardness (Ca2+ = 200 ppm, Mg2+= 96 ppm)

( )

Temporary hardness (due to HCO3- = 976 ppm)

( )

Permanent hardness = Total – temporary hardness = 900 - 800 = 100 ppm.

.................................................................................................................................. Q. Calculate the temporary, permanent and total hardness of water in ppm having the following

composition. Ca(HCO3)2 = 4ppm, Mg(HCO3)2 = 6 ppm, CaSO4 = 8ppm, MgSO4 = 10 ppm and Na HCO3

= 3 ppm?

Note: Total hardness depends on the amount of calcium and magnesium salts. If

NaHCO3 is present in water, the temporary hardness increases at the expense of

permanent hardness therefore consider NaHCO3 when calculating temporary hardness.

Ans.

( )

∑

21 Chapter 7

Total hardness (due to Ca(HCO3)2 = 4ppm, Mg(HCO3)2 = 6 ppm, CaSO4 =

8ppm, MgSO4 = 10 ppm)

( )

Temporary hardness (due to Ca(HCO3)2 = 4ppm, Ca(HCO3)2 = 6 ppm,

NaHCO3 = 3ppm)

( )

Permanent hardness = Total – temporary hardness = 20.8 - 8.4 = 12.4 ppm.

.................................................................................................................................. Pblm: Calculate the hardness of given sample of water, if 50 ml of sample water gives endpoint with 20 ml of EDTA solution. 20 ml of standard hard water of concentration 0.02 M gives end point with 25 ml of EDTA solution. Molarity of Standard hardwater solution M1 = 0.02 M

Standardization (Finding the concentration of EDTA solution using standard hard water) V1 = 20 ml V2 = 25 ml Standard Hard water (pH 10) Vs EDTA (EBT indicator wine red to blue) M1 = 0.02 M M2 M1 V1 = M2 V2

Then molarity of EDTA M2 = ( M1 V1 ) / V2

= ( 0.02 x 20 ) / (25) = 0.016 M

Estimation of hardness V3= 50 ml V4 = 20 ml Sample water (pH 10) vs EDTA (EBT indicator wine red to blue) M3 =………… M4= M2 Molarity of sample water M3 ? M3 V3 = M4 V4

Then M3 = ( M4 V4 ) / V3

= ( 0.016 x 20 ) / (50) = 0.0064 M Hardness in ppm is Molarity x 100000 ppm = 0.0064 x 100000 = 640 ppm

consider NaHCO3 when

calculating temporary

hardness.

22 Chapter 7

Q. Calculate the hardness of sample solution obtained by mixing 200 ml 0.1 M CaCl2 and 400 ml 0.2 M MgSO4. Solution A 200 ml 0.1 M CaCl2

Volume in litre 200/1000 = 0.2 L No: of moles = Molarity X Volume in litres = 0.1 X 0.2 = .02 moles Solution B 400 ml 0.2 M MgSO4

Volume in litre 400/1000 = 0.4 L No: of moles = Molarity X Volume in litres = 0.2 X 0.4 = .08 moles Total Volume of final solution 0.2 L + 0.4 L = 0.6 L Final molarity of CaCl2 = No of moles of CaCl2 / final Volume in litres

= 0.02 / 0.6 = 0.0333 M

Final molarity of MgSO4 = No of moles of MgSO4 / final Volume in litres = 0.08 / 0.6 = 0.1333 M

Total Hardness is due to CaCl2 and MgSO4. Tot. Hard = Molarity of CaCl2 . Valence of CaCl2 . 50000 mg/l + Molarity of MgSO4 . Valence of MgSO4 . 50000 mg/l = 0.0333 x 2 x 50000 + 0.1333 x 2 x 50000 = 333+ 1333 = 1666 ppm.

---------------------------------------------------------------------------------------------------------------- Pblm: Find the BOD of water sample containing 60 mg of carbohydrate (CH2O) per litre Ans : CH2O + O2 → CO2 + H2O Mol. Wt of CH2O = 12 + 1+1 + 16 = 30 g Mol. Wt of O2 = 32 g

30 g of CH2O 32 g of O2

Therefore, 1 g of CH2O

g of O2

Then, 60 mg/L of CH2O

g of O2 = 64 ppm.

BOD = 64 ppm.

