Environmental Quality Management

Environmental Quality Management

Małgorzata Szlachta, Ph.D. Eng.

Małgorzata Szlachta, Ph.D. Eng.

Environmental Chemistry (Water) -

Environmental Chemistry (Water) -

laboratory

laboratory

What is Environmental

Chemistry?

Environmental chemistry is the study of the

sources, reactions, transport, effects, and fates
of chemical species in water, soil, and air
environments.

Reference: S. M. Manahan, Environmental
Chemistry.

 

What is water?

“

Water

(wo’tar,wot’ar) n. 1.

A clear colorless, nearly

odorless and tasteless liquid, H

2

O, essential for most

plant and animal life and the most widely used of all
solvents. Melting point 0°C, boiling point 100°C,
specific gravity (4°C) 1.000, weight per gallon (15°C)
8.337 pounds. 2. Any of various forms of water such
as rain. 3. Any body of water such as a sea, lake,
river, or stream.”

Reference: W. Morris, The American Heritage Dictionary of the
English Language.

“

Water

is the dilute aqueous

solution/suspension of inorganic and organic
compounds that constitutes various types of
aquatic systems.”

Reference: V. Snoeyink and D. Jenkins, Water
Chemistry.

Distribution of Earth’s
Water

The regional distribution of the

884 million people

not using

improved drinking water sources in 2008, population (million)

Reference: WHO/UNICEF, Monitoring Programme for Water Supply and
Sanitation.
Report: Progress on Sanitation and Drinking Water, 2010 Update.

Safe drinking water

Revision

There are two basic ways to express the mass
concentration of dissolved species (solutes) in
solution:

1.

w/v basis

(weight/volume) - a concentration

in
units of mass of solute in a unit volume of
solution,

2.

w/w basis

(weight/weight) - weight of solute

in a given weight of solution.

Mass concentration

• mg/L = mass of substance (mg)

volume of solution (liter)

• ppm(parts per million) = mass of substance (mg)

mass of solution

(kg)

 
If the density of the solution is known:
 

• density of solution, ρ = mass of solution (kg)

volume of solution (liter)

 

• Concentration in ppm (mg/kg) = concentration in

(mg/L)

x 1/ρ (L/kg)

Equivalents and normal
concentration

The expression of solute concentration as
equivalents/liter, that is,

normal concentration

,

is based on a definition that is related to the type
of reaction in which the solution constituents are
involved.

Equivalent weight (EW)

definition, widely used

in water chemistry based on the charge of ion or
the number of hydrogen/ hydroxyl ions transferred
in an acid-base reaction.

The

equivalent weight

is defined as:

 

and the number of equivalents per liter -

normality

,

is defined as:

 
 

Definitions of equivalents and normality are used for

determining the number of eq/L of charge units and the
number of eq/L of species that participate in precipitation-
dissolution reactions.

molecular weight(MW)

Equivalent weight(EW)

ioncharge

=

massof substanceperliter

Normality

EW

=

Example
 Find the normality of the following solutions:

1. 120 mg CO

32-

/L, given that CO

32_

participates in the precipitation

reaction:
Ca

2+

+ CO

32-

→ CaCO

3

2. 155 mg Ca

3

(PO

4

)

2

/L, given that Ca

3

(PO

4

)

2

participates in the

dissolution reaction:
Ca

3

(PO

4

)

2

→ 3Ca

2+

+ 2PO

43-

 

 

Solution

1. The molecular weight (MW) of CO

32-

is 60 g/mole.

 
 

gramMW 60g/ mole

GramEW

30g/ eq 30mg/ meq

ioncharge 2eq/ mole

120mg/ L

Normality

4meq/ L

30mg/ meq

=

=

=

=

=

=

2. The MW of Ca

3

(PO

4

)

2

is 310 g/mole.

According to reaction Ca

3

(PO

4

)

2

forms six positive and six negative

charges, therefore:
 

310g/ mole

GramEW

51.6g/ eq 51.6mg/ meq

6eq/ mole

155mg/ L

Normality

3meq/ L

51.6mg/ meq

=

=

=

=

=

The

equivalent weight

of a substance in acid-base

reactions is defined as:
"the weight of a substance that will either replace
one H

+

(hydrogen ion/proton) in an acid, provide one

H

+

for reaction, or react with one H

+

to form an

acid."

 
where „n” is the number of hydrogen or hydroxyl
ions that react.

molecular weight(MW)

Equivalent weight(EW)

n

=

Example
Find the normality (meq/L) of the following

solutions:

 1. 40 mg HCl/L, with respect to the reaction:

HCl + NaOH → NaCl + H

2

O

2.

