Consider a raindrop in a polluted atmosphere with an SO_2 mixing ratio of 300 pp

Question

Consider a raindrop in a polluted atmosphere with an SO_2 mixing ratio of 300 ppb. What is the concentration of aqueous SO_2 in the droplet. If we assume there is no oxidation chemistry what is the total amount of dissolved sulfur and the pH of the droplet. If the atmospheric H_2O_2 mixing ratio is 100 ppb would you expect the actual concentration of dissolved sulfur to be lower or higher. Use the following: Total pressure: 101324 Pa Henry's law constant for SO_2 in equilibrium with water is 1.2 times 10^-5 ML -1 Pa_1 (aqueous SO_2) H_2O. SO_2 H^+ + HSO_3- K_a = 1.7 times 10^-2

Explanation / Answer

The SO2 , HO2 and O3 concentrations in cloud droplets and precipitation are assumed to be in equilibrium with the local gas phase concentration and are carried as implicit fields, i.e., when required in the source terms, they are computed as a function of the gas phase concentrations. these liquid phase concentrations are calculated using Henry's law

[M] = KHpM (mol L-1)

where pM is the partial pressure of species M

[M] = 1.7*10-2 *101324 = 1.7*103 = 9.51 ppb

the equilibrium hydrogen ion concentration in cloud droplets and precipitation is calculated from the assumption that HSO3- is the domimant form of aqueous S in cloud water and precipitation , and using charge balance,

[H+] = 0.5[2[SO4-2]-[NH4+]+((2[SO4-2]- [NH4+])2+ 4Ks + 4Kw)0.5]

Kh = 3.013(105) e3168.6/T

After SO2 is absorbed into cloud drops (H(298 K) = 1.2 M atm-1), it ionizes (step 3) to form bisulfite (HSO3-) and sulfite (SO32-) ions by the reactions have the equilibrium constants:

HSO2= [SO2.H2O ]/pSO2

Ks1 = [H+][HSO3-]/[SO2.H2O]

Ks2 = [H+][SO32- ]/[HSO3-]

Kw = [H+][OH-] ; This is called the ion product of water: Kw = 10-14 M2 @ 298 K.

Let total sulfur that is in the IV oxidation state in the aqueous solution be written as:

[S(IV)] = [SO2.H2O] + [HSO3-] + [SO32- ]

Writing all of the concentrations as a combination of constants and [H+]:

[S(IV)] = HSO2 pSO2 {1 + Ks1/[H+] + Ks1 Ks2 /[H+]2}

Therefore, we can write total dissolved sulfur as:

[S(IV)] = HS(IV)* pSO2

The effective Henry's Law coefficient is:

HS(IV)* = HSO2 {1 + Ks1/[H+] + Ks1 Ks2 /[H+]2}

the total dissolved sulfur is dependent on the pH.

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