Section 3

Environmental

Chemistry

1

Environmental Chemistry

• Definitions

• Chemical Reactions

➔ Stoichiometry

➔ Photolytic Reactions

• Enthalpy and Heat of Reaction

• Chemical Equilibria

➔ pH

➔ Solubility

• Carbonate Systems

2

Introduction

• Almost every pollution problem has some chemical basis

➔ From a chemical transformation or

reaction or from the chemical properties

of waste products

• Some problems involving chemical reactions:

➔ Greenhouse gases

➔ Ozone Hole

➔ Urban smog

➔ Acid deposition

➔ Water pollution

3

Definitions

• Atom

➔ 1 atomic mass unit (amu) = 1/12 the mass of

one carbon-12 atom (1.66053886 × 10–27 kg

• Proton (charge = +1, m = 1 amu)

• Neutron (charge = 0, m = 1 amu)

• Electron (charge = –1, mass ~0)

• Atomic weight

• Isotope

4

• Molecule

• Molecular weight

• Mole

➔ Number of moles = mass/molecular weight

➔ 1 g-mol = 6.022 × 1023 molecules

➔ 1 lb-mol = 2.7 × 1026 molecules

5

Chemical Reaction Stoichiometry

Balance those chemical reaction equations!

Example 1

A 1.67 × 10–3 M glucose solution (C6H12O6) is

completely biodegraded to carbon dioxide and

water. How much oxygen is required (mg/L)?

Balanced

equation:

C6H12O6 + O2 → CO2 + H2O

6

☛ From the amount of glucose we have in

moles, it takes six times that amount in

oxygen to decompose the glucose

If this is the amount of oxygen in one liter of

air, then the concentration in mg/ℓ is

This oxygen demand given by stoichiometry

is called the theoretical oxygen demand.

6 × 1.67 × 10−3 mol = 0.01 mol

0.01 molL

× 32 gmol

× 1000 mgg

= 320 mgL

7

Theoretical Oxygen Demand (TOD):

oxygen needed to fully oxidize a quantity

of organic material to carbon dioxide and

water

Biochemical Oxygen Demand (BOD):

oxygen required for oxidation of organic

wastes carried out by bacteria

BOD ~ TOD

8

Photodissociation

• Photodissociative reactions (also known as photolysis) perform key steps in atmospheric chemistry

• Photodissociation occurs when a molecule absorbs a photon of light

and decomposes

• “Photochemical” describes a reaction or set of reactions that derive at least

some of the energy needed for the reactions from sunlight

9

Energy in photons used in photochemistry

are related to the wavelength of the light:

E = energy of the photon (J)

h = Planck’s constant (6.6 × 10–34 J·s)

ν = frequency (cycles/s = Hz)

c = speed of light (3 × 108 m/s)

λ = wavelength (m)

E = hν =hcλ

10

Example

Energy per photon of light with wavelength

550 nm

E = hν =hcλ

E =hcλ

=6.6 × 10−34 Js( ) 3 × 108 m

s( )1

109m

nm⎛

⎝⎜

⎞

⎠⎟550 nm

= 3.6 × 10−19 J

11

Enthalpy

• We may need to determine the energy of a photon based on the enthalpy of the system

• Enthalpy (H) is determined by the internal energy (U) and the product of

the pressure (P) and volume (V)

For a process with

constant volume:

For a process with

constant pressure:

H = U + PV

ΔU = mcVΔT ΔH = mcPΔT

12

Heat of Reaction

(The zero superscript means the enthalpy is

measured at 1 atm and 298 K)

Endothermic reaction: ΔH0rxn is positive

Exothermic reaction: ΔH0rxn is negative

where ΔH0f is the heat of formation

ΔHrxn0 = ΔHf

0 Products∑ − ΔHf0 Reactants∑

13

Example

Using the enthalpy (ΔH0) of a reaction calculated

from the heats of formation (ΔH0f) of the reactants

and products, calculate the maximum wavelength

that can drive this photolytic reaction:

O3 + hν → O2 + OStandard enthalpies for the oxygen species

are given in Table 2.1 (text). Thus,

142.9 kJ/mol + energy of light = 0 kJ/mol + 247.5 kJ/mol

The energy of the photon would be 104.6 kJ/mol,

which corresponds a wavelength of 1.13 µm

14

Chemical Equilibria

• Examples where chemical equilibria are important:

➔ Acid/base reactions affecting pH

➔ Solubility products affecting precipitation

➔ Solubility of gases in water

Most liquid phase chemical reactions are

reversible, more or less:

aA + bB cC + dD

15

• When are chemical equilibria not important?

