ACID-BASE SPECIATION CALCULATIONS

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ACID-BASE SPECIATION CALCULATIONS Powered By Docstoc
					          TOPIC 3

  THE TERRESTRIAL
   ENVIRONMENT
  Required reading: Chapter 3 in
Andrews et al. (1996) or Chapters 4
  and 5 in Andrews et al. (2004)

                                      1
• Processes at the Earth’s surface are largely
  the readjustment of unstable, high-pressure
  and -temperature minerals to oxidizing,
  acidic, low-temperature conditions, i.e.,
  weathering.
• To fully understand the process of
  weathering, we need to understand bonding
  and structure of silicate minerals.


                                                 2
     CRYSTAL CHEMISTRY
• See Chapter 7 in Faure; Boxes 3.3 and 3.4 in
  Andrews et al. (1996); Section 2.32 and Box 4.2 in
  Andrews et al. (2004)
• Cations - elements that give up their valence
  electrons completely to form positively charged
  ions.
• Anions - elements that acquire a complete valence
  shell of electrons to become negatively charged
  ions.
                                                   3
                IONIC BONDS

• Ionic bonds - bonds formed via the electrostatic
  attraction between oppositely charged ions.
                     +           -
                           r
                           ZZ
• Coulomb’s law        F
                          4 0 r 2



0 = permittivity of free space = 8.84x10-9 farads m-1
                farad = coulomb volt-1
• Most minerals are held together via ionic bonds.
                                                     4
        COVALENT BONDS
• Covalent bonds - bonds formed by sharing of
  electrons.
• Many gaseous species and organic compounds
  contain covalent bonds, e.g., N2.
• A few minerals, such as diamond and graphite,
  contain covalent bonds.
• Sharing may not be equal. That is, covalent bonds
  may have some ionic character, and vice versa.

                                                      5
  MEASURE OF IONIC/COVALENT
        CHARACTER
• Electronegativity () - measures the ability of an
  atom in a compound to attract electrons to itself.
• Metals have low  values and non-metals have
  high  values.
• We use differences in electronegativity to
  determine degree of ionic character of a bond.
• Example: CsF Cs = 0.7; F = 4.0
                    F - Cs = 3.3
            > 92% ionic; < 8% covalent
                                                       6
7
8
OTHER EXAMPLES
 ZnS Zn = 1.6; S = 2.5
 S - Zn = 2.5 - 1.6 = 0.9
 19% ionic; 81% covalent

     H2     H = 2.1
       H - H = 0
 0% ionic; 100% covalent
                                           Cl
                              Cl
                                   C
 CCl4 C = 2.5; Cl = 3.0     Cl           Cl
 Cl - C = 3.0 - 2.5 = 0.5
 6% ionic; 94% covalent                9
10
   WHAT DIFFERENCE DOES THE
    TYPE OF BONDING MAKE?
• All physical and chemical properties of a
  compound depend on the character of the bonds.
• Example: Solubility in water
   Ionic bonded compounds have high aqueous solubilities
             NaCl      Cl - Na = 3.0 - 0.9 = 2.1
                   67% ionic, very soluble             (-)
                 C(diamond)        C - C = 0         O
                      0% ionic, insoluble          H       H
                                                   (+)     (+)
• Example: Physical properties
   Ionic bonded solids tend to be isotopic, covalent bonds -
     anisotropic.
                                                               11
   IONIC CRYSTAL STRUCTURES
• Ions form a crystal such that they are
  “closest packed”.
• This is a consequence of Coulomb’s law.
• Ions arrange themselves such that interionic
  distances are minimized.
• If we assume ions are hard, incompressible
  spheres (like billiard balls), we can use the
  concept of radius ratio as a key to
  explaining crystal structures.
                                              12
       COORDINATION NUMBER

• Coordination number - Number of anions
  surrounding a given cation.
• The coordination number is well predicted by the
  ratio of the radius of the cation to the radius of the
  anion (Rc/Ra).
• As the radius ratio increases, more anions can be
  fit exactly around a cation.
• Coordination numbers of 5, 7, 9, 10 and 11 are
  very rare in mineral structures.
                                                       13
14
                      Derivation of the
                      radius ratio for three-
                      fold coordination
    -
                                     = 60°
                                             Ra
                                 sin  
                                           Rc  Ra
    +   R+
         a
                                 sin  = 0.866
             R                              Ra
                                0.866 
              c

                                         Rc  Ra
-       Ra        -
                             Ra = 0.866Ra + 0.866Rc
                               0.134Ra = 0.866Rc
                              Rc/Ra = 0.134/0.866
                                     = 0.155
                                               15
                        Derivation of the radius
                        ratio for four-fold
                        coordination (square
                        planar)
-                   -
                                  = 45°
                                          Ra
                              sin  
                                        Rc  Ra
    +
                               sin  = 0.707
        R                                 Ra
        a   +                 0.707 
                R                      Rc  Ra
                c


-                   -     Ra = 0.707Ra + 0.707Rc
        Ra
                            0.293Ra = 0.707Rc
                           Rc/Ra = 0.293/0.707
                                  = 0.414
                                            16
                IONIC RADII
• Determined from X-ray studies of interatomic
  distances between ions in crystals.
• Ions are not really hard spheres. They can be
  deformed by electrical charges surrounding them.
  Thus, ionic radius is dependent on CN.
• Example: Na
  CN                 4     6      8      9
  radius (Å)        1.07 1.10 1.24 1.40
• Ionic radii are important for radius ratio
  arguments. Also, ions with similar radii can
  substitute for one another in minerals.
                                                     17
18
            TRENDS IN IONIC RADII
• The ionic radii of isoelectronic series decrease with increasing
  atomic number for both cations and anions.
• The radii of ions with the same charge in a group increase with
  increasing atomic number, i.e., downward in the periodic table.
• The radii of ions of the same element decrease with increasing
  positive charge and increase with increasing negative charge.
• The radii of ions with charges of +2 and +3 among the transition
  metals of the fourth period decrease with increasing atomic
  number, implying a contraction of the electron cloud as the 3d
  orbitals are filled.
• Ions of different elements may have similar radii: Na+, Ca2+,
  Cd2+, Y3+, Tl3+, REE3+, Th4+ ~ 1.00 Å (CN = 6); Hf4+, Zr4+ ~
  0.80 Å (CN = 8); Fe3+, Co3+, Ni3+ ~ 0.65 Å (CN = 6).
                                                              19
       CRYSTAL STRUCTURES

• We can describe ionic crystal structures as
  cation coordination polyhedra, outlined by
  the location of the anions around the
  cations. The nature of this polyhedron can
  be determined by the radius ratio rule.



                                                20
STRUCTURE OF HALITE (NaCl)




                             21
               PAULING’S RULES
These rules rationalize crystal structures of minerals.
1. A coordination polyhedron of anions is formed about
   each cation, the cation-anion distance being
   determined by the sum of the two radii, and the
   coordination number of the cation being determined
   by the radius ratio.
2. In a stable ionic structure, the valence of each anion,
   with changed sign, is exactly or nearly equal to the
   sum of the strengths of the electrostatic bonds to it
   from adjacent cations. [This rule states that crystals
   should be electrostatically neutral over very short
   distances.]                                            22
               PAULING’S RULES
                 (CONTINUED)
3. The presence of shared edges and especially of shared
   faces in a coordinated structure decreases its stability;
   this effect is large for cations with high valency and
   small coordination number. [Essentially, this says that
   any two cations tend to be as far from each other as
   possible. When polyhedra do share edges or faces, the
   cations have to be fairly close together, and the
   polyhedra are thus distorted due to the mutual
   repulsion of positively charged cations.]

                                                        23
               PAULING’S RULES
                 (CONTINUED)
4. In a crystal containing different cations, those with
   large valence and small coordination number tend not
   to share polyhedron corners, edges, or faces with each
   other. [This also reflects the dislike of cations for each
   other.]
5. The number of essentially different kinds of
   constituents in a crystal tends to be small. [This rule
   can be restated as follows. The structural environment
   of a given cation or anion tends to be the same
   throughout a crystal; i.e., only one type of
   coordination is likely for a given element.]           24
APPLICATION OF PAULING’S RULE
         2 TO HALITE
• Each Na+ ion is surrounded by 6 Cl-, so
  bond strength of Na-Cl bond is 1/6.
• Each Cl- ion in turn has 6 Na-Cl bonds
  reaching it, so 6 x 1/6 = 1 = -1.
• So the crystal is electrostatically neutral
  because the sum of bond strengths of all
  Na-Cl bonds reaching a single Cl- ion is
  equal to the absolute value of the charge of
  that ion.                                    25
PEROVSKITE STRUCTURE
               For Ca2+, CN = 12;
               for Ti4+, CN = 6.
               Each O2- is bound
               to 2 Ti4+ and 4
               Ca2+.




