Lesson Roadmap
To Read (Optional)	Chapter 2, Principles and Applications of Aquatic Chemistry, by F. M. M. Morel and J. Hering, 1993 Langmuir, 1997, Chapter 1 and 3 (Langmuir et al., 1997)
To Do	Finish take home practice 1 Homework questions 1 and 2

1.0 Aqueous complexation reactions

Example 1.1: A closed carbonate system. Imagine we have a closed bottle with clean water except with a head space filled with ${CO}_{2}$ gas. The system therefore only has carbonate species in the water phase, the pertinent reactions are as follows:

$\begin{aligned} H_{2} {CO}_{3}^{0} & \Leftrightarrow H^{+} + {HCO}_{3}^{-} K_{a 1} = \frac{a_{H^{+}} a_{{HCO}_{3}^{-}}}{a_{H_{2} {CO}_{3}^{0}}} = 10^{- 6.35} \\ {HCO}_{3}^{-} & \Leftrightarrow H^{+} + {CO}_{3}^{2 -} K_{a 2} = \frac{a_{H^{+}} a_{{CO}_{3}^{2 -}}^{2 -}}{a_{{HCO}_{3}^{-}}} = 10^{- 10.33} \\ H_{2} O & \Leftrightarrow H^{+} + {OH}^{-} K_{w} = a_{H^{+}} a_{O H^{-}} = 10^{- 14.00} \end{aligned}$

These are the speciation reactions between carbonate species and water species such as $H^{+}$ and ${OH}^{-}$ . All these reactions are aqueous speciation reactions and occur fast. All Ks are equilibrium constants at standard temperature and pressure conditions. Note that here we are not imposing the condition for charge balance. Reactions in Example 1.1 are examples of aqueous complexation reactions. These reactions occur ubiquitously in water systems and have significant impacts on water chemistry. For example, ligands, dissolved organic carbon (DOC), and other anions can form complexes with cations, including metals, to prevent cations from precipitation, therefore increasing the mobility of cations in water systems. The carbonate complexation and dissociation reactions with $H^{+}$ in example 1.1 works as a buffering mechanism to prevent the pH of water from changing rapidly.

1.1 Reaction thermodynamics

Reaction equilibrium constants $K_{e q}$ . Here we briefly cover fundamental concepts in reaction thermodynamics. For a more comprehensive coverage, readers are referred to books listed in reading materials [e.g., (Langmuir et al., 1997)]. A chemical reaction transforms one set of chemical species to another. Here is an example,

α A + β B É χ C + δ D

The reactants A and B combine to form the products C and D. The symbols $α, β, χ, and δ$ are the stoichiometric coefficients that quantify the relative quantity of different chemical species during the reaction in mole basis. That is, $α$ mole of A and $β$ mole of B transform into $χ$ mole of C and $δ$ mole of D. The Symbol $" É "$ indicates that the reaction is reversible and can occur in both forward and backward directions. According to reaction thermodynamics, the change in the Gibbs free energy $Δ G_{R}$ during the reaction can be expressed as:

Δ G_{R} = Δ G_{R}^{0} + R T \ln (\frac{a_{C}^{χ} a_{D}^{δ}}{a_{A}^{α} a_{B}^{β}})

Here, $Δ G_{R}^{0}$ is the change in reaction free energy under the standard condition (293.15 K and 10⁵Pa); R is the ideal gas constant (1.987 cal/(mol·K)); T is the absolute temperature (K); and $a_{A}, a_{B}, a_{C}, a_{D}$ are the activities of the species A, B, C and D, respectively. The derivation of this equation goes to the heart of reaction thermodynamics theory. In any particular aqueous solution, we define ion activity product (IAP) for the reaction (1) as follows:

I A P = \frac{a_{C}^{χ} a_{D}^{δ}}{a_{A}^{α} a_{B}^{β}}

The reaction reaches equilibrium when the forward and backward reactions occur at the same rate, the point at which the change in the reaction Gibbs free energy $Δ G_{R}$ reaches zero. At reaction equilibrium, the IAP equals to the equilibrium constant $K_{e q}$ :

K_{e q} = {(\frac{a_{C}^{χ} a_{D}^{δ}}{a_{A}^{α} a_{B}^{β}})}_{e q}

where $a_{A}, a_{B}, a_{C}, a_{D}$ are the activities of the species A, B, C and D at equilibrium, respectively. In general, larger $K_{e q}$ means larger reaction tendency to the right direction in the equation as written in (1). The $K_{e q}$ is a constant for a particular reaction at any specific temperature and pressure conditions. Although they look very similar, IAP is the ion activity product under any point along the reaction progress, while $K_{e q}$ is the IAP only at one particular point during the reaction, i.e., at equilibrium. We use the saturation index (SI) to compare IAP and $K_{e q}$

§ I = \log_{10} (\frac{I A P}{K_{e q}})

When SI =0, the reaction is at equilibrium; when SI < 0, the reaction proceeds to the right (forward); when SI > 0, the reaction proceeds to the left (backward).

