WEEN 442
ENVIRONMENTAL ENGINEERING
Water Chemistry-Basic properties of water
• Physical Properties
1. Mass density (ρ)=Mass per unit volume(kg/m3)

Varies with temperature

Dissolved impurities change the density in direct proportion to
their concentration and density

Its common to ignore the effect of the impurities on density when
they are in low concentration.

When in high concentration, their effects should not be ignored
2. Specific Weight (γ
)=Weight(Force)/unit volume (KN/m3)
Gamma
The specific weight of a fluid is related to its density by the acceleration of
gravity, g through the following relation
γ= ρg
• Physical Properties
3. Specific gravity=Ratio of mass density (ρ) or specific weight of water γ at
any given temperature to the mass density (ρo) or specific weight of
water at 3.98 oc (γo)
ρo =1000 kg/m3 while γo =9.81KN/ m3

4.
Viscosity = resistance of fluid to movement.
The resistance is caused by an internal friction.
The higher the friction, the harder it is to pump the liquid. Viscosity is
represented in 2 ways
i.
Dynamic viscosity or absolute viscosity (µ) and has dimensions of
mass per unit length per unit time(Pa.s)
• Physical Properties
ii. Kinematic Viscosity (ν)= µ/ρ (m2/s)
It is an alternative numerical expression the viscosity of a fluid
Often used to evaluate the friction coefficient for flow in pipes
• Impurities in solution
Substances can exist in water in one of three classifications
1. Suspended Solids = they are particles of organic or inorganic materials suspended
in water. They are large enough to settle out of solution or be removed by filtration.
Methods employed in the removal of suspended solids
 Sedimentation
Filtration
Centrifugation
2. Colloidal Solids/Particles = they are particles of organic or inorganic materials
suspended in water.
o They are smaller than suspended solids.
o They are prevented from settling out of solution by surface charges which prevent
them from coalescing together.
o surface charges exists only on very small particles.
o They are too small to be removed by sedimentation or by normal filtration
o Colloidal particles exhibit the Tyndall effect i.e. when light passes through a liquid
containing colloidal particles, the light is reflected by the particles.
o Colloidal particles cause turbidity in water.
o Turbidity is the relative measure of the degree to which a colloidal suspension reflects
light at a 90o angle to the entrance beam
o Turbidity is a relative measure. A given sample is compared to one of the various
standards that exist for turbidity
o The most common standard is Nephelometric turbidity unit (NTU).
o For any given particle size, the higher the turbidity, the higher the concentration of
colloidal particles.
• Impurities in solution
Methods employed in the removal of colloidal solids
Filtration through small pore membranes
High force centrifugation
3.
Dissolved Solids = they can be simple atoms or complex molecular
compounds. Dissolved solids form a single phase with the liquid. They
cannot be removed from solution without an accompanying phase
change. Methods employed in the removal of dissolved solids

Distillation- either the liquid or the dissolved solid is changed from a
liquid phase to a gas phase to achieve separation
• Impurities in solution
 Precipitation- the dissolved solid combines with another dissolved
chemical (in a chemical reaction) to form a new chemical in solid
phase, thereby separating from the liquid.
Adsorption – the dissolved solid adsorbs onto a solid particle and
thereby separates from the liquid
Liquid extraction-a dissolved solid is moved from one liquid into a
second liquid for which the solid has a higher affinity.
Ionic pore sized membranes- a membrane with pore sizes in the
ionic size range of the dissolved solid can separate the solids from
solution through a high pressure filtering process
• Color
 Color is a combination of dissolved and colloidal materials.
It’s widely used in environmental engineering because it can be
measured
It’s difficult to distinguish between dissolved and colloidal color i.e.
color caused by dissolved solids vs color caused by colloidal solids
The most common cause of color in water is the presence of
complex organic compounds that originate from the decomposition
of organic matter.
A common source of color is the degradation of soil humus, which
produces humic acids. Humic acids impart a reddish brown color to
water.
Humic acids have molecular weights between 800 – 50000, the
lower range being dissolved while the higher range is colloidal
Color is measured by the ability of the solution to absorb light
True color is a term used to describe color after turbidity has been
removed
• Chemical units
 Solutes in solution are often analysed by weight. Their weight is
expressed as
weight percent and
Milligram per liter