59 mg H

3

PO

4

/L, with respect to the reaction:

H

3

PO

4

→ 2H

+

+ HPO

42-

3.

45 mg CO

32-

/L, with respect to the reaction:

CO

32-

+ 2H

+

→ H

2

CO

3

 

Solution
 
1. One H

+

reacts per HCl, therefore:

gramMW 36.5g/ mole

GramEW

36.5g/ eq 36.5mg/ meq

n

1eq/ mole

40mg/ L

Normality

1.1meq/ L

36.5mg/ meq

=

=

=

=

=

=

2. Two H

+

react per H

3

PO

4

, therefore:

 

gramMW 98g/ mole

GramEW

49g/ eq 49mg/ meq

n

2eq/ mole

59mg/ L

Normality

1.2meq/ L

49mg/ meq

=

=

=

=

=

=

3.

Two H

+

react with each CO

32-

, therefore:

 

gramMW 60g/ mole

GramEW

30g/ eq 30mg/ meq

n

2eq/ mole

45mg/ L

Normality

1.5meq/ L

30mg/ meq

=

=

=

=

=

=

 

Concentration (meq/L) → Concentration

(mg/L)

C (meq/L) x EW (mg/meq) → C (mg/L)

massof substance

Equivalents

EW

=

Example
What is equivalent for 2 g of Ca

2+

?

 
Solution

2

2

2

MWof Ca

40g/ mole

40g/ mole

EW

20g/ eq

2eg/ mole

2gCa

EW

0.1eqCa

20g/ eq

+

+

+

=

=

=

=

=

Example
What is the EW of sodium sulfate Na

2

SO

4

?

 
Solution
Na

2

SO

4

→ 2Na

+

+ SO

42-

 

2

4

MWof Na SO

142.1g/ mole

142.1g/ mole

EW

71.1g/ eq

2eg/ mole

=

=

=

Example
What is concentration of 80 g/m

3

of aluminium ion Al

3+

in

meq/L and in eq/L?
 
Solution
 

3

3

3

Al

C

80g/ m 80mg/ L

MWof Al

30g/ mole

30g/ mole

EW

10g/ eq 10mg/ meq

3eg/ mole

C(mg/ L)

80mg/ L

8meq/ L 0.008eq/ L

EW(mg/ meq) 10mg/ meq

+

+

=

=

=

=

=

=

=

=

=

Water quality

parameters

Colour

Colour

in water is caused by:

• dissolved minerals,
• dyes,
• humic acids which cause a brown-yellow to

brown-black colour,

• coloured wastes, including dyes or pulp and

paper plants,

• the presence of iron, manganese and

plankton.

• Water colour caused by dissolved or colloidal

substances that remain in the filtrate after filtration
through
a 0.45 μm filter membrane is defined as

true colour

.

• Apparent colour

is the term applied to coloured

compounds in solution together with coloured
suspended matter.

Colour is measured spectrophotometrically at a

wavelength

between 450 and 460 nm using glass cells with

path

lengths of 10, 30 or 50 mm, with platinum-cobalt

solution

as standards.

Equipment

•Spectrophotometer

•Glass cells with a light path length equals 5 cm

•0.45 μm filter membrane without organic binder

Method
1.Turn on the instrument and let it warm up
according to manufacturer’s instructions.
2.Filter water sample using 0.45 μm filter
membrane to remove particular matter.

Colour
determination

3. Rinse the blank measuring cell with distilled

water, refill it and place cell at first position in
the spectrophotometer.

4. Rinse the second measuring cell with filtered

water sample, refill it and place cell at second
position in the spectrophotometer.

5. Choose the proper calibration curve, press

AUTOZERO and START.

6. Read colour directly from the instrument, the

result in mgPt/L will be displayed.

 

(2120 C. Spectrophotometric – Single-Wavelength Method,

Standards methods for the examination of water and
wastewater, 21 Edition, 2005)

UV absorbance at 254 nm

Some organic compounds present in natural waters,
such as humic substances and various aromatic
compounds, strongly absorb ultraviolet. Therefore

UV absorption

is a useful surrogate measure of

selected

organic constituents

.

Strong correlations may exist between UV absorption
and both colour and organic carbon content. UV-
absorbing organic matter in a water sample absorb
UV light in proportion to their concentration.

UV254 determination

UV absorption is measured at 253.7 nm but often
rounded off to 254 nm. Samples are filtered through
0.45 µm filter membrane to remove suspended and
colloidal particles.

Equipment
•Spectrophotometer
•Quartz cells with a light path length equals 5 cm
•0.45 μm filter membrane without organic binder

Method
1.Turn on the instrument and let it warm up

according to manufacturer’s instructions.