Dependence of $K_{e q}$ on temperature and pressure. Values of $K_{e q}$ are a function of temperature and pressure. According to reaction thermodynamics, the $K_{e q}$ dependence on temperature follows the Van’t Hoff equation under constant pressure:

{(\frac{\partial (\ln K_{e q})}{\partial T})}_{P} = \frac{Δ H_{r}^{o}}{R T^{2}}

Here $Δ H_{r}^{o}$ is the standard state enthalpy change of the reaction (cal/mol). If the reaction is exothermic $Δ H_{r}^{o} < 0), K_{e q}$ decreases with increasing T; if the reaction is endothermic $(Δ H_{r}^{o} > 0), K_{e q}$ , increases with increasing T. For example, $K_{e q}$ of quartz dissolution increases with temperature, meaning its solubility increases with temperature increase. The heat capacity at constant pressure, $Δ C p_{r}^{o}$ (cal/mol K), is defined as follows:

Δ C p_{r}^{o} = {| \frac{\partial (Δ H_{r}^{o})}{\partial T} |}_{P}

If

Δ H_{r}^{o}

is independent of temperature, the integrated form of equation (6) is the following:

- \log K_{e q, 2} = - \log K_{e q, 1} + \frac{Δ H_{r}^{o}}{4.576} (\frac{1}{T_{2}} - \frac{1}{T_{1}})

Here

K_{e q, 1}

and

K_{e q, 2}

are the equilibrium constants at T ₁ and T ₂, respectively. Typically T ₁ is

25^{\circ} C

because the equilibrium constants of many reactions are measured and available at

25^{\circ} C

. For reactions only involve aqueous ions, this equation is applicable if T ₂ is within 10 to 15 degree C difference from

25^{\circ} C

. If the reaction enthalpy is not a constant and

Δ C p_{r}^{c}

is non-zero however is constant, we have a more general integrated form as follows:

- \log K_{e q, 2} = - \log K_{e q, 1} + \frac{Δ H_{r}^{o}}{4.576} [\frac{1}{T_{2}} - \frac{1}{T_{1}}] - \frac{Δ C_{p}^{o}}{1.987} [\frac{1}{2.303} (\frac{T_{1}}{T_{2}} - 1) - \log \frac{T_{1}}{T_{2}}]

Similarly, pressure dependence of

K_{e q}

can be expressed by:

{(\frac{d (\ln K_{e q})}{d p})}_{T} = \frac{Δ V_{r}^{o}}{R T^{2}}

Here, $Δ V_{r}^{o}$ is the molar volume change of the reactions under the standard condition (cm³/mol). Typically, the effect of pressure is important when the reaction involves gas (e.g. ${CO}_{2 (g)}$ ) and when there is a significant difference in the molar volume of reactants and products.

If the $Δ V_{r}^{o}$ is a constant and does not depend on pressure, the integrated form the above equation is:

\frac{\ln K_{e q, 2}}{\ln K_{e q, 1}} = \frac{Δ V_{r}^{o} (P_{2} - P_{1})}{R T}

Where $K_{e q, 1}$ and $K_{e q, 2}$ are the equilibrium constants at P1 and P2, respectively. Typically, P1 is at atmospheric condition at 1 bar.

The standard thermodynamic properties (e.g. $H_{r}^{o}$ and $V_{r}^{o}$ ) of most compounds can be found or calculated from CRC handbooks (e.g., Handbook of Chemistry and Physics (Haynes, 2012), and the NIST Chemistry WebBook [1]. Readers are also referred to the standard geochemical database Eq3/6 for values of equilibrium constants (Wolery et al., 1990).

