Because the heterogeneous processes described in Chapter 4 are so fast, we can consider aqueous chemistry to be equilibrium chemistry for periods longer than a few seconds. Recall that the development and dissipation of clouds is typically tens of minutes, and sometimes more. So equilibrium aqueous chemistry should be very applicable to issues such as acid rain.
5.1 Aqueousphase
equilibria
In the aqueous phase, several chemical species rapidly dissociate into ions. The time constant achieving equilibrium between the adsorbed gas and its ions in solution is typically less than 1 ms. Thus, we can assume equilibrium for our calculations.
5.1.1 Water.
Water ionizes in to the hydrogen ion, H^{+}, and the hydroxide ion, OH^{}. In reality, H^{+} quickly attaches to another water to form the hydronium ion, H_{3}O^{+}.
At equilibrium, H_{2}O Û H^{+} + OH^{} (or H_{2}O + H_{2}O Û H_{3}O^{+} + OH^{1})
_{}
K_{w}^{’ }= 1.82 x 10^{16} M at 298 K. But [H_{2}O] = 55.5 M = 55.5 mole/L, which is much larger than the concentration of the ions and is essentially constant. We multiply by [H_{2}O] to get the ion product of water: K_{w}’[H_{2}O]
K_{w} = 1.0 x 10^{‑14} M^{2} at 298 K. In pure water, [H^{+}] = [OH^{‑}].
The pH of water is defined as pH =  log_{10}[H^{+}], and is 7.0 at 298 K. When pH is less than 7, the solution is acidic. When pH is greater than 7, the solution is basic.
5.1.2. Carbon
Dioxide / Water Equilibrium.
CO_{2} hydrolyzes in water to form CO_{2}·H_{2}O, which is sometimes called CO_{2}(aq). The following equilibria are established:
CO_{2}(g) + H_{2}O Û CO_{2}·H_{2}O
CO_{2}·H_{2}O Û H^{+} + HCO_{3}^{} (bicarbonate ion) K_{c1}
HCO_{3}^{} Û H^{+} + CO_{3}^{2} (carbonate ion) K_{c2}
Now that we have the equations, we can set up the equilibria. and then solve the simultaneous equations:
_{}
We want to get all of these in terms of things that we can actually measure.
_{}
We are interested in several quantities, including the total dissolved carbon:
_{}
More carbon is dissolved than we would expect from Henry’s Law alone. We can determine an effective Henry’s Law constant, which will be greater than the actual Henry Law Constant:
_{}
Note that the effective Henry’s Law Constant is greater than the Henry’s Law Constant. This difference results from the carbon that is dissolved in forms other than [CO_{2}∙H_{2}O], which is in equilibrium with the gasphase CO_{2}. We see that the effective Henry’s Law constant is dependent on pH through the hydronium ion concentration, [H^{+}] = [H_{3}O^{+}], as can be seen in following figure. The total amount of dissolved carbon is a strong function of pH, or [H^{+}]. The effective Henry’s Law Constant for a pH of 8 is more than 10,000 times larger than H_{CO2}.
