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 phaseAqueous-phase chemistry: heterogeneous chemistry occurring in or on particles (aerosols, fog droplets, cloud droplets, etc)

Bulk solutions Cloud/fog

droplets

Aerosol particles

Not too different

Aerosols may have high ionic strengths

Considerable uncertainty applying equilibrium and rate constants obtained from dilute solutions in the lab to atmospheric particles

Can also exchange material b/w phases (large reservoir

in gas phase)

Can be very different!

AQUEOUS PHASE REACTION MECHANISM

STEP 1: Diffusion to the surface

STEP 4: Chemical Reaction

STEP 2: Dissolution

STEP 3: Diffusion in aqueous phase

XX X

XX+Y ?

STEP 2’: Ionization (for some species), VERY fast

A+ + B-

SOLUBILITY AND HENRY’S LAW

Henry’s Law: Distribution of species between aqueous and gas phases (for dilute solutions)

2 2

2

( )

[ ]A A

A

A g H O A H O

A H OK H

p

HA = Henry’s Law ConstantUnits 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↓

STEP 2

A A

A A

P N,R 0.0821Latm

mo/ mole / K

RT V

H H RT

les

L

Can use ideal gas law to obtain the “dimensionless” Henry’s Law Constant:

298

1 1exp

298

HH T H

R T

THE ROLE OF LIQUID WATER

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 10 0.05-0.5 5x10-8 – 5x10-7 2-6

rain 500-5000 0.1-1 10-7 – 10-6 4-5

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)

Consider, the distribution factor of a species:

A AA A A

A

H P[A(aq)]H L H RTL L

[A(g)] P / RT

molesAinsolution

molesAinair

f

=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

Generally, L~10-6, then fA =1 for HA~4x104 M/atm. If HA << than this, most of A in gas phase

All of gas in solution:

Ama

2Ox

A

H

moles

L

P 1[A]

RT L

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:

A A Aa m2

BX B Xa m m

mA = molality [moles A/kg solvent] = molal activity coefficient = f(ionic strength of solution, I)zi = charge on each ion (i)

Challenge: calculate activity coefficients in the multi-component, high ionic strength solutions characteristic of atmospheric aerosols

For example, use Debye-Hückel limiting law:2

i ilog 0.5085z I

STEP 2

2i i

i

1I z m

2

IONIZATION REACTIONS

The most fundamental ionization reaction:H2O(l) ↔ H+(aq) + OH-(aq)

' 16

2

14 2

[ ][ ]1.82 10 , 298

[ ]

[ ][ ] 1 10 , 298

w

w

H OHK M at K

H O

K H OH M at K

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 = basic7 = 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…

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):2

2

2SO

SO

3a1

2

23

a23

[SO (aq)]H

P

[H ][HSO ]K

[SO (aq)]

[H ][SO ]K

[HSO ]

Solve for equilibrium concentrations of bisulphite and sulphite:

2 2

2 2

a1 SO SOa1 23

a1 a2 SO SO2 a2 33 2

K H PK [SO (aq)][HSO ]

[H ] [H ]

K K H PK [HSO ][SO ]

[H ] [H ]

With fast equilibria often group: [S(IV)]=[SO2(aq)]+[HSO3

-]+[SO32- ] all have same oxidation state

2 2

2

a1 a1 a2SO SO 2

*S(IV) SO

K K K[S(IV)] H P 1

[H ] [H ]

H P

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.

STEP 2’

H*≥H

S(IV) SOLUBILITY AND COMPOSITION DEPENDS STRONGLY ON PH

[Seinfeld & Pandis]

2 2

2

a1 a1 a2SO SO 2

*S(IV) SO

K K K[S(IV)] H P 1

[H ] [H ]

H P

STEP 2’

SOLVING THE SULFUR DIOXIDE / WATER EQUILIBRIUM

2 2

2 2

2 2

2 SO SO

a1 SO SO3

a1 a2 SO SO23 2

[SO (aq)] H P

K H P[HSO ]

[H ]

K K H P[SO ]

[H ]

From equilibrium we had:

Add the electroneutrality equation: [H+]=[OH-]+[HSO3-]+2[SO3

2- ]

2 2 2 2

3w a1 SO SO a1 a2 SO SO0 [H ] K K H P [H ] 2K K H P

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:

2 2 2 2

3 2 2w a1 SO SO a1 a2 SO SO 40 [H ] K K H P [H ] 2K K H P 2[SO ][H ]

STEP 2’

OTHER ACID/BASE EQUILIBRIA…

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[SO4

2-]+[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

CO2 dissolving in a drop (same as in ocean):CO2(g)

CO2.H2O

CO2.H2O HCO3

- + H+

HCO3- CO3

2- + H+

HCO2 = 3x10-2 M atm-1

Kc1 = 9x10-7 M

Kc2 = 7x10-10 M

OCEAN

electroneutrality: [H+]= [HCO3

-]+2[CO32-]+[OH-]

Can express in terms of K’s and [H+]Find at 283K, PCO2=350ppm, pH=5.6 (rain slightly acidic)

STEP 2’

2 2

2 2

T 22 2 2 3 3

c1 c1 c2CO CO 2

*CO CO

[CO ] [CO H O] [HCO ] [CO ]

K K KH P 1

[H ] [H ]

H P