1. ATS 621 Fall 2012 Lecture 18 Atmospheric Aerosols / Visibility Wrap-Up
Aqueous Phase Chemistry Part I

2. PM Welfare Effects: Visibility and Haze
Visual Range: 250 km
Visual Range: 25 km

3. PM10 Levels and Visibility in Beijing
WHO Guideline: 50 ug/m3
averaged over 24 hrs
8 ug/m3
12 July
26 ug/m3 15 July
32 ug/m3
20 July
104 ug/m3
5 August
191 ug/m3
7 August
278 ug/m3
10 August
(slide courtesy J. Volckens)

4. Visibility Human ability to see is a function of
Characteristics of object observed
Amount and distribution of light
Concentration of suspended particles and gases
Particles & gases scatter & absorb light causing
Appearance of haze
Decrease in contrasts
Change in perceived color (absorption / scatter of certain λ’s)
Scattering usually has a greater effect on visibility than absorption does
2 & 3 affect contrast between 1 & 4

5. Rayleigh scattering (isentropic) applies to gases, very small particles

6. Haze particle

7. Contrast that can be discerned by typical viewer
Koschmieder equation I I0
Assumes:
Homogeneous atmosphere
Uniform sky brightness
Viewing horizontally
Neglected curvature of Earth
Contrast that can be discerned by typical viewer
This equation relates visual range to the total extinction coefficient

8. Components of the extinction coefficient, σext
σext = σRayleigh + σabs,gas + σabs,part + σscat,part
Visible light: 0.4 μm < λ < 0.7 μm
Consider components:
σRayleigh
Due to gases (small particles also scatter this way); scattering α λ-4
-> shorter wavelengths scatter more (blue sky in clean conditions; red sky at sunset)
At sea level and 0.45 μm , σRayleigh ~ 0.03 km-1
-> Lv = 130 km (81 miles)
σabs,gas
NO2 is the only important absorber – absorbs most strongly at blue λs
-> colors plumes yellow, red or brown
σabs,part
Absorption = f(composition, size distribution)
-> soot is most important (5-40% of total visibility reduction)
-> overall soot is ~3x as efficient as sulfate, nitrate or organic aerosol species in terms of visibility reduction per unit mass of airborne particles
-> new data suggest organic species contribute to light-absorbing carbon (“brown carbon”) – prevalent in wood smoke

9. Scattering by particles
σscat,part
Particle scattering is ~60 – 95% of visibility reduction
Scattering importance (generally): sulfate > organic carbon > nitrate
- Particle WATER CONTENT is very important!
Most complicated to compute
Very small particles act like gases
Very large particles (Dp >> λ) extinguish light according to 2x their cross-sectional area (the “extinction paradox” – light scattered at very small angles to the beam is still seen when close to the object, but not when at a distance)
Intermediate-sized particles (“Mie scattering”) have the most effective light scattering mechanism
Extinction cross section can be up to ~4 – 5 times the cross sectional-area
0.1 – 1 μm are of similar λ to visible light: the “accumulation mode”

10. Scattering efficiency of a spherical particle
m=1.5
Extinction paradox: large particles extinguish 2 x their cross sectional area!
 But remember NUMBERS are very small
m= i
Diameter in microns

11. Scattering by particles
The scattering coefficient due to N particles, each of radius r, is
Can compute this for each ‘section’ of a size distribution to get the overall sscat
Similar for absorption by particles and total extinction due to particles
In practice, we often use “mass scattering efficiencies” and “mass absorption efficiencies” to compute overall extinction from measured mass concentrations

12. Effect of refractive index
Note per unit mass, large particles account ~0.6 m2 g-1
Mass scattering efficiency

13. Effects of relative humidity on visibility
More aerosol mass is present (in the form of condensed water in the aerosol phase)  visibility is reduced
The SIZES of the particles shift  generally to a more efficient scattering diameter, but can also grow OUT of the most effective scattering range if already very large
The refractive index DECREASES  this decreases the extinction coefficient, but effect is much smaller than the other effects

14. Aerosol water contents (Kreidenweis et al., ERL, 2008)
Relative Humidity

15. Effects of relative humidity on visibility
More aerosol mass is present (in the form of condensed water in the aerosol phase)  visibility is reduced
The SIZES of the particles shift  generally to a more efficient scattering diameter, but can also grow OUT of the most effective scattering range if already very large
The refractive index DECREASES  this decreases the extinction coefficient, but effect is much smaller than the other effects

