Chemistry of Natural Waters

ENGI 9628 – Environmental Laboratory

Faculty of Engineering & Applied Science

Friday, July 3, 2009

Lecture 5: The Chemistry of Natural

Waters

ENGI 9628 – Environmental Engineering Laboratory 2

Hydrologic Cycle


Importance of Water• All life forms on earth depend on water. Each human being

consumes several L water per day to sustain life.

• 97% is seawater: unsuitable for drinking and most agriculture.

• ¾ of the fresh water is trapped in glaciers and icecaps.

• Lakes and rivers are one of the main sources of drinking water.

• Although only 10% of the world’s population in 2000 lived under conditions of water stress and scarcity, expect to rise to 38% by 2025.

• Important to understand the types of chemical activity that prevail in natural waters.


Major Water Quality Issues

• Acidification – acid rain, acid mine drainage• Eutrophication – nutrient overload• Metals e.g. Pb, Cd, Mg, Al, etc.• Organic contamination• – Organic contaminants e.g. DDT, PCBs• – Organic matter e.g. sewage, manure• Pathogens• Salinization• Sediment load• Diversions, dams

ENGI 9628 – Environmental Engineering Laboratory

Concentration

• Solutions ‐‐ Reporting Concentrations and Solubility

• Concentration can be expressed as:– Mass/Volume units– temperature dependent!– Mass/Mass units


Concentration (mass/volume)

• Molarity (M, mol/L)

– Number of moles of solute per litre of solution

• Normality (N, eq/L)

– Equivalents of solute per litre of solution

– Also expressed as meq/mL


Normality

• The weight of one equivalent depends on the type of reaction considered

eq = M/nM = Molar massn = number of protons donated= no. of moles of H+ or OH‐ produced per mole of acid/base= total change in oxidation number of a compound= charge on an ion


Concentration (Mass/Mass)

• Modality (m, mol/kg)• – number of moles of solute per kilogram of solvent• Mole Fraction (XA)• – Number of moles of a component divided by the totalnumber of moles of all components in the solution• Weight Percent (%)• – Mass of solute per 100 g of solution• Parts per million (ppm)• – Grams of solute per million grams of solution (or mg/L)• – Can also use ppb or ppt


Equilibrium Constant

• K = Equilibrium Constant

aA bB cC dD+ → +

[ ] [ ][ ] [ ]

c d

a b

C DKA B

=


Solubility

• E.g. gypsum2 2

4 4( )CaSO S Ca SO+ −→ +2 2

4[ ][ ]Ksp Ca SO+ −=2 2

4[ ] [ ]Ca SO+ −=


Ion Activity Product

• • IAP = Ksp Saturated solution

• • IAP < Ksp Unsaturated solution

• • IAP > Ksp Supersaturated solution


Water Chemistry

• Two common reaction categories: Acid base reactions and oxidation-reduction (redox) reactions.

• Acid base and solubility phenomena predominantly control the concentrations of dissolved inorganic ions, such as carbonate.

• Redox reactions dominate the organic matters.


Oxidation Reduction: Dissolved Oxygen

• The most important oxidizing agent in natural waters is dissolved oxygen, O2.

• Each of the oxygen atoms in O2 is reduced from zerooxidation state to -2 state.

→++ −+ 4e4HO2

→++ −4eO2HO 22


Dissolved Oxygen, Cont’d

• The concentration of DO in water is small and therefore precarious from ecological point of view.

• The dissolution process

• The equilibrium constant is the Henry’s Law constant KH

)(dissolvedO(gas)O 22 ⇔

2O

2H Pressure Partial

)(dissolvedOK =


DO: Problem 1

• Confirm by calculation the value of 8.7mg/L for the solubility of Oxygen in water at 25oC. (given that KH=1.3x10-3 mol/L atm at 25 oC, PP O2= 0.21atm)


DO: Thermal Pollution

• River and lake water that has been artificially warmed can be considered to have undergone Thermal Pollution. Why?

• Gas solubility decreases with increasing temperature.

• Warm water contains less oxygen than cold water. To sustain life, most fish species require at least 5 ppm of DO.

