Class time: 4:45 PM - 6 PM on Monday and Wednesday.
The web notes serve as a text. As a backup text, try Environmental Chemistry (latest edition) by Stanley Manahan
Instructor: Ronald K. Stoessell, Room 1034, 280-6795,
rstoesse@uno.edu or ronlondi@charter.net
50% of the grade will be from 2 tests (midterm and final) and the remaining 50% will be from the homework assignments. The midterm and the final will be take-home tests. Graduate students will have to learn how to use MINTEQA2 and for their final homework assignment, pick a pollutant metal use the program to study the metal mobility in aqueous solutions as a function of pH and pe conditions. Undergraduates can do this final assignment for extra points.
The purpose of this course is to provide a practical background in basic geochemical principles that can be applied to environmental problems. These same principles can also be applied to geologic problems at other than earth-surface conditions. In general, at higher temperatures and pressures, much of the necessary geochemical data needed to apply the geochemical principles is not tabulated and has to be calculated. This is covered in another course Geology 4659 (Geochemical Thermodynamics).
The course will make use of computer programs to solve geochemical problems and the use of EXCEL spreadsheets to do simple modeling of pollutant transport. Some simple chemical analyses will be done to illustrate important environmental chemical reactions, e.g., pH, Eh, dissolved O2, alkalinity, hardness, nutrient concentrations.
Environmental geochemistry is the application of chemical principles to predicting the fate of organic and inorganic pollutants at the earth's surface and in the atmosphere.
Metal pollutants are treated using equilibrium with solids and aqueous solutions. Class members can use the EPA computer program MINTEQA2 to delineate pH and Eh conditions under which various metals are mobile and can pose a threat to humans.
Metal retardation factors will be calculated using mass constraints predicted by various algorithms such as the Freundlich, Langmuir, and Power Functions. Some simple spread-sheet modeling with Excell may be done for metals undergoing cation-exchange in plumes.
Organic processes discussed will include nutrient removal from agricultural wastes and municipal and home sewage systems. The sequence of reactions governing ozone depletion in the stratosphere will be described and environmental problems such as acid rain, global warming, smog, and acid mine drainage will be covered. Time permitting, a survey will be made of the fate of various industrial (e.g., PCBs and dioxin) and agriculture organic pollutants (e.g., DDT and the organophosphates).
Simple chemical analyses such as measuring pH and Eh (by electrodes), alkalinity (by pH titration), dissolved oxygen (by titration and by electrode) and conductivity (by electrode), as well as anions and cations (using the liquid ion chromatograph) will also be covered.
The necessary chemical background for students consists of the freshman chemistry courses required for the undergraduate major in geology.
A pollutant is a substance present in greater than natural concentration as a result of human activity and having a net detrimental effect upon its environment.
The hydrosphere includes all water in the earth's crust.
The lithosphere includes the outer portion of the earth's mantle and the earth's crust in geology; however, in environmental sciences, the term is used to refer to the minerals, organic matter, water, and air which make up the surface of the solid earth, in particular the soil.
The atmosphere is the envelope of gases surrounding the Earth.
The biosphere includes all living organisms and their surrounding environment.
A phase is simply a homogenous substance with uniform composition and properties, e.g., the atmosphere, sea water, a mineral grain. The concept of a phase becomes less clear on a microscopic scale near phase boundaries, because there is always a variation in composition near phase interfaces.
A solution is generally a liquid phase composed predominantly of one component, the solvent (e.g., water) and less abundant components, the solutes (e.g., dissolved salts). A solution can also be a solid or a gas phase, e.g., a solid solution (e.g., halite, the solvent, with minor impurities such as sodium bromide, a solute) or a mixture of gas components (e.g., the atmosphere). The term fluid can refer to either a liquid or gas phase.
Anions - negatively charged ions or complexes in aqueous solutions. An anion forms by reduction through acceptance of electrons. The major anions and their charges are OH-, Br-, Cl-, F-, HCO3-, HS-, NO3-, CO32-, SO42-, PO43-.
Cations - positively charged ions or complexes in aqueous solutions. A cation forms by oxidation through donation of electrons. The major cations and their charges are H+, K+, Na+, Ca2+, Fe2+, Mg2+, Al3+, Fe3+. The important class of pollutants known as heavy metals generally exist in solution as cations.
Complexes - molecules that can be anions, cations, or neutral in charge. For example, SiO2 combines with 2H2O in aqueous solution as a neutral aqueous molecule in the form of H4SiO4. Another example is Al(OH)4-, a negatively charged complex consisting of an Al3+ cation and 4(OH)- anions joined together.
Electrical Balance - In a solution, the charges contributed by the cations have to balance those contributed by the anions. Elemental analyses are frequently done which list the concentrations of the major components without determining if they exist in solution by themselves or as part of different complexes. For example, an analysis will only determine total Na concentration and not the proportion that actually exists in solution as Na+, NaClo, NaHCO3o, NaCO3-, etc. However, the overall electrical balance can still be estimated by summing the electrical charges in the elemental analyses using the dominant charge for each of the elemental components. For example all of the sodium is given a +1, all of the chloride is given a -1, all of the sulfate is given a -2, etc. The overall electrical balance gives an idea of the accuracy of the solution analyses. Without sophisticated corrections, errors in the computed electrical balance will arise for elements existing in solution with multiple oxidation states, e.g., Fe2+ and Fe3+ or with different complexes of hydrogen ions, e.g., HS- and H2S. The former produces an error because only the total of an element is generally measured and you would need to know the fraction in each oxidation state for the correct electrical balance. The latter produces an error because the pH, not total H+, is measured and most of the H+ is not included in the electrical balance except indirectly in the aqueous complexes. For example, you would need to know if most of the inorganic carbon is CO32- or HCO3- to account for it in the electrical balance. In general these are minor errors but to be precise, a computer program can partition elements into different oxidation states and into different concentrations of complexes with hydrogen species for an accurate computation of electrical balance.
Always do a simple hand computation of the electrical balance of analyses from commercial laboratories. A poor electrical balance often means that the data should be disregarded. A perfect electrical balance means that one of the components was determined from the electrical balance and not by analysis. This is often the case for either Na+ or Cl-.
In solutions, the fundamental concentration unit in is the mole fraction Xi; in which for j components, the ith mole fraction is
(1,1) Xi = ni/(n1 + n2 + ...nj),
where the number of moles n of a component is equal to the mass of the component divided by its molecular weight. Note that a substance composed of only one substance will have a mole fraction of unity for that substance. In an aqueous solution, the mole fraction of water, the solvent, is always near unity. In solids that are nearly pure phases, e.g., limestone, the mole fraction of the dominant component, e.g., calcite, will be near unity. In general, only the solutes (minor components) in a liquid solution and gas components in a gas phase will have mole fractions that are significantly different from unity.
Other more commonly used concentration units for the solutes are given below:
per cent, parts per hundred
ppt, parts per thousand
ppm, parts per million
ppb, parts per billion
ppt, parts per trillion
These units are often assumed to be in terms of mass (weight) unless otherwise stated. Note that one mg/kg is one ppm where a mg is 1/1000th of a gram and kg is 1000 grams. Similarly, one ug/g is also one ppm where a ug is 1/1,000,000th of a gram. One ug/kg is a ppb.
Because one liter of water weighs approximately one kg, mg/liter units of solution are nearly equal to ppm units. The precise equivalence is obtained by dividing by the density p:
(1,2) ppmi = [mgi/liter]/p
where the solution density is in grams/cm3. Older texts will substitute specific gravity for density in the above equation. The specific gravity is the ratio of the solution density to that of the density of pure water at 4oC. Since the density of pure water at 4oC is 1 gram/cm3, the specific gravity is equal to the solution density when expressed in metric units of g/cm3.
Note that a ml (1/1,000th of a liter) is equal to a cm3. The units of ppt and ppm are commonly used for the concentrations of solutes in aqueous solutions such as sea water. Trace components are represented in the ppb and ppt range.
The units of g/cm3 are used in diffusion. This unit is nearly equivalent to weight fraction for an aqueous solution since one cm3 of water approximates one gram.
Other concentrations units are described in terms of the number of moles of a components.
The molarity of i, Mi, is the moles of component i per liter (1000 cm3 or 1000 ml) of solution.
The formality of i, fi, is the moles of component i per kg of solution.
The molality of i, mi, is the moles of component i per kg of solvent (e.g., water). Molality is used only in aqueous solutions. In making thermodynamic calculations, the units used are molality for all aqueous components.