 To perform stoichiometric calculations, the weight percent is converted into
common units
𝑊
Weight percent P=
𝑋100
𝑊+𝑊0
Where P= percent of substances by weight
W= weight in grams of substance
W0=weight in grams of solution
Results are given in milligram per liter (mg/l)
It is assumed that the substance under consideration does not
change the density of water. Given that 1ml of water weighs 1
gram, then
• Chemical units
Given that 1ml of water weighs 1 gram, then
1𝑚𝑔
𝑙
=
1𝑚𝑔
1000𝑔
=
1𝑚𝑔
106 𝑚𝑔
= 1𝑝𝑝𝑚
the weight percent is then determined as
𝑊
1𝑚𝑔
10−3 𝑔
P=
𝑋100 =
X100 = 3 𝑋100 =
𝑊+𝑊0
𝑙
10 𝑔
1mg/l= 1𝑋10−4 % and 1%=10,000mg/l
1𝑋10−4 % 𝑜𝑟
Example:
A water treatment plant produces 100kg of dry matter of sludge per day.
That sludge is removed from the treatment process in 10m3 of water. What
is the actual percent solids and the approximate mg/l concentration,
assuming that the density of water does not change.
• Chemical units
Solution: density of water=1000kg/m3, therefore 10m3 will weigh 10,000kgs
Actual % solids P will be calculated as
100𝑘𝑔𝑠
P=
100𝑘𝑔𝑠+10000𝑘𝑔𝑠
concentration in mg/l=
𝑊
P=
𝑋100
𝑊+𝑊0
=
𝑋100 =0.99%
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑𝑠
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
=
100𝑘𝑔𝑠𝑋106 𝑚𝑔 𝑘𝑔
=10000
10𝑚3 𝑋1000𝑙 𝑚3
• To work with chemical reactions, weight concentrations are converted to
molarity or normality.
• A mole is 6.02*1023 molecules of a substance
• Chemical reactions are expressed in integral numbers of moles
• A mole of a substance has a relative weight called its molecular weight
(MW) and is equal to the sum of the atomic weights of the constituting
atoms
• Molarity is the number of moles in one liter of solution, therefore a 1Molar
(1M) solution has 1 mole of substance per liter of solution.
• Molarity is related to mg/l as follows
mg/l=Molarity(Moles/l)X molecular weights
𝑔
𝑚𝑜𝑙𝑒
∗
103 𝑚𝑔
𝑔
• Equivalent weight (EW) of a substance is its molecular weight divided by
the number of electrons (n) transferred in redox reactions or the number of
protons (hydrogen ions) transferred in acid-base reactions. The value of n
is dependent on the reaction.
• In a precipitation reaction, n is the valence of the element under
consideration.
• For compounds, n is equal to the number of hydrogen ions that would be
required to replace the cations e.g. for 𝐶𝑎𝑐𝑜3 , 2 hydrogen ions would be
required to replace the calcium, therefore n=2.
• In oxidation/reduction reactions n is equal to the change in oxidation
number that the compound undergoes in the reaction.
• Equivalent weights are frequently used in water softening calculations.
• Normality (N) is the number of equivalent weights per liter of solution.
• Normality (N) is related to Molarity (M) as follows
N=Mn where n=number of electrons/hydrogen ions/valency
Example:
Commercial sulfuric acid 𝐻2 𝑆𝑂4 is often purchased as a 93 weight percent
solution. Find the mg/l of 𝐻2 𝑆𝑂4 , the Molarity and Normality of the solution.
Sulfuric acid has a specific weight of 1.839
Solution 1liter of water weighs 1000g
1 liter of 100% pure 𝐻2 𝑆𝑂4 will weigh 1000g(1.839)=1839g
at 93% 1 liter of commercial 𝐻2 𝑆𝑂4 will weigh 0.93(1839g)=1710g
expressed as
1.7∗106 𝑚𝑔
𝑙
the molecular weight of 𝐻2 𝑆𝑂4 =
(2H+S+4O)=(2*1+(1*32)+(4*16))=98g/mole
Molarity=
𝑔 𝑙 𝑜𝑟 𝑚𝑔 𝑙
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 (𝑔 𝑜𝑟 𝑚𝑔)
=
1710𝑔 𝑙
98𝑔 𝑚𝑜𝑙𝑒
=
17.45𝑚𝑜𝑙𝑒𝑠
𝑜𝑟
𝑙
17.45𝑀
Normality (N) ; 𝐻2 𝑆𝑂4 can give up 2 hydrogen ions and therefore
n=2equivalents/mole.
N=Mn therefore N=17.45moles/l* (2equivalents/mole)
=34.9 equivalents/l
Example 2;
Calculate the weight of sodium bicarbonate 𝑁𝑎𝐻𝐶𝑂3, necessary to make a
