Exam Atmospheric Chemistry and Physics June 4, 2014 at 8:00-12:00

You may use: Calculator, Table booklet (Tefyma or similar), Equation sheet of the course Answer each question (1-5) on separate sheets.

1. Aerosol and hygroscopic growth

a) Which are the two main natural sources of aerosol particles, and which is the main human activity behind particles in the atmosphere? (1 p)

b) Which of these sources produce mainly particles smaller or larger than 1 µm diameter, respectively? (1 p)

c) Compare these three sources with respect to hygroscopic growth (water uptake). Discuss their importance in cloud formation with respect to particle size and composition. (1 p)

d) Why is hygroscopic growth of aerosol particles important for the climate? (1 p) e) A volcanic eruption induces an aerosol layer in the stratosphere of the earth with the

globally averaged optical depth δ = 0.05. The backscattered fraction β = 0.2. Calculate the radiative forcing induced by the eruption. (2 p)

2. The atmosphere

a) Which is the fundamental difference between the troposphere and the stratosphere, making them regarded as different parts of the atmosphere? (1p)

b) Is vertical motion most common in the troposphere or the stratosphere? Why? (1 p) c) The global average height of the tropopause is 14 km. Calculate the pressure at that

altitude, assuming constant atmospheric temperature of 250 K. (1 p) d) Assume that the global air flow from the troposphere to the stratosphere is equal to the

opposite flow (from the stratosphere to the troposphere), being 5.9 1017 kg/year in each of the two directions. Calculate the residence times of air in the troposphere and the stratosphere. The pressure at the stratopause is 1 hPa. (2 p)

e) Explain why the residence times of stratospheric and tropospheric air differ. (1 p) 3. Acidification Sulfur is emitted to the atmosphere mainly as the gas sulfur dioxide (SO2). The sources are both natural and anthropogenic (resulting from human activities). SO2 is then oxidized in the gas and aqueous phases to eventually cause acidification.

a) What are the most important natural and anthropogenic sources of sulfur? (1p) b) What are the most important SO2 oxidation agents (molecules) in the gas and aqueous

phases? Briefly describe how these oxidation agents form in the atmosphere. (1p) In order to be oxidized in the aqueous phase, the gas SO2 must first be dissolved in water according to the following reactions: (1) SO2(g) SO2(g)⋅H2O (K1 = 1.2 M⋅atm-1) (2) SO2(g)⋅H2O HSO3

- + H+ (K2 = 1.3⋅10-2 M) (3) HSO3

- SO32- + H+ (K3 = 6.3⋅10-8 M)

Here, the Ki are equilibrium constants. c) Will the reactions 1-3 above result in a permanent acidification of the aqueous solution?

Motivate your answer. (1p) Oxidation of SO2 in the aqueous phase – and thereby the acidification – mainly proceeds via the bisulfite ion (HSO3

-), and its aqueous phase concentration [HSO3-] is therefore important.

d) Calculate, using reactions 1-3 above, an expression for the mole fraction of bisulfite (HSO3

-) out of the total amount of S(IV) dissolved in water. Then use this expression to calculate the mole fraction of bisulfite at pH = 3, 4.5 and 7. (3p)

4. Stratospheric ozone a) Describe – using Figure 1 below – how emissions of halocarbons (freoner) at the surface of

Earth can affect the stratospheric ozone layer. Write the (few) reaction steps that lead to catalytic ozone loss for different compounds X, where X can be the chlorine radical as shown in the figure. Also summarize the net reaction showing the net ozone loss. (2 p)

Figure 1

NOx can also destroy ozone catalytically. The most important reactions for maintaining the radical balance within the ClOx- and NOx-families in the stratosphere are: (1) Cl + O3 ClO + O2 (k1 = 7.9⋅10-12 cm3 ⋅ molecules-1 ⋅ s-1) (2) ClO + NO Cl + NO2 (k2 = 2.7⋅10-11 cm3 ⋅ molecules-1 ⋅ s-1) (3) NO + O3 NO2 + O2 (k3 = 1.8⋅10-15 cm3 ⋅ molecules-1 ⋅ s-1)

