1. AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON
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)

2. OUTLINE
BASIC CHEMISTRY
See:
Seinfeld & Pandis: Appendix A
Wark and Warner: Chapter 1
Finlayson-Pitts & Pitts: Chapter 1

3. ATOMS – MOLECULES – IONS – RADICALS
ELEMENTS (atoms)
Atomic number = No. Protons = No. Electrons
Atomic weight ≃ No. Protons + No. Neutrons
Mole (gram) = 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

4. 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. = (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

5. Water self-ionizes H₂O = H⁺ + OH⁻ [H ⁺][OH⁻] = 1
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. = = 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

6. Hydrogen chloride HCl → H⁺ + Cl⁻ 3. 65 g of HCl gas bubbled through 1
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.

7. WEAK ACIDS EXAMPLE Nitrous acid If [HONO] = 1
WEAK ACIDS EXAMPLE Nitrous acid 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: pH = 1.65

8. PROBLEM LEFT FOR THE STUDENT Acetic acid is CH₃COOH pKa = 4
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 M soln? (4.5; 34%) What fraction of the acids are ionized?

9. AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON
LECTURE 2
AOSC 434
AIR POLLUTION
RUSSELL R. DICKERSON

10. 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)

11. UNITS OF CONCENTRATION
Ia GAS-PHASE
Atoms Molecules Radicals
He Ar N₂ O₂ CO₂ O₃ H₂CO CCl₂F₂ OH HO₂
monatomic 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

12. 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

13. 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

14. 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: 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.

15. 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 ppm. Above is an example of an irreversible reaction. There are also reversible reactions. EXAMPLE Equilibrium of ammonium nitrate NH₃ + HNO₃ ↔ Ammonium nitrate is a solid, and thus has a concentration defined as unity.

16. Number density nX [molecules cm-3]
Proper measure for
calculation of reaction rates
optical properties of atmosphere
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 = x 1019 cm-2.

17. 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

18. 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, Check out also

19. GAS HENRY’S LAW CONSTANT (M / atm at 298 K)
OXYGEN
O₂ x 10⁻²
OZONE
O₃ x 10⁻³
NITROGEN DIOXIDE
NO₂ x 10⁻²
CARBON DIOXIDE
CO₂ x 10⁻²
SULFUR DIOXIDE
SO₂
NITRIC ACID (effective)
HNO₃ x 10⁺⁵
HYDROGEN PEROXIDE
H₂O₂ x 10⁺⁴
HYDROPEROXY RADICAL
HO₂ x 10³
ALKYL NITRATES
(RONO₂)

20. 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

21. Today’s barometric pressure is 993 hPa = 993/1013 atm = 0. 98 atm
Today’s barometric pressure is 993 hPa = 993/1013 atm = 0.98 atm. Thus the partial pressure of CO₂ is In water CO₂ reacts slightly, but [H₂CO₃] remains constant as long as the partial pressure of CO₂ remains constant.

22. H+ = 2.3x10-6 → pH = -log(2.3x10-6) = 5.6 We know that: and THUS
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.

23. 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?

24. 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: (atm)
Ambient temperature: T(K)
LET:
be the concentration of X in the aqueous phase in moles/m³
be the concentration of X in the gas phase in moles/m³
Where is the aqueous concentration in M, and is the partial pressure expressed in atm. We can find the partial pressure from the mixing ratio and total pressure.

25. For the aqueous-phase concentration: units: moles/m³ = moles/L(water) x g(water)/m³(air) x L/g For the gaseous content: units: moles/m³ =

26. Notice that the ratio is independent of pressure and concentration
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.

27. What is the possible pH of water in a high cloud (alt
What is the possible pH of water in a high cloud (alt. ≃ 5km) that absorbed sulfur while in equilibrium with 100 ppb of SO₂? 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. This SO₂ will not stay as SO₂•H₂O, but participate in a aqueous phase reaction, that is it will dissociate.

28. 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? If most of the [H⁺] comes from SO₂•H₂O dissociation, then 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

29. Because SO₂ participates in aqueous-phase reactions, Eq
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.

30. 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