Unit08a-aqgeochem
Transcript of Unit08a-aqgeochem
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Unit 08a : Advanced Hydrogeology
Aqueous Geochemistry
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Aqueous Systems
In addition to water, mass exists in thesubsurface as:
Separate gas phases (eg soil CO 2) Separate non-aqueous liquid phases (eg
crude oil) Separate solid phases (eg minerals
forming the pm) Mass dissolved in water (solutes eg Na +,
Cl-)
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Chemical System in Groundwater
Ions, molecules and solid particles in waterare not only transported.
Reactions can occur that redistribute massamong various ion species or between thesolid, liquid and gas phases.
The chemical system in groundwatercomprises a gas phase, an aqueous phaseand a (large) number of solid phases
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Solutions
A solution is a homogeneous mixture whereall particles exist as individual molecules or
ions. This is the definition of a solution. There are homogeneous mixtures where the
particle size is much larger than individualmolecules and the particle size is so smallthat the mixture never settles out.
Terms such as colloid, sol, and gel are usedto identify these mixtures.
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Concentration Scales
Mass per unit volume (g/L, mg/L, g/L)is the most commonly used scale for
concentration Mass per unit mass (ppm, ppb, mg/kg,
g/kg) is also widely used
For dilute solutions, the numbers arethe same but in general:mg/kg = mg/L / solution density (kg/L)
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Molarity Molar concentration (M) defines the
number of moles of a species per litre ofsolution (mol.L -1)
One mole is the formula weight of asubstance expressed in grams.
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Molarity Example
Na 2SO 4 has a formula weight of 142 g A one litre solution containing 14.2 g of
Na 2SO 4 has a molarity of 0.1 M (mol.L -1) Na 2SO 4 dissociates in water:
Na 2SO 4 = 2Na + + SO 42-
The molar concentrations of Na + andSO 42- are 0.2 M and 0.1 M respectively
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Seawater Molarity Seawater contains roughly 31,000 ppm of NaCl
and has a density of 1028 kg.m -3. What is themolarity of sodium chloride in sea water?
M = (m c/FW) * rwhere m c is mass concentration in g/kg;
r is in kg/m 3; andFW is in g.
Formula weight of NaCl is 58.45 31 g is about 0.530 moles Seawater molarity = 0.530 * 1.028 = 0.545 M
(mol.L -1)
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Molality
Molality (m) defines the number ofmoles of solute in a kilogram of solvent(mol.kg -1)
For dilute aqueous solutions attemperatures from around 0 to 40 oC,
molarity and molality are similarbecause one litre of water has a massof approximately one kilogram.
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Molality Example
Na 2SO 4 has a formula weight of 142 g One kilogram of solution containing 0.0142 kg
of Na 2SO 4 contains 0.9858 kg of water. The solution has a molality of 0.101 m(mol.kg -1)
Na 2SO 4 dissociates in water:
Na 2SO 4 = 2Na + + SO 42- The molal concentrations of Na + and SO 42-
are 0.202 m and 0.101 m respectively
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Seawater Molality Seawater contains roughly 3.1% of NaCl. What
is the molality of sodium chloride in sea water?m = (m c/FW)/(1 TDS)
where m c is mass concentration in g/kg;TDS is in kg/kg andFW is in g.
Formula weight of NaCl is 58.45
31 g is about 0.530 moles Average seawater TDS is 35,500 mg/kg (ppm) m = (31/58.45)/ (1- 0.0355) = 0.550 mol.kg -1
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Molar and Molal
The molarity definition is based on thevolume of the solution. This makes molarity a
temperature-dependent definition. The molality definition does not have a
volume in it and so is independent of anytemperature changes.
The difference is IMPORTANT forconcentrated solutions such as brines.
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Brine Example
Saturated brine has a TDS of about 319 g/L Saturated brine has an average density of
1.203 at 15o
C The concentration of saturated brine istherefore 265 g/kg or 319 g/L
The molality m = (265/58.45)/(1-0.319)) is
about 6.7 m (mol.kg-1
) The molarity M = (265/58.45)*1.203 is about
5.5 M (mol.L -1)
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Equivalents
Concentrations can be expressed inequivalent units to incorporate ionic
chargemeq/L = mg/L / (FW / charge)
Expressed in equivalent units, the
number of cations and anions in diluteaqueous solutions should approximatelybalance
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Partial Pressures
Concentrations of gases are expressedas partial pressures.