O2 (4) NO2 + hν NO + O3 (k4 = 1⋅10-2 s-1, λ< 420 nm) The following concentrations are found in a parcel of air in the stratosphere: [O3] = 0.6 ppmv, [NOx] = 0.3 ppbv, [ClOx] = 1.0 ppbv. The air density at this altitude is 3.5·1018 molecules·cm-3. Assume steady state conditions for both Cl and NO. b) Show that [Cl] << [ClO] and [NO] << [NO2] for the air parcel in question.

Tip: Start with the chlorine radicals, and also remember that [NOx] = [NO] + [NO2]. (2 p) c) What are the lifetimes for the radicals Cl and NO considering reactions 1-4 above? Is the

assumption regarding steady state conditions for both Cl and NO reasonable? Motivate your answer. (2 p)

5. Tropospheric ozone Figure 2 describes how hydrocarbons (RH) are oxidized in the atmosphere, contributing to the formation of ozone. This involves many rather complex reactions. However, the formation of ground-level ozone can be condensed and summarized using the photostationary equilibrium for ozone, given as:

hν NO2 + O2 NO + O3 (1) a) Use the photostationary equilibrium (1) and the reactions shown in the Figure to explain why

hydrocarbons are needed for the formation of a photochemical smog and high concentrations of tropospheric ozone. (2p)

The Figure also shows the single most important tropospheric chemistry radical cycle. b) What determines whether a compound is radical or not? (1p) c) Which of the gas phase compounds in the Figure are radicals? (1p) d) What prevents this radical chain to continue forever, producing ever more ozone. Give one

example of a reaction that will end the radical chain. (1p) e) Name at least one harmful trace gas that is oxidized mainly by the radicals formed in the

Figure. What is the harmful effect for the trace gas in question in terms of negative impacts on human health and/or the environment, and give the end product of the oxidation process. (1p)

Figure 2. Schematic figure of how hydrocarbons (RH) are oxidized in the atmosphere,

contributing to the formation of ozone.

Answers to exam questions of June 4, 2014 1. Aerosol and hygroscopic growth a) Natural: Sea-spray and windblown erosion particles (dust) Anthropogenic: combustion b) Natural sources (mechanical generation): >1 µm Combustion: mainly < 1 µm c) Water solubility and large size => low critical supersaturation => good cloud condensation nucleus (CCN). Sea-spray (sodium chloride): soluble and large; Windblown dust (mineral): insoluble and large; Combustion: soluble and small; => Sea-spray the best CCN d) Soluble particles grow with increased relative humidity causing increased scattering of solar radiation, thus increasing the direct climate effect of the aerosol. The hygroscopic properties of particles also affect cloud formation and thus also the indirect climate effect of aerosol. e) The albedo of an aerosol layer is given by Aa = βδ, and the total albedo of two layers is obtained by AT = A0 + Aa(1-A0)2, where A0 = 0.28 is the albedo without the aerosol layer from the volcanic eruption. The albedo change ∆A = AT – A0 = Aa(1-A0)2. The radiative forcing is defined by ∆F = Fin,2 – Fout,2 in the system after the change of albedo (textbook pp 133 – 135). In this case we have changed the albedo, implying that the we have the same incoming radiation in the reference system (before the eruption) => Fin,2 = Fin,1 = Fout,1, were the latter is true because the reference system is at equilibrium. Hence, we can express ∆F = Fout,1 – Fout,2. Using the climate model of the textbook (Fig. 7-12): ∆F = FSA0/4 + (1 – f)σT0

