Lesson 1a: aqueous solutions at a fixed RH 
1^{st} Calculation 

Note: the above should be entered on the "simple" calculations page of Model III (http://www.aim.env.uea.ac.uk/aim/model3/model3a.php). 
System Pressure:  ambient pressure in atmospheres (usually fixed at 1 atm). 
Volume:  the assumed volume, in cubic metres, containing the aerosol mass. 
T:  ambient temperature in Kelvin. This is input by the user, except for Model III where temperature is fixed at 298.15 K. 
RH:  the ambient relative humidity, expressed as a fraction. This is either set by the user or determined from the partitioning of a known amount of water between the vapour and condensed phases. The equivalent partial pressure of water (pH2O) is also listed here. 
Scroll right so you can see the "Molality" and "Mole Frac." columns. The "Molality" column is the concentration in moles of species per kg of liquid water and the "Mole Fraction" column is the concentration in moles of species per total moles. Remember that this concentration and mole fraction are within the particles and refer only to the aqueous phase portion of the system. Consequently, the concentration (molality) of water is always 55.51 which is just the number of moles of water in 1 kg of water.
Returning now to the "Molality" column, the concentrations of ammonium (NH4^{+}) and nitrate (NO3^{−}) are both about 0.32 mol kg^{1}, which is the same as saying that the molality of ammonium nitrate in the particle is 0.32 mol kg^{1}.
The activity coefficients for the aqueous phase species given under "Act. Coeff." are on a mole fraction basis. The reference states for these activity coefficients (f), that is the compositions for which they are unity, are pure water for water itself, and infinite dilution in water for the ions. In this example the activity coefficient of water (1.001) is very close to one, but the activity coefficients of the ions are rather lower (0.6402). You will find that the activity coefficients of ions can deviate by quite large amounts from unity even in dilute solutions.
pH2O = fH2O × xH2O × pH2O^{o},
where fH2O is the mole fraction activity coefficient of water (1.001 in the "Act. Coeff." column), xH2O is the mole fraction of water (0.9886 in the "Mole Frac." column), and pH2O^{o} is the vapour pressure of pure water (0.031258 atm) which is calculated internally by the model and not shown on the output.
Remember that the relative humidity (RH) in a vapour phase is simply equal to the actual partial pressure of water divided by the equilibrium partial pressure of water over pure water at the system temperature. Thus, from the equation above:
RH = pH2O/pH2O^{o} = fH2O × xH2O,
which tells us that, for a system at equilibrium, the aqueous phase water activity is the same as the relative humidity.
The calculation of equilibrium partial pressures of NH3 and HNO3 from activities in the aqueous phase is straightforward:
pHNO3 = aH^{+} × aNO3^{−} / KH(HNO3)
pNH3 = aNH3 / KH'(NH3)
In the above equations KH and KH' are Henry's law constants and the activity of each aqueous species, denoted by the prefix a, is equal to its mole fraction multiplied by its activity coefficient. For example, the activity of H^{+} is equal to 0.2259E6 × 0.7246 = 1.6369E7.
Suppose we had not determined the concentrations of NH3 and H^{+} in the ammonium nitrate particle. In this case it would not be possible to obtain the individual equilibrium partial pressures of NH3 and pHNO3, but we can still calculate the equilibrium partial pressure product pNH3 × HNO3 (1.46E19 atm). The equilibrium constant for this product is related to the aqueous activities and Henry's law constants of the two gases, and the acid dissociation constant (Ka) of the ion NH4^{+}:
pNH3 × pHNO3 = aNH4^{+} × aNO3^{−} × Ka(NH4^{+}) / (KH'(NH3) × KH(HNO3)),
where:
Ka(NH4^{+}) = aH^{+} × aNH3 / aNH4^{+}
KH'(NH3) = aNH3 / pNH3
KH(HNO3) = aH^{+} × aNO3^{−} / pHNO3
The activity of each aqueous species, denoted by the prefix a, is equal to its mole fraction multiplied by its activity coefficient, as before. Notice how the activities of neither H^{+}(aq) nor NH3(aq) appear in the above equation for the partial pressure product, which is therefore independent of both quantities.
If you do the calculation described here on the "Comprehensive" input page for Model III (http://www.aim.env.uea.ac.uk/aim/model3/model3b.php), and select the option to switch off NH4^{+} dissociation, the value of the pressure product will be listed but not the equilibrium partial pressures of NH3 and HNO3.
2^{nd} Calculation 

Note: the above should be entered on the "simple" calculations page of Model III (http://www.aim.env.uea.ac.uk/aim/model3/model3a.php). 
The mole fraction activity coefficients of H^{+} and NO3^{−} are 0.7383. These are higher than for NH4^{+} and NO3^{−} at the same relative humidity, and most of the difference is because the HNO3 solution at 99% relative humidity is more dilute.
pH2O = fH2O × xH2O × pH2O^{o},
where fH2O is equal to 1.001 and xH2O (0.9894) is the mole fraction of water.
The output lists a partial pressure of 1.805E8 atm for HNO3, which is calculated from the Henry's law reaction given in Part 1:
KH(HNO3) = aH^{+} × aNO3^{−} / pHNO3
The mole fraction Henry's law constant is equal to 853.1 atm^{1} at 298.15 K, thus pHNO3 = (0.005316 × 0.7383)^{2} / 853.1 = 1.805E8 atm. The model output shows that this partial pressure is equivalent to 0.7379E6 moles in the system volume of 1 m^{3}.
You should now proceed to Lesson 1b, or return to the main page for this lesson.