• Consequently, their survival in warm water can be problematic.


Oxygen Demand

• The most common substance oxidized by DO in water is Organic Matter of biological origin (dead plant matter and animal wastes).

• The reaction process

• Oxidation state of carbon increases from 0 to +4, it is oxidized, the state of O2 decreases from 0 to -2.


Oxygen Demand: Problem 2

• Show that 1L water saturated with oxygen at 25oC is capable of completely oxidizing 8.2 mg of CH2O.


Oxygen Demand

• Water that is aerated by flowing in shallow streams and rivers is constantly replenished with oxygen.

• Stagnant water or that near the bottom of a deep lake is usually almost completely depleted of oxygen because of the reaction with OM.

• Biological Oxygen Demand (BOD): the capacity of the organic and biological matter in a sample of natural water to consume oxygen, a process catalyzed by bacteria present.

Biochemical Oxygen Demand (BOD)Concept

an indirect measure of organic‐matter concentration in water bodies

most widely used for measuring organic pollution in wastewater & surface water

High BOD concentration = high organic‐matter concentration = poor water quality

BOD Kinetics

If we introduce some organic matter into water

==> Initial amount of organic matter ==> initial BOD (BODinitial)

==> Bacteria keep decomposing organic matter

==> Amount of organic matter keeps decreasing

==> remaining BOD keeps decreasing (BODremaining)

BOD = amount of oxygen required (by bacteria) to decompose organic waste to carbon dioxide and water.

Define: BODutilized = BODinitial − BODremaining

==> BODutilized(at time t) = BODinitial − BODinitial10 –Kt

==> BODutilized(at time t) = BODinitial(1 − 10 –Kt)

BODutilized at day 5 = BODinitial(1 − 10 –5K)

BODutilized at day 10 = BODinitial(1 − 10 –10K)

Right figure shows the usual curve for BOD and is actually BODutilized. Also shown is the 5 day BOD.

BODremaining(at time t) = BODinitial10 –Kt

where K = reaction constant (day ‐1); and t = time (day).

Example: The 5‐day BOD for some wastewater

has been found to be 200 mg/L. With reaction

constant K = 0.1 day ‐1, find

10‐day BOD and BODinitial?

Measurement of BOD

Decomposition of organic matter is a slow process

Standard way to measure BODutilized during the first 5 days (at 20 °C)

==> named "5‐day BOD" or BOD5

==> employed by the standards of many countries, e.g. Canada, USA and EU.

Implication of BOD measurement:

‐ BOD5 for domestic sewage = several hundreds mg/L

‐ BOD5 for industrial sewage = several thousands mg/L

Therefore, when the sewage is discharged to water

==> quick depletion of oxygen => would kill fish

20 days ==> decompose 95 to 99% of organic matter

5 days ==> decompose 60 to 70% of organic matter

IMPACT FACTORS ‐ (1) Temperature

Temperature ==> affect biochemical reaction rates (and thus reaction constant K)

==> different BOD values

K value at temperature different from 20 oC:

where θ = temperature coefficient (between 1.056 to 1.135).

KT = K20θ(T‐20)

Example: The initial BOD concentration of a sewage is 300 mg/L. Reaction constant K is 0.1 day‐1 at 20 oC. Calculate BOD5 for this sewage at 30 oC (θ = 1.056).

IMPACT FACTORS ‐ (2) Effect of Nitrification

Organic nitrogen in sewage

Converted to ammonia during decomposition

Oxidation of ammonia to nitrite and nitrate requires oxygen

Second‐stage BOD (nitrogenous oxygen demand)

At 20 oC, reproduction rate of nitrifying bacteria ==> very slow (need 6 to 10 days to exert measurable oxygen).