4 + fσT14 – (FSAT/4 + (1 – f)σT0

4 + fσT14) = FSA0/4 – FSAT/4 =

= – FS∆A/4 = – FSAa(1-A0)2/4 = – 1370*0.2*0.05*(1 – 0.28)2/4 = – 1.78 W/m2 2. The atmosphere a) The temperature change with altitude. Troposphere: temperature decrease with altitude; Stratosphere: temperature increase with altitude b) The troposphere, because the temperature increase with altitude in the stratosphere (inversion) induces a strong counteracting force against vertical motion => very stable conditions c) The vertical pressure profile can be obtained from the barometric law: p(z) = p0 exp(-z/H), where H = RT/(Mag), p0 = the average pressure at the surface (984 hPa), z = the altitude, R = the gas constant, T = the temperature, Ma = average atmospheric molar mass and g = the gravity. p(14 km) = 145 hPa. d) The pressure at any given layer is caused by the weight of the overlying air: pA = mg => m =pA/g. The mass of the troposphere is given by the mass of the atmosphere minus the mass of the air above the tropopause: mT = (p0 – ptp)A/g = (98400 – 14500)4π(6.37 106)2/9.81 = 4.36 1018 kg and for the stratosphere mS = (ptp – psp)A/g = 7.48 1017 kg Residence times: τT = mT/Fout = 4.36 1018 / 5.9 1017 = 7.4 years; τS = 1.3 years e) The mass of the troposphere is much larger than the stratosphere and the mass flows in both directions are equal. Therefore more time is needed to exchange the tropospheric air.

3. Acidification a) Natural sources: The ocean (DMS), volcanoes. Antropogenic: Combustion of sulphur-containing fossil fuels, (also smelters that extract metals from sulfide ores). b) Gas phase: OH radical, formed in photochemical smog (photolysis of O3). Aqueous phase: H2O2, formed via termination of HO2 radicals; O3, formed in photochemical smog (photolysis of NO2) c) No, since these reactions are all reversible. There is no oxidation taking place, and the oxidation number remains as S(IV) thoughout reactions 1-3. When the water is removed (for instance when cloud droplets evaporate), the sulfur will repartition back to the gas phase. d)

( )[ ]( )[ ]

( )[ ] ( )[ ]2

2

12222

2

221 SO

SO

pKOHgSOp

OHgSOgSO

OHgSOK ⋅=⋅⇒⋅

=⋅

=

[ ] [ ]

( )[ ] [ ] ( )[ ][ ] [ ]++

−+− ⋅⋅

=⋅⋅

=⇒⋅⋅

=H

pKKH

OHgSOKHSOOHgSO

HHSOK SO2122223

22

32

[ ] [ ][ ] [ ] [ ]

[ ] [ ]2321332

33

23

32

++

−−

−

+− ⋅⋅⋅=

⋅=⇒

⋅=

H

pKKKHHSOKSO

HSOHSOK SO

Let

[total-S(aq)] denote the total aqueous pahse concentration of all S(IV)-containing molecules

and ions.

[ ] [ ] [ ] [ ]−− ++⋅=− 23322 )()( SOHSOOHgSOaqStotal

[ ] [ ] [ ] =

⋅⋅⋅+

⋅⋅+⋅=−

++ 232112

122

2)(

H

pKKKH

pKKpKaqStotal SOSO

SO

[ ] [ ]

⋅++⋅⋅=

++ 2322

1 12 H

KKHKpK SO

[ ][ ]

[ ]

[ ] [ ][ ] [ ]

[ ][ ]+

+

++

++

+−

++=

⋅++

=

⋅++⋅⋅

⋅⋅

=−

⇒

HK

KH

HKKKH

K

HKK

HKpK

HpKK

aqStotalHSO

SO

SO

3

2

322

2

2322

1

21

3

1

1

1)(

2

2

Calculate this mole fraction at pH = 3, 4.5 and 7. Use pH= −log([H+]) to get [H+]. At pH = 3, mole fraction of HSO3

- = 0.93 At pH = 4.5, mole fraction of HSO3

- = 1.00 At pH = 7, mole fraction of HSO3

- = 0.61.