1M solution. What will be the normality of the solution
Solution:
calculate the molecular weight of 𝑁𝑎𝐻𝐶𝑂3, MW=84 (Na=23; H=1;
C=12, O=16)
mg/l=Molarity(Moles/l)X molecular weights
=(1Mole/l)X
84𝑔
𝑚𝑜𝑙𝑒
103 𝑚𝑔
𝑔
𝑔
𝑚𝑜𝑙𝑒
∗
103 𝑚𝑔
𝑔
=84000
𝐻𝐶𝑂3− gives or accepts only one proton, therefore n=1 and
Normality=Molarity=1equivalent/l
Example 3;
Calculate the equivalents of each of the following 𝐶𝑜32− , 𝐶𝑎𝐶𝑂3, 𝐶𝑎2+
Solution:
Equivalent weight (EW)=
𝐴𝑡𝑜𝑚𝑖𝑐 𝑜𝑟 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡
𝑛
Units of EW are grams/equivalent(g/eq) or
milligrams/milliequivalent(mg/meq);
For Calcium n=valency or oxidation state in water=2
Atomic weight of calcium=40.08
therefore
40.08
(EW)=
2
=
20.04𝑔
𝑜𝑟
𝑒𝑞
𝑚𝑔
20.04
𝑚𝑒𝑞
For the carbonate ion, the oxidation state is 2 (NB. The base 𝐶𝑜32− , can
potentially accept 2 hydrogen ions), therefore n=2
Molecular weight =60.01 (12.01+3(16))
Therefore
60.01
(EW)=
2
=
30𝑔
𝑜𝑟
𝑒𝑞
𝑚𝑔
30
𝑚𝑒𝑞
Example 3;
Calculate the equivalents of each of the following 𝐶𝑜32− , 𝐶𝑎𝐶𝑂3, 𝐶𝑎2+
For the calcium carbonate 𝐶𝑎𝐶𝑂3, n=2 (it would take 2 hydrogen ions to
replace the cations 𝐶𝑎2+ to form carbon acid 𝐻2 𝐶𝑂3
Molecular weight =60.01+40.08=100.09
Therefore (EW)=
100.09
2
=
50.04𝑔
𝑜𝑟
𝑒𝑞
50.04
𝑚𝑔
𝑚𝑒𝑞
Chemical reactions
• There are four principle types of reactions of importance in environmental
engineering.
Precipitation
Acid-Base reactions
Ion association
Oxidation/reduction
• Dissolved ions can react with each other to form a solid compound
• This phase change reaction of dissolved to solid is a precipitation reaction
𝐶𝑎2+ + 𝐶𝑜32− ↔ 𝐶𝑎𝐶𝑂3 (𝑠)
• Acid-base reactions a special type of ionization where a hydrogen ion is
either added or removed from solution e.g.
 Addition of hydrochloric acid into water adds a hydrogen ion
HCL ↔𝐻+ + 𝐶𝑙 − ; 𝐻𝐶𝐿 + 𝐻2 𝑂↔𝐻3 𝑂 + 𝐶𝑙 −
 Addition of NaOH removes a hydrogen ion from water
NaOH+ 𝐻2 𝑂 ↔𝑁𝑎+ + 2𝐻2 𝑂 + 𝑂𝐻 − ;
• Ions can also exist in water as complexes with other ions. Formation of
dissolved complexes are ion-association reactions e.g. Mercuric chloride.
It can exist as a neutral compound of as its constituent ions
𝐻𝑔2+ + 2𝐶𝑙 − ↔ 𝐻𝑔𝐶𝑙2
• Often, the soluble complex has a charge and is itself an ion e.g. metal ion
complexes
𝐴𝑙3+ + 𝑂𝐻 − ↔ 𝐴𝑙𝑂𝐻 2+
• Oxidation/reduction reactions involve valency changes and the transfer of
electrons e.g. when iron metal corrodes, it releases electrons
𝐹𝑒 0 ↔ 𝐹𝑒 2+ + 2𝑒 −
• If one element releases electrons, another must be available to accept the
electrons, therefore in corrosion
2𝐻 + + 2𝑒 − ↔ 𝐻2 (𝑔)
Precipitation reactions
• All complexes are soluble in water to some extent. Some are very soluble
e.g. sodium chloride (Nacl) while others are sparingly soluble e.g. silver
chloride (𝐴𝑔𝐶𝑙2 ).
• When a solid compound is placed in distilled water, some of the compound
will go into solution. At some time (t), no more compound will dissolve and
equilibrium will be achieved.
• The time to reach equilibrium ranges from seconds to years. The solubility
reaction is written as follows
𝐴𝑎 𝐵𝑏 (𝑠) ↔𝑎𝐴𝑏+ + 𝑏𝐵𝑎− 𝑒𝑥𝑎𝑚𝑝𝑙𝑒 𝐶𝑎3 (𝑃𝑂4 )2 (𝑠) ↔ 2𝐶𝑎2+ + 2𝑃𝑂43−
• The product of the activity of the ions is always constant for a given
compound at a given temperature. (the activity is approximated by the
Molar concentration)
• This constant is called the solubility constant 𝐾𝑠 = 𝐴 𝑎 𝐵 𝑏 𝑤ℎ𝑒𝑟𝑒
=
𝑚𝑜𝑙𝑎𝑟 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
• Ks values are often reported as pKs, where pKs=-logKs. The constant
applies equally to reactions that involve precipitation.
Example
What concentration in mg/l of phosphates 𝑃𝑂43− would be in solution at
equilibrium with 𝐴𝑙𝑃𝑂4 (𝑠)