Lectures 1 & 2
AOSC 434
AIR POLLUTION
RUSSELL R. DICKERSON
Topics requested
Climate/Pollution (5)
Remote Sensing (3)
Epi/Health, Observations, Local AQ, Policy (2 each)
Fires, Developing world, Oceans (1 each)
I.
OUTLINE
BASIC CHEMISTRY
See:
Seinfeld & Pandis: Appendix A
Wark and Warner: Chapter 1
Finlayson-Pitts & Pitts: Chapter 1
ELEMENTS (atoms)
Atomic number = No. Protons = No. Electrons
Atomic weight ≃ No. Protons + No. Neutrons
Mole (gram) = 6.023 x 10²³ molecules/g mole
ATOMS – MOLECULES – IONS – RADICALS
Radicals – no charge, but unpaired electron. They exist in both gaseous
and aqueous phases.
e.g. OH, HO₂, NO₃
Very reactive
Hydroxyl not hydroxide; NO₃ not nitrate ion
STOICHIOMETRY
(mass balance)
C + O₂ → CO₂
NO + NO + O₂ → 2 NO₂
Remember the concept of valance?
AQUEOUS CHEMISTRY
(cloud water, aerosols → acid percep.)
Basic Units: mole/liter = M (molar)
EXAMPLE: KNO₃
Mwt. = 39 + 14 + 3(16) = 101 g/mole
Put 101 g in a bottle and fill to 1.00 liter with water
Makes 1.00 M KNO₃
Also [K+] = 1.00 M;
[NO₃⁻] = 1.00 M
Water self-ionizes
H₂O = H⁺ + OH⁻
[H ⁺][OH⁻] = 1.0 X 10⁻¹⁴
pH = -log[H ⁺] = -log[H₃O⁺]
pH = 7.0 (pure H₂O)
pOH = -log[OH⁻] = 14 –pH
(Look at your freshman chemistry text)
FURTHER EXAMPLES: sodium hydoxide
NaOH → Na⁺ + OH⁻
Mwt. = 23 + 16 + 1 = 40 g/mole
Add 40 g NaOH (1.0 mole) to a flask, fill to 1.00 liter with water
[Na⁺] = [OH⁻] = 1.00
[H⁺] = 1.0 x 10⁻¹⁴ [OH⁻]
pH = 14; pOH = 0.0
Hydrogen chloride
HCl → H⁺ + Cl⁻
3.65 g of HCl gas bubbled through 1.00 liter of water →
[H⁺] = 0.1 pH = 1.0
(Stomach acid pH = 1.0, this will dissolved flesh, bones, and iron)
Titrate 1.0 M HCl with 1.0 M NaOH produce warm salt water:
Na⁺ + OH⁻ + H⁺ + Cl⁻ → Na⁺ + Cl⁻ + H₂O + HEAT
“Acid” rain has a pH = 3-5, due mostly to sulfuric and nitric acids.
Rain falling in “clean” air has a pH = 5.6. This acidity is due to CO₂,
but we need to consider weak acids and Henry’s Law to calculate show
why this is true.
WEAK ACIDS
EXAMPLE
Nitrous acid
+
HNO 2 = H + NO
2
K a = 5 ´10 -4
[H + ][NO-2 ]
Ka =
HONO
If [HONO] = 1.0 M you can assume that most of the HONO remains
undissociated, that is unionized. You can also assume for reasonably
high acid concentrations that the H⁺ produced by HONO is >> [H⁺]
from the autoionization of water. To a good approximation:
[H  ]  [NO 2 ]  [HONO] Ka
  2.2  10  2
pH = 1.65
PROBLEM LEFT FOR THE STUDENT
Acetic acid is CH₃COOH
pKa = 4.75
What is the pH of a 1.0 M soln? (2.38; 0.4%)
How about a 0.0001 M soln? (4.5; 34%)
What fraction of the acids are ionized?
LECTURE 2
AOSC 434
AIR POLLUTION
RUSSELL R. DICKERSON
TODAY’S OUTLINE
Ia. Chemistry (Concentration Units):
1. Gas-phase
2. Aqueous-phase
Ib. Atmospheric Physics
