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AQUEOUS PHASE CHEMISTRY

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

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AQUEOUS PHASE CHEMISTRY

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  1. AQUEOUS PHASE CHEMISTRY MODIS, NASA’s Blue Marble Project Clouds cover 60% of the Earth’s surface! Important medium for aqueous phase chemistry

  2. 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

  3. 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

  4. STEP 2 SOLUBILITY AND HENRY’S LAW 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: Note: HA↑ as T↓ Can use ideal gas law to obtain the “dimensionless” Henry’s Law Constant:

  5. 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) THE ROLE OF LIQUID WATER 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

  6. STEP 2 NON-IDEAL SOLUTIONS 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

  7. STEP 2’ IONIZATION REACTIONS 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…

  8. STEP 2’ ACIDIC/BASIC IONIZATION REACTIONS, EXAMPLE: SO2 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*≥H H* =“effective” of “modified” Henry’s Law constant Solubility of S(IV) increases as pH increases. Effect of ionization in solution is to increase the effective solubility of the gas.

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

  10. STEP 2’ SOLVING THE SULFUR DIOXIDE / WATER EQUILIBRIUM 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)]) If other species are present need to modify electroneutrality equation, for example with sulfate:

  11. STEP 2’ CO2(g) CO2.H2O CO2.H2O HCO3- + H+ OTHER ACID/BASE EQUILIBRIA… CO2 dissolving in a drop (same as in ocean): HCO2 = 3x10-2 M atm-1 OCEAN electroneutrality: [H+]= [HCO3-]+2[CO32-]+[OH-] Kc1 = 9x10-7 M • 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 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

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