Acid Rain

Given is an acid rain sample with the following composition:

0.02 mM HNO3 (nitric acid)
0.03 mM H2SO4 (sulfuric acid)
0.01 mM HCl (hydrochloric acid)
0.02 mM NH3 (ammonia)

We are looking for the pH of this solution for three different scenarios:

  1. without CO2 gas exchange
  2. with CO2 gas exchange (‘open system’)
  3. with CO2 gas exchange and in contact with calcite

Scenario 1

We start with pure H2O (button New) and switch to molar units (activate checkbox Mol). This and all subsequent tasks are easily solved with the reaction module (button Reac). Please enter the four components as depicted in the right screenshot.

Click on Start. The solution is immediately shown:

pH = 4.11

All other parameters are presented in the subsequent panels, some of them are summarized below. In this example, the three acids are partially compensated by NH3.

Scenario 2

All proceeds like in scenario 1, but before you click on Reac you have to select the option “Open CO2 System” in the main input window as shown here. The default value (pCO2 = 3.408) corresponds to the normal conditions in the atmosphere. We adopt it.

The calculation will show, that the pH of 4.11 does not change in the ‘open system’. Nonetheless, the solution now contains carbonate species, mainly in form of CO2(aq).

Scenario 3

All proceeds like in scenario 2, but before you click on Reac you should incorporate at least 1 mmol/L calcite. For this purpose activate the checkbox Minerals in the main input window and enter an initial amount of clacite, say 1 mmol/L. An example is given here.

The calculation will show that the dissolution of 0.56 mmol/L enhances the pH significantly:

pH:     4.11   =>   8.21

[Note: The initial amount of calcite remains in the mineral list untill a new input water is taken, either via the button New or button Open. Then it is set to zero again.]


Parameter OS 1 Units Scenario 1 Scenario 2 Scenario 3
(with CO2) (with Calcite)
pH - 4.11 4.11 8.21
sulfate S(6) mM 0.03 0.03 0.03
chloride mM 0.01 0.01 0.01
ammonium N(-3) mM 0.0139 0.0139 0.0133
nitrite N(3) mM 0.0243 0.0243 0.0267
nitrate N(5) mM 0.0018 0.0018 <0.0001
DIC C(4) mM - 0.0134 1.03
Ca mM - - 0.56
Selected species (ions):
NH4+ N(-3) mM 0.0139 0.0139 0.0123
NH3 N(-3) mM <0.0001 <0.0001 0.0011
HNO2 N(3) mM 0.0028 0.0028 <0.0001
NO2- N(3) mM 0.0214 0.0214 0.0267
NO3- N(5) mM 0.0018 0.0018 <0.0001
CO2(aq) C(4) mM - 0.0133 0.0133
HCO3- C(4) mM - <0.0001 1.00
CO3-2 C(4) mM - <0.0001 0.0087

1 oxidation state or valence state of the corresponding element

Nitrogen Speciation and Redox Reactions

The initial solution contains 0.04 mM nitrogen (= 0.02 mM HNO3 + 0.02 mM NH3) in two redox states: N(5) and N(-3). The total amount of nitrogen does not change in the reactions, but what is changed is the equilibrium distribution of the nitrogen species. Thus in the equilibrium state nitrogen occurs in three redox states: N(5), N(3), and N(-3).

The transition from one to another oxidation state is legitim. However, it can be allowed or forbitten by the user. The default case allows the transitions:

N(5)   <=>   N(3)
N(5)   <=>   N(-3)

But this can be changed by Settings in the top menu bar of the main window.

[last modified: 2013-08-05]