The equivalents of i, ei, is generally
the moles of charge of i in molarity, formality, or molality units. If the equivalents
refer to a particular reaction, than it refers to the equivalents of charge needed
for the reaction to be completed in terms of molarity, formality, or molality
units./P>
Multiplication of the above mole concentration units by one thousand will convert them to milli units, e.g., mMi, mfi, mmi, and mei.
Note that molarity is defined in terms of volume, a liter of solution. Any concentration unit involving volume will change with temperature and pressure, because volume is a function of those parameters. For this reason, geochemists prefer not to use molarity units in aqueous solutions; however, chemists prefer molarity units.
(1,3) fi = ppmi/(1000 Mwi) where Mwi is the molecular weight of i.
(1,4) mi = [ppmi/(1000 Mwi)] [1/(1 - tds/1,000,000)]
where tds is the total dissolved solids in ppm in the solution.
(1,5) Mi = [ppmi/(1000 Mwi)] p
The concentration units are not absolute amounts of a component in the solution but are relative amounts, having been normalized to the total mass or volume of the solution or solvent. Hence, changing the amount of a solution does not change the concentration of any of the components. Concentration units, as well as temperature T and pressure P, are intensive parameters that do not depend upon the total mass of a solution. Volume is an example of an extensive parameter, one that depends upon the total mass of the solution.
Use the formulas given above with the seawater compositon given below (Pytkowicz and Kester, 1971, for sea water of 19000 ppt chlorinity) in a spreadsheet program such as Excel to convert from ppm to mg/l to mmolality and mmolarity. For the density of seawater use 1.025 g/ml. Check to see if the electrical balance in mequivalents is zero.
Cl- 18,971 ppm
Na+ 10,565 ppm
SO42- 2,660 ppm
Mg2+ 1,269 ppm
Ca2+ 404.2 ppm
K+ 391.4 ppm
HCO3- 142.3 ppm
Br- 65.99 ppm
B(OH)3 26.1 ppm
Sr2+ 7.79 ppm
F- 1.3 ppm
Li+ 0.19 ppm
A scientist looks at a sample of sea water in limestone and predicts that 100 mg of calcite are dissolving per liter of sea water and that 50 mg of dolomite formed in the process. Sea water has an expected Ca, Mg, and HCO3 concentration of 412, 1210, and 146 mg/l, respectively, as total concentrations. What are the concentrations of Ca, Mg, and HCO3 in the altered sea water that are consistent with the scientist's prediction.
If two endmember solutions I and II are mixed, the conserved (unreactive) concentrations of component a and component b in the mixture can be expressed as a function of the solution fraction of I, fI by
and ppmb,mix = ppmb,IfI + ppmb,II(1-fI)
Combining the two equations yields to eliminate f
(2,1) ppma,mix = (m1)ppmb,mix + m2
where m1 = (ppma,I - ppma,II)/(ppmb,I - ppmb,II)
and m2 = -(ppma,I - ppma,II)/(ppmb,I - ppmb,II)ppmb,II + ppma,II
and m1 and m2 are the slope and intercept of a straight line in which ppma,mix is plotted versus ppmb,mix.
Note that a linear relation is also obtained from equation (2,1) by dividing by ppmb,mix. A straight line results from plotting the ratio of the the components concentrations is plotted versus the concentration reciprocal of the component in the denominator.
(2,2) ppma,mix/ppmb,mix = m2/ppmb,mix + m1
where the slope is now m2 and the intercept is m1.
An isotopic ratio such as 87Sr/86Sr can also be plotted versus
the reciprocal of the concentration of total Sr, not just 86Sr to obtain a straight
mixing line. The relationship follows from combining the following two mass equations to eliminate fI
ppmSr,mix = ppmSr,IfI + ppmSr,II(1-fI) with [87Sr/86Sr]mix =
[87Sr/86Sr]IfI[ppmSr,I/ppmSr,mix]
+ [87Sr/86Sr]II(1-fI)[ppmSr,II/ppmSr,mix]
The resulting equation is ()mix = (m3)/(ppmSr,mix) + m4 where
m3 = ppmSr,IppmSr,II{[87Sr/86Sr]II - 87Sr/86Sr]I}/(ppmSr,I - ppmSr,II) and
m4 = {ppmSr,I[87Sr/86Sr]I - ppmSr,II[87Sr/86Sr]II}/(ppmSr,I - ppmSr,II)
The 87Sr/86Sr ratio is a useful indicator for the age of marine-related brines because this ratio has undegone a "more or less" steady increase since the mid-Jurassic in sea water. Marine-related brines initially have the ratio of sea water at the time of formation as a pore fluid. The ratio increases if the brine moves through clastic units and picks up 87Sr from the decay of 87Rb. Hence, the ratio can give a minimum age of the initial formation of the brine.
In Louisiana, pollution sometimes involves saline contamination of ground-waters that are used for irrigation. The saline contamination is often high sodium and chloride concentrations on the order of that found in sea water, approximately 10 to 30 ppt. These salinities can result from mixing dilute ground waters having a salinity of about 1 ppt with brines having concentrations of 100 ppt or more, e.g., oil field brines leaking up well bores. Determining which brines are the pollution sources is used to determine responsibility for cleanup.
Conservative-mixing of unreactive components can be used to pinpoint the brine compositions that are the source of the pollution. Because mass, not volume, is conserved in mixing, the data should be in units such as ppm (mass basis) or molality, not mg/liter. As shown above, the concentrations of two conserved components, plotted versus each other, fall on a straight line if they are the result of mixing. The concentrations in the polluted waters can be used to fit a linear line which can then be extrapolated to brine salinities. Then real brine compositions can be tested against the extrapolated brine composition to determine if they could be the pollution source. Conserved aqueous components in mixing are usually Na, Cl, Br, I, and certain isotopic ratios. However, as discussed above, isotopic ratios of an element have to be plotted versus the reciprocal of the total concentration of the element to produce a straight mixing line. An example of an application of the above procedure to determine the salinity source polluting fesh water is given by Stoessell (1997, Ground Water, 35, 409-417).
For example plots of Cl versus Br in ppm of the contaminated waters can be used to derived a linear mixing line which can then be extrapolated to brine compositions. Isotopic ratios of 87Sr/86Sr can also be plotted versus the reciprocal of total Sr in ppm to produce a linear mixing line which can be used to extrapolate to brine compositions. Also, it is always best to plot in linear space rather than log space because the actual fit of the data to a best-fit line is much easier to judge. Researchers have always used log-log plots when they want their data to look better than it really is. Another trick used by some researchers is to fit a best-fit linear line to two sets of clustered data: one in the polluted waters and one from actual brine samples that they hope to prove are the source of the pollution. The data will fit to a straight line with a correlation coefficient close to unity, because the data is clustered at two separated salinities.
The equilibrium constant K for a reaction can be compared with the reaction quotient Q of the activities (thermodynamic concentration units) of reaction components to test if the reaction is at equilibrium. The equilibrium constant is related to the mass-action expression described below.
The generalized chemical reaction below has reactants on the left being transformed into products on the right,
aA + bB ==> cC + dD.
The absolute values of the reaction coefficients for A, B, C, and D are a, b, c, and d, respectively. The reaction coefficient of a reactant is negative because the reactant is being destroyed and the reaction coefficient for a product is positive because the product is being produced. The components can be in different phases, e.g., carbon dioxide in the atmosphere can be reacting with calcium carbonate in limestone and with sea water, or all within the same phase, e.g., H+ reacting with CO32- to form HCO3- in ground water. If all the reaction components are within the same phase, a good assumption is that the reaction is at equilibrium. Notable exceptions are reactions involving redox (oxidation and reduction) reactions in aqueous solution which are generally not in equilibrium.
The reaction rate can be increased by increasing the activities (concentrations) of the reactants. Once equilibrium is reached, the reaction can be reversed by increasing the activities (concentrations) of the components on the right side of the reaction, driving the reaction to the left. This reversal would make the components on the right side, the reactants, and those on the left side, the products.
The mass-action expression or law
K = Q where
(3,1) Q = (aC)c(aD)d(aA)-a(aB)-b
Frequently, the law is written in log units because the activities are often very small numbers.
log K = log Q where
(3,2) log Q = c log aC + d log aD - a log aA - b log aB = log [(aC)c(aD)d]/[(aA)a(aB)b]
where K is the equilibrium constant for the reaction as written with the reactants on the left and the products on the right. The product of terms on the right and left sides of the equation is called the reaction quotient Q and is only equal to K at equilibrium. Hence, measuring Q at equilibrium is one way to determine K; however, K can also be predicted from thermodynamics. In Q, each term is raised to the power of its reaction coefficient which are positive for products and negative for reactants. ai is the activity which is the product of the concentration ci and a proportionality coefficient called the activity coefficient. Although ci is measured experimentally, the activity coefficient has to be computed. Note that chemists use the pK, which is the negative log of K, i.e., -log K.