Pblm: 100 ml sewage water is diluted to 500 ml using dilution water; the initial DO was 7.5 ppm. The Do level after 5 days of incubation was 3.5 ppm. Find the BOD of sewage. Ans. Before incubation 100 ml sewage diluted to 500 ml and DO measured was 7.5 ppm is 7.5 mg/l 500 ml sewage DO measured 7.5/2 mg (Ie is in 100 ml sewage) 1000 ml sewage contains (7.5/2) x 10 = 37.5 ppm DO After Incubation 100 ml sewage diluted to 500 ml and DO measured was 3.5 ppm is 3.5 mg/l 500 ml sewage DO measured 3.5/2 mg (Ie is in 100 ml sewage) 1000 ml sewage contains (3.5/2) x 10 = 17.5 ppm DO The DO consumed = DO before incubation – DO after incubation = 37.5 – 17.5 = 20 ppm.

23 Chapter 7

Or BOD = DO difference X dilution factor

dilution factor = final volume / initial sewage volume BOD = (7.5 – 3.5 ) x 500/100 = 4 x 5 = 20 ppm Q. In an EDTA experiment, the following values are obtained. Calculate the different types of hardness. i) 20 ml of std. hard water (10 g CaCO3/L) = 25 ml EDTA ii) 50 ml hard water sample = 25 ml EDTA iii) 50 ml boiled water sample = 14 ml EDTA Ans Molarity of Std. Hard water(M1) 10 g CaCO3 in 1 L = wt per litre / mol wt

= 10/100 = .1 M

Standardization (Finding the concentration of EDTA solution using standard hard water) V1 = 20 ml V2 = 25 ml Standard Hard water (pH 10) Vs EDTA (EBT indicator wine red to blue) M1 = 0.1 M M2 M1 V1 = M2 V2

Then molarity of EDTA M2 = ( M1 V1 ) / V2

= ( 0.1 x 20 ) / (25) = 0.08 M

Estimation of Total hardness V3 = 50 ml V4 = 25 ml Sample water (pH 10) vs EDTA (EBT indicator wine red to blue) M3 =………… M4= M2=0.08 M Molarity of sample water M3 ? M3 V3 = M4 V4

Then M3 = ( M4 V4 ) / V3

= ( 0.08 x 25 ) / (50) = 0.04 M Hardness in ppm is Molarity x 100000 ppm = 0.04 x 100000 = 4000 ppm

Estimation of Permanent hardness V5 = 50 ml V6 = 14 ml Sample water (pH 10) after boiling vs EDTA (EBT indicator wine red to blue) M5 =………… M6= M2=0.08 M Molarity of boiled sample water M5 ? M5 V5= M6V6

Then M3 = ( M6 V6) / V5

= ( 0.08 x 14 ) / (50) = 0.0224 M Hardness in ppm is Molarity x 100000 ppm = 0.0224 x 100000 = 2240 ppm Permanent Hardness = Total – Permanent = 4000-2240 = 1760 ppm

24 Chapter 7

Q. 100 ml sample water required 13.5 ml of 0.02 N EDTA solution for hardness

estimation. Another 100 ml sample water was boiled and removed the precipitate by

filtration. This filtrate required 6 ml of EDTA. Calculate temporary hardness

Soln Estimation of hardness V3 = 100 ml V4 = 13.5 ml Sample water (pH 10) vs EDTA (EBT indicator wine red to blue) N3 =………… N4= 0.02 N Normality of sample water N3 ? N3 V3 = N4 V4

Then N3 = ( N4 V4 ) / V3

= ( 0.02 x 13.5 ) / (100) = 0.0027 N Hardness in ppm = Normality x 50000 ppm = 0.0027 x 50000 = 135 ppm

Estimation of permanent hardness V5 = 100 ml V6 = 13.5 ml Sample water (pH 10) vs EDTA (EBT indicator wine red to blue) N5 =………… N6= 0.02 N Normality of sample water after boiling N5 ? N5 V5 = N6 V6

Then N5 = ( N6 V6 ) / V5

= ( 0.02 x 6 ) / (100) = 0.0012 N Permanent Hardness in ppm is Normality x 50000 ppm = 0.0012 x 50000 = 60 ppm

Temp. Hardness = total – perm = 135-60 = 75 ppm.

Q. A water sample is having 400 ppm permanent hardness and 100 ppm temporary hardness. If 42

mg/L NaHCO3 is added to this water calculate the new permanent and temporary hardness.