2.Filter water sample using 0.45 μm filter

membrane to remove the colloidal matter.

3.Rinse the blank measuring cell with distilled

water, refill it and place cell at first position in
the spectrophotometer.

4.

Rinse the second measuring cell with filtered
water sample, refill it and place cell at second
position in the spectrophotometer.

5.

Set wavelength to 254 nm, press AUTOZERO and
START.

6.

Read absorbance directly from the instrument.

Calculations

where:
Abs - absorbance measured
d - quartz cell path length, cm

1

Abs 100

UV245, m

d

-

�

=

Turbidity

Turbidity

in water is a measure of the

cloudiness

and is

caused by:
• suspended and colloidal matter such as clay, silt,

finely divided organic,

• inorganic matter,
• plankton,
• other microscopic organisms.

Turbidity can be correlated with

suspended solids

,

but only for waters from the same source. In such
cases, a simple turbidity measurement may replace
the complex time-consuming suspended solids test.

The instrument used for turbidity measuring is called

nephelometer

or

turbidimeter

. Turbidity also may

be determined by a visual comparison test with
standard turbidity suspensions.

Turbidity is measured in

Nephelometric Turbidity

Units (NTU)

.

Equipment
• Turbidimeter
• Sample cell

Method
1.

Turn on the instrument and let it warm up
according to manufacturer’s instructions.

2.

Gently agitate the water sample.

Turbidity
determination

3. Wait until air bubbles disappear and pour

the sample into the cell.

4. Dry the cell and place it in the turbidimeter,

then close the cover and wait for a while.

5. Read turbidity directly from the instrument,

the result in NTU will be displayed.

(2130 B. Nephelometric method, Standards methods for the

examination of water and wastewater, 21 Edition, 2005)

Solids

The solids content of water is one of the most
significant quality parameter. The amount, size and
type of solids depend on the specific water.

Untreated sewage wastewater may have organic
particulate matter, including food scraps of size
range in millimetres, while a purified drinking water
may have particles in the size range
10

-6

mm.

Solids size:

Solids general classification:

Solids classification:
• Total solids, TS = SS + TDS
• Suspended solids, SS
• Total dissolved solids, TDS = TS - SS
• Total volatile solids, TVS
• Volatile suspended solids, VSS

Total solids

defined as residue left in the vessel/dish after

sample evaporation and drying in an oven at a defined
temperature.

Total solids = sum of

total suspended solids

(the portion

of total solids retained on a filter paper) and

total

dissolved solids

(the portion that passes through a filter

paper).

Volatile fraction

is the weight loss on ignitron of dried

residue of total, suspended or dissolved solids.

Solids determination

Equipment
• Graduated cylinder, 100 ml
• Evaporating dish, 100 mL
• Standard filter paper
• Funnel
• Analytical balance
• Steam bath
• Drying oven

Total solids (TS)

Method

1.Weigh empty evaporating dish; recorded mass

mark as “a”.

2.Using the graduated cylinder measure 50 mL of

well-mixed water sample and pour into the
preweighed dish.

3.Evaporate the sample to dryness on a steam

bath.

4. Next place the evaporating dish in the drying oven

and dry sample for 1 hour at 105˚C.

5. Cool the dish in desiccator to balance the

temperature.

6. Using the analytical balance weigh the

evaporating dish with dried deposit; recorded
mass mark as “b”.

Calculations
 

where:
a – mass of empty evaporating dish, mg
b – mass of evaporating dish with dried deposit,

mg

V – volume of sample, mL

Total dissolved solids
(TDS)

Method
1.Weigh empty evaporating dish; recorded

mass mark as “a”.

2.A well-mixed water sample filter through a

standard filter paper.

3.Using the graduated cylinder measure 50 mL

of filtrate and pour into the preweighed dish.

4.Evaporate the sample to dryness on a steam

bath.

5. Next place the evaporating dish in the drying

oven and dry sample for 1 hour at 105˚C.

6. Cool the dish in desiccator to balance the

temperature.

7. Using the analytical balance weigh the

evaporating dish with dried deposit;
recorded mass mark as “b”.

Calculations

where:
a – mass of empty evaporating dish, mg
b – mass of evaporating dish with dried

deposit, mg

V – volume of sample, mL

Suspended solids (SS)

Calculations

Alkalinit
y

Alkalinity

is a measure of water capacity to

neutralize acids (sometimes referred as the acid
neutralization capacity).

Similarity, acidity is a measure of the base
neutralizing capacity.

