Chemistry👨‍🔬

1. HARDNESS OF
WATER,DETERMINATION &
ALKALINITY OF WATER

2. CONTENTS
 WATER :- AN INTRODUCTION
 IMPURITIES OF WATER
 ANALYSIS OF WATER
 HARDNESS OF WATER
 TYPES OF HARDNESS
 DETERMINATION OF HARDNESS
 UNITS OF HARDNESS
 INDUSTRIAL APPLICATION OF WATER
 ALKALINITY OF WATER : INTRODUCTION
 ALKALINITY : - FORMULA ,CALCULATIONS OF ALKALINITY

3. WATER :- AN INTRODUCTION
 Water is essential for the existence of human beings,
animals and plants
 Though 80% of the earth’s surface is occupied with water,
less than 1% of water is available for ready use.
 Water is a universal solvent .

4. Sources of Water
A. Surface Water : -
a) Flowing water : -
 Streams and Rivers
b) Still water : -
 Lakes , ponds and reservoirs.
B. Underground Water : - wells , springs
C. Rain Water
D. Sea water

5. IMPURITIES OF WATER
 Suspended Impurities :- : Inorganic/Organic impurities (Clay,
soil, sand/Animal and vegetable matter).
 Colloidal Impurities : - Fe, Al(OH)3 & Animal matter.
 Micro-organisms : - Bacteria, fungus, virus etc
 Dissolved Impurities : - Oxygen, Carbon dioxide, Carbonates,
Sulphates, Iron, Calcium, Magnesium etc.

6. Sources of Impurities in Water-
 Gases are picked up from the atmosphere by rainwater .
 Decomposition of plant and animal residues introduce
organic impurities in water .
 Water takes impurities when it comes in contact with
ground , soil or rocks .
 Organic impurities comes in contact with sewage or
industrial waste .

7. ANALYSIS OF WATER
 Hardness
 Alkalinity
 Total dissolved Solid
 Turbidity Value
 Dissolved Oxygen
 Chloride ions
 Biological Oxygen Demand
 Chemical Oxygen Demand
 Metal Ions : - Mn, Pb,As

8. HARD WATER
SOFT WATER

9. Hardness of Water
 Hardness of water is that characteristic, which prevents
lathering of soap.
 Originally ,it was defined as the soap-consuming capacity of
water sample .
 Hard water consumes lot of soap because of the pressure of
certain salts of Ca, Mg and other heavy metal ions like
Al3+,Fe3+ and Mn2+ In it

10.  A sample of hard water , when treated with soap does
not produce any lather ,but on the other hand , forms
insoluble white scum of precipitate , which do not
possess any detergent action. It is due to the formation
of insoluble soaps of calcium and magnesium.
 Typical reactions of soap with calcium chloride and
magnesium sulphate are show below:
 2C17H35COONa + CaCl2 (C17H35COO)2Ca + 2NaCl
 (sodium stearate) (Calcium stearate)
 2C17H35COONa + MgSO4 (C17H35COO)2Mg + 2Na2SO4
 (sodium stearate) (magnesium stearate)

11.  Hardness may be defined as the soap destroying power of
water. The consumer considers water hard if large amount
soap is required to produce lather.
The hardness in water is caused mainly by 4 dissolved
compounds:
These are:
1) Calcium Bicarbonate.
2) Magnesium Bicarbonate.
3) Calcium Sulphate
4) Magnesium sulphate
 The presence of any of these compound produces hardness.
There are others which are of less importance.
 Chlorides and nitrates of calcium and magnesium can also
cause hardness but they occur generally in small amounts.
 Iron, manganese, aluminium compounds also cause
hardness but as they generally are present in such small
amount it is a customary not to consider them in connection
with hardness.