➔ Gas phase reactions—reversible,

but low concentrations of the reactants means they may not

interact enough to make the reversibility meaningful

16

If the forward and reverse reactions are

proceeding at the same rate, the system is in

equilibrium

— constant concentrations of species

For ,

Concentrations must be expressed in

moles/liter = M

Molecules may dissolve to form ions:

cations (+), anions (–)

aA + bB cC + dD

K =C[ ]c D[ ]d

A[ ]a B[ ]b≡ Equilibrium Constant

17

Dissociation of Ions

K is a dissociation or ionization constant:

A2B 2A+ + B2−

H2SO4 2H+ + SO42−

K =A+⎡⎣ ⎤⎦

2B 2−⎡⎣ ⎤⎦

A2B[ ]

18

Acid-Base

Since this K can range over several orders

of magnitude, it is more convenient to use

logarithmic notation

pX = −log X

X = 10−pX

pH = − log H+⎡⎣ ⎤⎦

H+⎡⎣ ⎤⎦ = 10

−pH mol

19

Pure water dissociates:

A special equilibrium relationship applies

to water:

In neutral water,

Then,

H2O H+ + OH−

Kw = H+⎡⎣ ⎤⎦ OH−⎡⎣ ⎤⎦ = 1 × 10−14 (at 25°C)

H+⎡⎣ ⎤⎦ = OH−⎡⎣ ⎤⎦

Kw = H+⎡⎣ ⎤⎦ OH−⎡⎣ ⎤⎦ = H +⎡⎣ ⎤⎦2

= 1 × 10−14

20

➯ pH = 7 for a

neutral solution

H +⎡⎣ ⎤⎦ = 10−7

Kw = H+⎡⎣ ⎤⎦ OH−⎡⎣ ⎤⎦ = H +⎡⎣ ⎤⎦2

= 1 × 10−14

21

• Significance of pH to environmental science:

➔ Indicates whether a solution is

acidic (pH < 7) or basic (pH > 7)

➔ Life is very sensitive to pH,

particularly aquatic life (affects biodiversity of this group)

➔ pH affects solubility of gases and

solids, which affects effects of acidity in aquatic systems

➔ Industrial wastes may have extreme pH levels, requiring neutralization

22

Example

Household ammonia has pH = 11.9 at 25°C.

What is [H+} and [OH–]?

pH + pOH = 14, so pOH = 14 – 11.9 = 2.1

H +⎡⎣ ⎤⎦ = 10−pH molL

= 10−11.9 molL

= 1.26 × 10−12 molL

OH−⎡⎣ ⎤⎦ = 10−pOH molL

= 10−2.1 molL

= 7.94 × 10−3 molL

23

Solubility Product

— describes dissolution of solids and

precipitation of components

Ksp = A[ ]a B[ ]b ≡ solubility product

solid aA + bB

K = A[ ]a B[ ]b solid[ ]−1

The solid portion is in a different phase than the

solutes, and its “concentration” is irrelevant to

the equilibrium. We combine [solid] with the

equilibrium constant:

24

• In actual situations, solid may or may not be present

➔ [ions] > Ksp, solid is present or will

subsequently form

➔ [ions] < Ksp, more solid can dissolve

in the solution

25

Example

Fluoride (F–) in water from CaF2 dissolution:

Given that there is solid present, what will

the equilibrium concentration of F be?

CaF2 Ca2+ + 2F−

Ksp = Ca2+⎡⎣ ⎤⎦ F −⎡⎣ ⎤⎦2= 3 × 10−11

26

For each mole of Ca2+, there will be 2 moles

of F–.

