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

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

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 http://news.bbc.co.uk/2/hi/in_pictures/7506925.stm (slide courtesy J. Volckens)

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

Rayleigh scattering (isentropic) applies to gases, very small particles http://www.philiplaven.com

Haze particle http://www.atm.damtp.cam.ac.uk/people/mgb/aniso.html

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

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

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”

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=1.5 + 0.1 i Diameter in microns

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

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

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

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

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

UPTAKE OF WATER BY AEROSOLS

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

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

http://views.cira.colostate.edu/web/

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

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

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

A cloud drop….

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

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

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 Constant @ 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:

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 0.05-0.5 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 500-5000 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

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

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…

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.

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

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:

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 7.3.4  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