If Q is less than K (or log Q/K < 0), the reaction is moving from left to right. If Q equals K (or log Q/K = 0), the reaction is at equilibrium. If Q is greater than K (or log Q/K > 0) than the reaction is going from right to left, and the reaction is moving in the reverse direction from written.
For determining the direction of the reaction, it doesn't matter if the reaction was actually reversed from the way it was written. For example if Q > K, the reaction is moving right to left. If we reverse the reaction as written so the right side becomes the left side and vice versa, we also have to change the values of Q and K. Their new values are the reciprocals of their old values. The new values would give Q < K, predicting the reaction moving from left to right, consistent with reversing the way the reaction was written.
K is predicted from thermodynamics or measured experimentally. The value of K depends upon the conventions used to define the activities of the components. These conventions are based on the definition of the standard state chemical potential of each component (see below) in which the activity relates the actual state of a component to its standard state. K varies with pressure P and/or temperature T if the standard states of the components in the reaction move with P and/or T. In geochemistry, K varies with pressure P and temperature T for all reactions except those involving only gas components. The K for reactions involving only gas components varies with temperature but not pressure because the standard states for gas components are always held at 1 bar P. As shown in thermodynamics, if the effect of a variable such as pressure isn't included in the standard state of a component it is included in its activity. See the definition of activity below for gas components for use in the reaction quotients.
In thermodynamics, K is computed using the following relations based on the definition of the DGr, the Gibbs Free Energy change, in a Reaction:
(3,3) DGr = DGor + 2.303 RT log Q where
(3,4) DGor = Sniµoi and
(3,5) log Q = Snilog ai
where DGor is the standard state change in Gibbs Free Energy of the reaction at the P and T of interest. ni and µoi are the reaction coefficient (positive for a product and negative for a reactant) and the molar standard state Gibbs Free Energy (also called the standard state chemical potential) for the ith reaction component. µoi values are tabulated in reference books at 25oC and 1 bar pressure. R and T are the gas constant and absolute temperature (at 25oC and 1 bar R = 1.9872 cal/molo K and 298.15oK at 25oC). Although not covered in this course, various algorithms exist for calculating the standard state chemical potential values at different pressures and temperatures for use in computing equilibrium constants. This is covered in Geology 4659, Geochemical Thermodynamics.
At equilibrium,
(3,6) DGr = 0; hence, (3,7) log K = - Sniµoi/ (2.303 RT)which is the relation used to compute K if Q cannot be measured at equilibrium.
The values of µoi correspond to the energy a component has as a result of bond energies, heat content, and spatial location of the atoms on a lattice. These values are from experimental measurements of energy changes in reactions in which the component is formed from the elements. By convention, the µoi of each element in its stable form at 25oC and 1 bar is set to zero. This allows the change in free energy at 25oC and 1 bar in the reaction forming the component from the elements to be assigned to the component. In aqueous solutions, two additional conventions are needed, setting µoi for H+ and for e- to zero at 25oC and 1 bar. The use of these conventions is justified because we always are calculating the equilibrium constant from a free energy change across a reaction which is mass balanced, allowing the effect of the conventions to cancel out. These conventions and how to use them are covered in Geology 4659, Geochemical Thermodynamics.
From the above relations we can write by substitution:
(3,8) DGr = 2.303 RT log (Q/K)which is zero at equilibrium when Q = K.
The expression log (Q/K) is often called the saturation index and is negative if the reaction goes to the right and positive if the reaction goes to the left.
Note the simple form of the mass-action law. The expression Q contains the product (not the sum) of the activities of the components taking part in the reaction. Each activity is raised to the power represented by its reaction coefficient. This coefficient is positive for products (because the products are being produced) and negative for reactants (because the reactants are being destroyed).
(3,9) Q = P(aini)
The mass-action law is important if the K for a reaction is known. By computing Q and comparing it with K, the reaction can be tested to determine if equilibrium exists. If Q is not equal to K then there is no equilibrium. However, all reactions proceed in the direction of equilibrium, so the future reaction path can be predicted.
Whenever reactions are added together to form a new reaction, the equilibrium constant K for the new reaction is simply the product of the individual equilibrium constants of the reactions. In log space, the log of the equilibrium constant for the addition of reactions is simply the sum of the logs of the equilibrium constants for the individual reactions.
Subtracting a reaction means reversing the right and left sides of a reaction and then adding the reversed reaction to another reaction. If a new reaction is formed from subtracting one reaction from another reaction, the equilibrium constant of the new reaction is simply the quotient of the two reactions in which the denominator contains equilibrium constant of the reaction that is being subtracted. In log space, the log of the equilibrium constant for the first reaction minus the log equilibrium constant for the reaction being subtracted is the log of the equilibrium constant of the new reaction.
The definitions given below are the common ones used in geochemistry. Whenever, an equilibrium constant is used to evaluate if equilibrium exists, the user must use the activity definitions that are consistent with the conventions used to define the equilibrium constant. Any equilibrium constant given in the literature should list those conventions. Unfortunately, different conventions are sometimes used by chemists, geochemists, and chemical engineers, generating confusion if the conventions are not listed.
For an aqueous component in a solution,
(3,10) ai = migi
where mi is the molality of the ith component and gi (gamma) is the activity coefficient discussed later.
For a solid component in a solid phase or a liquid component in a non-aqueous solution,
(3,11) ai = Xili
where Xi is the mole fraction of the ith component in the phase and li (lamba) is the activity coefficient discussed later.
For a gas component in a gas phase,
(3,12) ai = XiciP
where Xi is the mole fraction of the ith component in the gas phase, ci (chi) is the activity coefficient discussed later, and P is the total gas pressure. Remember that K will not vary with pressure for a reaction involving only gaseous components. The reason is that pressure is already included in the activity expression of all the components.
Chemists generally use the same conventions as geochemists and the same activity definitions, with one exception. Their K values do not vary with pressure for liquids and solids. The result is that pressure variation is included in activity coefficient computations in these phases.
The activity coefficients gi, li, and ci each relate the activity to a concentration unit. In thermodynamics, the activity coefficient corresponds to a measure of the interactions of the ith component with other components in the phase. The activity coefficient with the concentration unit in the activity term allows for a correction to the Gibbs free energy of the component from what it would be in its standard state. The standard state is the pure substance for a solid and for liquids, a pure 1 bar ideal gas (obeys ideal gas law) for a gas component, and an infinitely dilute-behaving solution for dissolved aqueous components. The activity coefficient is one when the component is behaving as though it is in its standard state (in which there are no interactions with other components). In general the activity coefficient of the solvent component making up a nearly pure phase, e.g., water in an aqueous solution or a mineral that is nearly pure can be regarded as having unit activity coefficients. However, the activity coefficients of dissolved ionic salts are rarely one.
The activity coefficient of a charged aqueous component is generally calculated by a modified Debye-Hückel expression. For components in solids, gas phases, other non-aqueous solutions, various empirical and semi-theoretical expressions are used. The actual calculation of activity coefficients usually represents the biggest error in using mass-action laws. Examples of activity-coefficient expressions will be given in class. The Debye-Hückel equation and its modifications for use in higher ionic strength solutions are:
(3,13) log gi = -Azi2I1/2 for very dilute solutions, I < 10-2.3
(3,14) log gi = -Azi2I1/2/(1 + BaoiI1/2) for I < 10-1
(3,15) log gi = -Azi2I1/2/(1 + BaoiI1/2) + miC for I < 1
and the modification known as the Davies Equation which is generally written as
(3,16) log gi = -Azi2[I1/2/(1 + I1/2) - 0.3I]
where A and B are solvent (water) parameters which vary with pressure and temperature and have values of 0.5101 and 0.3285 x 108, respectively, at 25oC and 1 bar. C and aoi are fit parameters for the ith component and zi is its charge. In general, aoi ranges from 2 x 10-8 to 8 x 10-8 and depends upon the size of the ith ion. I is the ionic strength of the total aqueous solution which can be defined as
(3,17) I = 0.5(mjzj2 + mj+1z2j+1 + ... for all aqueous species).
When a salt is dissolved in water, the concentrations can refer to the stoichiometric concentrations resulting from dissolution or the true concentrations which take into account new species or complexes formed between ions in solution. If I is computed using the stoichiometic or total molalities of each component, then I is called a stoichiometric ionic strength. If I is computed using the true or "free" molalities of all species (including complexes) in solution, then I is called a true ionic strength.