The partial pressure of a gas in amixture is the pressure that would beexerted by the gas if it occupied thevolume alone.
Atmospheric CO 2 has a partial pressureof 10 -3.5 atm or about 32 Pa.
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Mole Fractions
In solutions, the fundamental concentrationunit in is the mole fraction Xi; in which for jcomponents, the ith mole fraction is
Xi = n i/(n 1 + n 2 + ...n j),
where the number of moles n of a componentis equal to the mass of the componentdivided by its molecular weight.
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Mole Fractions of Unity
In an aqueous solution, the mole fraction ofwater, the solvent, is always near unity.
In solids that are nearly pure phases, e.g.,limestone, the mole fraction of the dominantcomponent, e.g., calcite, will be near unity.
In general, only the solutes in a liquid solutionand gas components in a gas phase will havemole fractions that are significantly differentfrom unity.
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Structure of Water
Covalent bonds between H and O 105 o angle H-O-H Water molecule is polar Hydrogen bonds join molecules
tetrahedral structure
Polar molecules bind to chargedspecies to hydrate ions in solution
105 o -
+
+
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Chemical Equilibrium
The state of chemical equilibrium for a closedsystem is that of maximum thermodynamic
stability No chemical energy is available to
redistribute mass between reactants andproducts
Away from equilibrium, chemical energydrives the system towards equilibriumthrough reactions
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Kinetic Concepts
Compositions of solutions in equilibrium withsolid phase minerals and gases are readily
calculated. Equilibrium calculations provide no
information about either the time to reachequilibrium or the reaction pathway.
Kinetic concepts introduce rates and reactionpaths into the analysis of aqueous solutions.
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Reaction Rates
After Langmuir and Mahoney, 1984
Mineral Recrystallization
Solute-Solute
Hydrolysis of multivalent ions (polymerization)
Adsorption-Desorption
Mineral-Water Equilibria
Secs Mins Hrs Days Months Years Centuries My
Gas-Water
Solute-Water
Reaction Rate Half-Life
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Relative Reaction Rates
An equilibrium reaction is fast if it takesplace at a significantly greater rate than thetransport processes that redistribute mass.
An equilibrium reaction is slow if it takesplace at a significantly smaller rate than thetransport processes that redistribute mass.
Slow reactions in groundwater require akinetic description because the flow systemcan remove products and reactants beforereactions can proceed to equilibrium.
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Partial Equilibrium Reaction rates for most important reactions
are relatively fast. Redox reactions are oftenrelatively slow because they are mediated by
micro-organisms. Radioactive decayreactions and isotopic fractionation areextremely variable.
This explains the success of equilibrium
methods in modelling many aspects ofgroundwater chemistry. Groundwater is best thought of as a partial
equilibrium system with only a few reactions
requiring a kinetic approach.
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Equilibrium Model
Consider a reaction where reactants A and B react toproduce products C and D with a,b,c and d being therespective number of moles involved.
aA + bB = cC + dD For dilute solutions the law of mass action describes
the equilibrium mass distributionK = (C) c(D)d
(A)a(B)b where K is the equilibrium constant and (A),(B),(C), and
(D) are the molal (or molar) concentrations
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Activity
In non-dilute solutions, ions interact electrostaticallywith each other. These interactions are modelled byusing activity coefficients ( g) to adjust molal (or molar)concentrations to effective concentrations
[A] = ga(A) Activities are usually smaller for multivalent ions than
for those with a single charge The law of mass action can now be written:
K = gc(C) c gd(D)d = [C]c[D]d ga(A)a gb(B)b [A]a [B]b
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Debye-Hckel Equation
The simplest model to predict ion ion activitycoefficients is the Debye-Hckel equation:
log gi = - Az i2(I)0.5
where A is a constant, z i is the ion charge, and I is theionic strength of the solution given by:
I = 0.5 SMiz i2
where (M i) is the molar concentration of the ith
species The equation is valid and useful for dilute solutionswhere I < 0.005 M (TDS < 250 mg/L)
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Extended Debye-Hckel Equation
The extended Debye-Hckel equation is usedto increase the solution strength for whichestimates of g can be made:
log gi = - Az i2(I)0.5
1 + Ba i(I)0.5 where B is a further constant, a i is the ionic
radius This equation extends the estimates to
solutions where I < 0.1 M (or TDS of about5000 mg/L)
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More Activity Coefficient Models
The Davis equation further extends theionic strength range to about 1 M
(roughly 50,000 mg/L) using empiricalcurve fitting techniques
The Pitzer equation is a much more
sophisticated ion interaction model thathas been used in very high strengthsolutions up to 20 M
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Monovalent Ions
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.001 0.01 0.1 1 10
Ionic Strength
A c
t i v
i t y
C o e
f f i c i e n
t
Debye-Huckel
Extended
Davis
Pitzer
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Divalent Ions
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.001 0.01 0.1 1 10
Ionic Strength
A c
t i v
i t y
C o e
f f i c i e n
t
Debye-Huckel
Extended
Davis
Pitzer
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Activity and Ionic Charge
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.001 0.01 0.1 1 10
Ionic Strength
A c
t i v
i t y
C o e
f f i c i e n
t
Debye-Huckel
Extended
Davis
Pitzer
Monovalent
Divalent
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Non-Equilibrium
Viewing groundwater as a partial equilibriumsystem implies that some reactions may notbe equilibrated.