4. Stratospheric ozone a) Ozone can be destroyed in catalytic processes, which means that the compound X that is responsible for the ozone loss is recycled and not consumed. (1) O3 + X → O2 + XO (2) O + XO → O2 + X Net reaction ((1) + (2)): O + O3 → 2O2 A different (but also common) set of reactions is: (1) O3 + X → O2 + XO (3) O3 + XO → 2O2 + X Net reaction ((1) + (3)): 2O3 → 3O2 X can be various molecules: X = OH : OH (XO=HO2) X = Cl (XO=ClO) X = NO : NO (XO=NO2) X = Br (XO=BrO) The figure shows how Cl destroys ozone catalytically. The Cl radicals originate from halocarbons (freoner) that contain one or more chlorine atoms. In the troposphere, the halocarbons are very stable compounds and have a long lifetime. They can therefore slowly penetrate into the stratosphere, where they are photolyzed by UV radiation. For instance:

CF2Cl2 + hν CF2Cl + Cl (initiation, UV radiation required) (ClOx – family: Cl and ClO, both are radicals) b) Steady state condition for Cl:

k2[ClO][NO] = k1[Cl][O3]

[ ][ ]

[ ][ ]

[ ][ ] =<=⇒

31

2

31

2

OkNOk

OkNOk

ClOCl x 1.7⋅10-3 mol/mol.

Here we have used the fact that [NOx] = [NO] + [NO2], and thus [NOx] > [NO]. Steady state condition for NO:

k4[NO2] = k3[NO][O3] + k2[ClO][NO]

[ ][ ] [ ] [ ] [ ] [ ] =+

≈+

=⇒xClOkOk

kClOkOk

kNONO

233

4

233

4

2

0.1 mol/mol.

Since we have shown already that [Cl] << [ClO], then [ClO] ≈ [ClOx] = 1.0 ppbv. Note that you need to give the concentrations in units of molecules·cm-3 and not in ppb or ppm. c) The lifetime for Cl molecules in the stratosphere (MCl=[Cl], LCl loss rate of Cl):

[ ][ ][ ] [ ]3131

1OkOClk

ClLM

Cl

ClCl ===τ = 0.06 s

Similarly, the lifetime for NO:

[ ][ ][ ] [ ][ ] [ ] [ ]332332

1OkClOkONOkClONOk

NONO +

=+

=τ = 10 s

These lifetimes are short in comparison with the processes which control the formation and loss of Cl and NO (mainly UV radiation varying on a diurnal basis). This means that new steady state concentrations can be achieved rapidly (within approximately 3τ), balancing production and loss.

5. Tropospheric ozone a) By reacting with the OH radical (detergent of the atmosphere), hydrocarbons (RH) produce

organic oxy (RO) and peroxy radicals (RO2) that are needed to propagate the radical chain involving also the HOx radicals (OH and HO2) in the Figure. By doing so, they help shifting the photostationary equilibrium (1) to the right side by converting NO to NO2. (UV light is also needed to photolyze NO2 and O3.

b) A radical has one unpaired electron (an odd number of electrons)

c) The radicals in the Figure are

OH, HO2, RO, RO2, NO, NO2. d) The radical chain is terminated, which means that two radicals react to produce one non-radical compound. (This will happen more frequently as the radical concentrations increase, since their lifetime is typically very short meaning that they will tend to react first with non-radical compounds in propagation reactions. Alternatively, sunlight (UV) is lacking (night-time) or the supply of RH compounds that can be oxidized is depleted.) The important termination steps (only one of them needs not be mentioned):

OH + NO2 + M HNO3 + M HO2 + HO2 H2O2 + O2 RO2 + HO2 ROOH + O2

e) Only one example is needed. The main radical formed is the hydroxyl radical OH, the “detergent of the atmosphere”. CO is lethal to humans and animals since it prevents the uptake of oxygen in our blood cells. End product: Oxidized by OH to CO2. CH4 is a powerful greenhouse gas, even more potent than CO2. End product: Oxidized by OH to HCHO, CO and eventually CO2. SO2 is dangerous to humans (affects respiratory system) and causes acidification of ecosystems, soils and waters via dry and wet deposition. End product: Oxidized (in gas phase) by OH to sulphuric acid H2SO4.