The importance of the distinction is that the ionic strength is used to compute activity coefficients of aqueous species. If we use the true molality of a species such as Ca2+ in solution, then we use the true ionic strength (based on all true molalities) to compute the "free" activity coefficient. On the other hand, if we use the stoichiometric molality of Ca2+, then we use the stoichiometric ionic strength (based on all stoichiometric molalities) to compute the stoichiometric activity coefficient. The resulting activity of Ca2+ must be the same following either convention, hence the stoichiometric activity coefficient is always less than the "free" activity coefficient.
For use in the Debye-Hückel equation, use the "TOTAL" ionic strength computed for the entire solution using all dissolved salts. A common mistake among students is to use only the ionic strength contribution of a component in the Debye-Hückel equations in computing its activity coefficient.
If a solid is composed predominantly of one component, that component can usually be assumed to have an activity coefficient of unity, leading to an activity of unity because the mole fraction is near unity. Similarly, water in an aqueous solution can usually be assumed to have an activity coefficient of unity and an activity of unity if the solution is not a brine. In reality, for calculating osmotic pressures and other parameters, the activity coefficient of water cannot be assumed to be unity. However, for use in the mass-action law, the activities of nearly pure solids and water in aqueous solutions (not brines) are often set equal to unity.
For a noncharged species, e.g., SiO2 or H4SiO4, CH4 aq the aqueous activity coefficient can often be computed from a linear relationship with the solution ionic strength:
(3,18) log gi = kiI
where ki can be thought of as a salting-out coefficient due to the fact that the solubility of a neutral species frequently decreases with increasing ionic strength I. The solubility decrease is related to the decrease in free water molecules caused by the hydration of the dissolved salts. This decrease in solubility is input in the mass-action law by the increase in activity caused by the increase in gi.
Stoessell (1982, Geochim. Cosmochim. Acta, 46, 1327-1332) measured the
activity coefficient of methane in various single salt solutions at 25oC,
showing the above equation represented the methane activity coefficient
but with a different salting-out constant for each different type of salt
solution, e.g., NaCl versus CaCl
The pH is defined as
(3,19) pH = -log aH+
in an aqueous solution where the aH+ refers to the activity of the free (uncomplexed) H+ ion. In field samples, the pH must be measured immediately, before gas exchange takes place with the atmosphere. The temperature of the sample at the time of pH measurement would be recorded, so the pH can be back-computed for the temperature of the in situ sample.
The definition of pH relates to our concept of acids and bases in aqueous solutions. An acidic solution has a higher concentration of hydrogen ions H+ than of OH- ions. These two ions are related through the disassociation reaction with water H2O.
H2O = H+ + OH- where KH2O = aH+aOH-/aH2O is the mass-action law.
At 25oC and 1 bar (near atmospheric pressure), KH2O is 10-14.
The activity of water in most aqueous solutions (except brines) can be assumed to be unity, leading to
10-14 = aH+aOH- for the mass-action law.
From the mass-action law, in a neutral solution the H+ and OH- activities must each be 10-7. This neutral solution is neither acid nor basic and has a pH of 7 (-log aH+ = 7). Acidic solutions have a pH lower than 7 and basic solutions have a pH greater than 7.
The pH can be measured with a glass electrode or calculated from any mass-action expression for a reaction at equilibrium which involves the hydrogen ion. All non-oxidation-reduction reactions involving only aqueous components can generally be assumed to be at equilibrium. For example, the reaction between inorganic carbon species in an aqueous solution
HCO3- = CO32- + H+ can be used to compute the pH. The mass action law is
KHCO3(-) = aCO3(2-)aH+/aHCO3(-).
where pH = -log aH+ = -log KHCO3-+ log aCO3(2-) + log aHCO3(-).
The activities can be computed by measuring the concentrations of CO32- then be computed using the KHCO3(-) value.
Definition of pe and Eh
The pe is defined as
(3,19) pe = -log ae-
where ae- represents an aqueous electron activity in solution. In reality, free aqueous electrons do not exist in a solution; however, we can still use the concept. The pe is a measure of how oxidizing or reducing a solution can be. An oxidizing solution can be considered to have a deficiency in aqueous electrons, promoting the giving up of electrons to the solutions by ions being oxidized. A reducing solution can be considered to have an excess of aqueous electrons, promoting the acceptance of electrons by ions being reduced. Hence an oxidizing solution has a low ae- (positive pe) and a reducing solution has a high ae- (negative pe).
pe can be calculated from the activities of reaction components in any "half cell" oxidation-reduction reaction. The term "half-cell reaction" implies that the electrons released or consumed are not cancelled out in the overall reaction. A major problem is the general lack of equilibrium in oxidation-reduction (redox) reactions, combined with the necessity of accurate measurements of ion concentrations in different oxidation states. The latter is particular difficult, because exposure to the atmospheric oxygen during sampling and measurement will change the elemental proportions in different oxidation states.
An example of oxidation and reduction in an aqueous solution is the oxidation of ferrous iron with molecular oxygen.
Fe2+ + 0.25O2 + H+ = Fe3+ + 0.5H2O
The reaction is actually the sum of two half-cell reactions. Either one of the half-cell reactions can be used to compute pe. For example, the oxidation half-cell reaction, involving the oxidation of aqueous iron,(becomes more positive), is
The reduction half-cell reaction involves the reduction of
aqueous oxygen (becomes more negative).
0.25O2 + e- + H+ = 0.5H2O, and pe could also be
computed using the mass-action law for this reaction.
Note that the K value in the mass-action law is for a particular reaction, hence K will have a different value for the oxidation half-cell reaction (for Fe) then for the reduction half-cell reaction (for O2); however, the pe calculated from each half-cell reaction should be identical.
The pe calculation from half-cell reactions assumes equilibrium for the overall oxidation-reduction reaction, because the K values hold only for equilibrium. Frequently, the calculated pe values are different using different half-cell reactions. There are many oxidation-reduction reactions occurring in a solution, and pe values computed from the different half-cell reactions of the various oxidation-reduction reactions may not agree for the same solution, indicating a lack of equilibrium. Hence, the calculated pe values are often difficult to apply to the solution!
A different approach is to use an inert platinum electrode to
measure the half-cell voltage of the aqueous solution relative to the
oxidation of 0.5H2 to H+ in the standard hydrogen electrode
SHE. The standard hydrogen electrode consists of hydrogen gas
being bubbled under a pressure of 1 bar through an HCl aqueous solution
in which H+ has unit activity. By convention, the voltage of
SHE is set to zero, so the measured cell voltage (at zero current flow)
is assigned to the aqueous solution connected through it by the salt bridge.
The measured cell voltage is called the Eh. The platinum electrode
contributes no voltage but is simply a conductor for the flow of electrons.
In practice, the SHE half cell is too hard to maintain for
making measurements. Instead, the cell voltage is measured across an
external circuit between the platinum electrode in the sample solution
and a reference electrode, connected by a salt bridge to the solution.
The Ehref of the reference electrode relative to oxidation
in SHE has to be added to the measured voltage EMF to compute Eh
for the solution. Ehref is 0.2444 volts at 25oC and
1 bar for the saturated (4.16 M KCl) calomel electrode, corresponding to the reduction of
0.5Hg2Cl2 to Hgo + Cl-, relative
to oxidation in the standard hydrogen electrode. The saturated Ag-AgCl
reference electrode has a voltage of 0.1986 volts (199 mv) for the reduction of AgCl to Ag(s)
+ Cl- at 25oC and 1 bar.
(3,20) Eh = EMFmeasured + Ehref
By thermodynamic convention, the voltage corresponding to SHE is zero and the electron activity in SHE is unity. When these conventions are combined with other conventions used in the half-cell reactions, the Eh can be shown to be related to the pe by the following relationship.
(3,21) pe = FEh/(2.303RT)
where F is the Faraday constant (96,485 J/V/mole), R is the gas constant (8.314 J/mole) and T is the temperature in degrees Kelvin (298.15oK = 25oC).
However, measuring the Eh with a platinum electrode is not always accurate, because the surface of the electrode can be poisoned upon contact with the solution, or the redox reactions are not equilibrium so the measured value doesn't correspond to the actual Eh. If the solution contains the Fe redox couple (e.g., Fe2+ and Fe3+, or the Mn redox couple or significant HS-, then the solution generally gives a reliable Eh reading. In particular, the presence of HS- always results in a negative Eh and is a good indicator for anaerobic conditions, e.g., low dissolved O2.
For example, under what pe and Eh conditions would sulfate and aqueous sulfide be at equilibrium at a pH of 7.