In order to quantitative numbers, we must determine the ion concentrations. We can do this by assuming that the solution started out neutral and that the total positive ions exactly balance the total negative ions:
[H^{+}] = [OH^{}] + [HCO_{3}^{}] + 2[CO_{3}^{2}]
Using the equilibrium relations derived before, we can rewrite this expression in terms of equilibrium coefficients, the Henry’s Law Constant, and the partial pressure of CO_{2}:
_{}
For a temperature of 298 K and a CO_{2} mixing ratio of 350 ppmv, the pH of a solution of water and dissolved CO_{2} will be 5.6. Because atmospheric CO_{2} is always there, a pH of 5.6 is considered the pH of pure rainwater.
5.1.3. Sulfur dioxide / water equilibrium
After SO_{2} is absorbed into cloud drops (H(298 K) = 1.2 M atm^{1}), it ionizes (step 3) to form bisulfite (HSO_{3}^{}) and sulfite (SO_{3}^{2}) ions by the reactions:
(1) SO_{2}(g) + H_{2}O <==> SO_{2}^{.}H_{2}O
(2) SO_{2}^{.}H_{2}O <==> H^{+} + HSO_{3}^{ }; HSO_{3}^{ }is the bisulfite ion
(3) HSO_{3}^{} <==> H^{+} + SO_{3}^{2 }; SO_{3}^{2 }is the sulfite ion
(4) H_{2}O <==> H^{+} +
These reactions have the equilibrium constants:
H_{SO2}= [SO_{2}^{.}H_{2}O ]/p_{SO2}
K_{s1} = [H^{+}][HSO_{3}^{}]/[SO_{2}^{.}H_{2}O]
K_{s2} = [H^{+}][SO_{3}^{2 }]/[HSO_{3}^{}]
K_{w} = [H^{+}][OH^{}] ; This is called the ion product of water: K_{w} = 10^{14} M^{2 }@ 298 K.
The concentrations of the dissolved species are:
[SO_{2}^{.}H_{2}O] = H_{SO2} p_{SO2}
[HSO_{3}^{}] = K_{s1}[SO_{2}^{.}H_{2}O]/[H^{+}] = H_{SO2} K_{s1} p_{SO2}/[H^{+}]
[SO_{3}^{2 }] = K_{s2}[HSO_{3}^{}]/[H^{+}] = H_{SO2}K_{s1}K_{s2} p_{SO}^{2} / [H^{+}]^{2}
Let total sulfur that is in the IV oxidation state in the aqueous solution be written as:
[S(IV)] = [SO_{2}^{.}H_{2}O] + [HSO_{3}^{}] + [SO_{3}^{2 }]
Writing all of the concentrations as a combination of constants and [H^{+}]:
[S(IV)] = H_{SO2} p_{SO2} {1 + K_{s1}/[H^{+}] + K_{s1} K_{s2} /[H^{+}]^{2}}
Therefore, we can write total dissolved sulfur as:
[S(IV)] = H_{S(IV)}* p_{SO2}
The effective Henry's Law coefficient is:
H_{S(IV)}* = H_{SO2 }{1 + K_{s1}/[H^{+}] + K_{s1} K_{s2} /[H^{+}]^{2}}
Note that H*_{S(IV)} is always greater than H_{S(IV)}. Note also that the total dissolved sulfur is dependent on the pH.
The fraction of S(IV) that is SO_{2}∙H_{2}O, HSO_{3}^{}, and SO_{3}^{2} is given in the next figure. Note that the bisulfite ion is the most important for much of the range of cloud water. Recall that atmospheric CO_{2} creates cloud water with a pH of 5.6. The form of the sulfur becomes quite important when we think about reactions of sulfur compounds in the aqueous phase.
To find the pH of the solution, we can assume once again assume electroneutrality. Thus,
[H^{+}] = [OH^{}] + [HSO_{3}^{}] + 2 [ SO_{3}^{2 }]