Influence of Nitrification

APPLICATION of BOD:measure waste loading to treatment plants

evaluate efficiency of treatment systemsdetermine relative O2 requirements of treated effluents & polluted waters

==> useful in

‐ sewage treatment & water quality management

‐ sizing of waste treatment facilities ‐measuring efficiency of treatment processes

(5) Oxygen Sag Curve

when sewage is discharged into a receiving water, two simultaneous actions follow:

decomposition of organic matter by bacteria => consume O2

reaeration from atmosphere ==> bring O2 to water

==> their combination produces an oxygen sag curve

initial stage

at the point where [DO] = minimum

beyond minimum point

This sequence is called "natural self‐purification of water"

==> DO curve drops (i.e. rate of O2 consumption by bacteria > rate of reaeration with atmosphere)

==> rate of consumption = rate of reaeration

==> rate of consumption < rate of reaeration (DO level eventually returns to normal)


Chemical Oxygen Demand (COD)• A fast determination of oxygen demand can be made by

evaluating the Chemical Oxygen Demand (COD) of a water sample.

• The basis for the COD test is that nearly all organic compounds can be fully oxidized to carbon dioxide with a strong oxidizing agent under acidic conditions.

• For many years, potassium permanganate (KMnO4) has been used.

• Other oxidizing agents such as ceric sulfate, potassium iodate, and potassium dichromate have been used to determine COD.

• Of these, potassium dichromate (K2Cr2O7) has been shown to be the most effective: it is relatively cheap, easy to purify, and is able to nearly completely oxidize almost all organic compounds.


• Dichromate ion, can be dissolved as one of its salts, such as Na2Cr2O7 or Na2Cr2O7 in sulfuric acid, this mixture is a powerful oxidizing agent.

• It is this mixture, rather than O2, that is used to ascertain COD values.

• Moles of O2 is 6/4=1.5 times that of the moles of dichromate, since the latter accepts 6 electrons per ion whereas O2 accepts only 4

Chemical Oxygen Demand (COD)


COD: Problem 3• A 25-mL sample of river water was titrated with

0.001M Na2Cr2O7 and require 8.3 mL to reach the endpoint. What is the chemical oxygen demand (COD) in mg of O2 per liter of the sample?(0.001 M = 0.001 mole/Liter)


• Dichromate is such a strong oxidant that it oxidizes substances that are very slow to consume oxygen in natural waters.

• In another word, Dichromate oxidizes substances that would not be oxidized by O2 in determination of BOD.

• Therefore, the COD is slightly greater than BOD.• In many water pollution cases, the O2 required by OM is

higher than DO. If the water is not aerated, the O2 will be soon depleted, and fish will die.

• Wastewater need to be treated to reduce BOD.

Chemical Oxygen Demand (COD)


Acid‐base Chemistry: Carbonate System

• Natural waters, even when “pure” contains significant quantities of dissolved carbon dioxide CO2, and the anions it produces (CO3

2-, HCO3-), as

well as Calcium and Magnesium cations (Ca2+,Mg2+).• The pH of the natural waters is rarely exactly 7.0, the

value expected for pure water.• Equilibrium (carbonate, bicarbonate)


Calcareous Waters• Natural waters that are exposed to limestone are

called calcareous waters. • The limestone rocks are largely made up of

calcium carbonate, CaCO3. • Almost insoluble, a small amount dissolves when

water passes over it.

• Dissolved carbonate acts as a base, producing bicarbonate ion and hydroxide ion, OH-

−+ +⇔ 23

23 COCa)(CaCO s

−−− +⇔+ OHHCOOHCO 3223


Reaction Among Three Phases

Air

Water

Rock, Soil or Sediments

CaCO3 (solid)

Ca2+

CO32- + H2O OH- + HCO3

-

+

H2CO3 H+ + HCO3-

CO2 (gas)

H2O


Water in Equilibrium with Solid CaCO3

• Consider a hypothetical case: water is in equilibrium with excess solid CaCO3, all other reaction are of negligible importance.

• Equilibrium constant in this case is Ksp, the solubility product, equals to the product of the concentration of ions.

• Ksp=4.6x10-9 (25oC), S=[Ca2+]=[CO32-]

S2=4.6x10-9, S=6.8x10-5 M, or 6.8x10-5 mol/L

]][CO[CaK 23

2sp

−+=

Chemistry of Natural Waters

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