SO4+2 + 9H+ + 8e- = HS- + 4H2O
in which log K = 27.86
Hence, log pe = -3.94 (Eh = -233 mV) if the a(SO4+2) = a(HS-)
In general, any reaction taking place completely within a phase can be considered to be at equilibrium. The major exceptions are the oxidation-redox reactions discussed above within an aqueous solution. Reactions taking place wholly within a phase are usually at equilibrium because the reaction takes place rapidly.
Reactions taking place between phases can usually not be assumed to be at equilibrium. Exceptions are the generally fast reactions involving exchange of gaseous components from a gas phase to an aqueous phase or the adsorption and desorption of components between an aqueous solution and a solid surface. For example, the transfer of CO2 between the atmosphere and the surface ocean waters or the exchange of cations between clay interlayers and ground water usually reach equilibrium.
The importance of reactions being at equilibrium is the advantage of being able to use the mass-action law to predict equilibrium concentrations. However, even if reactions are not at equilibrium, we can generally test them to see how close they are to equilibrium and evaluate what changes in component concentrations are to be expected. The mass-action laws are extremely useful because all reactions will tend towards equilibrium if given enough time.
The saturation of one component in a phase represents an equilibrium between the component in two different phases. For example, the saturation of silver chloride AgCl in water represents an equilibrium of solid AgCl with the aqueous solution. The reaction can be written as
AgCl = Ag+ + Cl- where the ions are in solution and the compound is a separate solid phase. The mass-action law is
KAgCl = aAg+aCl-/aAgCl where the activity of the solid AgCl can be assumed to be unity if the solid AgCl is essentially a pure AgCl phase with only minor impurities. At 25oC and 1 bar, KAgCl equals 10-9.74. Thus,
10-9.74 = aAg+aCl-
Assuming activity coefficients of unity for Ag+ and Cl-, AgCl would dissolve in pure water until each of the ions had a molality of 10-4.87, a close approximation of zero concentration. AgCl is very insoluble. The precipitation of AgCl can be used to measure aqueous Cl- in solution.
What is the 25oC and 1 bar calcite solubility and pH in water in equilibrium with the atmosphere with a mole fraction of CO2 of 10-3.5. The aqueous species are listed below in solution. The µio values at 25oC and 1 bar are listed below.
| Component | µio in cal/mol |
|---|---|
| Ca2+ | -132,300 cal/mol |
| CaCO3o | -258,470 |
| CaHCO3+ | -274,330 |
| HCO3- | -140,260 |
| CO32- | -126,170 |
| H2CO3 | -148,940 |
| H2O | - 56,687 |
| OH- | - 37,594 |
| H+ | 0 |
| CO2 gas | - 94,254 |
| CaCO3 calcite | -269,800 |
Write the reactions for the complexes and the solubilities
of calcite and CO2 in terms of the following components:
H+, HCO3-, Ca2+, and
H2O. Write out the corresponding mass-action equations
for the reactions. Write an electrical solution balance.
The electrical solution balance equation can be simplified by substitution of the mass action equations so that only the molality of H+ is a variable plus the activity coefficients of the various components. Initially assume the activity coefficients are one. Solve for the molality of H+ by trial and error on a spreadsheet. The calculate component molalities from the mass-action expressions with this molality. Use these to calculate an ionic strength and calculate activity coefficients for the components with the Davies equation. Again solve for the molality of H+ in the electrical neutrality expression, using the new activity coefficients. Then recalculate the molalities from the mass action equations and repeat the process if there is any significant change in the molalities. The solubility of calcite is equal to the molality of Ca2+ + CaHCO3+ + CaCO3o.
Dissolve 0.001 moles of CaSO4 in 1 kg of water containing 0.001 moles of anhydrite, CaCl2. Several important species are listed below with their µio at 25oC and 1 bar. Cl- does not form complexes significantly so its molality will remain at 0.002. Calculate the true molalities of the species listed below. Use the Davies equation below for aqueous activity coefficients of charged species.
log gi = -Azi2[I1/2/(1 + I1/2) - 0.3I] where I is the true ionic strength. At 25oC and 1 bar, A = 0.5085. Assume neutral species have an activity coefficient of 1.
| Component | µio in cal/mol |
|---|---|
| Ca2+ | -132,300 cal/mole |
| SO42- | -177,750 |
| CaSO4o | -310,270 |
Calculate the equilibrium constant for the formation of
CaSO4o in the reaction below using the relation
log K = -DGr/2.303RT
where R = 1.987 cal/oK/mol and T = 298.15oK:
CaSO4o = Ca2+ + SO42- where K = [mCa(2+)mSO4(2-)/mCaSO4(o)][gCa(2+) gSO4(2-)/gCaSO4(o)]
log K is found to equal 0.16 or K = 1.45.
The procedure uses two mass balance equations for the total number of moles of aqueous Ca and aqueous sulfate:
nCa = 0.002 = mCa(2+) + mCaSO4(o)
nSO4 = 0.001 = mSO4(2-) + mCaSO4(o)
Initially neglect CaSO4o, i.e., use the stoichiometric (total) Ca and SO4 molalities to calculate the ionic strength and then calculate gi, the activity coefficients for Ca2+ and SO42-. Use these stoichiometric activity coefficients together with the stoichiometric molalities in the above equilibrium quotient to calculate mCaSO4(o) and then use the mass balance equations to calculate mCa(2+) and mSO4(2-). If the molalities differ significantly from the previous step then repeat the procedure until they agree without significant difference. When repeating the procedure use the new ionic strength with the new molalitites to recalculate the activity coefficients.
This is the same problem as before but now we want to dissolve CaSO4 (anhydrite) until it is at equilibrium with the solution. We want to know the solubility of anhydrite in this solution.
The only additional information needed is the anhydrite equilibrium reaction. We assume anhydrite is pure so its activity is one.
CaSO4 = Ca2+ + SO42- where K = [mCa(2+)mSO4(2-)][gCa(2+)gSO4(2-)]
and µio of anhydrite, CaSO4, is -315930 cal/mol, so K = 0.000049008.
In this case we can solve directly for mCa(2+) in the above reaction quotient because mCa(2+) = mSO4(2-) + 0.001. We use the stoichiometric molalities to calculate the stoichiometric ionic strength and then the activity coefficients. We substitute for mCa(2+) in the reaction quotient which produces a quadratic equation and solve for mSO4(2-). We then use the new molalities to calculate mCaSO4(o) and a new "true" ionic strength and calculate "true or free" activity coefficients. The process is repeated, solving again for mSO4(2-). The new molalites are compared with the old molalities to see if they have changed significantly. If so, the calculation is repeated. The major error in this process is that the Davis equation is not accurate at high ionic strengths (above 1) and the true solubility is much less because of a large increase in activity coefficients at high ionic strengths.
Aluminum solubility is usually controlled in soils by gibbsite precipitation (Al(OH)3). The precipitation reaction is written as
Al(OH)3 gibbsite + 3H+ = Al3+ + 3H2O, where Kgibbsite = aAl3+/(aH+)3
where the activities of water and the pure solid gibbsite can be assumed to be unity (one). The solubility of aluminum is limited by the maximum value that the activity of Al3+ can have in the above equation, such that the ratio of aAl3+/a3H+ remains below Kgibbsite. With increasing activity of Al3+, eventually gibbsite will precipitate. However, aluminum also form aqueous complexes with hydroxides and maintains equilibrium with these complexes, e.g.,
Al3+ + nH2O = Al(OH)n3-n + nH+, where KAl(OH)n(3-n) = (aAl(OH)n)(3-n)(aH+)n/aAl3+
If the activity of Al3+ increases, more Al(OH)n3-n forms, helping to keep the activity of Al3+ below the solubility of gibbsite. Determine the solubility of gibbsite in (sum of the molality of aqueous Al species) as a function of pH (8 pH values: 3, 4, 5, 6, 7, 8, 9, 10). Use the following standard state Gibbs free energies of formation at 25oC and 1 bar to calculate the equilibrium constants and use the Davis Equation to calculate activity coefficients. The OH- data is given so its concentration can be calculated to help compute the ionic strength of the solution for use in the Davis Equation. Note that R equlas 1.9872 cal/mol K at 25oC (298.15oK).
| Component | µio in cal/mol |
|---|---|
| Al3+ | -116,551 |
| Al(OH)2+ | -166,477 |
| Al(OH)2+ | -215,511 |
| Al(OH)3o | -263,050 |
| Al(OH)4- | -312,094 |
| H2O | - 56,678 |
| H+ | 0 |
| OH- | -37,572 |
| Al(OH)3 gibbsite | -276,018 |
To do this problem you simply calculate the activity of Al3+ in equilibrium with gibbsite at each pH value and then use the the equilibrium reactions linking Al(OH)n3-n with Al3+ to calculate the activities of each Al(OH)n3-n species. The sum of all the molalities of all these Al species would be the solubility of gibbsite at each pH point. However, at this stage, you have only calculated the sum of the activities. By assuming the activity coefficients are one, each activity becomes a molality and can be used to calculate an ionic strength. This ionic strength can be used to compute the activity coefficients of each of the species and they can be inserted into the activity expression to calculate molalities of each of the species. The molalities of the aluminum species can then be summed to get the solubility of gibbsite.