Alkalinity in most natural waters is predominantly
due to

bicarbonate (HCO

3-

), carbonate (CO

32-

),

and

hydroxide (OH

-

)

and may vary depending on

the pH.

The amount of acid required to react with OH

-

, CO

32-

and

HCO

3-

is defined as total alkalinity.

Total alkalinity (eq/L) = [OH

-

] + 2[CO

32-

] + [HCO

3-

] –

[H

+

]

 
where concentrations on right side are molarities (known as
molar concentration in mole/L).

Alkalinity is measured by titrating with sulphuric acid

(H

2

SO

4

)

or hydrochloric acid (HCl).

• For

pH > 11

, the added H

+

reacts with OH

-

.

• For

8.3 < pH < 11

, H

+

reacts with CO

32-

, producing

HCO

3-

.

• For

4.5 < pH <8.3

, H

+

reacts with HCO

3-

, producing

H

2

CO

3

.

• Inflection point at

pH 8.3

corresponds to equivalence

point for conversion of CO

32-

to HCO

3-

.

• Inflection point at

pH 4.5

corresponds to equivalence
point for conversion of HCO

3-

to H

2

CO

3

.

For water samples with pH > 8.3 the titration is
made in two steps.

•The first step is to pH of 8.3.
This endpoint corresponds to the equivalence
point for
conversion of CO

32-

to HCO

3-

:

CO

32-

+ H

+

→HCO

3-

Phenolphthalein alkalinity

= [OH

-

] + [CO

32-

]

• The second step is to pH of 4.5.
This endpoint corresponds to the equivalence point for
conversion of HCO

3-

to H

2

CO

3

:

HCO

3-

+ H

+

→H

2

CO

3

Total alkalinity

(known as methyl orange alkalinity)

= [OH

-

] + 2[CO

32-

] + [HCO

3-

] – [H

+

]

For water samples with pH < 8.3 the titration is made

in one step and it is to pH of 4.5, so it’s total
alkalinity.

In water chemistry the alkalinity (HCO

3-

, CO

32-

and OH

-

) can

be express following the calcium carbonate system where
the concentration of a substance is given as

mg/L as

calcium carbonate (CaCO

3

).

CaCO

3

→ Ca

2+

+ CO

32-

Each mole of CaCO

3

yields 1 mole, or 2 equivalents of Ca

2+

,

therefore with respect to Ca

2+

:

 

3

100g/ mole

EWof CaCO

50g/ eq 50mg/ meq

2eq/ mole

=

=

=

Example
What is the total alkalinity (TA) in mg/L as CaCO

3

if a solution has a total alkalinity of 0.002 eq/L?

 
Solution
 
TA = 0.002 eq/L
EW

CaCO3

= 50 g/eq = 50 000 mg/eq

 
TA = 0.002 eq/L · 50 000 mg/eq = 100 mg/L as

CaCO

3

The procedure is for water samples with pH >

4.6.

 
Equipment and reagents
• Conical flask, 250-300 ml
• Graduated cylinder, 100 ml
• Burette with HCl 0.1 molar solution
• Methyl orange, 0.1 % water solution
• pH-meter with electrode

Alkalinity determination

Method
1.Using the graduated cylinder measure 100 mL of

water sample and pour into the flask.

2.Add a few drops of methyl orange and gently mix.
3.Titrate at room temperature with HCl 0.1 molar

solution. The end of titration can be seen by a
colour change from yellow to orange.

4.Record the amount of HCl taken for titration as

”a”.

Calculations

where:
a - amount of HCl 0.1 molar solution taken for

titration, mL

0.1 - conversion factor related with 0.1 molar HCl
50 - the gram equivalent weight of CaCO

3

1000 - conversion factor, mL to L
V - volume of examined water, mL

3

a 0.1 1000

Alkalinity,mgCaCO / L

50

V

� �

=

�

Hardness

of water sample is defined as a

measure of capacity of water to precipitate soap.
The two main divalent ions responsible for soap
precipitation are

calcium

and

magnesium

.

Total hardness

is defined as the sum of calcium

and magnesium concentrations expressed as
calcium carbonate in milligrams per liter
(mg/L as CaCO

3

).

Hardness

When hardness is greater than the sum of carbonate
and bicarbonate alkalinity, that amount of hardness
equivalent to the total alkalinity is called

carbonate

hardness

while the amount of hardness in excess of

this is called

non-carbonate hardness

.

The hardness of water samples may range from zero
to hundreds of milligrams per liter.