12. DISADVANTAGES OF HARDNESS OF
WATER:
 There is wastage of soap and detergents.
 It is unsuitable for cooking certain vegetables, dal and
meat. They take very long time to cook in hard water.
 With hard water clothes are not cleaned properly and
they do not have a long life.
 Temporary hard water on boiling leads to deposit of a
layer of calcium carbonate on inside walls of boilers
and kettles which is known as scaling or furring of
boilers.
 It is harmful for industrial purposes and also shortens
the life of pipes and fixtures in the industries.

13. DISADVANTAGES OF HARDNESS OF
WATER:
 It is harmful to the health as in certain cases it may lead to
diarrhea and other digestive disorders.
 Since hard water does not lather easily with soap, it wastes a
great deal of soap when it is used in washing. It therefore is
not economical to be used in washing.
 It is not advisable to use hard water in washing white fabrics
since it tends to stain white fabrics by making them appear
grey. More often than not when you use hard water to wash
your white clothes, you are going to see the clothes turning
grey after you have washed them. This is what hard water
often does to white fabrics.
 Hard water is not good for dyeing materials. This is why the
dyeing industry doesn’t use it to work.
 Hard water forms annoying limescales in containers such as
kettles, pots, pipes, etc.

15. 1.TEMPORARY HARDNESS
 It is due to the presence of dissolved Bicarbonates and
Carbonates of Calcium and Magnesium salts.
 Temporary Hardness is also known as Carbonate or
Alkaline Hardness.
 It is determine by Acid-base titration with Hydrochloric
acid and Sulphuric acid using Methyl orange &
Phenolphthalein as indicator.
TYPES OF HARDNESS-

16. 2. PERMANENT HARDNESS-
 It is due to the presence of dissolved chlorides and
sulphates of calcium, Magnesium, iron and heavy
metals.
 Some salts responsible for permanent hardness are :
CaCl2. MgCl2, CaSO4, MgSO4, FeSO4, Al2 (SO4)3 .
 It can not be removed by Boiling.
 It is known as Non alkaline or Non carbonate Hardness.
 Water classified on the basis of hardness
Classification soft Slight
hard
Moderately
hard
hard Very hard
PPM or Mg/l 0-17 17-60 60-120 120-180 180 and
above

17. Why hardness expressed in terms of calcium
carbonate equivalents ?
 The reason for choosing CaCO3 as the standard for reporting hardness of water
is the ease in calculation as its molecular weight is exactly 100.
 Moreover, it is the most insoluble salt that can be precipitate in water treatment.
 Suppose, a given water is hard due to CaCl2. The soap consuming capacity of this
hard water and a standard water containing CaCO3 can be understand by chemical
equation.
Thus, 111 parts by weight of CaCl2 would react with the same amount of soap as
100 parts by weight of CaCO3.

18. Units of Hardness
 1. PPM – Parts Per Million.
 It is defined as the number of parts by weight of calcium carbonate present per million (106)
parts by weight of water,
 1 PPM = 1 part of caco3 equivalents hardness in 106 parts of water.
 Milligrams per litre (mg/L): It is the number of milligrams of CaCO3 equivalent
hardness present per litre of water.
 1 mg / L. = 1 mg of CaCO3 eq. Hardness / L of water
 But 1 L of water weights = 1000 gms.
 = 1000 x 1000 mg.
 1 mg / L = 1 mg / 106 mg = 1 ppm.
 1 Mg/lit = 1 PPM
 2. Clarke’s degree (0Cl): It is the number of grains of CaCO3 equivalent hardness per gallon of
water. It is the parts of CaCO3 equivalent hardness per 70,000 parts of water.
 3. Degree French (0Fr): It is the parts of CaCO3 eq. Hardness per 105 parts of water.
 Relationship between units:
 1 PPm = 1 mg / L = 0.1 0Fr = 0.07 0Cl
 1 0Fr = 10 PPm = 10 mg / L = 0.7 0Cl
 1 0 Cl = 14.3 PPm = 14.3 mg/L = 1.43 0Fr

19. Estimation of water hardness by EDTA method
 The estimation of water hardness is done by complexometric
titration using standard EDTA as titrant and EBT as an
indicator.
 EDTA : - It stands for Ethylene Diammine tetra acetic acid is
tetraprotic acid. EDTA is a hexadented ligand which form a
claw like stucture with metal ions present in water.
 Disodium salt of EDTA is used for this titration because it is
soluble in water .