1

Exam Atmospheric Chemistry and Physics June 4, 2015 at 8:00-13:00

You may use: Calculator, Table booklet (Tefyma or similar), Equation sheet of the course Answer each question (1-5) on separate sheets.

1. The Atmosphere

a) The atmospheric concentration of CO2 is since a long time increasing. Which are the main reasons for the increase? (1 p)

b) Fig. 1 shows the change in CO2 concentration from year 1960 to present at Mauna Loa, Hawaii. What is causing the annual variation (årstidsvariation)? (1 p)

c) The yearly average concentration at Mauna Loa is a good estimate of the global yearly average. Why can we use data from one place for a global estimate? (1 p)

d) Use the data in Fig. 1 to calculate the increase of carbon mass in the atmosphere during the industrial era (i.e. the increase from pre-industrial concentration to the concentration year 2014). The pre-industrial CO2 concentration was 280 ppmv. (3 p)

Figure 1. The CO2 concentration in the air at Mauna Loa, Hawaii.

310

320

330

340

350

360

370

380

390

400

1960 1970 1980 1990 2000 2010

CO

2(ppmv)

Year

Mauna Loa CO2

Monthly average

Yearly average

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3

3. Aerosol-water interaction

a) An aerosol particle has taken up water, formed a droplet, and obtained its equilibrium size with respect to relative humidity (RH). What is the RH (in %) if the droplet diameter (D) is 400 nm? The particle consists of 1.12·10-18 kg sodium chloride (100 nm dry diameter). Assume that the dissociation () of NaCl is complete and that the temperature is 298 K. Also assume that the molar mass, surface tension and density of the solution equals that of water. [Water: Mw = 18.02 kg/kmol, σDw = 0.073 N/m, ρ = 998 kg/m3; Sodium chloride: MNaCl = 58.44 kg/kmol] (2p)

b) How will the droplet react if the RH increases slightly (by 1%)? Will the droplet still be in equilibrium state? (1p)

c) Make a sketch diagram that describes the relation between RH (> 95%) and droplet diameter. Mark the critical RH, and critical diameter. (1p)

d) How will the droplet react if the RH increases above the critical level of the droplet? Describe in general terms. Motivate. (2p)

4. Stratospheric ozone In the stratosphere, both O2 and O3 are photolyzed by UV-radiation (reactions (1) and (2) below). a) Why is photolysis of O2 not enough to protect life on Earth from hazardous UV-radiation from the sun? (1p) The reactions below are the most important for the production and destruction of stratospheric ozon. Here, O denotes O(3P). (1) O2 + h O + O (2) O3 + h O2 + O(1D) Cl + O3 ClO + O2 HO2 + HO2 H2O2 + O2 NO2 + O3 NO3 + O2

O3 + O 2O2

OH + O3 HO2 + O2 NO3 + NO2 + M N2O5 + M

OH + HO2 H2O + O2 O + O2 + M O3 + M

Cl + CH4 HCl + CH3 NO2 + O NO + O2

ClO + NO2 + M ClNO3 + M O(1D) + M O + M NO2 + OH +M HNO3 + M HO2 + O3 OH + 2O2 ClO + O Cl + O2

NO + O3 NO2 + O2

Use these reactions to write b) the Chapman mechanism for ozone production in the stratosphere (1p) c) the catalytic ozone loss via the HOx, NOx and ClOx families. (2p) d) How are HOx, NOx and ClOx radicals formed in the stratosphere? (1p) e) Which of the reactions above constitute termination steps for catalytic ozone loss propagated via the HOx, NOx and ClOx radicals? (1p)