Hardness calculations
 

 where M

2+

is divalent metallic ion (e.g. Ca

2+

, Mg

2+

)

2

3

3

2

2

3

3

M (mg/ L)

Hardness,mg/ LasCaCO

EWof CaCO (mg/ meq)

EWof M (mg/ meq)

Hardness,mg/ LasCaCO M (meq/ L) EWof CaCO (mg/ meq)

+

+

+

=

�

=

�

Example
Determine the hardness in mg/L as CaCO

3

of the water sample

contains calcium, magnesium and biocarbonate ions in

concentrations

given in the table.

Solution

Total hardness (TH) = Ca

2+

+ Mg

2+

= 270 mg/L as

CaCO

3

Carbonate hardness (CH) = HCO

3

-

= 300 mg/L as

CaCO

3

The titration method is used in order to detremine the

hardness

as sum of Ca

2+

and Mg

2+

ions.

Equipment and reagents
• Conical flask, 250-300 mL
• Graduated cylinder, 100 ml
• Burette with EDTA-Na, 0.01 molar solution
• HCl, 0.1 molar and 1+1 solution
• NaOH, 2.5 molar solution
• Ammonia, 25% solution
• Reagent ET and murexide

Hardness determination

Method

Ca

+2

1.

Using the graduated cylinder measure 100 mL of water
sample and pour into the flask.

2.

Add the same mL of HCl, 0.1 molar solution as used for
total alkalinity determination.

3.

Add a pinch of murexide and gently mix.

4.

Add 2 mL of NaOH and mix the sample.

5.

Titrate at room temperature with EDTA-Na solution. The
end of titration can be seen by a colour change from
pink to violet.

6.

Record the amount of EDTA-Na taken for titration as ”a”.

Mg

+2

1.After calcium determination, add to the same

sample 7 mL of HCl, 1+1 solution and mix until the
colour disappears.

2.Add 7 mL of ammonia solution, a pinch o ET

reagent and gently mix.

3.Titrate at room temperature with EDTA-Na solution.

The end of titration can be seen by a colour
change from violet to blue.

4.Record the amount of EDTA-Na taken for titration

as ”b”.

Calculations

2

2

2

2

2

2

3

a 0.1 1000 20

Ca ,mgCa / L

2.8 V

b 0.1 1000 12

Mg ,mgMg / L

2.8 V

Totalhardness,mg/ LasCaCO

Ca

Mg

+

+

+

+

+

+

� �

�

=

�

� �

�

=

�

=

+

�

where:
0.1 – conversion factor, 1 mL of EDTA-Na equals

0.1 hardness

number
1000 – conversion factor, mL to L
V – volume of water sample, mL
20 – gram equivalent weight of Ca

+2

12 – gram equivalent weight of Mg

+2

pH

pH

is defined as the negative log (base 10) of

the hydrogen ion concentration and is unitless:

 

pH = -log[H

+

]

 
Taking the negative log of K

W

= [H

+

] [OH

-

]:

-logK

w

= -log[H

+

]-log[OH

-

]

pK

w

= pH + pOH

where:

pH=-log[H

+

]

pOH = -log [OH

-

]

Since K

w

= 10

-14

at 25

o

C, it follows that

pK

w

= 14 at 25

o

C,

which means that:

pH + pOH = 14

The definition of

neutrality

is:

pH = 7 = pOH

 

Acidity:

[H

+

] > [OH

-

]

[H

+

] > 10

7

mole/L

pH < 7

 

Basicity:

[H

+

] < [OH

-

]

[H

+

] < 10

7

mole/L

pH > 7

pH scale:

pH determination

Measure the pH at room temperature with
properly calibrated pH-meter.

Equipment
•pH-meter with electrode
•Glass beaker, 25 mL
 

Method
1.Remove electrode from storage solution,

rinse with distilled water and dry.

2.Pour the water sample into a small beaker.
3.Place the pH electrode in the beaker.
4.Let the electrode to stabilize.
5.The result of pH value will be displayed.

Example

What is the pH and alklinity of groundwater sample with quality
parameters listed in the table. The water temperature is 15 °C.

Constituent

Concentration

mg/L

Ca

2+

190

Mg

2+

84

Na

+

75

Fe

2+

0.1

HCO

3

-

260

SO

4

2-

64

CO

3

2-

30

NO

3

-

35

Solution

Consider the dissociation of HCO

3-

ion: [HCO

3-

] ↔ [H

+

] + [CO

32-

]

 At 15°C K

2

= 3.72·10

-11

mole/L

 

3

3

3

2
3

2

4

3

MWof HCO

61g/ mole

260

HCO

4.26 10 mole/ L

1000 61

MWof CO

60g/ mole

30

CO

5 10 mole/ L

1000 60

-

-

-

-

-

-

=

�

�=

=

�

�

�

�

=

�

�=

= �

�

�

�

Reorganized the equilibrium equation:
 