1. Pressure
2. Atmospheric structure and circulation
A. pressure and temp. profiles
B. thermo diagram and stability
C. circulation (winds)
Ia. 1. GAS-PHASE
Atoms
He Ar
monatomic
N₂ O₂
Molecules
CO₂ O₃
Radicals
H₂CO CCl₂F₂ OH HO₂
diatomic
triatomic
polyatomic
UNITS OF CONCENTRATION
Mole Fraction – for ideal gas this is the same as volume fraction. Also
called mixing ratio, or volume mixing ratio.
fraction
[O₂] ~ 1/5
percent
[Ar] = 1%
[H₂O] = up to 4%
parts per million (10⁶)
[CH₄] = 1.9 ppm
parts per billion (10⁹)
[O₃] = 30 ppb
parts per trillion (10¹²)
[CCl₂F₂] = 100 ppt
ATMOSPHERIC CO2 INCREASE OVER PAST 1000 YEARS
Jacob: Intergovernmental Panel on Climate Change (IPCC) document, 2001
Concentration units: parts per million (ppm)
number of CO2 molecules per 106 molecules of air
CO2 CONCENTRATION IS MEASURED HERE AS MIXING RATIO
For an ideal gas these concentrations are constant regardless
temperature and pressure.
Ideal Gas Law: PV = nRT
For example if T₂ = 2T₁ and if dP = 0 then V₂ = 2V₁.
Meteorologists favor the ideal gas law for a kg of air:
pα = R’T
Where R’ has units of J kg-1K-1 and α is the specific volume
(volume occupied by 1 kg of air; Mwt ~29 g/mole).
If air in New York is brought to Denver (P = 83% atm) there will be
no change in the concentration of pollutants as long as the
concentration is expressed as a volume (molar) mixing ratio.
MASS PER UNIT VOLUME
Best for particles (solid or liquid)
Weigh a filter – suck 1.00 m⁻³ air through it – reweigh it
Change in weight is conc “dust” in mass/unit volume or μgm⁻³
EXAMPLE
If you find 10 μg/m³ “dust” of which 2 μg/m³ are nitrate (NO₃‾), how
much gas phase HNO₃, expressed as a mixing ratio, was there in the air
assuming that all the nitrate was in the form of nitric acid? We must
convert 2.0 μg/m³ HNO₃ to ppb:
2.0 10 6 /63
1000/22.4
gm 3 / g / mole
L / m 3 / L / mole
Remember, one mole of an ideal gas is 22.4 liters at STP.
STP = 0o C & 1.0 atm.
2.0 μg/m³ HNO₃ = 7.1 x 10⁻10 = 0.71 ppb
In general: 1.0 μg/m³ HNO₃ = 0.35 ppb
Notice that the concentration in μg/m³ changes with P and T of the air.
Mixing ratios area also good for writing reactions:
NO + O₃ → NO₂ + O₂
1 ppm + 1 ppm = 1 ppm + 1 ppm
Note: the [O₂] in air is not 1 ppm, rather it is 0.2 x 10⁶ + 1.0 ppm.
Above is an example of an irreversible reaction. There are also reversible
reactions.
EXAMPLE
Equilibrium of ammonium nitrate
NH₃ + HNO₃ ↔ NH4 NO3(S)
[NH 4 NO3 ]
Keq 
[NH3 ][HNO 3 ]
Ammonium nitrate is a solid, and thus has a concentration defined as
unity.
Number density nX [molecules
-3
cm ]
Proper measure for
• calculation of reaction rates
• optical properties of atmosphere
# molecules of X
nX 
unit volume of air