20. Ethylene Di-amine terta acetic acid

21. Eriochrome Black - T
 It is an azo dye used as metal ion indicator in
this titration.
 Chemical name is : -
 Sodium1-(1-hydroxy-2-naphthylazo)-6-nitro-2-
naphthol-4-sulphonate.
Disodium salt of EDTA is a water soluble chelating agent and is always preferred.
It is non- hygroscopic and a very stable sequestering agent (Ligands which form
water soluble chelates are called sequestering agents). There are cheating agents
that form water insoluble chelates with metal ions

22. Theory-
 The hard water is buffered to a pH value 10 using NH4OH –
NH4Cl buffer and a few drops of EBT indicator solution are
added. EBT forms a weak complex with metal ions, which
has a wine red color.
 EDTA first combine with metal ion to give very stable,
colorless and water soluble metal EDTA complex.
 After all the free metal ion are consumed, the next drop of
added EDTA solution displace the indicator, EBT from
metal – EBT complex
 At the equivalence point , there is change in colour from
wine red to blue , the total hardness is determined .

23. Reactions-

24. Role of buffer solution in determination of
hardness of water by EDTA
 Wine red colored metal Eriochrome black – T unstable complex formation,
as per reaction.
 Colorless metal –EDTA stable complex formation.
 Displacement of blue colored free eriochrome black T indicator.

25. Procedures
 Standardization of EDTA solution.
 Determination of total hardness of water.
 Determination of permanent hardness of water.

26. FORMULA
 𝑪𝒉𝒆𝒎𝒊𝒄𝒂𝒍 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 =
𝑴𝑶𝒍𝒂𝒓 𝑴𝒂𝒔𝒔
𝒏−𝑭𝒂𝒄𝒕𝒐𝒓
 𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒊𝒄𝒂𝒕𝒊𝒐𝒏 𝑭𝒂𝒄𝒕𝒐𝒓 =
𝟏𝟎𝟎
𝟐×𝑪𝒉𝒆𝒎𝒊𝒄𝒂𝒍 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝒐𝒇 𝒉𝒂𝒓𝒅𝒏𝒆𝒔𝒔 𝒑𝒓𝒐𝒅𝒖𝒄𝒊𝒏𝒈 𝑺𝒖𝒃𝒔𝒕𝒂𝒏𝒄𝒆
 𝑯𝒂𝒓𝒅𝒏𝒆𝒔𝒔 = 𝑨𝒎𝒐𝒖𝒏𝒕 𝒐𝒇 𝒔𝒂𝒍𝒕 × 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒊𝒄𝒂𝒕𝒊𝒐𝒏 𝑭𝒂𝒄𝒕𝒐𝒓
 𝑨𝒎𝒐𝒖𝒏𝒕 =
𝑯𝒂𝒓𝒅𝒏𝒆𝒔𝒔
𝑴𝒖𝒍𝒕𝒊𝒑𝒍𝒊𝒄𝒂𝒕𝒊𝒐𝒏 𝑭𝒂𝒄𝒕𝒐𝒓
 𝑯𝒂𝒓𝒅𝒏𝒆𝒔𝒔 = 𝑺𝒕𝒓𝒆𝒈𝒏𝒕𝒉 (𝒎𝒈/𝒍𝒊𝒕𝒓𝒆) × 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒊𝒄𝒂𝒕𝒊𝒐𝒏 𝑭𝒂𝒄𝒕𝒐𝒓
 n – Factor : - Least common multiple of valencies of cation and anion