16. UPTAKE OF WATER BY AEROSOLS

17. Effects of relative humidity on visibility
More aerosol mass is present (in the form of condensed water in the aerosol phase)  visibility is reduced
The SIZES of the particles shift  generally to a more efficient scattering diameter, but can also grow OUT of the most effective scattering range if already very large
The refractive index DECREASES  this decreases the extinction coefficient, but effect is much smaller than the other effects

18. Effect of refractive index
Mass scattering efficiency
n~1.4 commonly assumed

20. Estimated Annual Visual Range http://vista. cira. colostate
ATS 555 F10 Lecture 3

21. AQUEOUS PHASE CHEMISTRY
MODIS, NASA’s Blue Marble Project
Clouds cover 60% of the Earth’s surface!
And, they concentrate species into much smaller volumes than in air.
Important medium for aqueous phase chemistry

22. DEFINITIONS AND ISSUES
Heterogeneous chemistry: chemistry involving more than one phase
Aqueous-phase chemistry: heterogeneous chemistry occurring in or on particles (aerosols, fog droplets, cloud droplets, etc)
Can also exchange material b/w phases (large reservoir in gas phase)
Aerosols may have high ionic strengths
Not too different
Can be very different!
Aerosol
particles
Bulk
solutions
Cloud/fog
droplets
Considerable uncertainty applying equilibrium and rate constants obtained from dilute solutions in the lab to atmospheric particles

23. A cloud drop….

24. AQUEOUS PHASE REACTION MECHANISM
STEP 2’: Ionization (for some species), VERY fast
STEP 1: Diffusion to the surface
STEP 2: Dissolution X X X
 A+ + B-
STEP 4: Chemical Reaction
STEP 3: Diffusion in aqueous phase
X+Y ? X

25. AQUEOUS PHASE REACTION MECHANISM
STEP 2’: Ionization (for some species), VERY fast
STEP 1: Diffusion to the surface
STEP 2: Dissolution X X X
 A+ + B-
FAST
(not really, but we will treat as such)
FAST
STEP 4: Chemical Reaction
STEP 3: Diffusion in aqueous phase
X+Y ? X
Rate limiting
FAST

26. SOLUBILITY AND HENRY’S LAW
STEP 2
Henry’s Law: Distribution of species between aqueous and gas phases (for dilute solutions)
HA = Henry’s Law Constant
Units here are mol/L/atm OR M/atm
Some Henry’s Law Constants of Atmospheric Relevance:
Chemical Species
Henry’s Law 25°C (mol/L/atm)
HNO3
2.1x105
H2O2
7.5x104
HCHO
3.5x103
NH3
57.5
SO2
1.2 CO
9.6x10-4
Note: HA↑ as T↓
Can use ideal gas law to obtain the “dimensionless” Henry’s Law Constant:

27. THE ROLE OF LIQUID WATER
STEP 2
The liquid water amount affects the partitioning of species between the gas and the aqueous phase (esp for very soluble species)
L = liquid water content of the atmosphere (m3 of water / m3 of air)
Diameter (m)
L (cm3/m3)
L (m3/m3) pH
haze
10-5 – 10-4
10-11 – 10-10
1-8
clouds 10
0.1-1
10-7 – 10-6
3-6
fog
5x10-8 – 5x10-7
2-6
rain
4-5
Consider, the distribution factor of a species:
=1, there are equal amounts of A in each phase
<<1, A is predominantly in the gas phase
>> 1, A is predominantly in the aqueous phase
All of gas in solution:
Generally, L~10-6, then fA =1 for HA~4x104 M/atm.
If HA << than this, most of A in gas phase

28. NON-IDEAL SOLUTIONS STEP 2 Rain/Clouds = dilute
Haze/plume = concentrated
Henry’s Law
(approximate activities using concentrations)
Calculate activities (a):
Undissociated species A:
Species BX which dissociates:
mA = molality [moles A/kg solvent]
= molal activity coefficient = f(ionic strength of solution, I)
zi = charge on each ion (i)
For example, use Debye-Hückel limiting law:
Challenge: calculate activity coefficients in the multi-component, high ionic strength solutions characteristic of atmospheric aerosols

