Section 3Environmental

Chemistry

1

Environmental Chemistry• Definitions• Chemical Reactions➔ Stoichiometry

➔ Photolytic Reactions

• Enthalpy and Heat of Reaction• Chemical Equilibria➔ pH

➔ Solubility

• Carbonate Systems2

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 1A 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 ms( )

1109

mnm

⎛

⎝⎜

⎞

⎠⎟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 = ΔHf0 Products∑ − ΔHf0 Reactants∑

13

ExampleUsing 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 × 10−4molL

2s = F−⎡⎣ ⎤⎦ = 4 × 10−4molL

s = 2 × 10−4 molL

27

Solubility of Gases in WaterHenry’s LawDescribes 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/atmPatm = 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

33

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+ + CO32−

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 ) Ca

2+ + 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+ +CO32−

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+ + CO32−

H2O H+ + OH−

42

H+⎡⎣ ⎤⎦ = HCO3−⎡⎣ ⎤⎦ + 2 CO32−⎡⎣ ⎤⎦ + OH−⎡⎣ ⎤⎦

CO2•H2O[ ] = KHPCO2

HCO3−⎡⎣ ⎤⎦ =K1 CO2(aq )⎡⎣ ⎤⎦

H+⎡⎣ ⎤⎦=K1KHPCO2

H+⎡⎣ ⎤⎦

CO32−⎡⎣ ⎤⎦ =K 2 HCO3−⎡⎣ ⎤⎦

H+⎡⎣ ⎤⎦

OH−⎡⎣ ⎤⎦ =KwH+⎡⎣ ⎤⎦

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 +KwH+⎡⎣ ⎤⎦

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

CO32−⎡⎣ ⎤⎦ HCO3

−⎡⎣ ⎤⎦

44

H+⎡⎣ ⎤⎦ = HCO3−⎡⎣ ⎤⎦ + 2 CO32−⎡⎣ ⎤⎦ + 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