Column concentration =  n( z )dz
0
Proper measure for absorption of radiation by
atmosphere
Column concentrations are measured in molecules cm-2 , atm*cm, and
Dobson Units, DU.
1 atm*cm = 1000 DU = 2.69 x 1019 cm-2.
STRATOSPHERIC OZONE LAYER (Jacob’s book)
1 “Dobson Unit (DU)” = 0.01 mm ozone at STP = 2.69x1016 molecules cm-2
THICKNESS OF OZONE LAYER IS MEASURED AS A COLUMN CONCENTRATION
AQUEOUS-PHASE CHEMISTRY
HENRYS LAW
The mass of a gas that dissolves in a given amount of liquid as a given
temperature is directly proportional to the partial pressure of the gas above
the liquid. This law does not apply to gases that react with the liquid or
ionized in the liquid. See Finlayson p.151 or Chameides, J. Geophys. Res.,
4739, 1984. Check out also
http://www.henrys-law.org/#1
GAS
OXYGEN
O₂
OZONE
O₃
NITROGEN DIOXIDE
NO₂
CARBON DIOXIDE
CO₂
SULFUR DIOXIDE
SO₂
NITRIC ACID (effective)
HNO₃
HYDROGEN PEROXIDE
H₂O₂
HYDROPEROXY RADICAL
HO₂
ALKYL NITRATES
(RONO₂)
HENRY’S LAW CONSTANT
(M / atm at 298 K)
1.3 x 10⁻²
9.4 x 10⁻³
1.0 x 10⁻²
3.1 x 10⁻²
1.3
2.1 x 10⁺⁵
9.7 x 10⁺⁴
9.0 x 10³
1.3
HENRY’S LAW EXAMPLE
What would be the pH of pure rain water in Washington, D.C. today?
Assume that the atmosphere contains only N₂, O₂, and CO₂ and that rain
in equilibrium with CO₂.
Remember:
H₂O = H⁺ + OH⁻
[H⁺][OH⁻] = 1 x 10⁻¹⁴
pH = -log[H⁺]
In pure H₂O pH = 7.0
We can measure:
[CO₂] = ca. 370 ppm
Today’s barometric pressure is 993 hPa = 993/1013 atm = 0.98 atm. Thus
the partial pressure of CO₂ is
PCO2 = 370 *10-6 (0.98) = 3.63*10 4 atm
[CO2 ]aq  H  P(CO2 )  3.4  102  3.62  104
 1.23  105 M
In water CO₂ reacts slightly, but [H₂CO₃] remains constant as long as the
partial pressure of CO₂ remains constant.
CO2  H 2 O  H 2CO3
 H 2CO3  H   HCO 3
[H  ][HCO 3 ]
 Keq  4.3  107
[H 2CO3 ]
We know that:
and
THUS
5
[H 2CO3 ]  1.23  10 M
[H  ]  [HCO 3 ]
[H  ]  Ka * H2CO3
H+ = 2.3x10-6 → pH = -log(2.3x10-6) = 5.6
EXAMPLE 2
If fog water contains enough nitric acid (HNO₃) to have a pH of 4.7, can any
appreciable amount nitric acid vapor return to the atmosphere? Another way
to ask this question is to ask what partial pressure of HNO₃ is in equilibrium
with typical “acid rain” i.e. water at pH 4.7? We will have to assume that
HNO₃ is 50% ionized.
pH  log[H  ]
[H  ]  10  4.7  2 10 5
PHNO3  [HNO 3 ]aq /H
  2 10 5 /2.1 105
  9.0  10 11 atm
This is equivalent to 90 ppt, a small amount for a polluted environment, but
the actual [HNO₃] would be even lower because nitric acid ionized in solution.
In other words, once nitric acid is in solution, it wont come back out again
unless the droplet evaporates; conversely any vapor-phase nitric acid will be
quickly absorbed into the aqueous-phase in the presence of cloud or fog water.
Which pollutants can be rained out?
We want to calculate the ratio of the aqueous phase to the gas phase concentration
of a pollutant in a cloud. The units can be anything , but they must be the same.
We will assume that the gas and aqueous phases are in equilibrium. We need
the following:
Henry’s Law Coefficient: H (M/atm)
Cloud liquid water content: LWC (gm⁻³)
Total pressure: PT (atm)
Ambient temperature: T(K)
LET:
X aq be the concentration of X in the aqueous phase in moles/m³
X gas be the concentration of X in the gas phase in moles/m³
[X] aq  HPX
Where [X] aq is the aqueous concentration in M, and Px is the partial pressure
expressed in atm. We can find the partial pressure from the mixing ratio and
total pressure.
PX  [X] gas PT
For the aqueous-phase concentration:
Xaq  [X]aq LWC 103
units:
moles/m³ = moles/L(water) x g(water)/m³(air) x L/g
Xaq  H[X] gas PT LWC 103
For the gaseous content:
X gas 
units:
moles/m³ =
[X] gas

