Section IV Water Pollution

Water Quality:

Chemistry: Review:

Molecular weight: is the mass of a molecule. It is a sum of the combining element

[g/mole].

Equivalent weight: is the molecular weight divided by the number of positive or

negative electrical charges that resulting from compound dissolution [g/eq. weight]

Molarity (M): [mole/L]: is the number of mole per litter of solution

Molality [mole/kg]is the number of mole per kg of material

Normality (N) [g equ./l]: is the number of equivalent weight per litter of solution

It is N = M*n

ppm : part per million and equals to mg/l

Fe2 (SO4)3 has a MW = 400 [g/mole]

-since its dissolution produces 6 charge

thus its eq. weight = 400/6 = 66.7 [g/eq. weight]

Examples:

(1) H2SO4 + 2OH-= 2H2O + SO42-

Molecular weight of sulfuric acid (H2SO4) = (2*1) + (1*32) + (4*16) = 98.07 g/mol

Equivalent weight of sulfuric acid (H2SO4) = 49.03 g/equivalent of H+

Reasoning:

98.07 g/mol H2SO4 * (1 mole H2SO4 / 2 equivalents of H+) = 49.03 g/equivalent of H+

In this example, the magnitude of the equivalent weight of sulfuric acid is HALF of

that of the molecular weight. This is because according to the balanced chemical

reaction, one mole of sulfuric acid reacts with TWO equivalents of hydroxide.

Example 2:

NH4OH + H+ = H2O + NH4+

Molecular weight of ammonium hydroxide (NH4OH) = 35.00 g/mol

Section IV Water Pollution

Equivalent weight of ammonium hydroxide (NH4OH) = 35.00 g/equivalent of OH-

Reasoning:

35.00g/mol of NH4OH * (1 mole NH4OH / 1 equivalent of OH-) = 35.00 g/equivalent

of OH-

In this example, the magnitude of the equivalent weight of ammonium hydroxide is

the same as the molecular weight.This is beacuse according to the balanced chemical

reaction, one mole of ammonium hydroxide reacts with ONE equivalent of H+.

Example 3:

Calculate the equivalent weights of HNO3 (MW = 63) and Ga(OH)3

(MW = 121) in the following three reactions:

- (a) 3HNO3 + Ga(OH)3 → 3H2O + Ga(NO3)3

EW(HNO3) = 63/1 = 63g

EW[Ga(OH)3] = 121/3 = 40.3g

All 3 OH- enter the reaction. Note the 3 water formed and that no unused OH- are

found in the salt Ga(NO3)3.

- (b) HNO3 + Ga(OH)3 → H2O + Ga(OH)2(NO3)

EW(HNO3) = 63/1 = 63g

EW(Ga(OH)3] = 121/1 = 121g

Since only 1 H+ is available to react with the 3 OH- from the Ga(OH)3, only 1 OH-

reacts. Note also that only one water forms and two OH - are left unused in the salt

Ga(OH)2NO3.

- (c) 2HNO3 + Ga(OH)3 → 2H2O + Ga(OH)(NO3)2

EW(HNO3) = 63/1 = 63g

EW[Ga(OH)3] = 121/2 = 60.5g

Two OH- are used by the 2H+ provided by the 2 moles of HNO3. Note also that 2

moles of water form and 1 OH- is left unused in the product Ga(OH)(NO3)2.

Section IV Water Pollution

Example:

500 mg of anhydrous Ca(HCO3)2 is dissolved in 750 ml of water. What is the molar

concentration, normality and the concentration in ppm as expressed as CaCO3?

Mw of Ca(HCO3)2 = 162gm/mole

The equivalent weight = 162/2 = 81g/equivelant

Concentration of Ca(HCO3)2 in the solution = 500/0.75 =666.67mg/l (ppm)

No of mole = 500/162 = 3.08 mmole

Morality concentration = 3.08/0.75 = 4.12 m mole/L

Normality concentration = 500/(81*0.75) = 8.24 N

666.67mgL

81𝑔

𝑒𝑞𝑢𝑖𝑣𝑒𝑙𝑎𝑛𝑡

= 8.24𝑚𝑒𝑞𝑢𝑖

𝐿

Thus the concentration as CaCO3 = 8.24*50 = 412mg/l as CaCO3

Example:

what mass of CO2 would be produced if (100g) of butane (C4H10) is completely

oxidized into CO2 and water?