Biogeochemical reaction systems typically include both slow reactions with kinetic rate laws and fast reactions that are governed by reaction thermodynamics. Kinetic reactions include, for example, mineral dissolution and precipitation and redox reactions. Thermodynamically-controlled reactions are those with rates so fast that the kinetics does not matter for the problem of interest. In geochemical systems, these include, for example, aqueous complexation reactions that reach equilibrium at the time scales of milli-seconds to seconds. For these reactions, the activities of reaction species are algebraically related through their equilibrium constants, or laws of mass action, as shown in equation (4). As such, their concentrations are not independent of each other and should not be numerically solved independently. This necessitates the classification of aqueous species into primary and secondary species. The primary species are essentially the building blocks of chemical systems, whereas the concentrations of secondary species depend on those of primary species through the laws of mass action. As such, with the definition of primary and secondary species, a reactive transport code only needs to numerically solve the number of equations for the species that are independent of each other, which is essentially the number of primary species. The number of primary species is equivalent to the number of components in a system. The concentrations of secondary species can be calculated based on the concentrations of primary species using laws of mass action.

1.2 Primary and secondary species

Here we will go through a few examples on how we choose primary and secondary species using the tableau method. Details are referred to in chapter 2, Morel and Hering, 1993.

How do we categorize the primary and secondary species of the system? Here we go through several steps for the choice of primary and secondary species.

What are the species?

List of species (5): H⁺, OH^-, H₂CO₃⁰, HCO₃^-, CO₃^2-, (H₂O is typically included implicitly).
How many algebraic relationships do we have that define the dependence between activities of different species?

We have 3 fast aqueous reactions, which means that we have 3 laws of mass action, as shown in three expressions defining the equilibrium constant for each reaction.
How many primary species do we have and what are they?

Here we have 5 species in total and 3 dependencies (equations). So we should have 5-3 = 2 primary species. The primary species should be defined so that all other secondary species can be written in terms of primary species.

Can we choose H⁺ and OH^- as primary species? No, because we cannot write the carbonate species as combinations of H⁺ and OH^-,

Can we choose H⁺ and CO₃^2-? Yes. See the following table. The species in the top row are primary species. The first left column includes all species.

Species	H⁺	CO₃^2-	(H₂O)
H⁺	1<	0	0
OH^-	-1	0	1
H₂CO₃⁰	2	1	>0
HCO₃^-	1	1	0
CO₃^2-	0	1	0

We can write all species in terms of H⁺ and CO₃^2-:

$\begin{array}{l} H^{+} = H^{+} \\ {CO}_{3}^{2 -} = {CO}_{3}^{2 -} \\ {OH}^{-} = (H_{2} O) - H^{+} \\ H_{2} {CO}_{3} = 2 H^{+} + {CO}_{3}^{2 -} \\ {HCO}_{3}^{-} = H^{+} + {CO}_{3}^{2 -} \end{array}$

Here $O H^{-}, H_{2} {CO}_{3}^{0}, and {HCO}_{3}^{-}$ ;are secondary species.

Is the choice of primary species unique? No. As long as the secondary species can be written as combinations of primary species, the list is legitimate. For example, we can also use OH^- and ${HCO}_{3}$ as primary species in this example.

Species	OH^-	HCO₃^-	(H₂O)
H⁺	-1	0	1
OH^-	1	0	0
H₂CO₃	-1	1	1
HCO₃^-	0	1	0
CO₃^2-	1	1	-1

We can then write all species in terms of primary species:

\begin{array}{l} H^{+} = (H_{2} O) - {OH}^{-} \\ {OH}^{-} = {OH}^{-} \\ H_{2} {CO}_{3} = {HCO}_{3}^{-} + H_{2} O - {OH}^{-} \\ {HCO}_{3}^{-} = {HCO}_{3}^{-} \\ {CO}_{3}^{2 -} = {HCO}_{3}^{-} - H_{2} O + {OH}^{-} \end{array}

Similarly, (H⁺, HCO₃^-), (H⁺, H₂CO₃), (OH^-, CO₃^2-) are also legitimate choices for primary species. However (H₂CO₃, CO₃^2-), (HCO₃^-, CO₃^2-), are not. You can practice using these species to write the expression of secondary species.

Take home practice 1.1:

If we impose the charge balance condition in example 1, we will then an additional algebraic relationship: $H^{+} = {OH}^{-} + {HCO}_{3}^{-} + 2 {CO}_{3}^{2 -}$ In this case how many primary species will we have? How many secondary species? What are they?

Click here for the answer

In example 1, we have 5 species and 3 reactions. Now we have 1 more relationship through this charge balance condition. This means that we have 4 algebraic relationships in total. In this case we will then have 1 = 5-4 degree of freedom so that as long as we specify 1 condition for the system, the system is defined.

