Meteo 532 – Atmospheric chemistry

Problem Set 5 Solutions

Problem 5.1. Aqueous phase sulfur chemistry.

Consider a situation in which SO₂ has been drawn up into cloudy air and has come to equilibrium with S(IV) in the liquid cloud drops. Assume that the mixing ratio of SO₂ in the cloudy air in equilibrium is 50 ppbv and that the liquid water content is 1g m^-3.

a.) What is the pH of the cloud water? When you get the expression for [H⁺], look to see which term(s) dominates before attempting to determine the pH.

Assume electroneutrality:

[H⁺] = [OH^-] + [HSO₃^-] + 2 [SO₃^2-].

Or, using the derivation in the notes:

[H⁺] = K_w / [H⁺] + {H_SO2K_s1p_SO2} / [H⁺] + 2 H_SO2K_s1K_s2 P_SO2 / [H⁺]²

[H⁺]³ – { K_w + H_SO2K_s1p_SO2}[H⁺] + H_SO2K_s1K_s2 P_SO2 = 0.

We can solve this exactly, or we can look at the value of the different terms. Typically, we might expect [H⁺] to be about 10^-5. If we assume this, then we can determine the order of magnitude of each term:

10^-15 - {10^-14 + 1.24 1.3x10^-2 50x10^-9}10^-5 - {2 1.24 1.3 x10^-2 6x10^-8 50x10^-9} = 0

10^-15 – 8.06x10^-15 – 9.7x10^-17 ~ 0

The first two terms are the most significant by an order of magnitude. We drop the last term and get the expression for [H⁺]:

[H⁺] = { H_SO2K_s1p_SO2}^1/2 = {8.06x10^-10}^1/2 = 2.83x10^-5

Thus, pH = - log₁₀ [H⁺] = 4.55

b.) What is the total amount of S(IV) in the cloud water?

[S(IV)] = [SO₂∙H₂O] + [HSO₃^-] + 2 [SO₃^2-].

= H_SO2 p_SO2 + {H_SO2K_s1p_SO2} / [H⁺] + H_SO2K_s1K_s2 P_SO2 / [H⁺]²

= 1.24 50x10^-9 + 8.06x10^-10 / 2.83x10^-5 + 9.7x10^-17 / {2.83x10^-5}²

= 2.86x10^-5 M

c.) What fraction of sulfur remains in the gas-phase as SO₂ for T = 0^oC, p = 1010 hPa (1 atm), and LWC = 1 g m^-3? From this answer, what can you say about the solubility of SO₂?

The total S(IV) is partitioned between the gas and liquid phases. The greatest difficulty is getting the two in the same units. We choose moles liter of air.

c_gas = p_SO2 / RT = 50x10^-9 / {0.08205∙273} = 2.23x10^-9 moles / L(air)

c_liq = [S(IV)] 10^-6 LWC = 2.86x10^-5 10^-6 = 2.86x10^-11 moles / L(air).

f_liq = 2.86x10^-11 / {2.86x10^-11 + 2.23x10^-9} = 0.013, or about 1.3%

Very little of the S(IV) is in the liquid phase.

For this problem, use the constants:

H_s = 1.24 mol L^-1 atm^-1

K_s1 = 1.3 x 10^-2 mol L^-1

K_s2 = 6.0 x 10^-8 mol L^-1

K_w = 1.0 x 10^-14 mol² L^-2

H_o = 0.011 mol L^-1 atm^-1

Problem 5.2. Conversion of S(IV) to S(VI).

Assume that the mixing ratios for O₃, H₂O₂, and OH are 50 ppbv, 1 ppbv, 0.1 pptv respectively for the conditions in problem 1.

The rate equation for the reaction of S(IV) with aqueous O₃ is:

d[S(IV)]/dt = - (k_o[SO₂∙H₂O]+k₁[HSO₃^-]+k₂[SO₃^2-])[O₃∙H₂O]

where k₀=2.4x10⁴ M^-1 s^-1; k₁=3.7x10⁵ M^-1 s^-1; and k₂=1.5x10⁹ M^-1 s^-1.

a.) Calculate the gas-phase and liquid-phase rates of conversion for SO₂, assuming that the gas-phase concentrations of OH, H₂O₂, and O₃ are maintained at the specified values (an open system).

gas phase:

[M] = 2.69x10¹⁹ 273/273}{1010/1013} = 2.68x10¹⁹ molecules cm^-3.