[H^{+}] = K_{w}/[H^{+}] + H_{SO2} K_{s1} p_{SO2}/[H^{+}] + 2 H_{SO2 }K_{s1 }K_{s2} p_{SO2} / [H^{+}]^{2}
or
[H^{+}]^{3} – (K_{w} + H_{SO2} K_{s1} p_{SO2}) [H^{+}]  2 H_{SO2 }K_{s1 }K_{s2} p_{O2} = 0
By looking at the values for each of the terms, we can simplify this expression.
5.1.4. Ammonia /water equilibrium.
Ammonia is the major basic atmospheric constituent that enters aqueous solutions. It equilibrium equations are:
NH_{3} + H_{2}O Û NH_{3} ·H_{2}O
NH_{3} ·H_{2}O
Û
NH_{4}^{+} +
where NH_{4}^{+} is the ammonium ion. The equilibrium constants has the following equations and values:
_{}
With these two equilibrium equations, we can find the ammonium ion concentration:
_{}
Thus, the total dissolved ammonia [NH_{3}^{T}] is given by:
_{}_{}
We were able to make the last approximation because for K_{a1}[H^{+}]/K_{w} ~ 1.5x10^{5} 10^{7} / 10^{14} >> 1. Thus, for pH less than 8, [NH_{3}^{T}] ~ [NH_{4}^{+}]. For pH < 5, essentially all ammonia will be dissolved in the cloud water.
5.1.5. Nitric acid / water equilibrium.
After sulfuric acid, nitric acid is the most important acid that comes from pollution. It is through nitric acid that the oxidation processes that create ozone are linked to the processes that cause acid rain and aerosol formation. The equilibrium equations are:
HNO_{3} + H_{2}O Û HNO_{3}×H_{2}O
HNO_{3}×H_{2}O Û NO_{3}^{} + H^{+}
The equilibrium constants and their values are:
_{}
With these two equilibrium constants, we can find the nitrate concentration:
_{}
so that the total dissolved nitric acid is:
_{}
K_{n1} = 15.4 M at 298 K. Thus, K_{n1} / [H^{+}] >> 1. Thus [NO_{3}^{}] >> [HNO_{3}×H_{2}O]
The effective Henry’s Law constant is:
_{}
For a pH of 5, so that [H^{+}] = 10^{5} M, the effective Henry’s Law Constant is ~3x10^{13} M atm^{1}. Almost all of the nitric acid will be in the aqueous phase.
5.1.6. Hydrogen Peroxide / water equilibrium.
The equilibrium equations for hydrogen peroxide are:
H_{2}O_{2}(g) +H_{2}O Û H_{2}O_{2}×H_{2}O
H_{2}O_{2}×H_{2}O Û HO_{2}^{}_{ } + H^{+}
The partitioning between H_{2}O_{2}×H_{2}O and HO_{2}^{} is given by:
_{}
We see that almost all dissolved H_{2}O_{2} is in the form of H_{2}O_{2}×H_{2}O for pH values that are typical for the atmosphere (i.e., less than 5.6).
5.1.7. Ozone.
Ozone is not very soluble in water. It exists essentially entirely in the aqueous form of O_{3}∙H_{2}O.
5.1.8. List of constants.
The following table contains the Henry’s Law Constants and the Equilibrium Constants for the atmospheric constituents and equilibria discussed above. These equilibria are temperature dependent, but we will assume that the temperature is always near 298 K. for our discussions.
Selected Henry’s Law Constants and Equilibrium Constants at 298 K. 