                              26
APPLICATION OF PAULING’S RULE
       2 TO PEROVSKITE
• Ti-O bond strength = 4/6
• Ca-O bond strength = 2/12 = 1/6
• Total bond strength reaching each oxygen
  ion is:
      4/6 x 2 + 1/6 x 4 = 12/6 = 2 = -2
     (from Ti) (from Ca)         (charge on O)
• Thus, Pauling’s 2nd rule is satisfied.
                                                 27
   ARE THERE ANY PRACTICAL
   APPLICATIONS OF PAULING’S
            RULES?
• Of course!
• Bond strength calculations useful in
  determining cleavage planes in minerals.
  Minerals will break along planes of
  weakness (low bond strength).
• We can also use Pauling’s rules to predict
  likely structures for new minerals.
                                               28
        COVALENT BONDING IN
             MINERALS
• Sulfides are the most important group of
  covalently bonded minerals.
• Pyrite (FeS2) - similar structure to halite, with Fe
  on the Na sites, and S2 on Cl sites (see figure).
• Bond distance between S atoms is 2.10 Å, which
  is less than 2rionic = 3.12 Å for S2-! This implies
  strong covalent bonding.
• Fe = 1.8; S = 2.5;       Fe - S = 0.7
  12% ionic, 88% covalent

                                                         29
STRUCTURE OF PYRITE




                      30
  HYBRIDIZATION OF ORBITALS
• Fe-S2 bonds are a good example of
  hybridization.
• Fe is octahedrally coordinated by S2, but s,
  p and d orbitals by themselves do not allow
  octahedral coordination.
• However, if we combine two d, one s, and
  three p orbitals mathematically, we can get
  6 new, hybrid d2sp3 orbitals which point to
  the corners of an octahedron.
                                             31
        ORGANIC COMPOUNDS
• Mostly held together by covalent bonds.
• C = 2.5; H = 2.1;     C - H = 0.4
  4% ionic, 96% covalent
• Solid organic compounds consist of molecules,
  strongly bound intramolecularly by covalent
  bonds; the molecules are weakly bound to one
  another via van der Waals forces.
• Thus, organic solids tend to melt at lower
  temperatures than ionic solids.


                                                  32
     MANY SOLIDS CONTAIN BOTH
     IONIC AND COVALENT BONDS
Calcite (CaCO3)
• C-O bonds form a trigonal planar array in the CO32- ion.
• O - C = 3.5-2.5 = 1.0; 22% ionic, 78% covalent.
               O
•     Ofrom
      CC-O bonds form                           sp2 hybrids.
              O
• CO32- ions are distinct within the calcite structure.
• CO32- and Ca2+ are in turn bonded ionically.
• Calcite dissolution: CaCO3  Ca2+ + CO32-
                                                          33
          SILICATE MINERALS
• Si-O bonds: O - Si = 3.5-1.8 = 1.7; 51% ionic,
  49% covalent.
• Si tends to form tetrahedral SiO44- units.
• Tetrahedral bonding allowed by sp3 hybridization.
• Tetrahedral units then link by sharing corners to
  form all the silicate structures known.
• Only a limited number of ways to link these
  tetrahedra, so we only have only a few
  fundamentally different silicate structure types.
                                                  34
        SILICATE CLASSIFICATION
Class             Arrangement of Shared corners   Repeat unit   Si:O   Example
                  tetrahredra
Nesosilicates     Independent           0            SiO44-      1:4   Olivine
                  tetrahedra
Sorosilicates     Pair of               1           Si2O76-     1:3.5 Hemimorphite
                  tetrahedra
                  sharing corner
Cyclosilicates    Closed rings of       2            SiO32-      1:3   Tourmaline
                  tetrahedra
Inosilicates      Infinite single       2            SiO32-      1:3   Pyroxenes
                  chain of
                  tetrahedra
                  Infinite double      2.5          Si4O116-    1:2.75 Amphiboles
                  chains of
                  tetrahedra
Phyllosilicates   Infinite sheets       3           Si2O52-     1:2.5 Micas
                  of tetrahedra
Tektosilicates    Unbounded             4            SiO2        1:2   Quartz,
                  framework of                                         feldspars
                  tetrahedra                                                     35
SILICATE STRUCTURAL UNITS - I




                                36
SILICATE STRUCTURAL UNITS - II




                             37
IONIC SUBSTITUTION IN
      CRYSTALS
  Read Chapter 8 in Faure (1998)




                                   38
           SOLID SOLUTION
• Occurs when, in a crystalline solid, one element
  substitutes for another.
• For example, a garnet may have the
  composition: (Mg1.7Fe0.9Mn0.2Ca0.2)Al2Si3O12.
• The garnet is a solid solution of the following
  end member components:
Pyrope - Mg3Al2Si3O12; Spessartine - Mn3Al2Si3O12;
Almandine - Fe3Al2Si3O12; and Grossular -
  Ca3Al2Si3O12.
                                               39
          GOLDSCHMIDT’S RULES

1. The ions of one element can extensively replace those
   of another in ionic crystals if their radii differ by less
   than approximately 15%.
2. Ions whose charges differ by one unit substitute
   readily for one another provided electrical neutrality
   of the crystal is maintained. If the charges differ by
   more than one unit, substitution is generally slight.
3. When two different ions can occupy a particular
   position in a crystal lattice, the ion with the higher
   ionic potential forms a stronger bond with the anions
   surrounding the site.                                   40
   RINGWOOD’S MODIFICATION OF
      GOLDSCHMIDT’S RULES
4. Substitutions may be limited, even when the size and
   charge criteria are satisfied, when the competing ions
   have different electronegativities and form bonds of
   different ionic character.
This rule was proposed in 1955 to explain discrepancies
   with respect to the first three Goldschmidt rules.
For example, Na+ and Cu+ have the same radius and
   charge, but do not substitute for one another.


                                                      41
   COUPLED SUBSTITUTIONS
Can Th4+ substitute for Ce3+ in monazite (CePO4)?
Rule 1: When CN = 9, rTh4+ = 1.17 Å, rCe3+ = 1.23Å. OK
Rule 2: Only 1 charge unit difference. OK
Rule 3: Ionic potential (Th4+) = 4/1.17 = 3.42; ionic
  potential (Ce3+) = 3/1.23 = 2.44, so Th4+ is preferred!
Rule 4: Th = 1.3; Ce = 1.1. OK

But we must have a coupled substitution to maintain
  neutrality:
              Th4+ + Si4+  Ce3+ + P5+
                                                      42
But can Si4+ substitute for P5+ according to
  Goldschmidt’s rules?

Rule 1: When CN = 4, rSi4+ = 0.34 Å, rP5+ = 0.25 Å.
  OK
Rule 2: Only 1 charge unit difference. OK
Rule 3: Ionic potential (Si4+) = 4/0.34 = 11.76; ionic
  potential (P5+) = 5/0.25 = 20, so P5+ is preferred.
Rule 4: Si = 1.8; P = 2.1. OK



                                                     43
  OTHER EXAMPLES OF COUPLED
         SUBSTITUTION
Plagioclase: NaAlSi3O8 - CaAl2Si2O8
              Na+ + Si4+  Ca2+ + Al3+

Gold and arsenic in pyrite (FeS2):
                Au+ + As3+  2Fe2+

REE and Na in apatite (Ca5(PO4)3F):
              REE3+ + Na+  2Ca2+

                                         44
  INCOMPATIBLE VS. COMPATIBLE
       TRACE ELEMENTS
Incompatible elements: Elements that are too large
  and/or too highly charged to fit easily into common
  rock-forming minerals that crystallize from melts.
  These elements become concentrated in melts.
Large-ion lithophile elements (LIL’s): Incompatible owing to
  large size, e.g., Rb+, Cs+, Sr2+, Ba2+, (K+).
High-field strength elements (HFSE’s): Incompatible owing to
  high charge, e.g., Zr4+, Hf 4+, Ta4+, Nb5+, Th4+, U4+, Mo6+, W6+,
  etc.
Compatible elements: Elements that fit easily into rock-
  forming minerals, and may in fact be preferred, e.g.,
  Cr, V, Ni, Co, Ti, etc.                             45
Changes in element concentration in the magma during
  crystal fractionation of the Skaergaard intrusion:
                   Divalent cations




                                                 46
Changes in element concentration in the magma during
  crystal fractionation of the Skaergaard intrusion:
                   Trivalent cations




                                                 47
THREE TYPES OF TRACE-
ELEMENT SUBSTITUTION
   1) CAMOUFLAGE

     2) CAPTURE

    3) ADMISSION

                        48
           CAMOUFLAGE
• Occurs when the minor element has the
  same charge and similar ionic radius as the
  major element (same ionic potential; no
  preference.
• Zr4+ (0.80 Å); Hf4+ (0.79 Å)
• Hf usually does not form its own mineral; it
  is camouflaged in zircon (ZrSiO4)

                                             49
               CAPTURE
• Occurs when a minor element enters a
  crystal preferentially to the major element
  because it has a higher ionic potential than
  the major element.
• For example, K-feldspar captures Ba2+ (1.44
  Å; Z/r = 1.39) or Sr2+ (1.21 Å; Z/r = 1.65) in
  place of K+ (1.46 Å, Z/r = 0.68).
• Requires coupled substitution to balance
  charge: K+ + Si4+  Sr2+ (Ba2+) + Al3+
                                              50
              ADMISSION
• Involves entry of a foreign ion with an ionic
  potential less than that of the major ion.
• Example Rb+ (1.57 Å; Z/r = 0.637) for K+
  (1.46 Å, Z/r = 0.68) in K-feldspar.
• The major ion is preferred.



                                              51
WEATHERING PROCESSES




                       52
MECHANISMS OF
 WEATHERING
   • Dissolution
    • Oxidation
 • Acid Hydrolysis


                     53
    DISSOLUTION

Read Chapter 10 in Faure (1998)




                                  54
           DISSOLUTION
• The simplest weathering reaction is
  dissolution of soluble salts, e.g.

      CaSO4(anhydrite)  Ca2+ + SO42-

• Solubility - The total amount of a substance
  that will dissolve in a solution at
  equilibrium.
                                             55
         SOLUBILITY PRODUCT
The solubility of a mineral is governed by the
  solubility product, the equilibrium constant for a
  reaction such as:
         CaSO4(anhydrite)  Ca2+ + SO42-
The solubility product is given by:
                           aCa 2 aSO 2
                  K SP               4

                             aCaSO4
If anhydrite is a pure solid, then aCaSO4 = 1.
and in dilute solutions: aCa2+  [Ca2+] and aSO42- 
   [SO42-].                                            56
So we may write:
           KSP = 10-4.5  [Ca2+][SO42-]

What is the solubility of anhydrite in pure water?
If anhydrite dissolution is the only source of both
   Ca2+ and SO42-, then by stoichiometry:
                  [Ca2+] = [SO42-] = x
                        x2 = 10-4.5
             x = 10-2.25 = 5.62x10-3 mol L-1
              MWanhydrite = 136.14 g mol-1
  solubility = (5.62x10-3 mol L-1)(136.14 g mol-1) =
                         0.765 g L-1
                                                       57
    A MORE COMPLICATED SALT
           Al2(SO4)3(s)  2Al3+ + 3SO42-
                            2      3
                          a Al 3 aSO 2 
                K SP                   4

                         a Al2 ( SO4 )3 ( s )
Assume aAl2(SO3)3 = 1, aAl3+  [Al3+] and aSO42- 
  [SO42-]. Then:
            KSP = 69.19  [Al3+]2[SO42-]3
Let x = the total number of moles of Al2(SO4)3(s)
  dissolved. Then, by stoichiometry:
           [Al3+] = 2x and          [SO42-] = 3x     58
       69.19 = (2x)2(3x)3
          69.19 = 108x5
       x = 0.9147 mol L-1
   [Al3+] = 2x = 1.829 mol L-1
  [SO42-] = 3x = 2.744 mol L-1

x = (0.9147 mol L-1)(342 g mol-1)
    = 312.8 g L-1 Al2(SO4)3(s)




                                    59
             SATURATION INDEX
In a natural solution, it is not likely that [Ca2+] =
   [SO42-], for example, because there will be more
   than one source of each of these ions. In this case
   we use saturation indices to determine if the water
   is saturated with respect to anhydrite.
             KSP = 10-4.5  [Ca2+]eq[SO42-]eq
                IAP = [Ca2+]act[SO42-]act

Saturation
             
                Ca  SO 
                     2
                          act
                                2
                                4 act
                                        
                                          IAP
index                     KSP             K SP

                                                     60
Suppose a groundwater is analyzed to contain 5x10-2 mol
  L-1 Ca2+ and 7x10-3 mol L-1 SO42-. Is this water
  saturated with respect to anhydrite?
                   KSP = 10-4.5 mol2 L-2
   IAP = (5x10-2)(7x10-3) = 3.5x10-4 = 10-3.45 mol2 L-2
               = 10-3.45/10-4.5 = 101.05 = 11.22
 > 1, i.e., IAP > KSP, so the solution is supersaturated
  and anhydrite should precipitate.

If  = 1, i.e., IAP = KSP, the solution would be saturated.
If  < 1, i.e., IAP < KSP, the solution would be
   undersaturated; the mineral should dissolve.
                                                        61
            ANOTHER EXAMPLE
Suppose the drainage from a tailings pile contains
  5x10-3 mol L-1 SO42-, 10-4 mol L-1 Fe3+, 10-3 mol L-1
  K+, and has pH = 3. Is this water saturated with
  respect to the mineral jarosite, KFe3(SO4)2(OH)6?
                 KSP = 10-96.5 mol12 L-12
            IAP = [K+][Fe3+]3[SO42-]2[OH-]6
      [OH-] = Kw/[H+] = 10-14/10-3 = 10-11 mol L-1
IAP = (10-3)(10-4)3(5x10-3)2(10-11)6 = 10-85.60 mol12 L-12
         = 10-85.60/10-96.5 = 1010.898 = 7.91x1010
So the solution would be highly supersaturated.
                                                       62
        HOW MUCH SALT SHOULD
            PRECIPITATE?
Returning to the previous example, i.e., the groundwater
   with 5x10-2 mol L-1 Ca2+ and 7x10-3 mol L-1 SO42-,
   how much anhydrite should precipitate at equilibrium?
If x mol L-1 of anhydrite precipitate, then at equilibrium:
         [Ca2+] = 5x10-2 - x; [SO42-] = 7x10-3 - x
                 and [Ca2+][SO42-] = 10-4.5
              (5x10-2 - x)(7x10-3 - x) = 10-4.5
            x2 - (5.7x10-2)x + (3.184x10-4) = 0
      x1 = 5.07x10-2 mol L-1 ; x2 = 6.28x10-3 mol L-1
                                                       63
The first root is unreal because it would cause both
  final concentrations to be negative.
So, 6.28x10-3 mol L-1 of anhydrite precipitates, or
  (6.28x10-3 mol L-1)(136.1 g mol-1) = 0.855 g L-1
                           and
 [Ca2+] = (5x10-2) - (6.28x10-3) = 4.37x10-2 mol L-1

 [SO42-] = (7x10-3) - (6.45x10-3) = 7.2x10-4 mol L-1

 Check: (4.37x10-2)(7.2x10-4) = 3.146x10-5 ≈ 10-4.5


                                                       64
      THE COMMON-ION EFFECT
Natural waters are very complex and we may have
   saturation with respect to several phases
   simultaneously.
Example: What are the concentrations of all species
   in a solution in equilibrium with both barite and
   gypsum?
1) Species: Ca2+, Ba2+, SO42-, H+, OH-
2) Mass action expressions:
         CaSO4·2H2O  Ca2+ + SO42- + 2H2O
               KSP = [Ca2+][SO42-] = 10-4.6

                                                       65
               BaSO4  Ba2+ + SO42-
             KSP = [Ba2+][SO42-] = 10-10.0

                  H2O  H+ + OH-
                Kw = [H+][OH-] = 10-14

3) Stoichiometric constraint: [Ba2+] + [Ca2+] = [SO42-]

4) Charge-balance:
     2[Ba2+] + 2[Ca2+] + [H+] = 2[SO42-] + [OH-]

                                                      66
   Ca 
      2
            104.6
                      SO  2   Ba 
                                   2
                                          1010.0
                                                   SO   2
                            4                             4


                                               SO 
       10 4.6          10 10.0                 2
       
               
                SO4 
                  2
                        
                        
                                 SO4  
                                      2
                                                    4




               10-4.6 + 10-10.0 = [SO42-]2
           [SO42-] = (10-4.6)1/2 = 10-2.3 mol L-1
          [Ca2+] = 10-4.6/10-2.3 = 10-2.3 mol L-1
         [Ba2+] = 10-10.0/10-2.3 = 10-7.7 mol L-1
The least soluble salt (barite), contributes a
  negligible amount of sulfate to the solution. The
  more soluble salt supresses the solubility of the
  less soluble salt (common-ion effect).                       67
The solubility of gypsum is hardly affected by the
  presence of barite.
Solubility of barite alone:
                [Ba2+][SO42-] = 10-10.0
                    [Ba2+]2 = 10-10.0
                [Ba2+] = 10-5.0 mol L-1
Solubility of gypsum alone:
                 [Ca2+][SO42-] = 10-4.6
                     [Ca2+]2 = 10-4.6
                [Ca2+] = 10-2.3 mol L-1

                                                     68
     REPLACEMENT REACTIONS
We can also calculate [Ba2+]/[Ca2+] in equilibrium
 with both barite and gypsum.
   Ca 
      2
            104.6
                      SO 
                        2     Ba 
                                   2
                                         1010.0
                                                    SO 
                                                      2
                        4                             4

                 [Ba 2 ] 10 10.0
                           4.6  10 5.4
                 [Ca 2 ] 10
What would happen if a solution with [Ba2+]/[Ca2+]
  = 10-3 came into contact with a gypsum-bearing
  rock?
Barite will precipitate and gypsum will dissolve until
  [Ba2+]/[Ca2+] = 10-5.4.                            69
REDOX REACTIONS

Read Chapter 14 in Faure (1998)




                                  70
       REDOX REACTIONS
Oxidation - a process involving loss of
 electrons.
Reduction - process involving gain of
 electrons.
Reductant - a species that loses electrons.
Oxidant - a species that gains electrons.


                                              71
                                                  72
Recovery of Cu using scrap iron, Butte, Montana
     RULES FOR ASSIGNMENT OF
        OXIDATION STATES
1) The oxidation state of all pure elements is zero.
2) The oxidation state of H is +1, except in
  hydrides (e.g., LiH, PdH2), where it is -1.
3) The oxidation state of O is -2, except in
  peroxides (e.g., H2O2), where it is -1.
4) The algebraic sum of oxidation state must
  equal zero for a neutral molecule or the charge
  on a complex ion.
                                                73
      VARIABLE VALENCE ELEMENTS
•   Sulfur: SO42-(+6), SO32-(+4), S(0), FeS2(-1), H2S(-2)
•   Carbon: CO2(+4), C(0), CH4(-4)
•   Nitrogen: NO3-(+5), NO2-(+3), NO(+2), N2O(+1), N2(0), NH3(-3)
•   Iron: Fe2O3(+3), FeO(+2), Fe(0)
•   Manganese: MnO4-(+7), MnO2(+4), Mn2O3(+3), MnO(+2), Mn(0)
•   Copper: CuO(+2), Cu2O(+1), Cu(0)
•   Tin: SnO2(+4), Sn2+(+2), Sn(0)
•   Uranium: UO22+(+6), UO2(+4), U(0)
•   Arsenic: H3AsO40(+5), H3AsO30(+3), As(0), AsH3(-1)
•   Chromium: CrO42-(+6), Cr2O3(+3), Cr(0)
•   Gold: AuCl4-(+3), Au(CN)2-(+1), Au(0)
                                                            74
        BALANCING OVERALL REDOX
               REACTIONS
Example - balance the redox reaction below:
                 Fe + Cl2  Fe3+ + Cl-
Step 1: Assign valences,
                Fe0 + Cl20  Fe3+ + Cl-
Step 2: Determine number of electrons lost or gained by
  reactants.
                Fe0 + Cl20  Fe3+ + 2Cl-
                     
                3e-    2e-
Step 3: Cross multiply.
              2Fe + 3Cl20  2Fe3+ + 6Cl-             75
    HALF-CELL REACTIONS
The overall reaction:
             2Fe + 3Cl20  2Fe3+ + 6Cl-
may be written as the sum of two half-cell reactions:
            2Fe  2Fe3+ + 6e- (oxidation)
            3Cl20 + 6e-  6Cl- (reduction)
All overall redox reactions can be expressed as the
  sum of two half-cell reactions, one a reduction and
  one an oxidation.
                                                    76
Another example:
             C2H6 + NO3-  HCO3- + NH4+
        C-32H6 + N+5O3-  2HC+4O3- + N-3H4+
                 
        14e-     8e-
          8C2H6 + 14NO3-  2HCO3- + NH4+
        8C2H6 + 14NO3-  16HCO3- + 14NH4+
 8C2H6 + 14NO3- + 12H+ + 6H2O 16HCO3- + 14NH4+




                                             77
Final example - balance the redox reaction:
            FeS2 + O2  Fe(OH)3 + SO42-
        Fe+2S2-1 + O20  Fe+3(OH)3 + 2S+6O42-
                   
         15e-       4e-
          4FeS2 + 15O2  Fe(OH)3 + 2SO42-
         4FeS2 + 15O2  4Fe(OH)3 + 8SO42-
 4FeS2 + 15O2 +14H2O  4Fe(OH)3 + 8SO42- + 16H+
This reaction is the main cause of acid generation in
  drainage from sulfide ore deposits. Note that we get
  4 moles of H+ for every mole of pyrite oxidized!
                                                   78
          ACID-MINE DRAINAGE
 4FeS2 + 15O2 +14H2O  4Fe(OH)3 + 8SO42- + 16H+

Hard-rock and coal mining can expose pyrite to
   oxidation and can lead to acid mine drainage (AMD).
In acid mine drainage the pH can be as low as 0-2, or
   even negative.
Acidity causes increased solubility of Al and other toxic
   heavy metals.


                                                     79
   WHAT SULFIDES OTHER THAN
    PYRITE CAN CAUSE AMD?
Pyrrhotite?
  FeS + 9/4O2 + 5/2H2O  Fe(OH)3 + 2H+ + SO42-
Yes, oxidation of pyrrhotite can form acid, but not as
  much as pyrite.

Galena?
             PbS + 2O2  Pb2+ + SO42-
No, oxidation of galena does not produce acid!
                                                     80
    PYRITE OXIDATION - A TWO
         STAGE PROCESS
Pyrite oxidation can be modeled as a two-step
   process:
1) Oxidation of pyrite sulfur to sulfate:
   4FeS2 + 4H2O + 15O2  8H+ + 8SO42- + 4Fe2+
2) Oxidation of ferrous to ferric iron:
      4Fe2+ + O2 + 10H2O  4Fe(OH)3 + 8H+
Equal amounts of acid are formed in each step.


                                                 81
           ROLE OF BACTERIA
• Inorganic oxidation is very slow at low pH.
• At pH < 3.5, Fe oxidation is catalyzed by the
  bacterium Thiobacillus thiooxidans
  (Acidithiobacillus thiooxidans).
• At pH = 3.5 - 4.5, oxidation is catalyzed by
  Metallogenium.
• The bacteria utilize Fe oxidation as a source of
  energy. This is not an efficient process; 220 g Fe2+
  must be oxidized to produce 1 g of cellular carbon.
  This leads to large deposits of Fe(III) oxide.
• Presence of H2O also catalyzes Fe(II) oxidation.
                                                    82
        ORGANIC MATTER
• Oxidation of reduced organic matter in soils
  is also catalyzed by micro-organisms.
• This is important because CO2 is produced,
  increasing soil acidity.
• In biologically active soils, PCO2 may be 10-
  100 times atmospheric.


                                              83
HALF-CELL REACTIONS AND THE SHE
• Half-cell reactions involve electron transfer, and
  so give rise to electrode potentials.
• The reaction:      H2(g)  2H+ + 2e-        is
  arbitrarily assigned a standard electrode potential
  of E° = 0.00 V, i.e., the EMF of this electrode is
  0.00 V when pH2 = 1 atm and [H+] = 1 mol L-1.
• The above electrode is called the standard
  hydrogen electrode (SHE).
• All other electrode potentials are measured and
  reported relative to SHE.
                                                        84
          AN IMPORTANT
          RELATIONSHIP

            G  nE
                   o
                   r
                                    o

                                    Faraday constant =
  number of electrons transferred   23.06 kcal V-1


And by convention:
            Gf(e-) = Gf(H+) = 0

                                                    85
        THE NERNST EQUATION
Consider the half-cell reaction: Fe2+  Fe3+ + e-
The following equation applies under all conditions:
             Gr = Gr° + RT ln (IAP)
                 IAP = [Fe3+]/[Fe2+]
         Gr = Gr° + RT ln ([Fe3+]/[Fe2+])
   Gr = nE; Gr° = nE°;  = 23.06 kcal V-1

                          2.3025RT
Nernst Equation: E  E   o
                                     log( IAP)
                             n
                      0.0592
At 25°C:       EE 
                   o
                              log( IAP)
                         n                     86
At equilibrium Gr = 0, so E = 0. Also, at
  equilibrium IAP = K
                    0.0592
               E 
                 o
                           log K
                       n
Example: Zn + Cu2+  Zn2+ + Cu
                 0.0592
          EE   o
                         log( IAP)
                     n
                                 [ Zn2 ] 
                                 [Cu 2 ] 
            1.10  0.0296 log           
                                          
If we fix E, we fix the ratio [Zn2+]/[Cu2+] and vice
   versa. If [Zn2+] = [Cu2+] = 1.0 mol L-1, then E = E°
   = -1.10 V.
                                                     87
      HOW DOES THIS APPLY TO
        NATURAL WATERS?
We cannot measure an EMF unless we define a
  reference half-cell or electrode.
Define Eh as the EMF generated between an
  electrode in any state and the H2 electrode in the
  standard state (SHE)
                    Fe2+  Fe3+ + e-
                    H+ + e-  1/2H2
               Fe2+ + H+  Fe3+ + 1/2H2
For this reaction the Nernst Equation is:
                                                       88
          0.0592  [ Fe ] pH22 
                                3    1


 Eh  E 
       o
                log                 
             1       [ Fe2 ][H  ] 
                                    
                        [ Fe3 ]        But by definition
                        [ Fe2 ] 
     E o  0.0592 log           
                                          of the SHE, pH2 =
                                        1 atm and [H+] =
                                          1 mol L-1
                          [ Fe3 ] 
                          [ Fe2 ] 
     0.769  0.0592 log           
                                   
Eh is an environmental parameter, just like pH or
  temperature and is a characteristic of a given
  natural water.
        High Eh implies oxidizing conditions.
        Low Eh implies reducing conditions.
                                                       89
In the above example, if we can measure [Fe3+]/[Fe2+],
   then we can calculate Eh, and vice versa. But what
   controls what?
Eh is one of the most difficult parameters to measure.
   Redox reactions may not attain equilibrium in
   natural waters at low temperatures.
Consider a second redox reaction:
      H3AsIIIO30 + H2O  H3AsVO40 + 2e- + 2H+
                     0.059     [ H 3 AsO4 ][H  ]2
                                          0
          Eh  E 0        log              0
                       2           [ H 3 AsO3 ]
We don’t always get agreement among the Eh values
 calculated from different redox couples.
                                                     90
  STABILITY LIMITS OF WATER IN
          Eh-pH SPACE
Upper limit
           H2O(l)  2H+ + 1/2O2(g) + 2e-
         Eh = E0 + 0.0295 log (pO21/2[H+]2)
   E0 = Gro/(n) = -(-56.687)/(2·23.06) = 1.23 V
       Eh = 1.23 + 0.0148 log pO2 - 0.0592 pH
At the Earth’s surface, pO2 can be no greater than 1
  bar so
                Eh = 1.23 - 0.0592 pH
                                                       91
Lower limit:

               1/2H2(g)  H+ + e-
          Eh = 0 + 0.059 log ([H+]/pH21/2)
              Again, let pH2 = 1 bar.

                  Eh = -0.059 pH




                                             92
                                               T = 25oC
                                               pH = 1 bar
                                                 2
                                                                 Eh-pH diagram
             1                    O
                                      2
                                               pO = 1 bar
                                                 2
                                                                 depicting the
                              HO
                               2                                 limits of stability
                                                                 of liquid water.
Eh (volts)




             0

                              HO
                               2
                              H
                              2




             -1
                  0   2   4       6        8    10   12     14
                                      pH                                         93
Range of Eh-pH
conditions in
natural
environments
based on data of
Baas-Becking et al.
(1960) Jour. Geol.
68: 243-284.




             94
         HOW IS Eh MEASURED?
Measure the voltage between a Pt indicator electrode
  and a reference electrode.
A SHE is not convenient in the field, so we often use
  a secondary reference called a saturated calomel
  electrode (SCE):
              2Hg + 2Cl-  Hg2Cl2 + 2e-
Then, to correct to SHE, we need to add 0.244 V to
  the reading obtained with the SCE.
Difficult to measure Eh in the field:
   Electrode itself disturbs equilibrium of system.
   Non-equilibrium among different redox couples.
                                                      95
           Eh-pH DIAGRAMS
•   Show both redox and acid-base reactions.
•   Depict mineral stabilities and solubilities.
•   Depict the predominant aqueous species.
•   Important for understanding processes in
    environmental geochemistry, exploration
    geochemistry, corrosion science,
    hydrometallurgy, and a variety of other
    fields.
                                                   96
         THE SYSTEM Fe-H-O
Species Gf(kcal mol-1) Species Gf(kcal mol-1)
H2O(l)           -56.687 Fe3+                -1.12
Fe3O4            -242.60 Fe2+               -18.85
Fe2O3             -177.6




                                             97
         THE SYSTEM Fe-H-O
Fe/Fe3O4 boundary:
      4H2O(l) + 3Fe(s)  Fe3O4(s) + 8e- + 8H+
                          0.0592
                 Eh  E 
                        0
                                 log[ H  ]8
                             8
The activities of water, all pure solids, and the electron
  are unity by convention.
       Gr° = -242.60 - 4(-56.687)= -15.85 kcal
    E° = Gr°/(n) = -15.85 kcal/(8·23.06 kcal V-1)
                      = -0.0086 V
                Eh = -0.0086 - 0.0592pH
This plots below the water stability line, so Fe not
  stable in the presence of water.                       98
Fe3O4/Fe2O3 boundary:
       2Fe3O4 + H2O  3Fe2O3 + 2H+ + 2e-
                       0.0592
              Eh  E 0
                              log[ H  ]2
                          2
Gr° = 3(-177.6) - 2(-242.60) - (-56.687) = 9.087 kcal
  E° = Gr°/(n) = 9.087 kcal/(2·23.06 kcal V-1)
                       = 0.20 V
                          0.0592
               Eh  0.2         log[ H  ]2
                             2

                  Eh  0.2  0.0592 pH

                                                   99
Fe3O4/Fe2+ boundary:
        3Fe2+ + 4H2O  Fe3O4 + 8H+ + 2e-
                          0.0592      [ H  ]8
               Eh  E 0         log
                             2       [ Fe 2 ]3
Gr° = -242.60 - 3(-18.85) - 4(-56.687) = 40.698 kcal
  E° = Gr°/(n) = 40.698 kcal/(2·23.06 kcal V-1)
                      = 0.88 V Need to fix [Fe2+]
            0.0592      [ H  ]8        arbitrarily. We
Eh  0.88         log                  choose 10-5 mol L-1.
               2       [ Fe 2 ]3
            Eh  0.88  0.237 pH  0.089 log[ Fe 2 ]
  Eh  0.88  0.237 pH  0.089(5)  1.325  0.237 pH
                                                          100
Fe2O3/Fe2+ boundary:
       2Fe2+ + 3H2O  Fe2O3 + 6H+ + 2e-
                          0.0592      [ H  ]6
               Eh  E 0         log
                             2       [ Fe 2 ]2
 Gr° = -177.6 - 2(-18.85) - 3(-56.687) = 30.16 kcal
  E° = Gr°/(n) = 30.161 kcal/(2·23.06 kcal V-1)
                      = 0.65 V      Need to fix [Fe2+] at
            0.0592      [ H  ]6        same value as before,
Eh  0.65         log                  i.e., 10-5 mol L-1.
               2       [ Fe 2 ]2
           Eh  0.65  0.177 pH  0.0592 log[ Fe 2 ]
 Eh  0.65  0.177 pH  0.0592(5)  0.946  0.177 pH
                                                          101
   Fe2O3/Fe3+ boundary:
              Fe2O3 + 6H+  2Fe3+ + 3H2O
    Gr° = 2(-1.12) + 3(-56.687) - (-177.6) = 5.299 kcal
          Gro               5299 cal
log K                           1    1
                                                    3.88
       2.3025RT 2.3025(1.987 cal K mol )(298.15 K)
                                       [ Fe 3 ]2
                      K  10 3.88          6
                                        [H ]
                      2log [Fe3+] - 6log [H+] = -3.88
                         log [Fe3+] + 3pH = -1.94
Need to fix [Fe3+] at        -5 + 3pH = -1.94
same value as before,            pH = 1.02
i.e., 10-5 mol L-1.                                     102
Fe3+/Fe2+ boundary:
                Fe2+  Fe3+ + e-
                         0.0592     [ Fe 3 ]
              Eh  E 0         log
                            1       [ Fe 2 ]
        Gr° = -1.12 - (-18.85) = 17.73 kcal
    E° = Gr°/(n) = 17.73 kcal/(23.06 kcal V-1)
                     = 0.77 V
                        0.0592     [ Fe 3 ]
            Eh  0.77         log
                           1       [ Fe 2 ]
By definition, this boundary occurs where [Fe2+] =
  [Fe3+], so Eh = 0.77 V.

                                                     103
             1.2                                     T = 25oC            Eh-pH diagram for the system
                              O                      pH = 1 bar
                             HO
                               2                       2                 Fe-O-H, showing the stability
             1.0              2
                                                     pO = 1 bar          and solubility of hematite and
                                                       2

             0.8             Fe3+                    Fe = 10-5 M        magnetite.
             0.6


             0.4
Eh (volts)




                                        Fe2O3
             0.2        Fe2+

             0.0
                         HO
                          2
                        H
                         2
             -0.2


             -0.4                                    Fe
                                                       3O
                                                           4
             -0.6


             -0.8
                    0    2          4   6        8     10      12   14
                                            pH                                                   104
          THE SYSTEM S-H-O
Species Gf(kcal mol-1) Species Gf(kcal mol-1)
                                 -
H2O(l)           -56.687 HSO4              -180.69
H2S0               -6.66 SO42-             -177.75
HS-                 2.93 S2-                 20.51




                                             105
               S-H-O SYSTEM
  HSO4-/SO42- boundary:
                SO42- + H+  HSO4-
     Gr° = -180.69 - (-177.75) = -2.94 cal mol-1
          Gro               2940 cal
log K                           1   1
                                                    2.16
       2.3025RT 2.3025(1.987 cal K mol )(298.15 K)
                                        
                                  [ HSO4 ]
                K  102.16   
                               [ H  ][SO4  ]
                                          2


  By definition, this boundary occurs where [SO42-] =
    [HSO4-], so pH = 2.16.
                                                    106
  H2S0/HS- boundary:
                 H2S0  HS- + H+
        Gr° = 2.93 - (-6.66) = 9.59 cal mol-1
          Gro               9590 cal
log K                           1    1
                                                    7.03
       2.3025RT 2.3025(1.987 cal K mol )(298.15 K)

                            [ HS  ][H  ]
              K  107.03 
                               [H2S 0 ]
  By definition, this boundary occurs where [H2S0] =
    [HS-], so pH = 7.03.


                                                       107
   HS-/S2- boundary:
                     HS-  S2- + H+
           Gr° = 20.51- 2.93 = 17.58 cal mol-1
          Gro              17580 cal
log K                           1   1
                                                    12.89
       2.3025RT 2.3025(1.987 cal K mol )(298.15 K)
                                      2    
                         12.89     [ S ][ H ]
                K  10                    
                                       [ HS ]
   By definition, this boundary occurs where [HS-] =
     [S2-], so pH = 12.89.


                                                       108
H2S0/HSO4- boundary:
           H2S0 + 4H2O  HSO4- + 9H+ + 8e-
 Gr° = -180.69 - (-6.66) - 4(-56.687) = 52.718 cal mol-1
                       0.0592                   
                                  [ H  ]9 [ HSO4 ]
            Eh  E 0         log
                          8           [H2S 0 ]
    E° = Gr°/(n) = 52.718 kcal/(8·23.06 kcal V-1)
                      = 0.286 V
                        0.0592                   
                                   [ H  ]9 [ HSO4 ]
           Eh  0.286         log
                           8           [H2S 0 ]
                 Eh = 0.286 - 0.0665 pH

                                                       109
H2S0/SO42- boundary:
           H2S0 + 4H2O  SO42- + 10H+ + 8e-
 Gr° = -177.75 - (-6.66) - 4(-56.687) = 55.658 cal mol-1
                        0.0592     [ H  ]10 [ SO4  ]
                                                 2
             Eh  E 0         log
                           8           [H2S 0 ]
    E° = Gr°/(n) = 55.658 kcal/(8·23.06 kcal V-1)
                      = 0.302 V
                        0.0592                    
                                   [ H  ]10 [ HSO4 ]
           Eh  0.302         log
                           8            [H2S 0 ]
                  Eh = 0.302 - 0.0739 pH

                                                         110
HS-/SO42- boundary:
           HS- + 4H2O  SO42- + 9H+ + 8e-
 Gr° = -177.75 - (2.93) - 4(-56.687) = 46.068 cal mol-1
                        0.0592     [ H  ]9 [ SO4  ]
                                                2
             Eh  E 0         log
                           8           [ HS  ]
    E° = Gr°/(n) = 46.068 kcal/(8·23.06 kcal V-1)
                      = 0.249 V
                        0.0592                   
                                   [ H  ]9 [ HSO4 ]
           Eh  0.249         log
                           8            [ HS  ]
                 Eh = 0.249 - 0.0665 pH

                                                        111
S2-/SO42- boundary:
            S2- + 4H2O  SO42- + 8H+ + 8e-
 Gr° = -177.75 - (20.51) - 4(-56.687) = 28.488 cal mol-1
                        0.0592     [ H  ]8 [ SO4  ]
                                                 2
             Eh  E 0         log
                           8            [ S 2 ]
    E° = Gr°/(n) = 28.488 kcal/(8·23.06 kcal V-1)
                      = 0.154 V
                         0.0592                    
                                    [ H  ]8 [ HSO 4 ]
           Eh  0.1544         log
                            8             [ S 2 ]
                  Eh = 0.154 - 0.0592 pH

                                                         112
H2S0/S(s) boundary:
               H2S0  S(s) + 2H+ + 2e-
                Gr° = 6.66 cal mol-1
                       0.0592     [ H  ]2 [S ( s)]
            Eh  E 0         log
                          2          [H2S 0 ]
     E° = Gr°/(n) = 6.66 kcal/(2·23.06 kcal V-1)
                      = 0.144 V
                          0.0592     [ H  ]2
             Eh  0.144         log                  aS(s) = 1.0
                             2       [H2S 0 ]
      Eh = 0.144 - 0.0592 pH - 0.0296 log [H2S0]
             set S = [H2S0] = 0.1 mol L-1
                Eh = 0.174 - 0.0592 pH                       113
HS-/S(s) boundary:
                HS-  S(s) + H+ + 2e-
                Gr° = -2.93 cal mol-1
                        0.0592     [ H  ][ S ( s )]
             Eh  E 0         log
                           2          [ HS  ]
     E° = Gr°/(n) = -2.93 kcal/(2·23.06 kcal V-1)
                     = -0.0635 V
                           0.0592      [H  ]
            Eh  0.0635         log                  aS(s) = 1.0
                              2       [ HS  ]
      Eh = -0.0635 - 0.0296 pH - 0.0296 log [HS-]
              set S = [HS-] = 0.1 mol L-1
                Eh = -0.034 - 0.0296 pH                      114
HSO4-/S(s) boundary:
         S(s) + 4H2O(l)  HSO4- + 7H+ + 6e-
    Gr° = -180.69 - 4(-56.687) = 46.058 cal mol-1
                       0.0592                     
                                  [ H  ]7 [ HSO 4 ]
            Eh  E 0         log
                          6            [ S ( s )]
    E° = Gr°/(n) = 46.058 kcal/(6·23.06 kcal V-1)
                      = 0.333 V
                     0.0592
        Eh  0.333                          
                            log[ H  ]7 [ HSO4 ]       aS(s) = 1.0
                        6
     Eh = 0.333 - 0.0691 pH + 0.00987 log [HSO4-]
             set S = [HSO4-] = 0.1 mol L-1
                 Eh = 0.323 - 0.0691 pH                      115
SO42-/S(s) boundary:
           S(s) + 4H2O(l)  SO42- + 8H+ + 6e-
     Gr° = -177.75 - 4(-56.687) = 48.998 cal mol-1
                        0.0592     [ H  ]8 [ SO4  ]
                                                  2
             Eh  E 0         log
                           6           [ S ( s )]
    E° = Gr°/(n) = 48.998 kcal/(6·23.06 kcal V-1)
                      = 0.354 V
                      0.0592
         Eh  0.354         log[ H  ]8 [ SO4  ]
                                             2
                                                        aS(s) = 1.0
                         6
     Eh = 0.354 - 0.0789 pH + 0.00987 log [SO42-]
             set S = [SO42-] = 0.1 mol L-1
                 Eh = 0.344 - 0.0789 pH                       116
          1.2                                          T = 25oC            Eh-pH diagram for the
                                  O                    pH = 1 bar
          1.0
                                  2
                                 HO
                                                           2               system S-O-H, showing the
                                  2
                                                       pO = 1 bar
                                                           2               solubility of native sulfur.
                                                       S = 0.1 M
                     -

          0.8
                     HSO4


          0.6

                                          SO42-
Eh (mV)




          0.4


          0.2                     S(s)

          0.0             HO
                           2
                         H       H S0
                         2        2
          -0.2


          -0.4                                     HS -
                                                                    S2-
          -0.6


          -0.8
                 0           2        4   6        8      10   12     14
                                              pH                                                 117
                                                                 ∑As = 10-5 M
               1                                                 ∑S = 10-3 M                        Data from
                                                                 T = 80 ºC                          Alvord Basin
                       H3 AsO4                                                                      geothermal
                                                                                                    systems
                                             -
              .5                       H2AsO4
Eh (volts)




                                   As(OH)3                        --
                                                              HAsO4
               0
                                  Orpiment
                                                                                         ---
                                                                               AsO4

                                             Realgar
                                                                       -
                                                              As(OH)4
             ±.5
                                                                                    --
                       80ÊC                                                AsO2OH
                   0          2        4         6        8        10          12              14
                                                                                                        118
                                                     pH
            THE SYSTEM Cr-H-O
Species Gf(kcal mol-1) Species    Gf(kcal mol )
                                                 -1


H2O(l)           -56.687 Cr(OH)2+             -103.0
HCrO4-           -182.77 Cr2O3(s)            -252.89
       2-                       -
CrO4             -173.94 CrO2                 -128.0




                                               119
                Cr-O-H SYSTEM
  HCrO4-/CrO42-:
               HCrO4-  CrO42- + H+
     Gr° = -173.94 - (-182.77) = 8.83 kcal mol-1
          Gro               8830 cal
log K                           1    1
                                                    6.47
       2.3025RT 2.3025(1.987 cal K mol )(298.15 K)
                             [CrO4  ][H  ]
                                  2
             K  106.47              
                               [ HCrO4 ]
  By definition, this boundary occurs where [HCrO4- ]
    = [CrO42-], so pH = 6.47.

                                                    120
Cr(OH)2+/Cr2O3(s)
     1/2Cr2O3(s) + 2H+  Cr(OH)2+ + 1/2H2O(l)
     Gr° = -103.0 + 1/2(-56.687) - 1/2(-252.89)
                  = -4.90 kcal mol-1
          Gro               4900 cal
log K                           1   1
                                                    3.59
       2.3025RT 2.3025(1.987 cal K mol )(298.15 K)
                                        2
                                [Cr (OH ) ]
              K  10   3.59
                              
                                  [ H  ]2
               3.59 = log [Cr(OH)2+] + 2pH
             [Cr(OH)2+] = Cr = 10-6 mol L-1
                     3.59 = -6 + 2pH
                        pH = 4.80                   121
 CrO2-/Cr2O3(s)
         1/2Cr2O3(s) + 1/2H2O(l)  CrO2- + H+
        Gr° = -128.0 - 1/2(-56.687) - 1/2(-252.89)
                   = 26.79 kcal mol-1
          Gro              26790 cal
log K                           1   1
                                                    19.64
       2.3025RT 2.3025(1.987 cal K mol )(298.15 K)
                       19.64             
              K  10             [CrO ][ H ]
                                      2

                -19.64 = log [CrO2-] - pH
               [CrO2-] = Cr = 10-6 mol L-1
                    -19.64 = -6 - pH
                       pH = 13.64                     122
Cr(OH)2+/HCrO4-
     Cr(OH)2+ + 3H2O(l)  HCrO4- + 6H+ + 3e-
        Gr° = -182.77 - (-103.0) - 3(-56.687)
                 = 90.291 kcal mol-1
                      0.0592                     
                                 [ H  ]6 [ HCrO 4 ]
           Eh  E 0         log
                         3         [Cr (OH ) 2 ]
    E° = Gr°/(n) = 90.291 kcal/(3·23.06 kcal V-1)
                     = 1.3052 V
                       0.0592                     
                                  [ H  ]6 [ HCrO 4 ]
         Eh  1.3052         log
                          3         [Cr (OH ) 2 ]
                Eh = 1.3052 - 0.1184 pH
                                                        123
Cr2O3(s)/HCrO4-
    1/2Cr2O3(s) + 5/2H2O(l)  HCrO4- + 4H+ + 3e-
      Gr° = -182.77 - 1/2(-252.89) - 5/2(-56.687)
                 = 85.393 kcal mol-1
                     0.0592
            Eh  E 
                   0                          
                            log[ H  ]4 [ HCrO4 ]
                        3
    E° = Gr°/(n) = 85.393 kcal/(3·23.06 kcal V-1)
                      = 1.234 V
                         0.0592
            Eh  1.234                           
                                log[ H  ]4 [ HCrO4 ]
                            3
     Eh = 1.234 - 0.0789 pH + 0.0197 log [HCrO4-]
                Eh = 1.116 - 0.0789 pH                  124
Cr2O3(s)/CrO42-
     1/2Cr2O3(s) + 5/2H2O(l)  CrO42- + 5H+ + 3e-
      Gr° = -173.94 - 1/2(-252.89) - 5/2(-56.687)
                   = 94.22 kcal mol-1
                     0.0592
            Eh  E 0
                            log[ H  ]5[CrO4  ]
                                           2

                        3
     E° = Gr°/(n) = 94.22 kcal/(3·23.06 kcal V-1)
                      = 1.362 V
                        0.0592
           Eh  1.362         log[ H  ]5[CrO4  ]
                                              2

                           3
      Eh = 1.362 - 0.0987 pH + 0.0197 log [CrO42-]
                Eh = 1.244 - 0.0987 pH                125
CrO2-/CrO42-
        CrO2- + 2H2O(l)  CrO42- + 4H+ + 3e-
        Gr° = -173.94 - (-128.0) - 2(-56.687)
                 = 67.434 kcal mol-1
                       0.0592     [ H  ]4 [CrO4  ]
                                               2
            Eh  E 0         log            
                          3           [CrO2 ]
    E° = Gr°/(n) = 67.434 kcal/(3·23.06 kcal V-1)
                      = 0.975 V
                        0.0592     [ H  ]4 [CrO4  ]
                                                2
           Eh  0.975         log            
                           3           [CrO2 ]
                 Eh = 0.975 - 0.0789 pH
                                                        126
          1.2                                        T = 25oC           Eh-pH diagram for the
                            HCrO4       -            pH = 1 bar
          1.0
                                                       2
                                                                        system Cr-O-H, showing
                                                     pO = 1 bar
                                                       2                the solubility of Cr2O3(s).
          0.8                               H
                                                O
                                                 2   Cr = 10-6 M
                                            2O


          0.6
Eh (mV)




          0.4        Cr(OH)2+                         CrO42-

          0.2
                                        Cr2O3(s)
          0.0


          -0.2
                                H
                                    2O

          -0.4                  H
                                2                          CrO2-

          -0.6


          -0.8
                 0      2   4       6           8     10      12   14
                                        pH                                                     127
   ACID HYDROLYSIS

Optional reading: Chapter 12 in Faure
               (1998)



                                        128
         ACID HYDROLYSIS
Acid hydrolysis - Reaction between a mineral and
  acid weathering agents.
Example:          H2CO30

    CaCO3 + CO2(g) + H2O(l)  Ca2+ + 2HCO3-
such a reaction neutralizes acid in water.
Increased PCO2 causes increased calcite solubility.
Decreased PCO2 causes precipitation.
This equation is the basis of cave formations, e.g.,
  stalagmites and stalactites.
                                                       129
      HYDROLYSIS OF SILICATES
Congruent dissolution (no solid products)
            Mg2SiO4 (forsterite) + 4H2CO30
             2Mg2+ + 4HCO3- + H4SiO40
Incongruent dissolution (leaves alteration products)
      CaAl2Si2O8(anorthite) + 2H2CO30 + H2O(l)
      Ca2+ + 2HCO3- + Al2Si2O5(OH)4(kaolinite)
Thus, weathering reactions tend to lead to surface
  waters with near-neutral pH, HCO3- as the major
  anion, and Ca2+ and Mg2+ as major cations.

                                                       130
                    A SINK FOR CO2
By itself, the precipitation of carbonates is not a sink for CO2:
              CaCO3 + CO2(g) + H2O(l)  Ca2+ + 2HCO3-
However, acid hydrolysis reactions of silicates, e.g.,
    Mg2SiO4 + 4CO2(g) + 4H2O(l)  2Mg2+ + 4HCO3- + H4SiO40

CaAl2Si2O8 + 2CO2(g) + 3H2O(l)  Ca2+ + 2HCO3- + Al2Si2O5(OH)4

                    2NaAlSi3O8 + 2CO2(g) + 11H2O(l)
             2Na+ + 2HCO3- + Al2Si2O5(OH)4 + 4H4SiO40
Combined with the precipitation of carbonates:
              Ca2+ + 2HCO3-  CaCO3 + CO2(g) + H2O(l)
     Ca2+ + Mg2+ + 4HCO3-  CaMg(CO3)2 + 2CO2(g) + 2H2O(l)
result in the net loss of CO2 into the solid phase!    131
WEATHERING RATES

Read Chapter 15 (pp. 253-256) in
             Faure



                                   132
       CONTROLS ON MINERAL
        WEATHERING RATES
1) Temperature - consider the following chemical
   reaction:         A+BC
The rate of this reaction is given by the equation
                   rate = k[A]n1[B]n2
Arrhenius equation
                   k = k0 exp(-Ea/RT)
Higher temperatures lead to faster chemical
   reactions. A rule of thumb: an increase of 10°C in
   temperature generally leads to a doubling of
   reaction rate.
                                                    133
           tropical regions: Tmean annual  20°C
         temperature regions: Tmean annual  12°C
So there will be about a factor of 2 faster weathering
   rate in tropical as opposed to temperate regions.
2) Flow rate
   High flow rates of water also lead to greater
   weathering rates, because ions are more efficiently
   flushed from the system.
               dissolution rates  (1 - )n
Thus, mineral dissolution rates are much faster when
   solutions are undersaturated.
The weathering rate of more soluble minerals is
   especially dependent on flow rate.                134
3) Parent bedrock
   Weathering rates follow a series roughly the
   inverse of the order of original crystallization for
   igneous minerals.
               Bowen’s Reaction Series
     decreasing temperature of




                                                                      increasing rate of weathering
     magmatic crystallization


                                  olivine              Ca-feldspar




                                                                                                      Goldich’s series
                                 pyroxene              Na-feldspar
                                 amphibole
                                                         K-feldspar
                                   biotite

                                             muscovite
                                                                                                      135
                                              quartz
136
137
4) Mechanical factors
   Grinding and fracturing by mechanical processes
   leads to increased surface area and increased
   reaction rate.
5) Plants and bacteria
The influence of plants and bacteria can enhance
   weathering rates by factors of 100-1000 times.
   a) Increased CO2 partial pressure.
   b) Organic acids oxalic, citric, etc.
   c) Polysaccharides and biofilms.


                                                     138
SOLID PRODUCTS OF
   WEATHERING
Read Chapter 13 in Faure (1998)




                                  139
            SOLID PRODUCTS OF
               WEATHERING
The final stable products of weathering consist of quartz
  and clay minerals.
Clay minerals: Hydrous sheet silicates (phyllosilicates)
  with a grain size < 4 m.
Clays are constructed of two major structural
  components:
  1) Sheets of SiO44- tetrahedra sharing three oxygens
  with neighbors.
  2) Sheets of Al, Fe and/or Mg in octahedral
  coordination with O2- and/or OH-.
                                                     140
         DIOCTAHEDRAL VS.
          TRIOCTAHEDRAL
Dioctahedral - Only two out of three
  octahedral sites are occupied by trivalent
  ions.
Trioctahedral - All three out of three
  octahedral sites occupied by a divalent ion.



                                             141
       1:1 CLAY MINERALS
              Serpentine-Kaolin Group
Kaolinite - Al2Si2O5(OH)4

                   octahedral sheet
                                        hydrogen bonds
                    tetrahedral sheet

1) Cations cannot get between layers.
2) Solid solution is limited.

                                                     142
        2:1 CLAY MINERALS
micas, illite, smectite, chlorite
                     tetrahedral sheet

                     octahedral sheet

                      tetrahedral sheet


solid solution is quite common in the 2:1 clays.


                                                   143
                     ILLITE
Illite - A general term to describe clay-size, mica-
   type minerals. Generally the composition is
   similar to muscovite.
• One out of four Si4+ ions are replaced by Al3+ in
   the tetrahedral sheet. This leads to a strong net
   negative charge.
• Some octahedral Al3+ may be replaced by Fe2+ and
   Mg2+, which also leads to net negative charge.
• The charge is neutralized by large cations, usually
   K+, in the interlayer spaces.

                                                   144
ILLITE STRUCTURE



               Interlayer sites filled
               with K+. Strongly
 K+ K+ K+ K+   bonded, so cations
               cannot easily
               exchange with K+.




                                145
                   SMECTITE

Smectite - similar structurally to illite. However, the
  2:1 units are not as tightly bound. Water can
  penetrate the interlayer sites, causing them to
  swell. Cations such as H+, Na+, Ca2+ and Mg2+
  also can enter the interlayer sites.
Thus, the weak interlayer bonding makes smectites
  prone to replacement by other cations. This leads
  to a high cation exchange capacity (CEC).

                                                      146
        THE CHEMICAL INDEX OF
             ALTERATION
It is predominantly feldspars that weather to clays. We
   can thus base a measure of the degree of weathering on
   how far the composition is from that of an ideal
   feldspar.
During weathering, Al and Fe are insoluble as oxides or
   oxyhydroxides. Other cations and Si are quite soluble.
                          Al 2O3          
                 Al O  CaO*  Na O  K O   100
          CIA                            
                 2 3             2     2  
The concentrations are in molecular proportions. CaO* is
  CaO in silicates (excluding that in carbonates and
  phosphates).                                       147
• CIA values of  100% are typical of heavily
  leached materials such as topical laterites
  and bauxites.
• Kaolinite and gibbsite occur in well-
  drained, heavily leached soils.
• Smectites form in poorly drained soils.




                                           148
Figure 3.17 Andrews (1996); Fig. 4.18 Andrew (2004)   149
Figure 19.1: A. Variation of
the chemical composition of
saprolites representing
increasing intensity of
chemical weathering of
granitic gneisses from
Minnesota.
 B. Variation of the measured
abundances of minerals in the
saprolites shown above.
From Faure (1998).




                       150
            ION EXCHANGE
       Clay-OH + K+  Clay-OK + H+
Clays (smectites) can hold ions both on their
  surfaces, on their edges, and in interlayer
  sites.
Clays can be used as adsorbents, e.g., as
  backfill in nuclear waste repositories.
Natural clays in groundwater aquifers retard
  the migration of pollutants by adsorption.
Clay surfaces may act as catalysts.
                                                151
Figure 13.2: Stability of kaolinite, K-montmorillonite, and Na-
montmorillonite in the presence of amorphous silica. From Faure
(1998).




                                                            152
Figure 13.3: Stability of selected minerals in the system K2O-MgO-
Al2O3-SiO2-H2O-HCl at 25°C and 1 atm in the presence of
amorphous silica. From Faure (1998).




                          microcline


                                                phlogopite
              muscovite
                                       illite

              kaolinite
                                                             chlorite




                                                                 153
        CLAYS IN ENVIRONMENTAL
             REMEDIATION
• 2,3,7,8-tetrachlorodibenzo-p-dioxin (or more simply
  dioxin)            C l    O       C l



                      OC l
                      C l

   – one of the most toxic priority pollutants of EPA
   – nerve poison
   – no safe lower limit of environmental contamination
• Use clays to absorb dioxin, then heat to destroy the
  dioxin.
• Problem: smectite clays have high capacity to adsorb
  dioxin, but heating to T > 200°C causes dehydration of
  interlayer sites and conversion of smectite to illite.
                                                          154
Need to build pillars in the interlayers to keep the
  layers from collapsing.
A pillar is a thermally stable, large cation such as
  Al13O4(OH)283+.
Use of pillars is advantageous because they:
  1) prevent interlayer collapse
  2) increase internal surface area of interlayer sites
  3) height of pillar can be engineered.
Also need a surfactant to make clay more favorable
  for absorbing organic molecules (Box 3.11).
Sodium lauryl sulfate                             polar head
                                    O     (hydrophilic)
   hydrocarbon tail     +
                      ( C
                      ) N
                      C H
                      3
                      2S
                       O
                      11O
   (hydrophobic)
                                                155
                               O
   CHEMISTRY OF
CONTINENTAL WATERS
   Required reading: Chapter 5 in
           Andrews (2004)
Optional reading: Chapter 20 in Faure
               (1998)
                                        156
   CHEMISTRY OF CONTINENTAL
           WATERS
• The 20 largest rivers on Earth carry 40% of total
  continental runoff; the Amazon alone accounts for
  15% of the total. These 20 rivers give a good
  indication of global average river composition.
• There are three major characteristics:
   – Four metals dominate all freshwater, i.e., Ca2+, Na+, K+,
     Mg2+. All are present as simple ions.
   – Low overall abundance of ions.
   – The composition of freshwater is markedly different
     from continental crust.
                                                            157
• The relative proportions of Al and Fe to other
  metals are particularly different.
            Al3+ + 3H2O(l)  Al(OH)3(s)
            Fe3+ + 3H2O(l)  Fe(OH)3(s)
• Ions with low Z/r dissolve as simple ions, e.g.,
  Cl-, Na+, Ca2+, etc.
• Ions with intermediate Z/r precipitate as
  hydroxides or oxides.
• Ions with high Z/r form oxyanions and are
  soluble, e.g.,
         Mo6+ + 4H2O(l)  MoO42- + 2H+
                                                   158
Ratio of average elemental riverine particulate to dissolved
     concentrations vs. ratio of charge to ionic radius.




                                                        159
   Fig. 3.23 Andrews (1996); Fig. 5.2 Andrew (2004)
 VARIATIONS IN IONIC COMPOSITION OF
  CONTINENTAL WATER OCCUR FROM
          PLACE TO PLACE

Dissolved ion composition depends on:
   1) The composition of rainfall and dry deposition.
   2) Modification of atmospheric inputs by
     evapotranspiration.
   3) Varying inputs from weathering and organic
     matter decomposition in soils and rocks.
   4) Differential uptake in biological processes.


                                                   160
• When weathering inputs are low, sea spray, dust
  and anthropogenic gases control freshwater
  chemistry. The Na and Cl contents are a measure
  of sea-salt input.
• Where rainwater input is dominant, Na+ will be
  the dominant cation.
• Where weathering processes are important, the
  dominant cations will be controlled by rocks (but
  usually Ca2+ will be dominant).
• Evaporation causes Ca to be removed as CaCO3,
  so Ca decreases relative to Na.
• Bicarbonate (HCO3-) is the dominant anion owing
  to hydrolysis reactions.
                                                 161
Variation in the weight ratio
of Na/(Na + Ca) as a
function of total dissolved
solids and ionic strength for
surface waters.
The ratio Na/(Na + Ca) and
the total dissolved salts
concentration can be used to
discriminate rainwater vs.
weathering vs. evaporation
contributions to freshwater
compositions.
Fig. 3.24 Andrews (1996);
Fig. 5.3 Andrews (2004)


                       162
                 IONIC STRENGTH
                     I       2 ci Z i2
                          1

                                i

          Ion     mmol L-1       Ion       mmol L-1
          Na+       10           Li+        0.01
           K+        1          HCO3-        12
          Ca2+       2          SO42-        1.5
          Mg2+      0.5          Cl-          1

    I = 1/2[cNa+·12 + cK+·12 + cCa2+·22 + cMg2+·22 + cLi+·12
                + cHCO3-·(-1)2 + cSO42-·(-2)2 + cCl-·(-1)2 ]

I = 1/2[10·1 + 1·1 + 2·4 + 0.5·4 + 0.01·1 + 12·1 + 1.5·4 + 1·1]

                      I = 20.05 mmol L-1                       163
                 ALKALINITY

Alkalinity - The amount of acid (millieq L-1)
  required to neutralize all bases in a water sample
  (usually the amount of acid required to titrate to a
  pH < 4). In most waters this can be calculated as:
               [alk]  [HCO3-] + 2[CO32-]
This is an important quantity. The higher the
  alkalinity of a natural water, the greater capacity it
  will have to buffer the pH upon addition of acid
  via, e.g., acid rain or acid mine drainage.
                                                      164
     PROBLEMS ENGENDERED BY
          ACIDIFICATION
• Increased solubility of toxic metals. For example, Al
  is insoluble in most natural waters, but solubility
  increases at extremes of low and high pH. Al is toxic
  to fish and other organisms.
• Loss of forests owing to acid leaching of nutrients
  from soil.
• Acidification a significant problem in poorly
  buffered freshwaters in the Adirondacks,
  Scandinavia and Scotland. For example, in
  Adirondacks in 1930, the median pH of lakes was
  6.7. In 1975 it fell to 5.1.                        165
                       Solubility of ZnO as a function of pH
           2
                             2+
                        Zn
                                           Oversaturated
           0
                                                                  2-
                        +                                   Zn(OH)4
           -2       ZnOH
log Zn




           -4


           -6
                    Undersaturated
           -8
                                              -
                                       Zn(OH)3
          -10
                4           6      8              10   12             14
                                         pH                                166
                               Solubility of Fe(OH)3 as a function of pH
                     1


                     -1


                     -3                                oversaturated
log concentration




                                                        Fe(OH)3(s)
                     -5


                     -7
                              Fe2(OH)24+                FeOH
                                                                2+                    -
                                                                                Fe(OH)4
                     -9


                    -11                                                     +
                                                                     Fe(OH)2
                                              3+
                          0       2        4 Fe    6        8          10        12       14
                                                       pH
                                                                                          167
                                 Solubility of Al(OH)3 as a function of pH
                    1
                                          Al3+
                             AlOH2+
                                                         oversaturated
                    -1                                    Al(OH)3(s)
log concentration




                    -3
                              Al(OH)2+                                           -
                                                                         Al(OH)4
                    -5


                    -7                                               Al(OH)30

                                undersaturated                    undersaturated
                    -9


                         0            2          4   6        8     10      12        14
                                                         pH
                                                                                     168

				
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