Let the molar concentration of Ca2+ be

represented by s

Then

Ksp = 3 × 10−11molL

= Ca2+⎡⎣ ⎤⎦ F−⎡⎣ ⎤⎦2= s 2s( )2 = 4s3

and s = Ca2+⎡⎣ ⎤⎦ = 2 × 10−4 molL

2s = F−⎡⎣ ⎤⎦ = 4 × 10−4 molL

s = 2 × 10−4 molL

27

Solubility of Gases in Water

Henry’s Law

Describes how much gas can dissolve into

water (at equilibrium)

[X]aq = aqueous phase concentration of X, in

KH = Henry’s Law coefficient, in

PX = partial pressure of X

X[ ]aq = KH,XPXmolL

molL iatm

28

Ex. Sea level pressure is 1 atm; pressure in

Boulder CO is 0.8 atm (at about 6000 ft altitude)

At sea level, partial pressure of oxygen (O2) is

about 0.21 atm (concentration is 21%). In

Boulder, it is

(21%)(0.8 atm) = 0.17 atm

1 ppm = 10–6 atm at 1 atm

partial pressure = concentration in volvol

⎛

⎝⎜

⎞

⎠⎟ atmospheric pressure( )

29

Oxygen in the human body

Approximate partial pressure of oxygen in the

blood is 0.13 atm.

If we assume that pressure is related to altitude

by:

where H = 7 km

What altitude is within the human body’s

comfort limits?

Find altitude where partial pressure of oxygen is not less

than the blood’s partial pressure of oxygen

P z( ) = P 0( )e−zH

30

What altitude is within the human body’s

comfort limits?

Find altitude where partial pressure of oxygen is not less

than the blood’s partial pressure of oxygen

0.62 = e−

z7 km

Partial pressure of O2 = 0.21( )P z( ) = 0.21( )P 0( )e−zH

0.13 atm = 0.21 1 atm( )e−

z7 km

z = − ln 0.62( ) × 7 = 3.4 km ≈ 11000 ft

31

Henry’s Law constants are temperature

dependent.

Ex., the solubilities of CO2 and O2 roughly

double between 25 and 0°C

T (°C) KH,CO2 (M atm–1) KH,O2 (M atm–1)

0 0.076 2.2 × 10–3

10 0.053 1.7 × 10–3

20 0.039 1.4 × 10–3

25 0.033 1.3 × 10–3

32

ExampleLeave some water on a table outside on a

cold day in Denver (10°C, 1525 m)

How much CO2 will dissolve in the water

(in M) if its concentration in the

atmosphere is 360 ppmv?

KH,CO2 (10°C) = 0.0532 M/atm

Patm = 0.825 atm in Denver

[CO2] = 360 ppmv = 0.036%

PCO2=

360106 0.825 atm( )

= 2.97 × 10−4 atm

X[ ]aq = KH,XPX

CO2[ ]aq = 0.0532 M atm−1( ) 2.97 × 10−4 atm( ) = 1.58 × 10−5 M

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Carbonate Systems

• Carbonates are the largest reservoir of carbon on Earth

• Controls pH in natural systems

• Four important species:

➔ Carbonic acid H2CO3

➔ Bicarbonate ion HCO3–

➔ Carbonate ion CO32–

➔ Calcium carbonate CaCO3

34

Carbon dioxide dissolves in water to form

CO2(aq), also stated as (CO2·H2O)

This turns out to be H2CO3

This dissociates in water:

The carbonate ion serves as a carbon sink—

when it forms, bicarbonate is removed and

more carbon dioxide is allowed to dissolve

CO2•H2O(aq) H(aq)+ + HCO3 (aq)

−

HCO3− H+ + CO3

2−

35

An effective Henry’s Law constant can take

into account the effect of dissociation and

chemical loss

We also need to consider competing effects

found in natural systems, such as the

presence of limestone:

The equilibria are:

CaCO3(s ) Ca2+ + CO32−

CO2 + H2O K1

H+ + HCO3− K2

H+ + CO32−

CaCO3(s ) Ksp

Ca2+ +

H2O Kw

H+ + OH−

36

Equilibrium constants are:

K1 ⇒ CO2 +H2O H+ +HCO3−

K 2 ⇒ HCO3− H+ +CO3

2−

Ksp ⇒ CaCO3(s ) Ca2+ +CO32−

H+⎡⎣ ⎤⎦ HCO3−⎡⎣ ⎤⎦

CO2(aq )⎡⎣ ⎤⎦=K1 = 4.47 × 107 M

= 10−6.35 M

pK1 = − logK1 = 6.35

H+⎡⎣ ⎤⎦ CO32−⎡⎣ ⎤⎦

HCO−3⎡⎣ ⎤⎦

=K 2 = 4.68 × 10−11 M

= 10−10.3 M

pK 2 = − logK 2 = 10.3

Ca2+⎡⎣ ⎤⎦ CO32−⎡⎣ ⎤⎦ =Ksp = 4.57 × 10−9 M

37

How much carbonate vs. bicarbonate?