Equilibrium 
H (mole L^{1} atm^{1}) 
K (M) 
H_{2}O ↔ H^{+} + OH^{}


1.0 x 10^{14} 
CO_{2} + H_{2}O ↔ CO_{2}∙H_{2}O 
3.4 x 10^{2} 

CO_{2}∙H_{2}O ↔ H^{+} + HCO_{3}^{} 

4.3 x 10^{7} 
HCO_{3}^{} ↔ H^{+} + CO_{3}^{2} 

4.7 x 10^{11} 
SO_{2} + H_{2}O ↔ SO_{2}∙H_{2}O 
1.23 

SO_{2}∙H_{2}O ↔ H^{+} + HSO_{3}^{} 

1.32 x 10^{2} 
HSO_{3}^{} ↔ H^{+} + SO_{3}^{2} 

6.42 x 10^{8} 
NH_{3} + H_{2}O ↔ NH_{3}∙H_{2}O 
62 

NH_{3}∙H_{2}O ↔ NH_{4}^{+} + OH^{}^{} 

1.8 x 10^{5} 
HNO_{3} + H_{2}O ↔ HNO_{3}∙H_{2}O 
2.1 x 10^{5} 

HNO_{3}∙H_{2}O ↔ H^{+} + NO_{3}^{}^{} 

15.4 
H_{2}O_{2} + H_{2}O ↔ H_{2}O_{2}∙H_{2}O 
7.45 x 10^{4} 

H_{2}O_{2}∙H_{2}O ↔ H^{+} + HO_{2}^{} 

2.5 x 10^{12} 
O_{3} + H_{2}O ↔
O_{3}∙H_{2}O 
0.011 

5.2. Aqueous sulfur chemistry
The equilibrium processes descried above are reversible. If the partial pressure of SO_{2} decreases for some reason, like dilution or surface deposition, the sulfur in the aqueous phase will remain in equilibrium with the partial pressure and will thus decrease. The situation is quite different for more oxidized forms of sulfur, sulfates, the S(VI) compounds. These have a very low vapor pressure and remain for the most part in the liquid phase. Thus, the reactions that take S(IV) to S(VI) are essential for the production of acid rain.
S(IV) can be oxidized to S(VI) by a number of chemical species, including O_{3}, H_{2}O_{2}, dissolved oxygen catalyzed by metals, and dissolved NO_{2}.
Application to the atmosphere, which has clouds and fogs.
How do we apply liquidphase chemistry to the atmosphere? How do we compare the rates of processes that occur in the liquid phase to those that occur in the gasphase? Recall that the reaction rate is proportional to the concentration of the reactants. In the gas phase, we usually use the units molecules cm^{3}, while in the aqueous phase, we use the units M = moles/liter. Converting between these two can be a headache. However, for reactions or equilibria, it is the concentrations that matter. We can compare the gasphase and aqueous phase rates by considering the rate in a volume of air. To calculate the rate for the aqueous phase, we need to know the volume of liquid that is in that volume of air. This value is usually called the liquid water content (LWC). In cloud physics, this is usually given units of g m^{3}. There is another value that is also used: the cloud water mixing ratio, which is w_{L} = (volume of liquid)/(volume of air). The two are related. Since for water 1 m^{3} = 10^{6} g, the conversion between the two is:
LWC = 10^{6} w_{L}
What are typical values for w_{L} for different conditions? They are the following:
wet aerosol: 10^{12}  10^{9}
fog : 10^{9}  10^{7}
cloud : 5x10^{8}  3x10^{6} , or 0.05 – 3 g m^{3}.
Assume that a solute A is neither created nor destroyed. The distribution factor, f_{A}, is the ratio of the aqueous phase concentration and the gasphase concentration, in the same units (g/(Lof air)), or other units will do.
f_{A} = (mass in liquid / volume of air)/(mass in air / volume of air) = concentration in the liquid in the volume of air / concentration in the gas phase.
We will denote the concentration in the liquid in the volume of air with C_{A}(aq) and the concentration in the gasphase with C_{A}(g). Note that [A(aq)] is the concentration of A in the liquid in terms of moles/L(liquid).
So, C_{A}(aq) = [A(aq)] w_{L}, which has the units (moles/L(aq)) x (L(aq)/L(air)
C_{A}(g) is different from [A(g)] because it must have units of moles/L(air) if f_{A} is to be unitless.
Now, we can transform these relations using the Ideal Gas Law, to get the
gasphase amount in terms of the partial pressure,
p_{A} = C_{A}(g) RT implies the C_{A}(g) = p_{A} / RT.
We should make sure that we use R = 0.08205 atm L mol^{1} K^{1} and p_{A} in units of atm.
Further, we can transform the liquid phase concentration by using Henry’s Law:
