Lesson 3a: a gas/particle system at a fixed relative humidity

### Content

The partitioning of HNO3 and NH3 between vapour and condensed phases in a system containing HNO3 and NH4NO3 is examined. The model treatment of the equilibrium between the NH4+(aq) ion and and H+(aq) ions, and gas phase NH3(g), is explained.

### Entering the Data

The following example corresponds to 0.1 micromole of nitric acid and 0.1 micromole of ammonium nitrate partitioned somehow between particles and 1 m3 of atmosphere. These values are reasonable for locations polluted by ammonia and nitric acid. Fill in the input form in the other browser window as follows:

 1st Calculation Enter the values and select the options under the following headings: Ambient Conditions Relative humidity = 0.90 (i.e., 90%). Ionic Composition in Moles Hydrogen = 1E-7, Ammonium = 1E-7, Nitrate = 2E-7. Trace Gases There are no entries under this heading. Solid Phases There are no entries under this heading. Click on the "Run" button at the end of the page to do the calculation. Note:  the above should be entered on the "comprehensive" calculations page of Model III (http://www.aim.env.uea.ac.uk/aim/model3/model3b.php).

### Interpreting the Results

A liquid phase exists which contains the ions H+(aq), NH4+(aq) and NO3(aq), and very small amounts of OH(aq) and NH3(aq). There are no solids, but the gases HNO3(g) and NH3(g) are present in the vapour phase at partial pressures of 4.0099E-9 and 1.6339E-9 atm, respectively. The vapour phase contains, as HNO3 gas, almost 5× the amount of NO3 ion present in the aqueous phase – look at the "Moles" columns in the "Aqueous Phase" and "Gases" sections. This is not a coincidence: we designed the problem this way. About 30% of the total ammonium is in the particle phase and 70% is in the gas phase as NH3(g).

Notice that the amount of hydrogen ion that we entered in the system is 1E-7 moles, but the total moles of hydrogen in the result is equal to 1.68722E-7 moles, which is the sum of the 0.72473E-10 moles of aqueous H+ and 1.6865E-7 moles of HNO3 in the gas phase. This apparently exceeds the amount input. To explain this we look at the gas/solution equilibria, which are calculated as follows:

HNO3(g) = H+(aq) + NO3(aq)         (1)

NH3(g) + H+(aq) = NH4+(aq)         (2)

The second reaction is more easily understood as the combination of the following:

NH3(g) = NH3(aq)         (3a)

NH4+(aq) = NH3(aq) + H+(aq)         (3b)

Reaction (3a) is the Henry's law equilibrium of the ammonia gas, and reaction (3b) is the acid dissociation of NH4+(aq). Both equilibrium constants KH'(NH3) and Ka(NH4+) (which has a value of 5.6825E-10 mol kg-1 at 298.15 K) are defined in Lesson 1a. The equilibrium constant K for reaction (2) is given by:

K = aNH4+ / (pNH3 × aH+) = KH'(NH3) / Ka(NH4+)         (4)

We can see from the reactions above that a small amount of NH4+(aq) dissociates in solution to produce aqueous H+ ion and NH3(g) in the vapour phase. By combining the two reactions (3a) and (3b) into reaction (2) we eliminate the aqueous NH3. Thus, the species could have been omitted from this calculation and we would still have correctly predicted the ammonia partitioning into the vapour phase. You can verify that this is correct by repeating the calculation, but this time selecting the "Show Options" radio button (below where the moles of species are entered), and then checking the box to switch NH4+(aq) dissociation off.

To make sure the mole balances are correct first add up the total ammonia in the system. This is equal to 0.31281E-7 moles of NH4+(aq) plus 6.8719E-8 moles of NH3(g) and 0.76093E-15 moles of NH3(aq). The total is 1.0E-7 moles, which is the amount input. Next, consider the total hydrogen ion balance: this is equal to 0.72473E-10 moles of H+(aq) plus 1.6865E-7 moles of HNO3(g) minus 6.8719E-8 moles of NH3(g) = 1.0E-7 moles, which is the amount input. The reason we subtract the number of gas phase moles of NH3 in this balance is because reaction (3b) shows that for every molecule of NH3(g) produced by the dissociation of NH4+(aq) one molecule of additional H+(aq) ion is also generated.

The other equilibrium that can affect the hydrogen ion balance of the system is the dissociation of water itself:

H2O(l) = H+(aq) + OH(aq)         (5)

The value of the equilibrium constant is about 1E-14 mol2 kg-2 at 298.15 K. The activity of the hydroxide ion in the solution to be equal to aH2O × 1E-14 / aH+. In the present result the amount of OH(aq) is clearly negligible. It will only be important in an alkaline system or one very near neutral pH where the molality of the hydrogen ion is low (and the total hydrogen ion input by the user is at or close to zero).

Proceed to Lesson 3b to learn how the same system behaves at a lower relative humidity, or return to the main page for this lesson.