{d[S(IV)]/dt}_gas = -k_OH+SO2 [SO₂][OH] = 9.85x10^-13 0.1x10^-12 50x10^-9 {2.68x19¹⁹}²

= 3.54x10⁶ molecule cm^-3 s^-1

This fractional removal rate is ~1% per hour.

liquid phase

{d[S(IV)]/dt}_liq = -10^-6 LWC({(k_H2O2+S(IV)[H⁺][H₂O₂∙H₂O][HSO₃^-])/(1+K_HOOH[H⁺])} +

{(k₀[SO₂∙H₂O] + k₁[HSO₃^-] + k₂[SO₃^2-])[O₃∙H₂O]}

= -10^-6 1 ({(7.5x10⁷∙2.83x10^-5∙7.45x10⁴∙10^-9∙2.86x10^-5)/(1+13∙2.83x10^-5)} +

{(2.4x10⁴∙6.2x10^-8 + 3.7x10⁵∙2.86x10^-5 + 1.5x10⁹∙6.1x10^-8)∙0.011∙50x10^-9}

= -10^-6∙({4.52x10^-6}+{5.56x10^-8}) = 4.57x10^-12 moles / L(air) = 2.75x10⁹ molecules cm^-3

Since we started with 50 ppbv of SO₂ (1.3x10¹² molecules cm^-3), the initial liquid phase removal rate is 750% per hour.

b.) Compare the gas-phase and liquid-phase rates.

The liquid phase removal rate is almost a thousand times greater than the gas-phase rate for these conditions.

c.) Assume now that H₂O₂ starts out at 1 ppbv, but that the only source of H₂O₂ is by the reaction:

HO₂ + HO₂ ® H₂O₂ + O₂,

where the mixing ratio of HO₂ is 10 pptv. Using calculations and/or speculations to support your views, say what you think will happen to the oxidation rate of SO₂ with time.

Recall that one H₂O₂ molecule is consumed for every SO₂ molecule that is oxidized.

The production rate of HOOH is: P(HOOH) = k_HO2+HO2[HO₂]² = 2.07x10^-12 {10^-11∙2.68x10¹⁹}² = 1.45x10⁵ molecules cm^-3 s^-1. This rate is far smaller than the destruction rate of S(IV) by HOOH. Thus, the rate of S(IV) destruction will drop rapidly – in a matter of 10 or so minutes, and will remain at about ½ % of the rate of S(IV) destruction by ozone (3.3x10⁷ molecules cm^-3 s^-1).

Problem 5.3 True acidity of rain.

The pH of rain reported by monitoring agencies is based on analysis of rain samples that are collected weekly in buckets. The weekly collection schedule is fine for HNO₃ and H₂SO₄, which do not degrade; however, formic acid (HCOOH) is rapidly consumed by bacteria in the buckets and therefore escapes analysis. The Henry’s Law and acid dissociation equilibria for HCOOH are:

HCOOH(g) Û HCOOH(aq) H_hcooh = 3.7 x 10³ M atm^-2

HCOOH(aq) Û HCOO^‑ + H⁺ K₁ = 1.8 x 10^-4 M

If the monitoring agency reports a rainwater pH of 4.7, calculate the true pH of the rain. Assume 1 ppbv HCOOH in the atmosphere, a typical value for the eastern United States.

We assume that the pH due to sulfate and nitrate do not change as the formic acid is eaten up. The ion balance for formic acid is:

[HCOO^-] = K₁ H_HCOOH p_HCOOH / [H⁺]_true

So the true acidity is equal to the sum of the acidity when the formic acid was absent and the acidity lost when the formic acid was eaten up:

[H⁺]_true = [H⁺]_meas + [HCOO^-] = 2.0x10^-5 + 3.7x10³∙1.8x10^-4∙10^-9 /[H⁺]_true = 2.0x10^-5 + 6.66x10^-10 / [H⁺]_true

[H⁺]_true² – 2x10^-5 [H⁺]_true – 6.7x10^-10 = 0

or

[H⁺]_true = -{b ± (b² – 4ac)^1/2 }/2a = -{-2x10^-5 ± (4x10^-10 + 4∙1∙6.66x10^-10)^1/2}/2 = 3.77x10^-5 M

The true pH is 4.42, about 0.3 or 7% lower than the measured pH.

Problem 5.4. Aqueous sulfur and nitrogen chemistry.

An urban pollution plume comes into equilibrium with clouds that have a liquid water content of 1 g m^-3. The initial equilibrium values for SO₂ is 20 ppbv and for HNO₃ is 0.01 pptv. The atmospheric pressure is 1013 hPa and the temperature is 298 K. Assume no CO₂. We assume that all the HNO₃ in the aqueous phase is in the form of nitrate (NO₃^-), and that H_HNO3 = 2.1x10⁵ M/atm and that the equilibrium constant between HNO₃·H₂O Û NO₃^- + H⁺ is 15.4 M.

a) What is the equation for the ion balance?