C_{A}(aq) = [A(aq)] w_{L} = H_{A} p_{A} w_{L}
We will need to use the appropriate Henry’s Law Constant, which in many cases will be the effective Henry’s Law Constant.
Putting all of this together gives us the ratio of A that is in the liquid phase to A in the gas phase:
What we are really interested in is the fraction of A that is in the liquid phase, denoted as X^{A}_{aq}:
For LWC = 10^{9} l m^{3}, (R T LWC)^{1} = 4x10^{4} M atm^{1}. Thus, chemical species that have a Henry's Law constant, or effective Henry's Law constant much less than 4x10^{4} M atm^{1} will be mostly in the gasphase, as in the figure. As a result, in the absence of oxidation, most SO_{2 }will be in the gasphase and most HNO_{3} will be in the liquid.
So, now we can put the aqueous phase reactions in the context of the atmosphere.
For a dissolved species A,
S(IV) + A(aq) ® products
In the liquid phase, the rate equation for this reaction is simply:
The units for this reaction rate are M s^{1} = moles L^{1} s^{1}. To get this reaction rate in terms of liters of air, we must multiply by w_{L}. This will give us the reaction rate for A and S(IV) due to liquid phase the liquid phase reaction.
The reaction rate coefficient should have units of L mole^{1} s^{1}; the concentrations are M = moles L(aq)^{1}; LWC is g m^{3}; and w_{L} is (moles L(aq)^{1})/(moles L(g)^{1}).
If we want to express these equations in terms of per cent change per hour, we simply divide the third expression by the mixing ratio of SO_{2} and multiply by 100:
The lifetime is simply this 100 divided by this last equation, in units of hours.
We can also get the fraction converted per unit time. The fraction of sulfur converted per unit time is the conversion rate divided by the total amount of sulfur not yet converted, in both the gas and the liquid phase. The fractional rate is thus:
fraction converted per unit time = {d([S(IV)])/dt} / {p_{SO2}/RT + H*_{SO2} p_{SO2} w_{L}}
Now let’s look specifically at the oxidation of S(IV) to S(VI). We can see the reaction rates for typical amounts of different oxidants. The liquidphase oxidants, H_{2}O_{2} and O_{3}, are largely created in the gasphase and absorbed into the liquid. Note that O_{3} and H_{2}O_{2} are both formed by processes involving HO_{x} radicals. We should expect that the greatest oxidation will occur during the day, when oxidant concentrations are largest, and will be greater in the summer than the winter for the same reason.
Because the oxidation by HOOH is by far the largest for anything but very acidic or basic solutions, we will focus on it.
The reaction pathway is given by the equations:
HSO_{3}^{}+H_{2}O_{2} <==> SO_{2}OOH^{} + H_{2}O
SO_{2}OOH^{} + H^{+}<==> H_{2}SO_{4}
The specific rate equation is from Hofmann and Calvert (1985):
d([S(IV)])/dt =  {k_{HOOH+S(IV)}[H^{+}][H_{2}O_{2}(aq)][HSO_{3}^{}] }/{1 + K_{HOOH}[H^{+}]}
where k_{HOOH+S(IV)} = 7.5x10^{7} M s^{1}, and K_{HOOH} = 13 M^{1}. Note that we must determine [H^{+}], [H_{2}O_{2}(aq)] = H_{H2O2} p_{H2O2}. Note that we cannot use [S(IV)](aq), but must instead use [HSO_{3}^{}], which can be calculated from the equilibrium conditions.
This reaction rate is essentially independent of pH when we multiply all the terms together. The reason for this weak dependence is that k_{HOOH+S(IV)} is essentially independent of [H^{+}], {H_{2}O_{2}(aq)] is independent of [H^{+}], [HSO_{3}^{}] depends on 1/[H^{+}], and K_{HOOH}[H^{+}] << 1.
For p_{H2O2} = 1 ppb,
d[S(IV)]/dt = 300 mM h^{1} (ppbSO_{2})^{1} (700% SO_{2}(g) h^{1} (g water / m^{3} air)^{1}
The reactions on page 367 show that the immediate product is H_{2}SO_{4}, sulfuric acid.
What happens to it? It is ionized:
H_{2}SO_{4}(aq) Û HSO_{4}^{}_{ } + H^{+} K_{svi1} = 1000;
HSO_{4}^{} Û SO_{4}^{ 2} + H^{+} K_{svi2} = 1.02x10^{2}
H_{2}SO_{4}(g) Û H_{2}SO_{4}(aq) remains in liquidphase.
Therefore, we can determine the dominant forms of S(VI):