If we add an additional species Ca²⁺ in the closed carbonate system in example 1, we then have the following reactions in addition to those in Example 1:

$\begin{array}{l} {CaCO}_{3}^{0} \Leftrightarrow {Ca}^{2 +} + {CO}_{3}^{2 -} \\ {CaHCO}_{3}^{+} \Leftrightarrow {Ca}^{2 +} + {HCO}_{3}^{-} \\ {CaOH}^{+} \Leftrightarrow {Ca}^{2 +} + {OH}^{-} \end{array}$

How many species do we have in total?
How many dependencies if we know the equilibrium constants of all these reactions above?
What is the number of primary species?
What are the primary and secondary species? How many different sets of primary species can you come up with?

Click here for the answer

Now we have 9 species, with the additional 4 species of Ca²⁺, CaCO₃⁰, CaOH⁺, and CaHCO₃⁺.
Originally we have 3 dependencies. Now we have added 3 more dependencies through the above 3 reactions so we have 6 dependencies.
The total number of primary species is 9 - 6 = 3
An example is (Ca²⁺, H⁺, HCO₃^-). All other species can be written in terms of these three primary species. They are however not the only set of primary species.

1.3 Setting up CrunchFlow for aqueous complexation in closed, well-mixed system

In this subsection we will discuss setting up aqueous complexation calculations in CrunchFlow.

Example 1.1 RTM setup

We have a closed carbonate system as an example 1.1 with the total inorganic carbonate concentration (TIC) being 10^-3 mol/L. Please answer the following questions:

What are the equations to be solved in this system?
If the pH of the system is 7.0, what are the concentrations of all involved species?
Calculate the concentrations of all individual species at pH varying from 1~14, with pH interval 2. That is, calculate concentrations of all individual species at pH 2, 4, 6, 8, 10, 12, 14.
Plot TIC, H₂CO₃, HCO₃^-, CO₃^2-, H⁺, and OH^- as a function of pH.
Observing from the plot, under what pH range H₂CO₃, HCO₃^- and CO₃^2- dominate, respectively?

Assuming this is a dilute system so that the values of activities are the same as concentrations. If we do not impose the charge balance condition, the code solves the following 5 equations for the 5 species questions 0):

$\begin{array}{l} K_{a 1} = \frac{C_{H^{+}} C_{H C O_{3}^{-}}}{C_{H_{2} C O_{3}^{0}}} = 10^{- 6.35} \\ K_{a 2} = \frac{C_{H^{+}} C_{C O_{3}^{2 -}}}{C_{H C O_{3}^{-}}} = 10^{- 10.33} \\ K_{w} = C_{H^{+}} C_{O H^{-}} = 10^{- 14.00} \\ - \log_{10} C_{H^{+}} = 7.0 (pH is 7) \\ C_{T} = 1.0 \times 10^{- 3} = C_{H_{2} C O_{3}^{0}} + C_{{HCO}_{3}^{-}} + C_{C O_{3}^{2}} \end{array}$

If charge balance is imposed, with the following equation:

$Ch arg e balance: C_{H^{+}} = C_{{HCO}_{3}^{-}} + 2 C_{{CO}_{3}^{2 -}} + C_{{OH}^{-}}$

Then either the pH condition or the TIC condition should disappear so we do not over condition they system (6 equations for 5 unknowns).

Please watch the following 37 minute video, Lesson 1, Example 1

Click here for a transcript of the Example 1 video.

PRESENTER: Let's go through this example on aqueous complexation reaction, how to set this up with zero dimension, essentially a closed, well-mixed system in CrunchFlow. This is for lesson 1 in the aqueous complexation lesson.

So let's just go through the example. We a closed carbonate system as an example one, where we already talked about primary species, secondly species, what's the general principle of picking primary species and secondary species. These are very important concepts.

So here we have an example with a total inorganic carbonate contraction equals 10 to minus 3 mole per liter. TIC, 10 to minus 3 mole per liter. So essentially you would have, the question is, first of all, the pH of the system is 7.0, what are the concentrations of all involved species? We already talked about this. A carbonate system, we should have hydrogen, OH minus, and all the three carbonate species. You had you should have carbonic acid, bicarbonate, and carbonate. So the total concentration is 10 to minus 3 mol per liter. Let's set this up.

So again, I'm opening this folder. Let's open the folder. Lesson 1, CrunchFlow example. So again, you see the four different files that are required to have in order to run the simulation. You have executable. You have the library file, input file, and database. So what I have here is a template. so

There's some keyword blocks already there. Let's put this title lesson 1-- aqueous complexation. And then you have all these database files. It's already specified. These shouldn't be changing much.

And the output file, when you have time dimension, notice here that in lesson 1, because we only talk about reaction dynamics, there's no kinetics involved. So there's no need of involving the time dimension as well. So it's zero space dimension and no time dimension. It's as simple a system as you can get.