29. IONIZATION REACTIONS STEP 2’ The most fundamental ionization reaction:
H2O(l) ↔ H+(aq) + OH-(aq)
Ka = acid dissociation constant
(the larger the value, the stronger the acid, and thus the more acid is dissociated)
pKa = -log[Ka]
If pH > pKa a molecule is more likely to donate a proton (deprotonate)
Electroneutrality (charge balance): in pure water [H+]=[OH-]
pH = -log[H+]  the activity of H+
< 7 = acidic
> 7 = basic
7 = neutral
Some species (eg. O3) simply dissolve in water and do not undergo reactions.
But not always so straight-forward for acidic or basic gases…

30. ACIDIC/BASIC IONIZATION REACTIONS, EXAMPLE: SO2
STEP 2’
Illustrate with SO2 dissolved in a cloud drop:
[SO2(aq)]=HSO2PSO2  from Henry’s Law
However, SO2 is an acid in aqueous solution:
SO2(aq)+H2O(l) ↔H+(aq)+HSO3-(aq)
HSO3-(aq) ↔H+(aq)+SO32- (aq)
Acid dissociation constants (Ka1, Ka2):
Solve for equilibrium concentrations of bisulphite and sulphite:
With fast equilibria often group:
[S(IV)]=[SO2(aq)]+[HSO3-]+[SO32- ]  all have same oxidation state
H* =“effective” or “modified” Henry’s Law constant
H*≥H
Solubility of S(IV) increases as pH increases. Effect of ionization in solution is to increase the effective solubility of the gas.

31. S(IV) SOLUBILITY AND COMPOSITION DEPENDS STRONGLY ON PH
STEP 2’
S(IV) SOLUBILITY AND COMPOSITION DEPENDS STRONGLY ON PH
[Seinfeld & Pandis]

32. SOLVING THE SULFUR DIOXIDE / WATER EQUILIBRIUM
STEP 2’
From equilibrium we had:
Add the electroneutrality equation: [H+]=[OH-]+[HSO3-]+2[SO32- ]
If S(IV) is the only species in solution we can solve this for [H+], with one more piece of information (for example PSO2=1ppb, T=298K  pH=5.4, could then calc [S(IV)])
Sum(anions)=sum(cations)
If PSO2 = 1ppt, then pH=6.7  close to pure water because so little SO2  in the future save this for a class Q (ie. if the concentration of SO2 in the atmosphere goes down, do we expect the rainwater to be more or less acidic?)
To calc [S(IV)] either sum up individual or use H* expression
S(IV) increases with increasing pH, decreasing T or increasing PSO2
If other species are present need to modify electroneutrality equation, for example with sulfate:

33. OTHER ACID/BASE EQUILIBRIA…
STEP 2’
CO2 dissolving in a drop (same as in ocean):
CO2(g)
CO2.H2O
HCO2 = 3x10-2 M atm-1
OCEAN
electroneutrality:
[H+]= [HCO3-]+2[CO32-]+[OH-]
Kc1 = 9x10-7 M
CO2.H2O
HCO3- + H+
Can express in terms of K’s and [H+]
Find at 283K, PCO2=350ppm, pH=5.6
(rain slightly acidic)
Kc2 = 7x10-10 M
HCO3-
CO32- + H+
Ammonia (basic in solution):
NH3 (g) + H2O(l) ↔NH4OH(aq)
NH4OH(aq) ↔NH4++OH-
electroneutrality: [H+]+[NH4+]=[OH-]=kW[H+]-1
CO2 hydrolyzes, H* > H, so total amount of CO2 dissolved always exceeds that predicted by Henry’s Law alone for CO2 (for pH < 5, H*~H)  CO2 is not as strong an acid as SO2, so range of H* not as large (cannot pull in as much via ionization)
Can replace [OH-] with Kw/[H+]
Ammonia is basic in solution, probably most important neutralizing agent in the atm
(can do full ammonia derivation following S&P section  find for pH < 8, [NH3T(aq)]=[NH4+]
Salt definition: neutral compound formed by a union of an acid and a base
Salts (dissolution):
(NH4)2SO4=2NH4++SO42-
electroneutrality: [H+]+[NH4+]=2[SO42-]+[OH-]
What is the pH? If assume no exchange with the gas phase, then NH4+ equilibrates with NH3(aq). Then, [NH4+]< 2[SO42-], so [H+]>[OH-] and pH < 7