Ans: Total Hardness remains constant. Total = permanent + temporary = 400 + 100 = 500.

Hint NaHCO3 added will react with permanent hardness causing substance and convert it into

bicarbonates of Ca and Mg. Therefore temporary hardness increases and permanent hardness

decreases.

Increase in temporary hardness with the addition of 42mg/l NaHCO3 in terms of CaCO3 equivalents

is = 42 mg/l x

84 is molecular weight of NaHCO3.

= 25 ppm.

New temporary hardness = 100 + 25 = 125 ppm

New permanent hardness = total – new temporary hardness = 500-125 = 375 ppm.

25 Chapter 7

Q. 100 ml of water sample after reaction with fixed amount of acidified potassium dichromate

consumes 15 ml of Fe2+ solution (0.1 N). For a blank titration the ferrous solution consumed is 25

ml. Find COD of water sample.

Water sample taken = 100 ml

Volume of Fe2+ equivalent to oxidizable substance = 25-15 = 10 ml

Normality of Fe2+ solution = 0.1 N

Normality of water sample = (NV) of Fe2+ divided by Volume of water sample

= 0.1 x 10 / 100 = 0.01 N

Normality x equivalent weight of O x 1000 = COD in ppm

COD = 0.01 x 8 x 1000 = 80 ppm.

Q. What type of ligand is EDTA? Give the structure of EDTA.

EDTA is chelating ligand (Hexadentate) . It form thermodynamically very stable complex with

Calcium and magnesium 2+ ions.

Q. Why NH4OH — NH4Cl buffer solution is added in determination of hardness of water by EDTA

method?

M-EDTA complex is more stable at pH-10. NH4OH — NH4Cl solution is a basic buffer which

maintains the pH at 10. M-EBT complex formation is favourable at pH 10.

Q. What is the main advantage of RO process over ion exchange process.

ION Exchange Process (IX) vs Reverse Osmosis (RO)

IX:- Physical-Chemical Method of demineralization RO:- Physical method of demineralization IX:- Removes only ionic substances (excluding Na+) from water RO:- Removes not only ionic substances (including Na+), but also bacteria, organic substances etc.

26 Chapter 7

Osmosis

Osmosis is a naturally occurring spontaneous phenomenon. It is a process where water

from low salt concentrated region migrates to a strong saline solution. Examples of

osmosis are when plant roots absorb water from the soil and our kidneys absorb water

from our blood.

For example, if you had a container full of water with a low salt concentration and

another container full of water with a high salt concentration and they were separated

by a water permeable membrane, then the water with the lower salt concentration

would begin to migrate towards the water container with the higher salt

concentration.

Reverse Osmosis is the process of Osmosis in reverse . Whereas Osmosis occurs

naturally without energy required, to reverse the process of osmosis you need to apply

Pressure greater than osmotic pressure to the more saline solution. Flow of water is

reversed, that is from sea water side to pure water side. It holds back a majority of

contaminants like ionic contaminants (like Na+, Ca2+, SO42-), non-ionic contaminants

(like silica, organic matter, bacteria etc) .

Below is a diagram outlining the process of Reverse Osmosis. When pressure is applied

to the concentrated solution, the water molecules are forced through the semi-

permeable membrane and the contaminants are not allowed through.

27 Chapter 7

Chemicals That Remove Hardness (Very Important)

NaHCO3 Convert permanent hardness to temporary hardness (Total remain

same)

When lime (CaO) and soda ash (Na2CO3) are added, hardness-causing

minerals form nearly insoluble precipitates. Calcium hardness is precipitated

as calcium carbonate (CaCO3). Magnesium hardness is precipitated as

magnesium hydroxide (Mg(OH)2).

Calcium hydroxide, Ca(OH)2 - addition of calculated amount of Ca(OH)2 will

remove only temporary hardness from water.

Sodium hydroxide, NaOH - addition of caustic soda (NaOH) removes both

temporary and permanent hardness by precipitating the metal ions which

cause the hardness as insoluble hydroxides.

Sodium trioxocarbonate(IV), Na2CO3 (in form of crystals (i.e. washing soda)

and soda ash (anhydrous)) - addition of Na2CO3.10H2O (crystals) or

Na2CO3 (anhydrous) will remove both temporary and permanent hardness.

Borax, Na2B4O7. 10H2O - addition of borax removes both temporary and

permanent hardness.

28 Chapter 7