 

pH = - log [H

+

] = - log 3.17·10

-10

= 9.5

 

3

2

2

3

HCO

H

K

CO

-

+

-

�

�

�

�

� �

� �

�

�

�

�

= �

3

11

10

4

4.26 10 mole/ L

H

3.72 10 mole/ L

3.17 10

5 10 mole/ L

-

+

-

-

-

�

� �=

�

�

=

�

� �

�

2

3

3

Alkalinity HCO

CO

260 60 320mg/ L

-

-

=

+

=

+ =

Example
Given the following water quality analysis, determine the
unknown values:

Constituent

Concentratio

n

Ca

2+

40 mg/L

Mg

2+

?

Na

+

?

K

+

39.1 mg/L

HCO

3

-

?

SO

4

2-

96 mg/L

Cl

-

35.5 mg/L

Alkalinity

3 meq/L

Non-

carbonate

hardness

1 meq/L

Solution

• Use alkalinity to determine [HCO

3-

]

 
 MW of HCO

3-

= 61 g/mole

[HCO

3-

] = 183 mg/L

3

61g/ mole

EWof HCO

61g/ eq 61mg/ meq

1eg/ mole

-

=

=

=

3

HCO

Alkalinity 3meq/ L

61mg/ meq

-

�

�

�

�

=

=

• Use hardness to determine [Mg

2+

]

 
 

{

}

2

2

Total hardness Carbonate hardness noncarbonate hardness

Ca

Mg

3meq/ L

1meq/ L

+

+

=

+

+

=

+

2

2

2

2

Ca

Mg

4meq/ L

EW

EW

Mg

40

4meq/ L

20

EW

Mg

24.4mg/ L

+

+

+

+

� � �

�

� � �

�

+

=

�

�

�

�

+

=

�

�=

�

�

•

Use the anion-cation balance to determine [Na

+

]

Cation

s

Concentr

a-

tion

Equivale

nt mass

Concentr

a-

tion

Anion

s

Concentr

a-

tion

Equivale

nt mass

Concentr

a-

tion

mg/L

mg/meq

meq/L

mg/L

mg/meq

meq/L

Ca

2+

40

20

2

HCO

3

-

183

61

3

Mg

2+

24.4

12.2

2

SO

4

2-

96

48

2

K

+

39.1

39.1

1

Cl

-

35.5

35.5

1

Na

+

x

23

x/23

Total

5+x/23

6

Anions

Cations

x

6 5

23

Na

x 33mg/ L

+

=

= +

� �= =

� �

�

�

Conductivity

Conductivity

is a parameter used to measure the

ionic concentration and activity of a solution, and is
usually express in

µS/cm

.

Conductivity is an important parameter that is often
used for monitoring water quality. For instance
conductivity measurements are used for applications
such as determining the total dissolved solids (

TDS

,

mg/L) or

salinity

of sea water. It’s also used for

assessment the degree of mineralization of distilled
and deionized water.

Conductivity of aqueous solutions:

Conductivity determination

Measure the conductivity at room temperature
with properly calibrated conductivity-meter.

Equipment
•Laboratory conductivity meter with the special
conductivity cell
•Glass beaker, 25 mL
 

Method
1.Rinse the conductivity cell with distilled water

and dry.

2.Pour the sample into a small beaker.
3.Place the cell in the beaker.
4.Operate the conductivity-meter according to

manufacturer’s instructions.

5.Record the result in µS/cm.

Ammonia/nitrite/nitrate

Nitrogen forms - nitrate, nitrite, ammonia, are
of great interest in water chemistry because
they are components of the nitrogen cycle.

Ammonia

is present naturally in water bodies

and wastewaters. Its concentration is
considerably lower in waters - µg/L, than in
wastewaters - mg/L.

Nitrate

is an essential nutrient for many

photosynthetic autotrophs and in some cases has
been identified as the growth-limiting nutrient. It
occurs in trace quantities in surface water but
may attain high levels in some groundwater.

In fresh domestic wastewater nitrate is found in
small amounts but in the effluent of nitrifying
biological treatment plants nitrate may be
present in concentrations of up to 30 mg/ L.

Nitrite

is an intermediate oxidation state of

nitrogen, both in the oxidation of ammonia to
nitrate and in the reduction of nitrate.
Such oxidation and reduction may occur in
wastewater treatment plants, water
distribution systems and natural waters. Nitrite
can enter a water supply system through its
use as a corrosion inhibitor in industrial process
water.