The solubility of inert gases can be predicted using equilibrium between the gas phase, e.g., the atmosphere, and the aqueous solution. The term inert implies that other reactions such as hydrolysis (as in the case of dissolved CO2) are not taking place.
The mass-action expression for a gas component G in equilibrium with both a gas phase and an aqueous solution is
Ggas = Gaq, where KG = aG aq/aG gas = (mGgG)aq/ (XGcGP)gas
or mG aq =
XGP(KGcG gas/gG aq)
or expressed in terms of the Henry's law constant kH and the partial pressure
of the gas component of the gas phase XGP
(3,22) kH = mG aq/XGP or
(3,23) kH = (KGcG gas/gG aq)
Note that kH is only a constant if the activity coefficient ratio of the component between the gas phase and the aqueous solution is constant. Henry's law constants have been measured for a large number of gases in equilibrium with the atmosphere with different aqueous solutions. These constants can be used to predict the solubility of various inert gases in natural waters in equilibrium with the atmosphere. If a gas hydrolyses into other components, e.g., CO2, than the solubility is only that of the unhydrolyzed component and does not include HCO3- or CO32-.
An example of the use of Henrys Law constant can be made with the aqueous solubility of O2 in contact with the atmosphere. kH in units of mol atm/l is 1.28 x 10-3. This is a value measured for dry air in which the mole fraction of O2 is 0.2095 (also the volume fraction). Water vapor in saturated air has a mole fraction of 0.0313 so to correct the mole fraction of O2, multiply P by (1. - 0.0313). Note that kH values are almost always reported in molarity and atmosphere units, rather then in molality and bar units.
MO2 = XG(kH) = 0.2095(1-.0313)(1 atm)(0.00128 mol atm/l) = 0.00026.
Cation Exchange
The exchange of an aqueous component in solution with a component on the surface of a solid is a common reaction between clays and an aqueous solution as well as colloids and an aqueous solution. For example, the exchange of aqueous cadmium with calcium on the surface of a clay can be written as
Cd2+ + Ca-clay = Ca2+ + Cd-clay
where the mass-action law becomes
(3,24) K = aCa(2+)aCd-clay/ aCd(2+)aCa-clay
and ai-clay = Xi-clayli-clay where Xi-clay is the mole fraction of i on the exchange sites, e.g,
XCd-clay = mCd on clay/(molesCa on clay + molesCd on clay)
However, if the exchange involved cations of different charge, Xi-clay has been used as either the charge fraction or mole fraction of the exchange ion (as shown below). li-clay is the activity coefficient of i on the exchange sites. There is no way to compute or measure li-clay independently of ai-clay.
The ratio of the solution activities is usually called a partition coefficient Kp, e.g.,
(3,25) Kp = aCa(2+)/aCd(2+) = KaCa-clay/aCd-clay
In the literature, these partition coefficients can be defined in many different ways. For example, they may be defined in terms of concentration ratios rather than activity ratios, so the activity coefficients would be removed from the activities. They are sometimes defined for the direct exchange reaction of the component with itself on the solid. The important point is never use a partition or distribution coefficient without knowing how it is defined. The same is true of equilibrium constants.
If the reaction involved the exchange of a monovalent with a divalent cation, the reaction may be written with the Gapon convention, e.g.,
Cd2+ + Na2-clay = 2Na+ + Cd-clay
and the mass-action law becomes
(3,26) K = (aNa(+))2aCd-clay/
[aCd(2+)aNa(2)-clay]
where the mole fractions of Na2clay and Cd-clay are identical to the charge fractions of Cd2+ and Na+ on the exchange sites, e.g.,
charge fraction of Na on clay = molesNa on clay/ (molesNa on clay + 2 molesCd on clay)
= 2 molesNa(2)-clay/ (2 molesNa(2)-clay + 2 molesCd-clay) = mole fraction of Na2-clay
The power exchange function used in geochemistry is a modification of the above equilibirum constant and has a power exchange constant, Kp.e..
(3,27) Kp.e. = [(aNa(+))2 /aCd(2+)][XCd-clay/ XNa(2)-clay]nwhere n is a exponent to fit the data. Note that there are no activity coefficient terms for Na2-clay and Cd-clay in the expression so the constant is not a true constant but has to contain those terms which will vary with the clay composition.
An alternative procedure to the Gapon convention is to write the exchange reaction between the monovalent and bivalent cations as
Cd2+ + 2Na-clay = 2Na+ + Cd-clay
and the mass-action law becomes
(3,28) K = (aNa+)2aCd-clay/[aCd2+(aNa-clay)2]
where the Vanselow convention uses mole fractions and the Gaines-Thomas convention uses charge fractions, respectively, of Cd2+ and Na+. This produces confusion becasue the charge fractions are not equal to the mole fractions. Your instructor believes the Gapon Convention is the thermodynamically consistent way of handling heterovalent cation exchange (Stoessell, 1998, Clays and Clay Minerals, 46, 215-218).
Three exchange algorithms called isotherms are commonly used in which a component M exchanges with itself on sites on a solid: Distribution, Freundlich, and Langmuir. All of these neglect explicit inclusion of the competition by other components for the sites on the solid. The reaction can be written as follows:
M = Msolid where
(3,29) K = aM solid/aM = [XM solid/mM] [lM solid/gM]The first bracketed ratio can be measured experimentally and is a modified Distribution coefficient which is often used for prediction in other solutions. However, this ratio will be a function of the solution composition, so it cannot be generally considered a constant.
In practice, the isotherms are written with units for M of mg/kg of solid called SM and mg/liter of solution called CM. KD, the Distribution coefficient is defined as:
(3,30) KD = SM/CM
Note that the concentration of M cannot increase without changing the ratio which would change the value of KD. Hence, the implication is that there are an unlimited number of sites available for M to sorb on in order to prevent the ratio from increasing.
The Freundlich (isotherm) algorithm is a modification of the Distribution algorithm.
(3,31) KF = SM/(CM)n
where KF is the Freundlich coefficient and n lies between a range of 0.7 and 1.1. If n is one, the expression reduces to the Distribution algorithm.
Other times, experimental data can be fitted to the Langmuir adsorption algorithm (Langmuir isotherm) which has a maximum value of SM called Smax incorporated as a parameter into the algorithm.
(3,32) SM = SmaxKLCM /(KLCM + 1) where KL is the Langmuir coefficient.
In general, in modeling contaminant flow, either simple Distribution, Freundlich, or Langmuir isotherms are used. Experimental data are used to pick which of the best three equations fit the data. To determine the fit to the Distribution isotherm, plot S versus C and fit the experimental data to the equation of a straight line with 0 intercept. The slope of the fit equation is KD. To determine the fit to the Freundlich isotherm, plot log S versus log C and the intercept is log KF and the slope is n. To determine the fit to the Langmuir equation, plot S-1 versus C-1 and the intercept is (Smax)-1 and the slope is (SmaxKL)>-1.
The modified Power Exchange Function for this reaction:
2Na+ + Zn-Clay = Zn2+ + Na2-Clay,
is
Kpe = [MZn2+/(MNa+)2]
[XNa2-Clay/(XZn-Clay)]n
where the modification is to use molarities instead of aqueous activities. Note n is a fit
parameter and the X symbols refer to mole fractions or charge fractions on the clay exchange
sites. Because of the use of the Gapon convention, the mole fractions are equal to the charge
fractions.
Use values of Kpe and n of 0.32 and 0.80, respectively, and assume the clay has
100 meq of exchange sites per kg of clay. Assume a constant equilibrium molarity of
0.001 for Na and 10 different equilibrium Zn molarities of 1 x 10-7,
2 x 10-7, 3 x 10-7, ..., 1 x 10-6. Calculate the equilibrium
XNa2-Clay/(XZn-Clay) for each solution composition of
0.001 M Na paired with one of the Zn molarities.
Express this data as mg of Zn/kg of clay and mg of Zn/liter of solution. You will have to
make use of the total of 100 meq/kg of exchange sites on the clay. Plot the data and then
fit it to the 3 isotherms and comment on the fits.
