AQUEOUS PHASE CHEMISTRY MODIS, NASA’s Blue Marble Project Clouds cover 60% of the Earth’s surface! 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

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

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: 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) The most fundamental ionization reaction: H2O(l) ↔ H+(aq) + OH-(aq) 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. Others do, and in some cases, reaction with liquid water does not change the essential proportionality of the liquid phase [X] to the gas phase Px. For example, when formaldehyde dissolves in water it forms a gem-diol: CH2O (aq) + H2O (l) ↔ CH2(OH)2 (aq) Here [CH2(OH)2] ~ PCH2O 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” of “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