T 1
 22.4 10 3

273 PT 

L(X)/L(air )
L/mole  m 3 /L


X aq
X gas

1
3 T

/[X] gas
 H[X] gas PT LWC 10  22.4 10
273 PT 

3
X aq
X gas
 H  LWC 
T
 22.4  10 6
273
Notice that the ratio is independent of pressure and concentration. For a
species with a Henry’s law coefficient of 400, only about 1% will go into a
cloud with a LWC of 1 g/m³. This points out the need to consider aqueous
reactions.
What is the possible pH of water in a high cloud (alt. ≃ 5km) that absorbed
sulfur while in equilibrium with 100 ppb of SO₂?
SO 2  H 2 O  SO 2  H 2 O
[SO 2 ]  100ppb
PSO 2  [SO 2 ]PTotal  [SO 2 ]P5km
In the next lecture we will show how to derive the pressure as a function of
height. At 5km the ambient pressure is 0.54 atm.
PSO 2  100 10 9 0.54  5.4 10 8 atm
[SO 2 ]aq  HPSO 2
  7 10 8 M
This SO₂ will not stay as SO₂•H₂O, but participate in a aqueous phase
reaction, that is it will dissociate.
SO2  H 2O  H   HOSO2
The concentration of SO₂•H₂O, however, remains constant because more SO₂
is entrained as SO₂•H₂O dissociates. The extent of dissociation depends
on [H⁺] and thus pH, but the concentration of SO₂•H₂O will stay constant
as long as the gaseous SO₂ concentration stays constant. What’s the pH for
our mixture?


[H ][HOSO 2 ]
Ka 
[H 2O  SO 2 ]
If most of the [H⁺] comes from SO₂•H₂O dissociation, then
[H  ]  [HOSO 2 ]
[H  ]  K a [H 2 O  SO 2 ]  3  10 5
Note that there about 400 times as much S in the form of HOSO₂⁻ as in the
form H₂O•SO₂. HOSO₂⁻ is a very weak acid, ant the reaction stops here.
The pH of cloudwater in contact with 100 ppb of SO₂ will be 4.5
Because SO₂ participates in aqueous-phase reactions, Eq. (I) above will
give the correct [H₂O•SO₂], but will underestimate the total sulfur in
solution. Taken together all the forms of S in this oxidation state are called
sulfur four, or S(IV).
If all the S(IV) in the cloud water turns to S(VI) (sulfate) then the hydrogen
ion concentration will approximately double because both protons come off
H₂O•SO₄, in other words HSO₄⁻ is a strong acid.
This is fairly acidic, but we started with a very high concentration of SO₂,
one that is characteristic of urban air. In more rural areas of the eastern US
an SO₂ mixing ratio of a 1-5 ppb is more common. As SO₂•H₂O is
oxidized to H₂O•SO₄, more SO₂ is drawn into the cloud water, and the
acidity continue to rise. Hydrogen peroxide is the most common oxidant
for forming sulfuric acid in solution; we will discuss H₂O₂ later.
Summary
Lectures 1&2
Basic chemistry of aqueous phase – see freshman chem text
Ideal gas law
Unit conversion – you must know intuitively.
Column contents (molecules cm-2, atm cm, matm cm =DU)
Henry’s law [X]aq = HPX