Updated: December 5, 2001. Copyright © 2001 by Walt W. McNab, Concord, CA, U.S.A. All Rights Reserved.
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MULTISPECIES REACTIVE TRANSPORT IN GROUND WATER |
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Senior
Scientist Exponent-Failure Analysis Associates Oakland, California |
Lecture 2: Mass balance, mass action, and charge balance equations for
multiple species Introduction
In the next few lectures, weíll briefly examine the constraints on the concentrations of aqueous species that are in a state of thermodynamic equilibrium with one another and with mineral assemblages that may be present. These constraints include
The mathematical solution of such systems, in the form of predicted concentrations of n pertinent aqueous species, is ultimately based on solving n coupled algebraic equations in n unknown variables. Once we have developed an understanding of how the multispecies aquifer geochemistry is modeled in a zero-dimensional system (i.e., a system that is chemically uniform) we will move on, in subsequent lectures, to consider how the effects of solute transport are taken into account.
Basic concepts in equilibrium
relationships 
In the last lecture, we made a distinction between equilibrium and kinetic models. Here, we will focus on the equilibrium model. In this model, all reactions proceed immediately to a state of equilibrium that is dictated by thermodynamics. In other words, a single state of thermodynamic equilibrium is the end point the system ìwantsî to achieve. For thermodynamic equilibrium modeling, we donít care about how the system gets to that state; we only care about the end result. In essence, this end state corresponds to a state of minimal free energy in the system. Free energy considerations are a subject for a course in chemical thermodynamics. In this course, we will take for granted the validity of thermodynamic models and focus instead solving for the equilibrium concentrations of species in practical situations.
Geochemical speciation modeling is based upon components and species. Components refer to basic building blocks of matter in the system, with each component generally representing a chemical element present in the problem. Species, on the other hand, refer to physical entities present in the system and consist of a combination of one or more components.
Weíll tackle the subjects of components, species, and how they relate to one another by the use of an illustrative example. Consider the simple problem of the dissolution of calcite (CaCO3) into a closed system with an initial fixed quantity of total inorganic carbon (Figure 1). First, letís start with the species. The system contains, as an idealization, six distinct species, H+, OH-, Ca2+, along with species of inorganic carbon: CO32-, HCO3-, and H2CO3. In reality, other aqueous species will exist in such a system (e.g., CaHCO3+), but weíll ignore these for the time being for the sake of clarity. Calcite and H2O could also be regarded as species in an aesthetic sense. However, since the solution is assumed to be in equilibrium with calcite, the activity of calcite is set to one (see discussion below), so it cannot be treated as an unknown variable. In the case of water, the concentration of H2O in such a system will change by only a negligible amount, with its activity also at one (see discussion below), so it cannot be regarded as an unknown either.
The components that describe this system are limited to H+, H2O, Ca2+, and CO32-. As we shall see, this set of components is necessary and sufficient to constrain the mass balances and charge balance in the system. It is always possible to define another set of components if we wish, as long as the species present in the system can be described by linear combinations of
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Thus, mass
balance for the element carbon, in terms of species, is given by,
Next, because the electrical neutrality of the system must be preserved, a charge balance constraint exists on the species with non-zero electrical charges:
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In essence, the
charge balance equation replaces a mass balance equation on the element H,
inorganic carbon species, we note that,
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(Eq.-4)
and,
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(Eq.-5)
The dissociation reaction of water also plays an important role in this system,
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(Eq.--6)
The final reaction is provided in the equilibration between the aqueous phase
composition and calcite as given by Eq.-1.
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The ratios of products to reactants in each of the chemical reactions given above are related to one another through mass action equations. Mass action equations are written in terms of the product of the product activities divided by the product of the reactant activities. |
The proportionality constants associated with mass action expressions can be calculated based upon thermodynamic considerations involving the free energies of formations. The development of mass action expressions and the equilibrium constants are explained in many introductory chemistry textbooks and will not be addressed here.
For the reactions listed, these mass action equations may be given by,
8)
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(Eq.-9)
and,
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(Eq.10)
In mass
action expressions, a species activity is always raised to an exponent equal to
the stoichiometric coefficient (= the number of
moles participating in the reaction) of that species in the reaction.
This becomes apparent when the stoichiometry is greater than one (e.g.,
the 2 moles of H+ in Eq-5; the
exponent of 2 for {H+} in Eq.-8).