Dissolution-precipitation reactions arecertainly in the non-equilibrium category.
Departures from equilibrium can be detected
by observing the ion activity product (IAP)relative to the equilibrium constant (K) whereIAP = [C]c[D]d = products
[A]a[B]b reactants
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Dissolution-PrecipitationaA + bB = cC + dD
If IAP1) then the reaction isproceeding from right to left.
If the reaction is one of mineral dissolutionand precipitation IAP/K1 the system is supersaturated and is
moving towards saturation by precipitation
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Saturation Index
Saturation index is defined as:SI = log(IAP/K)
When a mineral is in equilibrium withthe aqueous solution SI = 0
For undersaturation, SI < 0 For supersaturation, SI > 0
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Calcite The equilibrium constant for the calcite dissolution
reaction is K = 4.90 x 10 -9 log(K) = -8.31 Given the activity coefficients of 0.57 for Ca 2+ and 0.56
for CO 32- and molar concentrations of 3.74 x 10 -4 and5.50 x 10 -5 respectively, calculate IAP/K.
Reaction: CaCO 3 = Ca 2+ + CO 32-
IAP = [Ca 2+][CO 32-] = 0.57x3.37x10 -4x0.56x5.50x10 -5 [CaCO 3] 1.0
= 6.56 x 10 -9 and log(IAP) = -8.18 {IAP/K}calcite = 6.56/4.90 = 1.34log{IAP/K}calcite = 8.31 - 8.18 = 0.13
The solution is slightly oversaturated wrt calcite.
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Dolomite The equilibrium constant for the calcite dissolution reaction
is K = 2.70 x 10 -17 and log(K) = -16.57 Given activity coefficients of 0.57, 0.59 and 0.56 for Ca 2+ ,
Mg2+ and CO 32- and molar concentrations of 3.74 x 10 -4,
8.11 x 10-5
and 5.50 x 10-5
respectively, calculate IAP/K. Reaction: CaMg(CO 3)2 = Ca 2+ + Mg 2+ + 2 CO 32-
Assume the effective concentration of the solid dolomitephase is unity
log[Ca2+
] = -3.67 log[Mg2+
] = -4.32 log[CO 32-
] = -4.51log(IAP)=log([Ca 2+][Mg2+][CO 32-]2)= -3.67-4.32-9.02= -16.31
log{IAP/K}dolomite = 16.57 17.01 = -0.44 The solution is undersaturated wrt dolomite.
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Kinetic Reactions Reactions that are slow by comparison with
groundwater transport rates require a kineticmodel
k1
aA + bB = cC + dDk2 where k 1 and k 2 are the rate constants for the forward (L to R)and reverse (R to L) reactions
Each constituent has a reaction rate:r A = dA/dt; r B = dB/dt; r c = dC/dt; r D = dD/dt;
Stoichiometry requires that:-r A/a = -r B/b = r C/c = r D/d
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Rate Laws
Each consituent has a rate law of theform:
r A = -k 1(A)n1 (B)n2 + k 2(C) m1 (D)m2 where n 1, n 2, m 1 and m 2 are empirical orstoichiometric constants
If the original reaction is a single step(elementary) reaction then n 1=a, n 2=b,m 1=c and m 2=d
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Elementary Reactions
Fe 3+ + SO 42- = FeSO 4+
d(Fe 3+)/dt = -k 1(Fe 3+ )(SO 42-) + k 2(FeSO 4+)
The reaction rate depends not only onhow fast ferric iron and sulphate are
being consumed in the forward reactionbut also on the rate of dissociation ofthe FeSO 4+ ion.