Everything is at a equilibrium because there's only aqueous complexation reactions. And these reactions are really fast. So we don't need output to put anything output. This is for when you have time dimension. We don't discretization because this is used when you have space dimension. We also don't need boundary condition, initial condition because we don't have either space and time dimension. We don't have transport. We don't have flow. We don't have no porosity.

So all you need is putting in primary species and secondary species. So what are the primary species we talk about? The primary species are the building blocks of the system. This needs to be, in this system that you have carbonate, you have the three CO2 species. You have the pH and everything. So you should have, let's see. You at least should have H plus.

We always use H plus as primary species because it's so important. And we can also put, because we need to put at least one of the carbonate species. Because otherwise, you wouldn't be able to build up other species as a secondary species. So then you also have, corresponding to the H plus, you should have hydroxide. You should have CO2 aq. You should have-- and you also should have carbonate, these three species.

Now you have five different species. So this is like the example that we were talking in this example one, closed carbonate system. You only have these three reactions. You have five species in total. So then you have two primary species and three secondary species. And all the secondary species can be expressed in primary species.

So we have all these species. Now it's saying that the question one is pH is 7.0. And we know the total inorganic carbonate species is 10 to minus 3 mole per liter. So let's look at it what shall we put in this. So we should have conditions.

The unit is mole per liter instead of mole per kilogram. Let's call that condition pH 7, maybe. Because later on, we'll be doing other pH conditions. So I think it's useful to have the condition named with our variables of pH.

When we specify pH, we more or less can already specify the pH condition, the hydrogen ion concentration. And another condition we need to do is for the other primary species which is bicarbonate. And here when we specify bicarbonate, this should be already specified in total concentration.

There are several choices in controls that you can use to specify what are different Let's just search condition to make sure. That's how I do geochemical condition. Let's get to the point. Let's see. Are there lots of conditions. Can't believe it. I think we're almost there. These are for the runtime conditions.

So this part, page 48, is talking about input file entry of primary species. You can look through what you're putting for the primary species. It has to be coming from the primary species block of the database, as we talked about last time. And the secondary species has to be coming from the secondary species block.

But you actually can specify, for example, one of the carbonate species, you can specify one of them. The code uses a basic switch technique. You can pick any one of the carbonate species as primary species. The code kind of knows and can transfer between are different-- which one you choose primary and which is secondary species.

So it's fine to either pick carbonic acid or bicarbonate or carbon, even when in the database, bicarbonate is in primary species while the other two are not in primary species. The code uses basic switching technique to switch between the two. So it's OK as long as you choose one of them.

Can't believe I still haven't found it. Let' see. These are all of our databases that we talked about last time. Aqueous species. This is the page you need, page 65. So type of constraint for concentrations, aqueous species. So this table is useful.

If you want to put constraint for total concentration, you can just-- let's say you want to put sodium concentration 0.001, you just put the total concentration. So you have the uses for when you have more balance on total aqueous or total aqueous plus absorbed concentration.

So by default, it's total concentration when you specify. If you want to specify individual species concentration, then you should have a species after the number. So here, it's saying the total concentration with sodium is 0.001. So the sodium, when it's in water, it can be Na plus 3 species and also Na carbide or NaOH. But they add up to be 0.001.

Now in the second choice, here it's essentially saying you would have Na. Only the free sodium has the concentration 0.001. And then other species are calculated based on this and their equivalent constants. And there's species activity you should specify sodium plus 0.001 and specify that's activity. So this differs. So if you have active coefficient equal to 1, these two wouldn't make much difference. But if you are in a highly concentrated solution, these two will make a difference.

You can specify pH, a specific number like what we just did. You can also specify concentration in terms of the equilibrium with the gas phase. For example, oxygen, if you want to specify aqueous oxygen as an equilibrium with gas oxygen at partial pressure of 0.20 atmosphere, this is what you do.

Or if you want to equilibrate your primary species with a mineral oxygen aq with pyrite. Or if you want specified charge balance, you can do sodium charge. That would ensure charge balance is specified. Charge balance is honored.

So this is what do we do. And then here, other things later that mainly talk about how it does these calculations. Let's ignore this. So let's go back to the input file. So here, essentially, this is in the same format as, for example, here. So that means we are specifying total concentration, which is consistent with the condition we're given here, 10 to minus 3 mole per liter.

So if we do that, then we should be able to calculate-- don't need another condition. Only need one condition. That's good. So let's run this. We know all these species are in the aqueous phase, are in the database. So we don't need to check that. Let's round that. So you'll be putting one lesson example 1.5.