Equilibrate different concentrations of NaH2PO4 in 25 ml of a NaCl solution containing 0.1 grams of bauxite of a given size. Put the bottles on a shaker overnight to equilibrate. Use 0.1, 0.2, 0.4, 0.8, 1.6, 3.2, and 6.4 mM NaH2PO4 solutions in a matrix of either 0.1 M, 0.055 M, and 0.01 M NaCl - so 7 bottles per ionic strength. Each student should do one of the NaCl solutions, i.e., pick one of the above concentrations. Measure the PO4 concentrations on the liquid ion chromatograph and plot the data as mM PO4 in solution versus mM PO4 absorbed per cm2 of bauxite. Note that the density of bauxite is about 2.4 g/cm3, and you have to assume a cube of a particular size and calculate the total surface area for your 0.1 gram of sample. Fit your data to a Langmuir or Freundlich isotherm. Compare your results to determine the effect of ionic strength.
Each student can do one of the anions. Place 0.50 grams of bauxite in 200 ml of 0.01 M NaCl and 0.00005 M of the sodium salt of the anion, e.g., either Na2SO4, NaH2PO4, or NaNO3. Let sit overnight to come to equilibrium. In the titration, add enough acid to lower the starting pH to about 4, let equilibrate for 10 minutes and take a 2 ml sample and filter it. Try not to remove any of the bauxite. Record how many ml of acid were added, how many ml of solution were removed and the equilibrated pH. Add base to raise the pH a half pH unit, let equilibrate for 10 minutes and take another 2 ml sample and filter it. Again try not to remove any of the bauxite. Record the equilibrated pH, the ml of base added, and the ml of sample taken. Repeat the process to bring the pH to about 10. Analyze your solutions on the liquid ion chromatograph and compute the amount of sorption on the solids from the concentration left in solution. You will have to correct for the volume changes due to addition of acid and for the volume changes due to removal of sample. Plot the amount of anion sorbed as a function of pH.
The retardation of a concentration front of a pollutant B can be computed using a mass balance across the concentration front. This method uses a flushing factor equal to the mass needed to advance the concentration front one pore volume divided by the difference in mass per pore volume of the pollutant between that in upstream pore fluid and in the downstream pore fluid.
The concentration front moves downstream by replacing downstream ground water with the upstream ground water and the downstream mass of pollutant on sediment sites with those in equilibrium with the upstream ground water. If the pollutant is absent downstream, the retardation of the front Rf is equal to the mass of pollutant (per pore volume) on the upstream solids plus in the upstream ground water divided by the amount of pollutant (per pore volume) in the upstream ground water.
The retardation factor of the front is written below in equation form, using the ratio of r/t, the ratio of the dry sediment density to the sediment porosity fraction, to normalize to one pore volume. The equation is simply a statement of the mass constraints governing the movement of the front. The retardation factor is the factor by which the ground water velocity should be divided to get the velocity of the concentration front.
Rf = [(CB,upstream - CB,downstream) + (SB,upstream - SB,downstream)] (r/t)/(CB,upstream - CB,downstream) or
(3,33) Rf = 1 + (SB,upstream - SB,downstream) (r/t)/ (CB,upstream - CB,downstream).
If the Distribution Coefficient holds, substitution into the above equation yields:
(3,34) Rf = 1 + KD(r/t)
The latter is the well known retardation factor presented in hydrology texts. It is almost always derived from the advection and sorption partial differential equation. That equation, which describes the movement of a dissolved component and is used in numerical modeling, is based on the continuity equation which is a statment of the conservation of mass over a finite volume element. As will be shown later, the partial differential approach is important in that the retardation of a particular composition can be determined if the appropriate algorithm describing sorption is known. However, since that algorithm is never known accurately in the field, the second equation above gives a very accurate calculation of the retardation factor, as recently shown by Stoessell (1999, Ground Water, 37, 701-705) in a comparison of retardation factors from published column experiments versus those predicted by the equation.
A sandstone has a porosity fraction of 0.25, and a density of 2.5 g/ml (equal to 2.5 kg/l).
There are 0.25 moles of sites for Ca and for Zn per kg of rock. Upstream of the concentration
front, there are 0.00001 moles of Zn and 0.001 moles of Ca per liter of solution. Downstream
of the concentration front, there are 0.001 moles of Ca and no Zn per liter of solution. The
partitioning of Zn and Ca on exchange sites on the sandstone is controlled by the following algorithm
in which M is molarity (moles per liter of solution) and X is the mole fraction on the exchange sites:
1.5 = (MZn/MCa)(XCa-rock/XZn-rock)1.4
What is the velocity of the concentration front if the ground-water velocity is 1 meter per year?
Use the mass balance by looking at the difference (per liter of ground water) in zinc in ground water (g.w.) and on the sandstone (s.s.) upstream of the concentration front from that in g.w. and on the s.s. downstream of the concentration front and devide this by the difference in zinc concentration per liter of g.w. upstream of the front from per liter of g.w. downstream of the front.
First, solve for the concentration of Zn on the s.s. upstream of the concentration front and reason
that since there is no zinc in the g.w. downstream of the concentration front, there must also be no zinc on the
downstream sandstone. You must use the relation that the sum of the mole fractions equals 1 on the exchange sites.
1.5 = (10-5/10-3)[(1-XZn-rock)/XZn-rock)]1.4
or XZn-rock = 0.027145
Using the symbols M and Sm, respectively, for moles of Zn per liter of g.w. and moles of Zn per kg of s.s, the
retardation factor Rf is:
Rf = 1 + (SmZn,upstream)
(r/t)/ (MZn,upstream).
Rf = 1 + [(0.027145)(0.25) moles Zn/kg ss][(2.5 kg s.s./liter s.s.)/0.25 liter g.w./liter s.s.)]
/[10-5 moles Zn/liter g.w.]
Rf = 6,787. So the concentration front is moving at (1/Rf)
g.w.vel] = (1/6,787)(1 m/yr) or 0.15 mm/yr. Not very fast!
A sandstone has a porosity fraction of 0.25, and a density of 2.5 g/ml (equal to 2.5 kg/l). In ten years a pollution front containing zinc has moved 0.5 meters downstream (from a point source of zinc) through the sandstone. Upstream of the front, the ground water has 5 mg/l of zinc in solution and 49.5 mg/kg of zinc on the sandstone. Downstream of the front there is no zinc in ground water and no zinc on the sandstone. What is the velocity in m/y of the ground water in the sandstone?
In the above question, the pollution point source is removed and unpolluted ground water now enters the 0.5 m of zinc-containing sandstone and flows through it. After 1 year what is the average concentration of sorbed zinc in mg/kg of sandstone, within the 0.5 m block of sandstone? There are two different approaches to answering this question. One way of answering this question requires the ground-water velocity and the other procedure doesn't.
The "so-called" octanol partition coefficient Kow gives the equilibrium distribution of the solute between equal volumes of an octanol solution and an aqueous fluid, measured at a particular temperature and pressure. The octanol solution is often used to simulate any nonpolar solvent, e.g., gasoline. Hence, Kow is used as the partition coefficient Kpc for doing partitioning calculations of a component between water and any organic phase.
(3,35) Kow = Xoc/Xaq
where Xoc + Xaq = 1.
The X fractions refer to the actual solute fraction of the total moles of solute in both phases, not their mole fractions within the two fluids. Because equal volumes were used, the mole fraction ratio equals the concentration ratio in units of mass per unit volume. For example, consider the exchange of n moles of benzene between equal volumes of octanol and water.
Kow = {nbenzene in octanol/
[nbenzene in octanol + nbenzene in water]}/
{nbenzene in water/
[nbenzene in octanol + nbenzene in water]}
= nbenzene in octanol/nbenzene in water
= [nbenzene in octanol/liter of octanol]/[nbenzene in water/liter of water]
= ratio of benzene concentrations (molarities) in octanol to water.
Note that Kow is equal to an apparent thermodynamic equilibrium constant for a chemical reaction passing a component between the two phases in which the constant contains the thermodynamic equilibrium constant and the ratio of activity coefficient terms. The concentration units in this apparent thermodynamic equilibrium constant are in terms of molarity/liter.
As an example in the use of this constant, suppose there was a leak of 1000 gallons of gasoline, containing 100 moles of benzene, into a ground-water aquifer. If the gasoline came into contact with 10,000 gallons of water, describe the partitioning of benzene between the two phases. Kow for benzene is 2. Assume the same value holds for partitioning benzene between gasoline and water.
The mass balance is nmoles benzene in gasoline + nmoles benzene in water = 100,
and Kpc = 2 = [nmoles benzene in gasoline/1,000 gallons]/ [nmoles benzene in water/10,000 gallons].