2C4H10+ 13O2 8CO2 + 10H2O

Solution : from the balanced chemical equation

Each 2 mole of butan produces 8 mole of CO2

- Determine the number of mole of butane No of mole = weight/MW = 100/58 = 1.724 molw

2 mole of CH4 8 mole of CO2

1.72 of CH4 X

Thus X= 1.72*8 / 2 = 6.8 mole of CO2 is produced

MW of CO2 = 44 g/mole

Thus the amount of CO2 is 6.8 * 44 = 303 g of CO2

Section IV Water Pollution

Example:

Commercial Sulphuric acid H2SO4 is often purchased as 93% solution. Find the

concentration (mg/l),the molarity and normality fo the solution. If the specific gravity

of H2SO4 = 1.839.

Solution:

If H2SO4concentration is 100% , thus its itsspecific gravity will be 1.839

but since the concentration is only 93%

thus the solution S. gravity = 0.93* 1.839 = 1.710

MWH2SO4 = 2*1 + 1*32+4*16 = 98g/mole

Assume one litter of the solution:

Thus it contain 1710 g/l

Thus the molarity = 1710

𝑔

𝑙

98𝑔

𝑚𝑜𝑙𝑒

= 17.45𝑚𝑜𝑙𝑒

𝑙 𝑜𝑟 17.45 𝑀

Normality : MWH2SO4 = 98/2 = 48

N= 1710/48 = 34.9 eq/l

Example:

find the weight of sodium bicarbonate [NaHCO3]that is required to prepare 1M

solution?

MWNaHCO3= 84g/mole

1mole/l= mass

84𝑔/𝑚𝑜𝑙𝑒 = mass = 84g/l = 84,000mg/l

Water quality is studied for the following reasons:

1- Determine the degree of pollution

2- Determine the required step for W.T. P (water treatment plant)

3- Proper design for treatment unit

4- Check the effluent of WTP in reference with the specified standards.

Characteristic of water:

1- Physical characteristics

2- Chemical characteristics

3- Microbiological characteristics

4- Radiological characteristics

Section IV Water Pollution

A- Physical characteristics: it is associated with water appearance and includes

Parameter Range for Drinking Water (DW)

Temperature 15:20 oC

Turbidity ≤ 1 𝑁𝑇𝑈 (Nephlometric turbidity unit)

Color Colorless

Taste and Odor No odor

B- Chemical Characteristics:

- can be evident be their reaction like hardness , PH residual elements

- it includes also toxic substances such as organic matters, Nitrate (NO3),

Cyanides (CN) and heavy metals.

- It controlled by maximum allowable pollutants (organic, non-organic)

concentrations

Parameter Maximum Range for Drinking Water

(DW)

PH 6.5-8.5

Acidity

Alkalinity

Dissolved Oxygen

Hardness >300 mg/l

Chlorides >250 mg/l

sulphates 150-200 mg/l

Nitrates >45 mg/l

Iron 0.1 – 0.3 mg/l

Solids - Total dissolved Solid (TDS) >500*

- suspended solids

*if there is no other source it my increased to 1500 mg/l

C- Microbiological characteristics

- Mainly caused by microorganism activities like bacteria, Viruses, protozoa

- E-coli and total coliform tests are used to determine the water pollution with

bacteria

- Their presence in water indicates pollution with human or animal fecal.

- Controlled by safe drinking water standards and wastewater effluent standard

- Usually the result presented by presence abcence test, most probable

number (MPN)

- For Filtered water Pseudomonas aeruginosa is tested , because it comes as a

result of unclean old filter.

Section IV Water Pollution

D- Radiological characteristics

- The water should not contain any radioactive materials, that cause chronic or

cancer diseases

Solids: Types of impurities in water:

1- Suspended Solids (SS):

It is the large particle in water with diameter between (10 -1 : 10-3mm) such as

clay, salt, sand.