5. Acidi The gas S

(1) S HSO2 is Haqueous p= 1 ppb atwater is aS(IV) in w”neutral” example win describ

Figur

a) Exb) Exc) De

S(be

d) Inre

e) Thaqwa

ification

SO2 dissolve

SO2(g) S

enry’s law cphase concet standard a

actually strowater is mupH in the a

with SO2 clbing how a g

re 4. The eff

xplain why xplain qualieduce an eq(IV) dissolve able to quan which formlevance (3-his phenomqueous phasaters. Expla

es in water a

SO2H2O

constant forentration of atmosphericongly depench greater t

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ffective Henfunc

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r SO2. At T=SO2 of 1.23

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4

o the equilib

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onstant H*S(I

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5

Solutions to exam in Atmospheric Chemistry and Physics June 4, 2015 1. The Atmosphere a) Fossil fuel combustion and deforestation, in particular in the tropics b) The growth period consumes CO2, whereas biological decay increases the concentration c) The residence time of CO2 is long compared to typical transport times in the atmosphere, implying that the concentration is similar everywhere, except close to strong sources. d) C = 398 – 280 = 118 ppmv = 118 10-6; mc = Mcnc = McnaC We need to calculate the number of moles of gas molecules in the atmosphere (na): The weight of the overlying air causes the pressure at a given level: mag = pA =4R2p => na = ma/Ma = 4R2p/(gMa) => mc = McnaC = McC4R2p/(gMa), where R is the radius of the Earth (6.37 106 m), g the gravitation (9.81 m/s2), Ma the average molar mass of air (29.0 kg/kmol) and p the average pressure ate the surface of the Earth (984 hPa). Insert the numbers: mc = 2.5 1014 kg 2. Climate a) The atmosphere absorbs radiation from the surface which then emits blackbody radiation at its temperature. The temperature decreases with altitude in the atmosphere, implying that less radiation is emitted to space at wavelengths where the atmosphere has strong absorption. b) In Fig 2b we see that the absorption of terrestrial radiation is small in the “atmospheric window” 8 – 13 µm in wavelength. The estimated surface temperature by comparing with blackbody radiation is approximately 320 K (47° C; North Africa) c) At the wavelengths where the atmosphere absorbs/emits radiation less radiation leaves the Earth. Therefore the temperature of the entire system is elevated to reach equilibrium with the incoming radiation from the sun. d) When the absorption of the atmosphere exceeds 100%, the probability for double or multiple absorption increases. As a result, the last emission of radiation before leaving the Earth will on average be at a higher altitude. This is also a colder part of the atmosphere, thus implying that less radiation leaves the earth. 3. Aerosol-water interaction a) Use the Köhler equation for the calculation. Result: 98.5% b) The droplet will become larger due to the increase in relative humidity (RH). It will still be in equilibrium state, because the RH is still lower than the critical relative humidity. c) Based on the Köhler equation, see Fig. 3 in Aerosol, Water and Clouds by Martinsson. d) The droplet will follow the Köhler curve until it reaches the maximum level (=critical RH). Up to the critical value, the droplet will stay in the equilibrium state. If the RH increases further and exceed the critical RH, the droplet will leave the equilibrium state. This means that the droplet will grow without limitation, the droplet activates. This is the process for cloud droplet formation.