Open and closed systems.
1. Open system.
1. Gas partial pressures are maintained at fixed values by resupply from new air or from photochemistry.
2. Not a very realistic condition.
2. Closed system.
Let us consider what happens in a closed system.
In an open system, [H_{2}O_{2}(aq)] = H_{H2O2 }p^{0}_{H2O2}. In this case, the aqueous phase H_{2}O_{2} is independent of LWC.
In a closed system, initial H_{2}O_{2} is split between the gas and aqueous phases, as the cloud forms:
[H_{2}O_{2}]_{total} = (p_{H2O2}/RT) + [H_{2}O_{2}(aq)]_{closed} w_{L}
where recall that w_{L} has units of L liquid / L air and R = 0.082 (atm L)/(mole K).
If we are in equilibrium, then Henry’s Law holds and:
[H_{2}O_{2}(aq)]_{closed} = H_{H2O2} p_{H2O2}
If initially there was no cloud, then the conservation equation for H_{2}O_{2} becomes:
p^{o}_{H2O2} = p_{H2O2} + [H_{2}O_{2}(aq)]_{closed} w_{L} RT
= [H_{2}O_{2}(aq)] {(1/H_{H2O2}) +w_{L} RT}
Combining these equations, we get:
[H_{2}O_{2}(aq)]_{closed} = (H_{H2O2} p^{o}_{H2O2}) / (1 + H_{H2O2} w_{L} RT)
Note that [H_{2}O_{2}(aq)] decreases as LWC or w_{L} increases. More water means that more H_{2}O_{2} can be accommodated in the aqueous phase. The open system is represented by the dashed lines. For the closed system, because p_{H2O2} is proportional to [H_{2}O_{2}(aq)], p_{H2O2} drops as [H_{2}O_{2}(aq)] drops. In addition, because p_{H2O2} is proportional to [H_{2}O_{2}(aq)], the fraction of H_{2}O_{2} that is in the aqeuous form [H_{2}O_{2}(aq)] increases as the LWC increases.
More sulfate is produced in the open system than the closed system, as in the figure below. This makes sense: the oxidants and SO_{2} are being constantly replaced and is thus able to continue the conversion of S(IV) to S(VI).
Low solubility gases, like ozone, maintain aqueous phase concentrations that are similar to that for open systems. For example, if w_{L} = 10^{6}, T = 300 K, then H_{O3} w_{L} RT = 0.0113 10^{6} 0.082 300 = 2.8 x 10^{7}, which is much smaller than 1 and can be ignored.
[O_{3}(aq)]_{closed} = (H_{O3} p^{o}_{O3}) / (1 + H_{O3} w_{L} RT) ~ (H_{O3} p^{o}_{O3})
We can do the same thing for dissociating species, such as HNO_{3}:
[HNO_{3}(aq)] +[NO_{3}^{}] = (H^{*}_{HNO3} p_{HNO3}) / (1 + H^{*}_{HNO3} w_{L} RT)
where
H^{*}_{HNO3} = H_{HNO3} {1 + K_{n1} / [H^{+}]}
Consider a cloud forming in clear air that contains SO_{2} and H_{2}O_{2}.
Global transport models suggest that gasphase oxidation and liquid phase oxidation are about equally important for global SO_{2} emissions, even though the rates of oxidation in cloud water are several 10’s of percent per hour, while gasphase oxidation is at most a few percent per hour. While clouds cover roughly half of Earth’s surface at any given time, they occupy a much smaller volume of the troposphere at any given time.
Once the sulfate aerosol is in cloud drops, it can come down as precipitation. The figure below depicts the processes that occur in a rain storm. The photochemical constituents are drawn up into the storm in the low level inflow. Aqueous phase conversion of SO_{2} to sulfate occurs. Rain out deposits some of this sulfate on Earth’s surface. What is not rained out can be delivered to the global free troposphere by the upper level outflow. This sulfate can then travel great distances as aerosol before it is brought down to lower levels by subsidence of it air parcel, where it can be deposited by dry or wet processes.
The pattern of sulfate deposition for the Eastern United States is shown in the following figure.
The figure below shows the annual variation of deposition for sites in the Eastern United States. There is a clear peak in deposition during the summer months. This peak is not due to increases in SO_{2} and NO emissions, which are roughly constant over the whole year. What could be the cause of this summertime increase?