-It cause the water turbidity

- Can be removed by filtration or sedimentation

-Turbidity measurement principle:

Section IV Water Pollution

2- Colloidal matters: its smaller particle size usually (10 -3 : 10-6mm) and can’t

be settled alone

- Can be removed by very high force centrifugation or using an additive

(coagulant)

3- Dissolved solids (D.S): it is the salts and dissolved organic matter

- Usually cause water problem like calcium and potassium carbonate and

bicarbonate

- Can be removed by distillation, adsorption , ion –exchange or liquid extraction

What is water conductivity? Ability of water to conduct

Electrical current

Max .value 5000 µS/cm

Exercise: At home: Bring a cup of water and dissolve the following:

1- One spoon salt (NaCl)

2- Single spoon of Sugar (C6 H12 O6)

3- Single spoon of very fine spice (any type)

4- Single spoon of fine sand

Describe to which W characteristic each one belong with justification?

TDS [ppm] = 1.56 of

conductivity [µS/cm]

Why? Sugar and salt

Section IV Water Pollution

Testing summary:

1- Dry known weight of water sample at 103 oC to 105oC to drive off water in the

sample.

2- The residue is cooled, weighed, and dried again at 550oC to drive off volatile

solids in the sample.

3- The total, fixed, and volatile solids are determined by comparing the mass of the

sample before and after each drying step. Where

Laboratory procedure to determine TS and TVS utSource: UMT

Total Suspended solid (SS) = w s

V

Volatile solid = 𝑤𝑠−𝑤𝑓

𝑉

Note: for water the each 1000ml= 1000g

Example:

If the following data is obtained from a 100ml of water sample ,calculate TSS and

VS?

filter mass = 1.5413g

Mass after drying at 105 oC= 1.5541g

Mass after igniting at 550oC= 1.5519g

Solution:

Suspended solid (SS) = 1.5541−1.5413

100/1000= 0.128 g/l

Section IV Water Pollution

Volatile solid = 𝑤𝑠−𝑤𝑓

𝑉 = 0.022 g/l

Fixed solid = 1.5519−1.5413

100/1000= 0.106 g/l

Dissolved Oxygen (DO):

- The source of D.O in water is photosynthesis and aeration

- It is one of important parameters to measure the water quality.

- It gives pleasure taste to water

- As the temp D.O

- If the D.O concentration decreases to less than 4mg/l all fish die

- If the D.O concentration is less than 2 mg/l all organism dies and the water is

called septic water

- Best D.O concentration is between 8-10 ppm. Optimum is 9ppm.

- The maximum naturally accrued is 14 mg/l.

Biochemical Oxygen demand (BOD) :

- is the quantity of oxygen that is used by microorganism to stabilize the wastewater,

Usually measured after 5 days.

Section IV Water Pollution

- a BOD test can be used to measure waste loadings to treatment plants, plant efficiency and the effects of a discharge on a receiving stream, and to control the

plant process.

- It is indicator for the required aeration amount.

- The main equation describes the process is:

DO + organic matter CO2 + biological growth

- Drinking water usually has a BOD of less than 1 mg/L

- Ordinary domestic sewage may have a BOD of 200 mg/L.

-Any effluent to be discharged into natural bodies of water should have BOD less than

30 mg/L.

Test Summary:

1- The sample is filled in an airtight bottle and incubated at 20 oC for 5 days.

2- The dissolved oxygen (DO) content of the sample is determined before and

after five days of incubation at 20°C

3- and the BOD is calculated from the difference between initial and final DO.

The initial DO is determined shortly after the dilution is made; all oxygen uptake

occurring after this measurement is included in the BOD measurement

Calculations:

BOD5 mg/l = (Initial DO - DO5) x Dilution Factor

Dilution Factor =Bottle Volume (300 ml)

𝑠𝑎𝑚𝑝𝑙𝑒 𝑉𝑜𝑙𝑢𝑚𝑒

Section IV Water Pollution

BOD At any time:

BODt = L(1-10-kt)

Where:

BODt : BOD at any time

L: ultimate bio-oxygen demand

k: oxygen decay constant, [day-1]

t: time, [days]

Example:

A BOD test is done by pipiting 5 ml of waste water into 300 ml testing bottle. If the

initial DO was 8.4 mg/l and the DO after 5-days of incubation at 20 oC was 3.7mg/l,

calculate the BOD and estimate the 20-days BOD value assuming the reaction decay

constant k = 0.1 day-1.

Solution:

Dilution factor = 300/5 = 60

BOD5 = (8.4 – 3.7)*60 = 282 mg/l ….*

b- to determine the BOD after 20 days:

from (*) calculate L

282 = L (1-10−0.1 ×5 ) L = 412 mg/l

Thus BOD20 = 412 (1-10−0.1 ×20 ) BOD20 = 407.8 mg/l

Chemical oxygen demand (COD):

-Measure the amount of organic compounds in water that can be oxidized by strong

oxidant like mixture of sulfuric and chromic acids.