—look for [CO32–] / [HCO3–]

Divide H+⎡⎣ ⎤⎦ CO3

2−⎡⎣ ⎤⎦HCO−

3⎡⎣ ⎤⎦ by H+⎡⎣ ⎤⎦

H+⎡⎣ ⎤⎦ CO32−⎡⎣ ⎤⎦

HCO−3

⎡⎣ ⎤⎦

1H+⎡⎣ ⎤⎦

=10−10.3

H+⎡⎣ ⎤⎦

CO32−⎡⎣ ⎤⎦

HCO−3

⎡⎣ ⎤⎦=10−10.3

10−pH= 10pH−10.3

38

Role of Organisms in Carbon Cycle

• Carbon enters oceans as carbon dioxide dissolution, then may be converted to carbonate or bicarbonate

• Then, certain organisms bind calcium to bicarbonate

➔ Calcium carbonate CaCO3 used to make

shells, coral, exoskeletons, etc.

➔ Parts of dead organisms collect on sea

floor and eventually become sedimentary

rock—largest carbon sink on Earth

39

40

ApplicationNatural acidity of rainwater

from dissolved carbon dioxide

Determine the pH of rainwater that has an

equilbrium amount of atmospheric carbon

dioxide dissolved in it. Actual pH may be

lower in polluted areas due to sulfuric, nitric,

or organic acids.

Get [H+] from an electroneutrality expression

(charge conserved)—insert each species from

dissociation expressions, equilibrium relation

for water, and Henry’s Law

41

CO2•H2O(aq) H(aq)+ + HCO3 (aq)

−

HCO3− H+ + CO3

2−

H2O H+ + OH−

42

H+⎡⎣ ⎤⎦ = HCO3−⎡⎣ ⎤⎦ + 2 CO3

2−⎡⎣ ⎤⎦ + OH−⎡⎣ ⎤⎦

CO2•H2O[ ] = KHPCO2

HCO3−⎡⎣ ⎤⎦ =

K1 CO2(aq )⎡⎣ ⎤⎦H+⎡⎣ ⎤⎦

=K1KHPCO2

H+⎡⎣ ⎤⎦

CO32−⎡⎣ ⎤⎦ =

K 2 HCO3−⎡⎣ ⎤⎦

H+⎡⎣ ⎤⎦

OH−⎡⎣ ⎤⎦ =Kw

H+⎡⎣ ⎤⎦

43

Substitute for the concentrations of

bicarbonate, carbonate, and hydroxyl

ions in the electroneutrality expression

so everything ends up in terms of [H+]:

Solving this for [H+] is not easy. Retry with a

simplified version by applying:

H+⎡⎣ ⎤⎦ =K1KHPCO2

H+⎡⎣ ⎤⎦+ 2

K1K 2KHPCO2H+⎡⎣ ⎤⎦

2 +Kw

H+⎡⎣ ⎤⎦

H+⎡⎣ ⎤⎦3− H+⎡⎣ ⎤⎦ K1KHPCO2 + Kw( ) − 2K1K 2KHPCO2 = 0

CO3

2−⎡⎣ ⎤⎦ HCO3−⎡⎣ ⎤⎦

44

H+⎡⎣ ⎤⎦ = HCO3−⎡⎣ ⎤⎦ + 2 CO3

2−⎡⎣ ⎤⎦ + OH−⎡⎣ ⎤⎦

H+⎡⎣ ⎤⎦ =K1 CO2(aq )⎡⎣ ⎤⎦

H+⎡⎣ ⎤⎦+

10−14 M2

H+⎡⎣ ⎤⎦

H+⎡⎣ ⎤⎦2= K1 CO2(aq )

⎡⎣ ⎤⎦ + 10−14 M2

CO2(aq )⎡⎣ ⎤⎦ = KHPCO2

≅ 1.3 × 10−5 M

· · · ➔ pH = 5.62

45