27. Find n-factor or valency factor

28.  Examples-
 A water sample has the following dissolved salts in mg/litre : - Mg(HCO3)2= 83 , CaSO4 =
124, MgCl2 = 84 , CaCl2 = 94 , NaCl = 50 , Urea=15.
 Calculate the temporary and permanent hardness in ppm of calcium carbonate equivalent.
Temporary Hardness =
Permanent Hardness =
CONSTITUENT n-Factor Chemical
Equivalent =
mol.wt./n-
factor
Multiplicati
on factor =
100 /
2*Chem.
Equi
(M)
Amount (A) CaCO3
equivalent
(mg/L) = A *
M
Mg(HCO3)2 2 146/2 100/146 83 56.85
CaSO4 2 136/2 100/136 124 91.18
MgCl2 2 95/2 100/95 84 88.42
CaCl2 2 111/2 100/111 94 84.69

29.  A hard water sample has following compositional data : -
Ca(HCO3)2=162 PPM , Mg(HCO3)2 = 73 PPM , CaCl2 = 111 PPM , MgCl2= 190
PPM , CaSO4 = 272 PPM , MgSO4 = 240 PPM, FeSO4.7H2O= 139 PPM
Calculate temporary , permanent and total hardness in degree clark and degree french .
CONSTITUENT n-Factor Chemical
Equivalen
t =
mol.wt./n
- factor
Multiplicatio
n factor =
100 /
2*Chem.
Equi
(M)
Amount (A) CaCO3
equivalent
(mg/L) = A
* M
Ca(HCO3)2 2 162/2 100/162 162 100
, Mg(HCO3)2 2 146/2 100/146 73 50
CaCl2 2 111/2 100/111 111 100
MgCl2 2 95/2 100/95 190 200
CaSO4 2 136/2 100/136 272 200
MgSO4 2 120/2 100/120 240 200
FeSO4.7H2O 2 287/2 100/287 139 47

30. NUMERICAL
 Calculate temporary and permanent hardness of a sample of
water contains following salts
 Mg(HCO3)2 = 80 mg/li
 MgCl2= 95mg/litre
 CaCl2 = 111mg/litre
 Ca(HCO3)2 = 162mg/litre

31. NUMERICALS
 A water sample contains 272 mg of calcium sulphate per
litre. Calculate the hardness in terms of calcium
carbonate equivalents.
 Calculate the hardness of water sample in which
calcium bi carbonate is dissolved as amount 73
mg/litre.
 How many grams of ferrous sulphate disssovled per litre
gives 210.5 ppm hardness .

32. Alkalinity of Water
 The total content of those substances in it which cause an
increased hydroxide ion concentration upon dissociation
or due to hydrolysis. Also called basicity PH ranges above 7
 Alkalinity is a measure of the ability of water to neutralize
the acids.
 The alkalinity of water is attributed to the presence of the
 Caustic alkalinity -due to hydroxide & Carbonate ions
 Temporary hardness -due to bicarbonate ions

33.  With respect to the constituents causing alkalinity in
water, the following situations may arise:-
 Hydroxide ions only
 Carbonate Ions only
 Bicarbonate ions only
 Hydroxide & Carbonate Ions together
 Carbonate & Bicarbonate ions together

34. Estimation of Alkalinity of water
 Alkalinity of water can be estimated by the titration named
as neutralization titration.
 Hydroxide ,carbonate and bicarbonate ions can be
estimated separately by titration against standard acid ,
using phenolphthalein and methyl orange as indicators.
 In this titration we get two different end points : -
 1st pink to colorless
 2nd yellow to red

37. Numerical Problem-
 A sample of water was alkaline to both P and M indicators 100 ml of this water sample
required 10 ml of N/50 acid for P end point and 10 ml acid to M end point on separate
titration .
 Determine type and extent of alkalinity
Vp = 10 ml
Vm= 10 ml
N= N/50
Volume = 100 ml

38.  A water sample is alkaline to both P and M . 200 ml of water sample on titration with
n/50 acid require 9.4 ml acid to P end point . And again titration it required 21 ml
acid for M end point
 Find type and extent of alkalinity

39.  THANK YOU