Concentration unit not recognized. Looks like we-- Let's just check on the In the condition, we have mole per liter. And so I search units. And it jumped to concentrations units. So there's different concentration units there.

So it should be always mole per liter, millimole per liter, or micromole per liter, or PPM. So mole per liter is not there. So if things are not working, let's go back to the menu again to see where We have to do mole per kilogram of water.

Now in dilute solution, it doesn't really matter if it's mole per liter or mole per kilogram water. It's the same thing. But if it's very concentrated, then you might need to do some conversion based on the activity coefficient and all that based on the density.

Here, so let's say we change it back to mole per kilogram, which is almost Note here, in dilute solution, which is what we have here, the mole per kilogram approximates mole per liter. If it's a concentrated concentration, this liter water versus kilogram water is different. So you need to use a density to convert between the two. Let's see. Let's run again.

pH 7.1, initialization condition. Speciation of initial boundary condition successfully completed. No aqueous kinetic block found. NZ. So that's what happens when you have no initial condition.

So let's specify, this means you will need to specify a discretization. Let's try that. This is how you run. You have no guarantee that you always get the right answer until you, for example, are really familiar with the code. Sometime you still make mistakes. So this is how you use it and learn. And I'm trying to show the process so eventually we'll get to it, how we can get it run.

So discretization, let's say you need to put one. For equivalence, it shouldn't really matter. Let's say we put units of-- you will need to put this condition in the initial conditions. And we do the speciation.

Or the other way you do it is specify this is only going to be speciation only. Let's do that. Let's just make sure we are running the right thing. Hm. Interesting. They might change the code-- let's see-- different from previous version.

There used to be a speciation only keyword. Speciate only, that's what. This is going to page 42. In the table is a list of all the keywords. So it's another way of getting fast hold of the keyword you need.

The example we have is only some dynamics, no timestamping. So you really don't need the initialization. So let's say we put speciate only with two. So then it should be running just a speciation reaction without necessarily setting up initial condition and everything. Let's try that. It's completed. That means it's run.

So always when you have input file, you there will be another corresponding output file. I talked about before, it's kind of anchoring what has been done the system. And actually for just aqueous speciation reaction, this is essentially the initialization process, going through initialization process.

And for just aqueous complexation, because we don't need timestamping, this is all it does. It's reading the total concentration, reading the pH, temperature, decide which one occurred, what condition in terms of temperature, pH conditioning. And then it's calculating the concentration of all the individual species.

Now if you look through this, these are the everything. It reads a number component, which is the primary species. Number secondary species is three. Number of gas, number of kinetic mineral, we don't have all of these. And it reads in the log K value of these aqueous complexation reactions.

So for OH it's with H plus. Log K is 13.99. This is CO2. So you want to make sure the log K values they're using is correct. If it's not correct, then you need to make sure they are the right number.

So here is I've conditioned it It said primary species are bicarbonate. And then it has the gas. The code needs initial gas to start the simulation. Total concentration constraint phase, so total concentration bicarbonate So it's reading everything correctly.

And then it does the speciation geochemical conditions. So speciation process is part of the initialization. Now it anchors all the geochemical conditions you have. Temperature, we didn't specify a porosity so the defined is 1. Saturation is 1. Liquid density, is this solid, answer no. Ion exchange solution pH is 7. Total charge is this.

So if you want to specify-- here we didn't really specify a charge balance. We can specify a charge balance if we want. We can do it in another example. Then the code essentially calculates the individual species, log activity, activity, H plus, bicarbonate, OH minus. So it has molarity of the different species. The activity, all these are pretty close to one.

Carbon is relatively small because it has two charges. So it is relatively small. Everything else is very close to one. It has activity. So now you can answer the question seeing how much were these concentration of individual species. As I said, answer the question one in the example.

So second question is asking you to do concentration of all individual at pH from 2, 6, 7, 8, 9. So essentially, it will be very similar to what you have in question one, except that you are going to do different pH conditions. So one way you can do it is you copy the condition and specify 2.0, for example. Then here you should have 2.0 instead of 7.0. 4.0. 6.0. And here should be-- OK, I'm sorry. 4, 6, 10, 12, and then 14.

So we have all the conditions. And essentially, you do this is one file, which is convenient. So if everything else is the same, only the pH is different, so if you run again, you see the effect of pH. Now when we do run this again, a new output file would replace old output file. But we still have that condition for 7.0 there. So this output will still be there. OK. It's done.