Ammonia (NH

4

+

)

determination

Equipment and reagents
• Standards, mg NH

4+

/100 mL

• Nessler tube, 100 mL
• Seignette salt
• Nessler reagent (K

2

HgI

4

)

Method
1.Pour 100 mL of well-mixed water sample

into the Nessler tube.

2.Add 1 mL of Seignette salt and 1 mL of

Nessler reagent.

3.Mix content of the Nessler tube by turning

it upside down and wait 10 minutes. Yellow
colour occurs and its intensity is
proportional to ammonia concentration.

4.

Compare the sample with standards by looking
vertically downward through the Nessler tube
toward a white surface placed at such angle
that light is reflected upward through the
column of liquid.

5.

Record the result in mg/L NH

4+

.

Nitrite (NO

2

-

) determination

Equipment and reagents
• Conical flask, 250-300 mL
• Spectrophotometer
• Glass cells with a light path length equals 5 cm
• Sulfanilamide acid
• α- naphtylamine solution

Method
1.

Pour 100 mL of water sample into the flask.

2.

Add 1 mL of sulfanilamide acid, mix and wait 5
minutes.

3.

Add 1 mL of α-naphtylamine solution.

4.

Mix gently the sample and wait 10 minutes. Red-
violet colour occurs and its intensity is
proportional to the nitrite concentration.

5.

Turn on the spectrophotometer and let it warm
up according to manufacturer’s instructions.

6. Rinse the blank measuring cell with blank

sample, refill it and place cell at first position
in the spectrophotometer.

7. Rinse the second measuring cell with water

sample, refill it and place cell at second
position in the spectrophotometer.

8. Set the proper calibration curve (a wavelength

520 nm), press AUTOZERO and START.

9. Read nitrite concentration in mg/L directly

from the instrument.

Nitrate (NO

3

-

)

determination

Equipment and reagent
• Nessler tube, 100 mL
• Evaporating dish, 100 mL
• Glass rod
• Pipet
• Steam bath
• Standards, mg NO

3-

/100 mL

• Phenolodisulfonic acid
• NaOH solution

Method
1.Pipet 10 mL of water sample into the

evaporating dish and using the steam bath
evaporate it to dryness.

2.Remove the dish from the steam bath, dry it

and cool it down.

3.Add 1 mL of phenolodisulfonic acid and

using the glass rod, distribute the acid on
the dish walls (the acid makes the deposit
dissolved).

4. Add a few mL of distillated water and pour

the dish contents into Nessler tube.

5. Pipet 5÷7 mL of NaOH solution to the tube

and fill it up to 100 mL with distillated
water. Yellow colour occurs and its intensity
is proportional to the nitrate concentration.

6. Mix the sample by turning it upside down.

7. Compare the sample with standards by looking

vertically downward through the Nessler tube
toward a white surface placed at such angle
that light is reflected upward through the
column of liquid.

8. The results express as mg/L NO

3-

taking into

account that 10 mL of water sample was used
for nitrate determination.

Chlorides

Chloride ions (Cl

-

)

are one of the major inorganic

anions in aquatic environment. The salty taste of
water is due to presence of chloride ions and
dependents on the chemical composition of water.

Waters containing 250 mgCl

-

/L may have a detectable

salty taste if the predominant cation is sodium. The
typical salty taste may be absent in waters containing
as much as 1000 mg Cl

-

/L when the predominant

cations are calcium and magnesium.

Chlorides determination

This method can be used to determine the Cl

-

ions

concentration of water from many sources such as
seawater, stream water or river water. The pH of
the water sample should be between 6.5 and 10.

Equipment and reagents
•Conical flask, 250-300 ml
•Graduated cylinder, 100 ml
•Burette with AgNO

3

solution

•Potassium chromate (K

2

CrO

4

) indicator

Method
1.

Using the graduated cylinder measure 100 mL of
water sample and pour into the flask.

2.

Add 1 mL of K

2

CrO

4

and gently mix.

3.

Titrate the sample with AgNO

3

solution. The end of

titration can be seen by a colour change from faint
lemon-yellow to brick-red.

4.

Record the amount of AgNO

3

taken for titration as

”a”.

Calculations

where:
a – amount of AgNO

3

solution taken for titration, mL

V – volume of water sample, mL
1000 – the conversion factor, mL to L
0.3 – factor related with the amount of AgNO

3

solution, which is necessary for Ag

2

CrO

4

formation

in 100 cm

3

of distillated water

(a 0.3) 1000

Chlorides, mgCl / L

V

-

-

�

=

Determination of organic content

of water

The determination of the organic content of water can

be

done by:

• Specific tests

are suitable for measurement the

concentrations of specific compounds.