With these two relations, we can solve for the two unknowns: nmoles benzene in gasoline and nmoles benzene in water.
10 moles of xylene are within 1000 liters of an organic solvent forming a liquid pool in contact with ground water in the subsurface. The partition coefficient of xylene between the organic fluid and water is 75. The partition coefficient is the concentration (moles/l) ratio of xylene in the organic fluid to that in water.
There are two scenarios. What will be the concentration of xylene in the organic fluid if 10,000 liters of ground water equilibrate with the organic solvent? What would be the final concentration of xylene in the organic fluid if instead of 10,000 liters equilibrating at once, 10 successive 1,000 liters of ground water equilibrated with the organic fluid and then moved on, leaving the organic fluid behind?
The conservation of mass equation is: ntotal xylene = [nxylene in org. sol. + [nxylene in water]
and the partition coefficient function is:
Kpc = 75 = [nxylene in org. sol./litersorg. sol.]/
[nxylene in water/literswater]
By substitution into the conservation of mass equation:
ntotal xylene = nxylene org. sol.[1 + (literswater/litersorg. sol.)1/75]
For the first situation: 10 = nxylene in org. sol.[1 + 0.1333]
Hence, nxylene in org. sol. = 8.824 moles
For the second situation, the xylene concentration is solved for a 1000 liter equilibration with water and a new total moles of xylene computed from that remaining in the 1000 liter organic solvent. The calculation is then repeated 9 more times.
The equation becomes: ntotal xylene = nxylene in org. sol(1.01333).
The initial step has ntotal xylene = 10. Hence, nxylene in org. sol.
= 9.868 moles which is then used in the next iteration as ntotal xylene
| # iteration | - total moles of xylene - | moles of xylene in org. sol. |
|---|---|---|
| 1 | 10 | 9.868 |
| 2 | 9.868 | 9.738 |
| 3 | 9.738 | 9.610 |
| 4 | 9.610 | 9.484 |
| 5 | 9.484 | 9.359 |
| 6 | 9.359 | 9.236 |
| 7 | 9.236 | 9.114 |
| 8 | 9.114 | 8.994 |
| 9 | 8.994 | 8.876 |
| 10 | 8.876 | 8.760 |
In nature, a pollutant such as xylene is eaten by bacteria. Bacterial decay usually follows first order kinetics in which
ln NXy/NoXy = -k(t-to)
where NXy is the moles of xylene at time t in days and k is the decay constant. Assume the half life of xylene is 90 days and the partitioning of 10 moles of xylene in the above exercise between 1000 liters of organic solvent and 1000 liters of ground water takes place over a 36 day time span. How much xylene is left in the organic phase after 360 days (about one year)in which ten 1000 liter equilibrations will have occurred.
Organic Chemistry Review
Organic compounds contain carbon atoms. Isomers are compounds having the same formulas but different structures.
Hydrocarbons are organic compounds containing carbon and hydrogen atoms. Each carbon atom has four electrons in the outer s and p orbitals, and can be used to make four bonds which can be in different combinations of single, double, and triple bonds. For the atoms being bonded, a single bond involves 2 electrons; a double bond, 4 electrons; and a triple bond, 6 electrons. Each hydrogen atom has only one electron to donate to make one bond. In the absence of radicals or carbonium ions, all four bonds of C are utilized in organic compunds, leading to 4 single bonds, or 2 single bonds and a double bond, or 1 single bond and a triple bond, or 2 double bonds.
The alkanes, alkenes, and alkynes exist as straight chains and as branching chains with attachments. They do not form rings.
| alkanes CnH2n+2 | alkenes CnH2n | alkynes CnH2n-2 |
|---|---|---|
| (parafins) | (olefins) | |
| only C-C bonds | one C=C bond | one C=C bond |
| methane CH4 | ||
| ethane C2H6 | ethene (ethylene) C2H4 | acetylene C2H2 |
| propane C3H8 | propene (propylene) C3H6 | propyne C3H4 |
| butane C4H10 | butene (butylene) C4H8 | butyne C4H6 |
| pentane C5H12 | pentene C5H10 | pentyne C5H8 |
| hexane C6H14 | hexene C6H12 | hexyne C6H10 |
| heptane C7H16 | heptene C7H14 | heptyne C7H12 |
| octane C8H18 | octene C8H16 | octyne C8H14 |
| nonane C9H20 | nonene C9H18 | nonyne C9H16 |
| decane C10H22 | decene C10H20 | decyne C10H18 |
Note that a diene has two C=C bonds, a triene has three C=C bonds, a tetraene has four C=C bonds and these are alkene modifications. A simple straight chain hydrocarbon diene has a formula CnH2n-2. Isoprene is the common name for 2-Methyl-1,3-butadiene which has a methyl (CH3) group attached to the second carbon atom and a double bond attached to the first and third carbon atoms.
Loss of a H atom leads to the formation of radicals which have an unshared electron, making them very reactive. Radicals do not carry a charge, i.e., they are not ions. Alkyl radicals are derived from alkanes by losing a H atom. Allyl radicals are derived from the alkenes by losing a H atom.
| alkyl radical |
|---|
| CnH2n+1 |
| methyl CH3 |
| ethyl C2H5 |
| propyl C3H7 |
| butyl C4H9 |
These include cycloalkanes and cycloalkenes, formed into rings from the chains of alkanes and alkenes. The rings differ from the aromatic hydrocarbon rings which have a resonance nature, such that the structure oscillates between two isomers. An example of a cyclic aliphatic hydrocarbon is cyclohexane (C6H12) which can be considered to have formed from hexane (C6H14) by removing the two end hydrogens and joining the two end carbons together.
Aromatic hydrocarbons form from derivatives of benzene rings C6H6 in which various radicals are attached to benzene rings through replacement of the hydrogens. Other aromatic compounds do not contain a benzene ring; however, they have similar chemical properties as benzene.
attachment of a single alkyl radical
methylbenzene (CH3)C6H5 (toluene)
ethylbenzene (C2H5)C6H5
propylbenzene (C3H7)C6H5
butylbenzene (C4H9)C6H5
attachment of a single allyl radical
styrene (C2H3)C6H5
attachment of two alkyl radicals - Note that isomers arise because of the different possible positions on the benzene ring of the two alky radicals.
dimethylbenzenz CH3-(C6H4)-CH3 (xylene)
ethylmethylbenzene or ethyltoulene CH3-(C6H4)-C2H5
If the benzene ring has a very complicated attachment than the benzene ring is considered the radical and named as a derivative of the attachment instead of vice versa. The C6H5 benzene radical is called a phenyl. Whenever benzene rings are attached together through an alkane, they are generally named as derivatives of alkanes.
diphenylmethane C6H5-(CH2)-C6H5
diphenylethane C6H5-(C2H4)-C6H5
Different classes of organic compounds are characterized by containing a specific functional groups. Often, the functional group is derived from a hydrocarbon by substitution of a particular radical group. Some of the characteristic functional groups are given below. R and R' stand for either aliphatic or aromatic groups.
An example of the reaction between a carboxylic acid and an alcohol to produce an ester (and releases a water molecule) is acetic acid + methanol = methyl acetate (ester) + water:
CH3(C=O)OH + CH3OH = CH3(C=O)O-CH3 + H2O
Nitrogen normally has three bonds, e.g., ammonia NH3, and can form a positively-charged group by losing an electron through the addition of a 4th bond to a proton, e.g., ammonium, NH4+. Nitrogen can also use all five electrons by forming 1 single bond (to R or OH), 1 double bond (to O), and 1 single bond with 1 ionic bond formed by donating an electron (to O).
Phosphorous usually forms three bonds but can also use the remaining pair of electrons, as a double bond with S or a mixture of 1 single bond plus 1 ionic bond caused by donating one of the electrons to O.
Sulfur can form one bond each to two carbons (sulfide or thioether) or one bond to a carbon and one to a hydrogen (mercaptan or thiol) leaving two pairs of unshared electrons. Sulfur can also form 4 bonds as a sulfoxide (sulfinic acid, H2SO3, and 6 bonds as a sulfone (sulfuric acid, H2SO4). When S uses two electrons to bond with O, the bonding consists of sharing one electron and donating one electron to form an ionic charge. When S forms a double bond with C or P it shares both electrons to form a conventional double bond.
acetonitrile CH3(C=N) + 2H2O = acetic acid CH3(C=O)OH + NH3.
Polymers consist of long chains of molecules, made up from monomers. In the examples below, the monomers are linked together by replacing the double bonds between carbon atoms in the alkene monomers with single bonds.
| polymer | monomer |
|---|---|
| polyethylene | ethylene C2H4 | </