So it should have given you the speciation of each condition. It's essentially reading all the different conditions. And then 7.0 is first. And it's there. It calculates 7.0. And then you have condition 2.0. 4.0. So here you see pH is this. You have another set of concentration and everything. You have another set of concentrations.

So it essentially gives you the concentration of different species under different pH conditions. And you can pull the numbers out in either Excel file or whatever to plot what is asked for this as a function pH.

Now what if the [INAUDIBLE] the function [INAUDIBLE]? That's question three. And you can see under what pH conditions do these different species dominate. Now what if we try to run this with charge balanced? You can see that in that condition, charge is not balanced. So let's do pH 7.0. Let's just do another example.

Instead of specifying-- let's call the pH 7.0 charge. And maybe we will see. You can almost see from the output right now at 7.0, if we do not specify charge balance, it's a negative charge. That means in order to have charge balance, you need to make a positively charged species to do the charge balance, to kind of balance out this.

Instead of doing bicarbonate, we should do the pH calculated from charge balance. Let's try that. And same thing for other situations. If you want to make sure charge is balanced, you need to make sure that it's the right species, either positive charge or negative charge, in order to balance. And how you get the clue is from when you did not run the charge balance, what is the total charge? is it positive?

For example, pH 2.0, if you do charge balance, you need to use-- this is already positive charge. So you need a negative charge species to make charge balance. Then when you do pH 2.0 for charge balance, you need to put this for an example there. Then you should put, let's call it again charge. And then you should have bicarbonate charge because it's positive charge already. Let's run it again to see how it works.

Look at output file. I'm curious how that will come up with-- so and again, all this. Let's just look at the top two. 7.0, you still have this. And then when you have some 7.0, you see now the total charge is very small. So charge is more or less balanced.

Now in the solution, pH will have to be almost 4.7 in order to have total carbonate being in the concentration of 10 to minus 3. You to have a solution pH of 4.7 to balance it out, which is very different from 7.0.

And of course then, the concentration of the different species will be also very different. Hydrogen, H plus, OH minus, bicarbonate, carbon, they are all very different now. And if you look at the pH 2, it's this. And with the pH 2 charge, again, this a much smaller number than the previous one. You will need very, very high concentration of carbonate in order to have charge balance with a pH 2.

Because at pH 2, it should be the carbonic acid being dominating species. And you don't have that. And so you need bicarbonate and carbon to charge it to maintain charge balance. And these concentrations are very small. So in order to balance it, you need to have very, very high concentration of inorganic carbon or bicarbonate to balance it out. So in order to have that, you need a very high total inorganic carbon.

So that is what this is. This is very high, unrealistically high concentrations. So this essentially, it's almost like saying at pH 2.0, it's very hard to get a concentration pH to 1.0 in a system that you only have inorganic carbon species. So I think now we are good with this example.

So there's a take-home Practice One essentially asking you to add calcium in the simulation and answer these questions. I think it'll be an interesting exercise. One, just to give you a hint, the only thing you need to do is having another total calcium concentration. And you can specify and you put this-- I'm going to change it to make you have it to be charged balanced. And you need to make sure charge is balanced in the system.

So let's do this. And then you have another homework assignment with question one, question two. So question one is still the carbonate system with an open system. So you have P-CO2 of this. So you can look at the manual different constraints you should make it with CO2 gas in the 10 to minus 3.5.

And then the other question two is for metal complexation in seawater. And you can specify different species and everything. Make sure you check which secondary species might be dominant

So I think this is it for now. And I'm going to close that. And we can talk about it again once you have finished the homework give you enough exercise to work on it. This finishes lesson 1.

Credit: Li Li @ Penn State University is licensed under CC BY-NC-SA 4.0 [2]

Take home practice 1.1 RTM Set up

We have a closed system with total inorganic carbonate concentration (TIC) being 10^-3 mol/L and the total Ca(II) concentration (summation of all Ca-containing species) being 10^-4 mol/L. This is the same system as in the example with addition Ca²⁺, CaHCO₃^-, CaCO₃⁰ species.

What do you think are the equations that are solved in this system?
If the pH is 7.0, what are the concentrations of all involved species?
Calculate the concentrations of all individual species at pH varying from 1~14, with pH interval 2. That is, calculate concentrations of all individual species at pH 2, 4, 6, 8, 10, 12, 14 (Hint: for each pH, you will need to have one "condition" block in the input file).
Plot TIC, H₂CO₃, HCO₃^-, CO₃^2-, H⁺, and OH^-, and Ca(II)-containing sepcies as a function of pH.
Observing from the plot, describe the top 2 dominant species under each pH condition.