Details of specific tests are in

Standard Methods (2005).

• Non-specific

tests to measure the overall

concentration of the organic content.

 

Tests for the overall concentrations include:

• BOD

(a biochemical test that uses microorganisms)

• COD

(a chemical test)

• TOC

(an instrumental test)

 

BOD - biochemical oxygen
demand

The

BOD

5

is the amount of dissolved oxygen used

up from the water sample by microorganisms as
they break down organic material at 20˚C over a 5-
day period. It measures the

readily biodegradable

organic carbon

.

The BOD

5

is arbitrarily set at 5 days and this may

not be long enough to determine the

BOD ultimate

(BOD

u

)

, which is again arbitrarily set at

20 days

.

BOD

u

≈ 2 x BOD

5

Clean waters have BOD

5

values of less than 1

mg/L.
Rivers are considered polluted if the BOD

5

is

greater than 5 mg/L.

The BOD

5

of municipal wastewaters ranges

from about 150 to 1000 mg/L, while for
industrial wastewaters (food industries) the value
may be several thousands.

COD - chemical oxygen
demand

The

COD

test measures the total organic carbon, with

the exception of some aromatics such as benzene
which are not oxidized in the reaction. The test
determines the amount of

oxygen needed to

chemically oxidize the organics

in water/

wastewater sample.

A strong chemical oxidizing agent is used to oxidize
the organics rather than using microorganisms as in
the BOD test. The oxidizing agent is potassium
dichromate in an acid solution.

COD is attractive as a test since it takes about 2
hours by comparison with 5 days for the BOD. A
disadvantage is that it tells nothing about the
rates of biodegradation.

In typical municipal wastewaters:

BOD ultimate (≈ BOD

20

) is ≈ equal (92%) to

COD

BOD

5

≈ 0.6 COD

Example
If bacterial cells are represented by the chemical

formula

C

5

H

7

O

2

N, determine the potential carbonaceous BOD.

Solution
As the cells require O

2

to stabilize them to end

products, first stoichiometrically balance the
equation:

C

5

H

7

O

2

N + 5O

2

→ 5CO

2

+ 2H

2

O + NH

3

stable end products as a result of oxidation

Therefore each mole of bacterial cells requires 5
moles of O

2

for oxidation:

 

 
BOD

u

= 0.92 COD = 0.92 · 1.42 = 1.31

 
If the bacterial cell concentration was, e.g. 1000
mg/L, then the potential BOD

u

= 1310 mg/L.

2

5 7 2

5moleof O

5 32

COD

1.42

1moleof C H O N 113

�

=

=

=

Dissolved oxygen
determination (DO)

Equipment and reagents
• Conical flask, 250-300 ml
• Graduated cylinder, 100 ml
• BOD bottles, 125 mL
• Burette with Na

2

S

2

O

3

solution

• Solutions of MnSO

4

, KJ, and H

2

SO

4

• Starch indicator

Method
1.Fill the bottle with the sample without

entraining the air.

2.Add 1 mL of MnSO

4

and 2 mL of KJ; a

precipitate will form.

3.Cap the bottle that insertion of the stopper

leaves no air bubbles in the bottle and mix it
by inverting several times.

4.Place the bottle in the dark place and allow

the precipitate to settle.

5. Add 1 mL of H

2

SO

4

.

6. Next cap the bottle and gently shake until

the reagent and precipitate have dissolved
(depending on the oxygen content of the
sample a clear yellow to brown colour will
appear).

7. Using the graduated cylinder measure 100

mL of sample and pour into the flask without
entraining the air.

8.

Titrate at room temperature with Na

2

S

2

O

3

solution

until the faint yellow colour will be seen.

9.

Then add 5 mL of starch indicator and continue the
titration until the blue colour disappears.

10.

Record the amount of Na

2

S

2

O

3

taken for titration as

”a”.

Calculations

where:
x – amount of Na

2

S

2

O

3

taken for titration (before

and

after starch addition), mL
1000 – conversion factor, mL to L
0.2 – conversion factors
V – volume of sample, mL

2

0.2 a 1000

DO, mgO / L

V

��

=

Literature

•

S. E. Manahan, Fundamentals of Environmental
Chemistry. Second Edition, Lewis Publishers, 2001.

•

V. L. Snoeyink, D. Jenkins, Water Chemistry. John
Wiley & Sons, 1980.

•

S. D. Faust, O. M. Aly, Chemistry of water
treatment. Lewis Publishers, 1998.

•

Standards Methods for the Examination of
Water and Wastewater. American Public Health
Association, American Water Works Association,
American Public Health Association, 21 Edition,
2005.