Note that in Eqs. 7-10, the square brackets that denote concentration according to our earlier convention have been replaced by squiggly brackets that denote the thermodynamic activity of the aqueous species. The concept of activity is needed for addressing the non-ideal behavior of solutes in real solutions, where electrostatic interactions, which depend on the total strength of the electrolyte, complicate the task of assigning an equilibrium constant to a reaction based solely on concentrations. As a result, the concept of species activity is introduced, which is related to the species concentration by the activity coefficient,
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For fairly dilute solutions (i.e., with total ionic concentrations less than
that of seawater),
strength, which may be expressed as,
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(Eq.-13)
As an aside, note that activities of water and CaCO3 in our example system are equal to one, so these are not taken into consideration in the mass action equations.
Solution techniques 
Given certain constraints on a
solution (i.e., fixed total amounts of component, a fixed pH, and/or specified
equilibrium with some mineral or gas phase), the concentrations of each of the
species modeled in the system can be quantified by solving the series of
simultaneous algebraic equations presented above (Eqs. 2-3, 7-10).
However, the system of equations is not linear, so special techniques
must be employed, especially when a large number of species and components are
involved in more general cases. For
realistic results in any but the most dilute of water compositions, activities
must be used in place of concentrations in the mass action equations.
These activities, in turn, depend on the final solution chemistry (Eqs.
11-12), so an iterative procedure is always required.
We
will explore some of the techniques used to solve the geochemical speciation
equations for large systems by numerical approximation in a subsequent lecture.
For the relatively simple carbonate equilibrium example provided in this
lecture, the small number of equations involved lends itself to a simple
solution scheme, particularly if the activity coefficients are all assumed to be
equal to unity (i.e., concentrations can replace activities in the mass
action expressions. A non-linear equation solver, such as that supplied with the
popular MathCad calculation software, can easily be used to demonstrate simple
calculations for such a system. A
set of MathCad worksheet calculations for this system is provided at the end of
this lecture for inspection.
PROBLEM 1:
Activity
coefficients. Suppose
you are given the concentrations of Ca2+ and
SO42- in a water of pH 7 to
be 5.1 x 10-3 mole/L each. What
is the ionic strength of the solution? Given
a value of 0.5085 for the A parameter in the Davies equation, what are the
activity coefficients for these two species?
Now, suppose that the mass action relationship for equilibrium between an
aqueous solution and the mineral gypsum, CaSO4×2H2O,
can be written as,
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QUESTION:
Using
the calculated activities, will gypsum be favored to precipitate from this
PROBLEM 2:
Carbonate
equilibria.
Given that pH = -log{H+}, what is the dominant species of
inorganic carbon, in terms of activity, among the three illustrated in this
lecture (H2CO3, HCO3-,
or CO32-) at pH 3?
At pH 7? At pH 11?
ANSWER
TO PROBLEM 2:
EQUILIBRIUM OF CALCITE IN WATER CONTAINING INITIAL TOTAL INORGANIC CARBON
You are now ready to continue to LECTURE 3: Oxidation and Reduction Processes.
You may e-mail me questions and comments.
Walt W. McNab
E-mail address: Walt McNab <WaltMcNab@prodigy.net>
From the definition of ionic strength, we see that,
I = 0.5 x [(0.0051 mol/L) x 4 + (0.0051 mol/L) x 4] = 0.02 mol/L
Now, from the Davies equation, we can estimate that g for both Ca2+ and SO42- is equal to 0.947. The logarithm of the activity products on the right side of the mass action equation is thus,
log(IAP) = 2 x log(g x 0.0051) = -4.63
This value is ever so slightly smaller than the logarithm of the equilibrium constant, -4.6, so that we may conclude that gypsum is not necessarily favored to precipitate from solution. If we ignore the activity corrections, we would calculate that,
log (IAP) = 2 x log(0.0051) = -4.58
This value slightly exceeds the logarithm of the equilibrium constant, so that we could conclude here that there are slightly favorable circumstances for precipitation of gypsum. At elevated concentrations, the activity coefficient factors become much more important. For example, if both ions are present at 0.1 mol/L, then I = 0.4 mol/L and g = 0.557.
ANSWER
From the mass action equations between the carbonate species, we see that,
Given the definition of pH, we can calculate that at pH 3,
{HCO3-}/{CO32-} = 2 x 107 ,
while
{H2CO3}/{CO32-} = 5 x 1010.
Clearly, these activity ratios indicate that
{H2CO3} >> {HCO3-} >> {CO32-}.
At pH 7,
{HCO3-}/{CO32-} = 2000
while
{H2CO3}/{CO32-} = 501.
In this case, the ratios suggest that {HCO3-} is now the dominant species, although barely so, exceeding {H2CO3} by only a factor of 4 (you can see this by dividing the first ratio expression by the second); {CO32-} is still quite insignificant.
At pH=10, however, and
{H2CO3}/{CO32-} = 5 x 10-4,
so that {CO32-} ~ {HCO3-}, whereas {H2CO3} becomes negligible.