6

4. Stratospheric ozone a) Photolysis of ozone (<320 nm) can proceed also in the UVB region where there is no photolysis of O2 (<240 nm). The stratospheric ozone layer thus protects us from the hazardous UVB radiation (280 nm < λ < 320 nm) which is not removed by ordinary oxygen O2. Answer b) The Chapman mechanism: (1) O2 + h O + O O + O2 + M O3 + M

(2) O3 + h O2 + O(1D) O(1D) + M O + M O3 + O 2O2

c) Catalytic ozone loss via HOx: OH + O3 HO2 + O2 HO2 + O3 OH + 2O2 (the latter reation does not follow the general scheme for a catalyst X which is XO + O ) Catalytic ozone loss via NOx: NO + O3 NO2 + O2 NO2 + O NO + O2 (XO + O ) Catalytic ozone loss via ClOx: Cl + O3 ClO + O2 ClO + O Cl + O2 (XO + O ) d) Formation of HOx: H2O + O(1D) 2OH Formation of NOx: N2O + O(1D) 2NO N2O originates from soil decomposition processes and is stable in the troposphere. NO is also emitted directly into the stratosphere by airplanes. Formation of ClOx: Photolysis of chlorine-containing CFCs that are i stable in the troposphere. e) Termination steps for catalytic ozone loss via HOx: OH + HO2 H2O + O2 HO2 + HO2 H2O2 + O2 Termination steps for catalytic ozone loss via NOx: NO2 + OH +M HNO3 + M NO2 + O3 NO3 + O2 and NO3 + NO2 + M N2O5 + M

Termination steps for catalytic ozone loss via ClOx: Cl + CH4 HCl + CH3 ClO + NO2 + M ClNO3 + M 5. Acidification a) Dissolution of CO2 at present atmospheric levels (380 ppm) results in a “neutral” and non-acidified pH of 5.6 in cloud droplets and rain water. Only a pH below this value is a sign of acidification. b) Dissolution of the gas SO2 in water shifts the equilibrium for SO2(g) to push more S(IV) into the aqueous phase than would otherwise be the case. The reactions are identical to those involving CO2 (exchange SO2 by CO2), and are given in the “Equation Sheet”. (1) SO2(g) SO2H2O (HSO2 = 1.23 Matm-1, given) (2) SO2(g)H2O HSO3

- + H+ (K2 = 1.310-2 M) (3) HSO3

- SO32- + H+ (K3 = 6.310-8 M)

7

The Ki are equilibrium constants. HSO2 is the ”traditional” Henry’s law constant for dissolution of S(IV) in water. c) [S(IV)(aq)] = [SO2H2O] + [HSO3

-] + [SO32-].

The ”effective” Henry’s law constant H*S(IV) for dissolution of S(IV) in water is defined as H*S(IV) = [S(IV)(aq)] / [SO2(g)]. The equilibrium reactions in a) give

2

2

22222

2

222 SOSO

SOSO pHOHSO

p

OHSO

gSO

OHSOH

H

pHK

H

OHSOKHSO

OHgSO

HHSOK SOSO 222222

322

32

2322332

33

23

32

H

pKKH

H

HSOKSO

HSO

HSOK SOSO

2322

2

2

322222

23322

1

)(

2

22

2

H

KK

H

KpH

H

pKKH

H

pHKpH

SOHSOOHSOIVS

SOSO

SOSOSOSOSOSO

The ”effective” Henry’s law constant H*S(IV) can thus be expressed as

2322

22

*)( 1

))(())((

2 H

KK

H

KH

p

aqIVS

gSO

aqIVSH SO

SOIVS

The factor

23221

H

KK

H

K

that follows the ”traditional” Henry’s law constant HSO2 explains how the pH of the aqueous solution affects the solubility of S(IV). Increased pH (lower [H+]) causes this factor to increase considerably, and more S(IV) can be dissolved in water. d) As HSO3

-. Oxidation in the atmosphere need therefore proceed via HSO3-.

e) If the solubility of gas phase S(IV) in water would have been determined only by the ”traditional” Henry’s law constant HSO2, then very little SO2(g) would have been dissolved in water, for instance in cloud droplets. As a result, nearly all oxidation of SO2 would have taken place in the gas phase (via OH). The lifetime for SO2(g) would have been considerably higher and the problems associated with acidification would have shifted further away from the source regions. Concentrations of SO2(g) would also have been higher close to the sources with negative consequences for human health.