Example 1.1 Files [3]

Practice 1.1 Solution [4]

1.4 Homework Assignment

Question 1

Metal complexation in seawater. The seawater composition is as follows:

Total Molar Composition of Seawater (Salinity = 35)
Component	Conc. (mol/kg)
pH [5]	8.1
Cl⁻ [6]	0.546
Na⁺ [7]	0.469<
Mg²⁺ [8]	0.0528
SO₄²⁻ [9]	0.0282
Ca²⁺ [10]	0.0103
K⁺ [11]	0.0102
Total Inorganic carbon (TIC) [12]	0.00206
Br⁻ [13]	0.000844
Ba²⁺ [14]	0.000416
Sr²⁺ [15]	0.000091
F⁻ [16]	0.000068

Please set up CrunchFlow to do the aqueous complexation calculation for the seawater system and identify the dominant species (the top 2 species with the highest concentrations) for the major metals (Na, Ca, Mg, Sr). Is charge balanced? If not, what species should you put to balance charge?
Try the keyword “database_sweep” (look it up in the manual). What are the dominant secondary species formed in seawater? Pick the dominant secondary species for each cation. What is the difference by including and not including the dominant secondary species?
A Pb-containing solution is accidentally added to a tank of seawater, resulting in a total concentration of all Pb-containing species being 10^-2 mol/kg. All other concentrations remain roughly the same. In calculation, please make sure charge is balanced.
- Redo the calculation and identify the dominant species for all major metals, including Pb.
- Note that the total concentrations of Ca and Pb are similar. Do they differ in their dominant species? If there are differences, which parameter leads to such differences?

Question 2 open carbonate system

Here we assume we have an open carbonate system instead of a closed system. This means H₂CO₃, or CO₂(aq), is in equilibrium with atmospheric CO₂ concentration at P_CO2 of 10^-3.5atm. Henry’s law constant for CO₂ dissolution is K_H = 10^-1.47 mol/L/atm. Suppose we have a solution with a given pH and H₂CO₃ is solely from gas dissolution, please answer the following questions:

(Hint: you will need to look into the CrunchFlow manual to know how to set up a solution in equilibrium with a gas phase at given pressure. Look up the table for "Types of Constraints: Aqueous Species" on page 65-66).

At pH=7, what are the concentrations of individual species (H₂CO₃, HCO₃^-, CO₃^2-)?
Calculate the concentrations of TIC, H₂CO₃, HCO₃^-, CO₃^2-, H⁺, and OH^-, under pH of 2, 4, 6, 8, and 10.
Plot the concentrations of all individual species as a function of pH.
Observing from the plot, what are the dominant species under different pH conditions.
How does this open system behave differently from the closed system in example 1.1?

Question 3 buffering effect of carbonate system

Imagine you have two closed bottles. One bottle has pure water at a pH of 7.0 (with only H+, OH-, and water). In the other bottle, you have carbonate water as those in example 1.1 at a pH of 7.0 and a total inorganic carbon concentration of 0.001 mol/L. Both systems are charge balanced. If I add 0.001 mol/L of Ca(OH)2, what is the new pH of the two systems when each system reaches their new equilibrium? How does the presence of carbonate species influence pH changes? Think ahead how the two systems might be different before you do the calculation, and check if your calculation confirms your hypothesis.

HW1 files and solution package [17]

Reference:

Haynes, W.M. (2012) CRC handbook of chemistry and physics. CRC press.
Langmuir, D., Hall, P. and Drever, J. (1997) Environmental Geochemistry. Prentice Hall, New Jersey.
Wolery, T.J., Jackson, K.J., Bourcier, W.L., Bruton, C.J., Viani, B.E., Knauss, K.G. and Delany, J.M. (1990) CURRENT STATUS OF THE EQ3/6 SOFTWARE PACKAGE FOR GEOCHEMICAL MODELING. Acs Symposium Series 416, 104-116.

References

Haynes, W.M., 2012. CRC handbook of chemistry and physics. CRC press.

Langmuir, D., Hall, P., Drever, J., 1997. Environmental Geochemistry. Prentice Hall, New Jersey.

Wolery, T.J., Jackson, K.J., Bourcier, W.L., Bruton, C.J., Viani, B.E., Knauss, K.G., Delany, J.M., 1990. Current Status of the EQ3/6 Software Package for Geochemical Modeling. Acs Symposium Series 416, 104-116.

Lesson 1: Aqueous Complexation

Learning Outcomes

Lesson Roadmap

Questions?