[H⁺] = [OH^-] + [HSO₃^-] + 2∙[SO₃^2-] + [ NO₃^-]

b) What is the initial pH of the clouds?

[H⁺] = K_w / [H⁺] + {H_SO2K_s1p_SO2} / [H⁺] + 2 H_SO2K_s1K_s2 P_SO2 / [H⁺]² + K_n1 H_HNO3∙p_HNO3 / [H⁺]

[H⁺]³ – { K_w + H_SO2K_s1p_SO2}[H⁺] + H_SO2K_s1K_s2 P_SO2 = 0.

We can solve this exactly, or we can look at the value of the different terms. Typically, we might expect [H⁺] to be about 10^-5. If we assume this, then we can determine the order of magnitude of each term:

10^-15 - {10^-14 + 1.24∙1.3x10^-2∙50x10^-9 + 15.4∙2.1x10⁵∙10^-14}10^-5 - {2∙1.24∙1.3 x10^-2∙6x10^-8∙50x10^-9} = 0

10^-15 – {8.06x10^-15 + 3.23x10^-13}– 9.7x10^-17 ~ 0

The first two terms are the most significant by an order of magnitude. We drop the last term and get the expression for [H⁺]:

[H⁺] = { H_SO2K_s1p_SO2 + K_n1 H_HNO3∙p_HNO3 }^1/2 = {8.06x10^-10 + 3.23x10^-8}^1/2 = 1.82x10^-4

Thus, pH = - log₁₀ [H⁺] = 3.74

c) What fraction of the acidity is due to S(IV) and what fraction is due to NO₃^-?

The fraction due to HNO₃ is: f₁ = 1.80x10^-4 / (1.82x10^-4) = 0.99, or 99%

Assume that all the S(IV) in both the aqueous and gas phase is reacted to S(VI) and all the NO_x is reacted to HNO₃. Assume that all the S(VI) is in the aqueous phase. Note that the amount of NO₃^- depends on [H⁺].

d) What is the new pH of the clouds?

From the notes, we know that essentially all the S(VI) is in the form of sulfate, SO₄^2-. Also, the total amount of sulfur in the liquid must be equal to the initial amount of SO₂ in the gas phase plus the few percent that are in the aqueous phase.

[SO₄^2-] = {p_SO2 / RT} / 10^-6 LWC

= 20x10^-9 / {0.08205∙273}/10^-6 = 8.93x10^-4 M in the liquid

This is more than 10 times larger than when all the nitric acid is in the liquid as nitrate. Thus, we need to recalculate the nitrate concentration assuming that all the pH is controlled by sulfate. We make this adjustment by comparing the partitioning before and after the sulfate is formed.

p_HNO3(new)∙{1 /RT + 10^-6∙LWC ∙K_n1 H_HNO3 / [H⁺](new)) = p_HNO3(old)∙{1/RT + 10^-6∙LWC ∙K_n1 H_HNO3 / [H⁺](old))

p_HNO3(new)∙{1/0.082∙273 + 10^-6∙3.23x10⁶ /8.93x10^-4) = 10^-14∙{1/0.082∙273 + 10^-6∙3.23x10⁶ /1.82x10^-4)

p_HNO3 = 4.91x10^-14 atm

Thus, the new amount of nitrate is p_HNO3(old)∙{1/RT + 10^-6∙LWC ∙K_n1 H_HNO3 / [H⁺](old)) - p_HNO3(new)∙(1 /RT) =

10^-14∙{10⁶/0.082∙273 + 3.23x10⁶ /1.82x10^-4) – 10⁶ ∙ 4.91x10^-14 /0.082∙273 =

-K_n1 H_HNO3 p_HNO3(new)/ [H⁺](new)) = 15.4∙2.1x10⁵∙4.91x10^-14/ 8.93x10^-4 = 1.78x10^-4 M

Thus,

[H⁺] = [OH^-] + 2∙[SO₄^2-] + [NO₃^-]

~ 2∙8.93x10^-4 + 1.78x10^-4 = 1.96x10^-3 M

The pH is 2.71

e) What fraction of the acidity is due to S(VI) and which fraction is due to NO₃^-?

The fraction due to nitrate is 1.78x10^-4/1.96x10^-3 = 0.09 (about 9%)

Is [NO₃^-] now different from its initial value?

[NO₃^-] changes a little while p_HNO3 increases by a factor of 5. However, there is so little of it to begin with that the change in nitrate is negligible.