-Most applications of COD determine the amount of organicpollutants found in

surfacewater.

- It indicates of the strength (degree of pollution) of industrial WW that are not

biodegradable

- the test is faster than BOD test.

Section IV Water Pollution

Water PH

- Very important parameter that affects treatment processes, especially coagulation, anddisinfection

- any unusual change may reflect a major event

H2O H+ + OH-1 ,

A- kw = [H+] [OH-1];;;; kw = 1 × 10−14

B- PH = -log[H+] [H+] = 10-PH

Alkalinity :

- The source of alkalinity in natural water is bicarbonates that formed in reactions in the soils through which the water percolates.

- It measures the capacity of water to neutralize acids (buffer capacity) in other

words its inherent resistance to pH change).

- river alkalinity valuesof up to 400 mg/l CaCO3

- it is determined by titration against standardized sulfuric acid (H2SO4)

- sorurce of alkalinity is bicarbonate (main) carbonat and hydroxde

CO2 + H2O H2CO3 (carbonic acid)

Section IV Water Pollution

H2CO3 H+ + HCO3-1 Bicarbonate ka1 = 4.3 × 10−7

HCO3-1 H+ + CO3-2 carbonate ka2 = 4.7 × 10−11

C- ka1 = 4.3 × 10−7 = [𝐻+] [𝐻𝐶𝑂3−1]

[H2CO 3]

D- ka2 = 4.7 × 10−11 = [𝐻+] [𝐶𝑂3−2]

[𝐻𝐶𝑂3−1]

E- Alkalinity = [HCO3-1]+2[CO3-2] + [OH-1] - [H+] …. Mole/l

-in a typical alkalinity problem , it is required to determine [H+]; [HCO3-1]; [OH-1];

[H2CO3]; [CO3-2]

Section IV Water Pollution

Example:

a water has alkalinity of 250mg/l as CaCO3 and a PH equals 7.5, determine [H+];

[HCO3-1]; [OH-1]; [H2CO3]; [CO3-2].

Solution:

1- PH = 7.5 = -log [H+] …

thus[H+] = 10−7.5 = 3.16 × 10−8 𝑚𝑜𝑙𝑒/𝑙

thus [H+] = 3.16 × 10−8 × 1 × 103 = 3.16 × 10−5 𝑚𝑔/𝑙

2- Kw = 10-14 = [H+][OH-1] .. thus [OH-1] = 10−14

3.16 × 10−8 = 3.16 × 10−7 𝑚𝑜𝑙𝑒/𝑙

[OH-1] = 3.16 × 10−7 𝑚𝑜𝑙𝑒

𝑙 ×

17𝑔

𝑚𝑜𝑙𝑒= 5.37 × 10−6 𝑔

𝑙= 5.37 × 10−3 𝑚𝑔/𝑙

3- Determine the equivalent number of mole :

No of Eq. mole = weight/ eq. weight

250 × 10−3𝑔/𝑙

50𝑔/𝑒𝑞 . 𝑚𝑜𝑙𝑒= 5 × 10−3 𝑚𝑒𝑞/𝑙

Thus

CaCO3 Ca + CO3-2

- Every 1 mole of CaCO3 produces 1 mole of CO3-2

- 100g/mole of CaCO3 produces 60 g/ mole of CO3-2

- thus 250mg/mole of CaCO3 will produce = 250*60/100 = 150 mg of CO3-2

[CO3-2] = 5 × 10−3 𝑒𝑞.𝑚𝑜𝑙𝑒

𝑙

Alkalinity = 5 × 10−3 𝑒𝑞.𝑚𝑜𝑙𝑒

𝑙= [HCO3-1]+2[CO3-2] + [OH-1] - [H+]…..A

ka1 = 4.3 × 10−7 = [𝐻+] [𝐻𝐶𝑂3−1]

[H2CO 3]

ka2 = 4.7 × 10−11 = [𝐻+] [𝐶𝑂3−2]

