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. 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:

5. 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

6. 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

7. 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…

8. 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.

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

10. 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:

11. 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