[𝐻𝐶𝑂3−1]…. Thus

𝐶𝑂3−2 = 4.7× 10−11

3.16× 10−7 𝑚𝑜𝑙𝑒 /𝑙* [𝐻𝐶𝑂3−1] =1.48 × 10−4[𝐻𝐶𝑂3−1]

Substitute in A

Section IV Water Pollution

5 × 10−3 𝑒𝑞.𝑚𝑜𝑙𝑒

𝑙= [HCO3-1]+2×1.48 × 10−4[𝐻𝐶𝑂3−1]+[OH-1] - [H+]

[HCO3-1]= 4.99×× 10−3𝑞.𝑚𝑜𝑙𝑒

𝑙

Hardness:

- it is the capacity of a water to destroy the lather of soap. Causes economical losses.

- hardness mainly caused by the calcium (Ca+2) and magnesium (Mg+2)ions

- It is expressed as mg/l CaCO3

CaCO3 [mg/l] Type of water

Up to 50 ppm as CaCO3 Soft water

51- 100 ppm as CaCO3 Moderately soft

101-150 ppm as CaCO3 Slightly hard water

151-250 ppm as CaCO3 Moderately hard water

> 250 ppm as CaCO3 Hard water

- Total Hardness:is the expression of the results of direct measurement (principally of calcium andmagnesium) expressed as mg/l CaCO3

- Calcium Hardness:is the expression of the results of the measurement of calcium only, as mg/l CaCO3.

- Magnesium Hardness: is the difference between total hardness and calcium

hardness is taken as the magnesiumhardness

- There are other types of hardness like Carbonate Hardness( is removed by heating) , Non-carbonate Hardness and Permanent Hardness

Ca+2 + CO3 CaCO3

Mg+2 + OH- Mg[OH]2

Molecular weight of CaCO3 = 1*40 + 12 + 3*16 = 100 g/mole

Equivalent weight = Mw/charge = 100/2 = 50 g/eq. weight

Example

: the result of water analysis show the following for cations :calcium 35.8; magnesium

9.9; sodium 4.6 and potassium 3.9 ppm. While the anions were chloride 7.1; HCO 3-1

131; SO4-2 26.4, draw a milliequivalent – per litter bar graph and express alkalinity

and hardness in unit of mg/l as CaCO3?

Section IV Water Pollution

element Conc. [mg/l] Equ. weight M eq/l

Major cations Ca+2 35.8 20 1.79 Mg+2 9.9 12.2 0.81 Na+ 4.6 23 0.20 K+ 3.9 39.1 0.10

Total cation 2.9

Major anions HCO3-1 131 61 2.15

SO4-2 26.4 48 0.55

Cl- 7.1 35.5 0.20 Total anion 2.9

Note: usually total anions equal total cations; remember the arrangement

Thus we have:

1.79 meq/l of Ca(HCO3)2

0.36 meq/l of Mg(HCO3)2

0.45 meq/l of MgSO4

0.1 meq/l of KCl

0.1 meq/l of Na2 SO4

0.1 meq/l of NaCl

thus total carbonate hardness = 1.79 +0.36 = 2.15 meq/l

2.15 meq/l * 50 mg/meq = 107.5 mg/l as CaCO3

- total non-carbonate hardness

= 0.45 * 50 = 22.5 mg/l as CaCO3

Example 2:

element Conc. [mg/l] Equ. weight m eq/l

Major cations Ca+2 29 20 1.45 Mg+2 16.4 12.2 1.34 Na+ 23 23 1.00 K+ 17.5 39.1 0.45

Total cation 4.24

Major anions HCO3-1 171 61 2.81

SO4-2 36 48 0.75

Cl- 24 35.5 0.68 Total anion 4.24

Draw milieq./l bar

graph

1.79 Ca++ 0.81 Mg++ 0.2 Na+ 0.1 K+

2.15 HCO3- 0.55SO4

-2 0.2 Cl-

Section IV Water Pollution

1.45meq/l of Ca(HCO3)2

1.34meq/l of Mg(HCO3)2

0.02 meq/l of Na(HCO3)

0.75meq/l of Na2 SO4

0.23 meq/l of NaCl

0.45 meq/l of KCl

Total hardness = 1.45 + 1.34 = 2.79

2.79* 50 = 139.5mg/l as CaCO3

Hardness is determined (tested) in Lab by tetrametric with EDTA

Section IV Water Pollution

Residual Chlorine :

Heavy Metals: