Conversions

1
Conversions

• Equivalent wt. Formula wt. / charge of ion
• Na 22.99g/1 22.99g
• Ca2 40.078g/2 20.039g
• meq/l mg/l / (equivalent wt. in g)
• 10 mg/l Na / 22.99 0.435 meq
• 10 mg/l Ca2 / 20.039 0.499 meq
• Molality (m) (mg/l 10-3) / (formula wt. in g)
• 10 mg/l Na 10-3 / 22.99 4.35 x 10-4 m
• 10 mg/l Ca2 10-3 / 40.078 9.98 x 10-4 m
• Molality (m) (meq/l 10-3) / (valence of ion)

2
WHAT CAN THERMODYNAMICS TELL US?

• In the context of the geochemistry of natural
  waters, thermodynamics can tell us
• Whether a mineral should dissolve in or
  precipitate from a solution of a given
  composition.
• What types of other reactions that control water
  chemistry (e.g., acid-base, oxidation-reduction)
  might occur.

3
THE MEANING OF EQUILIBRIUM

• A system at equilibrium has none of its
  properties changing with time, no matter how long
  it is observed.
• A system at equilibrium will return to that state
  after being disturbed, i.e., after having one or
  more of its parameters slightly changed, then
  changed back to the original values.
• Thermodynamically speaking, a system is at
  equilibrium when ?rG 0

4
STABLE VS. METASTABLE EQUILIBRIUM

• Stable equilibrium - System is at its lowest
  possible energy level.
• Metastable equilibrium - System satisfies
  criteria for equilibrium, but is not at lowest
  possible energy.

G
5

• Gibbs Free Energy, G
• A measure of a reactions ability to accomplish
  useful work, that is, energy released by the
  reaction that can be converted to work.
• - A portion of then energy involved in a chemical
  reaction is unavailable to do work.
• Really interested in the change in the Gibbs free
  energy, ?G, of a reaction
• ?Gr ?Hr - T ?Sr
• T temperature in K (C 273.15)
• ?Hr the change in enthalpy, or the heat
  transfer between a system and its surroundings
  for a process under constant pressure.
• Consider the formation of liquid water from
  gaseous oxygen and hydrogen
• H2(g) ½ O2(g) H2O(l)
• 25.5L 12.3L 0.018L
• 0 0 -68.315 kcal/mol ?H
  -68.315 kcal/mol
• highly exothermic
• ?Sr change in entropy (degree of randomness
  or disorder in the system)
• (e.g. liquid H2O ? Vapor H2O has a positive ?S)

6
THE GIBBS FREE ENERGY CHANGE OF REACTION

• Consider the reaction
• aA bB ? cC dD
• Where a,b, c and d are the molar amounts of
  compounds A, B, C and D, respectively. The Gibbs
  free energy change of reaction is written as
• ?? rG c?f GC d?f GD - a?f GA - b?f GB
• or most generally
• If ?? rG 0, the reaction is at equilibrium if
  ?? rG lt 0, the reaction will proceed to the
  right if ?? rG gt 0, the reaction will proceed to
  the left.

7
Gibbs Free Energy (Cont.)

• Example The following reaction, governs the
  oxidation of sulfide minerals exposed to the
  atmosphere
• 4FeS2 15O2 14H2O ? 4Fe(OH)3 8SO42- 16H
• We would write ? rG as
• ? rG 4? f G(Fe(OH)3) 8? f G(SO42-) 16? f
  G(H)
• - 4? f G(FeS2) -15? f G(O2) - 14? f G(H2O)
• Using data a geochemical table we calculate
• ? rG 4(-696.5) 8(-744.0) 16(0) - 4(-160.1)
  - 15(0) - 14(-237.1)
• -4,778.2 kJ mol-1
• Assuming that all the reactants and products are
  in their standard state, this shows that the
  above reaction is highly favored to proceed to
  the right. In other words, the reaction should
  proceed spontaneously as written.

8
Gibbs Free Energy (Cont.)

• But, what if the reactants and products are not
  in the standard state?
• Need to calculate an equilibrium constant

9
ACTIVITY AND ACTIVITY COEFFICIENTS

• In thermodynamic expressions, the activity takes
  the place of concentration. (an effective
  concentration)
• Activity and concentration are related
• ai ?iMi
• where ai is the activity, Mi is the concentration
  and ?i is the activity coefficient.
• In dilute solutions, ?i ? 1, so ai ? Mi. However,
  in concentrated solutions activity and
  concentration may be far from equal.

10
LAW OF MASS ACTION

• Consider the reaction
• aA bB ? cC dD
• Where a,b, c and d are the molar amounts of
  compounds A, B, C and D, respectively. At
  equilibrium it must be true that
• K (the equilibrium constant) is independent of
  concentration.
• Thus, if we, for example, increase aA, then to
  maintain equilibrium, the reaction must shift to
  the right so that the activities of the reactants
  decrease and the activities of the products
  increase, keeping K constant. This is an example
  of Le Chatliers Principle.

11
Law of Mass Action

• Le Chatliers Principle
• based on the law of mass action.
• It states that
• if a reaction is perturbed, then that reaction
  will proceed in a direction so as to lessen the
  effect of the perturbation.
• For example, if the reaction is initially in
  equilibrium, and we increase the concentration of
  reactant A, then the reaction must proceed to the
  right to decrease the activities of A and B, and
  increase the activities of C and D, and thereby
  keep the ratio of product and reactant activities
  constant.

12
HOW DO WE CALCULATE K?

• For the above reaction, we can write
• ??rG c?fGC d?fGD - a?fGA - b?fGB
• And we also write
• where R is the gas constant and equals 8.314 J
  K-1 mol-1 or 1.987 cal K-1 mol-1.

13
Example

• Suppose we have the reaction
• CaSO4(s) ? Ca2 SO42-
• For which we can write
• Note that, the activity of most pure solids can
  be taken equal to unity (i.e., aCaSO4(s) 1).
• If we increase the activity of Ca2, Le
  Chatliers Principle tells us that the reaction
  will shift to the left (anhydrite will
  precipitate) so that K will remain constant.

14
SOLUBILITY PRODUCT

• The equilibrium constant for a reaction of the
  type
• CaSO4(s) ? Ca2 SO42-
• is called a solubility product (KSP). The KSP can
  be calculated according to
• ?? rG ?f GCa2 ?f GSO42- - ?f GCaSO4(s)
• and

15
AN EXAMPLE CALCULATION OF THE SOLUBILITY PRODUCT

• Example Determine the solubility product of
  gypsum
• From the Geochemical tables we obtain
• ?f GCa2 -553.6 kJ mol-1
• ?f GSO42- -744.0 kJ mol-1
• ?f GCaSO4(s) -1321.8 kJ mol-1
• so
• ??rG -553.6 (-744.0) - (-1321.8) 24.2 kJ
  mol-1

16
Another Example

• Determine the Ksp of calcite at 25C
• CaCO3 ? Ca2 CO32-
• The activity of solid CaCO3 1
• From tables Species ?Gf (kJ/mol)
• Ca2 -553.6
• CO32- -527.0
• CaCO3 -1128.4
• ?Gr ??G f-products - ??G f-reactants
• ?Gr (-553.6) (-527.0) (-1128.4) 47.8
  kJ/mol
• Using log K - ?Gr /2.3026RT
• log Ksp -8.37 and K 10-8.37

17
Another Example

• Calculate the solubility product of mackinawite
  at 25C
• FeS(s) ? Fe2 S2-
• ??rG ?f GFe2 ?f GS2- - ?f GFeS(s)
• ??rG -90.0 (85.8) - (-93.0) 88.8 kJ mol-1

18
Yet Another Example

• Calculate the solubility product of dolomite at
  25C
• CaMg(CO3)2(s) ? Mg2 Ca2 2CO32-
• ?rG ?f GMg2 ?f GCa2 2?f GCO32- -
• ?f GCaMg(CO3)2(s)
• ??rG -455.5 (-553.6) 2(-527.0) - (-2161.3)
  
• 98.2 kJ mol-1

19
So, What can this tell us?

• Gypsum Ksp 10-4.24
• Calcite Ksp 10-8.37
• Mackinawite Ksp 10-15.6
• Dolomite Ksp 10-17.2
• Gypsum is more soluble than calcite
• Calcite is more soluble than mackinawite
• Mackinawite is more soluble than dolomite.
• Also, whether a solution is saturated with any of
  the above

20
Solubility Products
21
THE ION ACTIVITY PRODUCT (IAP)

• Consider once again the reaction
• CaSO4(s) ? Ca2 SO42-
• The equilibrium constant is expressed in terms of
  the activities of the reactants and products at
  equilibrium
• However, a real solution may or may not be in
  equilibrium. The ion activity product (IAP ) or
  reaction quotient (Q ) has the same form as the
  equilibrium constant, but involves the actual
  activities

22
THE SATURATION INDEX

• The saturation index (SI) is defined according
  to
• If IAP KSP, then SI 0, and the water is
  saturated with respect to the mineral.
• If IAP lt KSP, then SI lt 0, and the water is
  undersaturated with respect to the mineral.
• If IAP gt KSP, then SI gt 0, and the water is
  supersaturated with respect to the mineral.

23
APPLICATION

• 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?
  (ignore activity coefficients)
• KSP(anhydrite) 10-4.24 mol2 L-2
• IAP (5x10-2)(7x10-3) 3.5x10-4 10-3.45 mol2
  L-2
• In this case, SI gt 0, i.e., IAP gt KSP, so the
  solution is supersaturated and anhydrite should
  precipitate.

24
A SECOND APPLICATION

• A water contains the following 3.13x10-4 mol L-1
  Mg2
• 8.48x10-4 mol L-1 Ca2 and 2x10-6 mol L-1 CO32-.
  Determine whether this water is saturated with
  respect to dolomite (CaMg(CO3)2) (ignore activity
  coefficients).
• KSP(dolomite) 10-17.20 mol2 L-2
• IAP (8.48x10-4)(3.13x10-4)(2x10-6)2
• 1.06x10-18 10-17.97 mol2 L-2
• Because SI lt 0, the solution is undersaturated
  with respect to dolomite the mineral should
  dissolve.

25
THIS APPROACH CAN BE APPLIED TO ALL TYPES OF
REACTIONS

• Consider the acid-base reaction
• H2CO30 ? HCO3- H
• for which K 10-6.35. Which way should the
  reaction go if pH 7, a H2CO30 10-4 and a
  HCO3- 10-3?
• First, we must calculate the IAP. To do so, we
  recall that pH -log a H, so a H 10-7. Now
• and IAP gt K (10-6 gt 10-6.35). The reaction will
  shift to the left until IAP K.

26
VARIATION OF LOG K WITH TEMPERATURE

• The solubility product is determined at T 25C.
• K is a function of temperature and cannot be used
  at temperatures lt or gt 25C
• How do we then find solubility relationships at
  other temperatures?

27
VARIATION OF LOG K WITH TEMPERATURE

• The following is a generally valid relationship
• ?rG ?rH - T?rS
• If we assume that ?rH and ?rS are approximately
  constant (true over a limited temperature range),
  then because
• we can write

28
THE VANT HOFF EQUATION

• As an alternative, we can use the Vant Hoff
  equation
• If we assume again that ?rH is approximately
  constant, we can write the expression
• We can calculate ?rH and ?rS according to

29
Example 1

• Calculate the solubility product of anhydrite at
  60C.
• CaSO4(s) ? Ca2 SO42-
• First, we calculate ?rH and ?rS
• ??rH ?fHCa2 ?fHSO42- - ?fHCaSO4(s)
• ??rH -543.0 (-909.3) - (-1434.4) -17.90 kJ
  mol-1
• ??rS SCa2 SSO42- - SCaSO4(s)
• ??rS -56.2 18.5 - 107.4 -145.1 J K-1 mol-1
• Method 1

(17,900 J mol-1) (333.15 K)(-14,510 J K-1
mol-1) 2.303 (8.314 J K-1 mol-1 )
(333.15 K) -4.77 K 10-4.77
30
Example 1 (Cont.)

• Method 2 (the Vant Hoff equation)
• The differences in the results of these two
  methods are due to slight inconsistencies in the
  thermodynamic data.

31
Another Example

• Determine the solubility product of calcite at
  40C
• CaCO3 ? Ca2 CO32-
• From tables Species ?Hf(kJ/mol) ?Sf
  (kJ/mol)
• Ca2 -543.0 -0.0562
• CO32- -675.2 -0.0500
• CaCO3 -1207.4 0.09197
• ?Hr ?Hfproducts - ?Hfreactants
• ?Hr -543.0 -675.2 (-1207.4) -10.8
• ?Sr ?Sfproducts - ?Sfreactants
• ?Sr -0.0562 - 0.0500 (0.09197) -0.19817
• Using ln Keq ?Hr - T ?Sr / -RT
• Log K -10.8 (313.15)(-0.19817) /
  2.303(-8.3143 ? 10-3)(313.15) -8.56
• Log K -8.56
• K 10-8.56

32
Example 2 (cont.)

• So what is the point?
• For calcite K40 10-8.56 and K25 10-8.37
• Calcite is less soluble at 40C than at 25C
• Implications?

33
Example 2 (Cont.)

• Tufa deposited where cold spring water meets
  atmospheric conditions
• Scale forms from carbonate-rich water in boilers
• Other?

34
IONIC STRENGTH - I

• Recall that activity and concentration are
  related through the activity coefficient
  according to
• ai ?iMi
• Activity coefficients different from unity arise
  because of the interaction of ions as
  concentration rises.
• The degree of ion interaction depends on ionic
  charge as well as concentration.

35
IONIC STRENGTH - II

• Ionic strength (I ) is a quantity that is
  required to estimate activity coefficients. It
  takes into account both concentration and charge
• The calculation of ionic strength must take into
  account all major ions

36
Ionic Strength III

• Calculation example
• A river water has the following composition
• Calculate the Ionic Strength
• First, convert mg/l to molality

I ½(0.00584 22 0.0016 22 0.00475 12
0.00518 22) I 0.0276
37
DEBYE-HÜCKEL EQUATION

• Used to calculate activity coefficients for ions
  at ionic strengths lt 0.1 mol L-1.
• A, B are functions of temperature and pressure
  and are given in Geochemistry tables.
• ao is the distance of closest approach and it is
  a property of the specific ion.
• ionic charge, zi.

38
DEBYE-HÜCKEL PARAMETERS
39
DISTANCES OF CLOSEST APPROACH FOR SELECTED IONS
40
Example Calculation

• Given the analysis we used earlier to calculate
  the Ionic strength
• I 0.0276
• What is the activity of Ca2?

log ?i (- 0.5085 (2)2 (0.0276)1/2) / (1
(0.3281 x 10-8)(5 x 10-8) (0.0276)1/2) log ?i
-0.3379 / 1.2725 log ?i - 0.2655 ?i 0.543
41
THE DAVIES EQUATION

• Used to calculate activity coefficients for ions
  at ionic strengths lt 0.5 mol L-1.
• The value of A is the same as the one employed in
  the Debye-Hückel equation.
• The advantage over the D-H equation is that the
  only ion-dependent parameter is the ionic charge,
  zi.

42
ACTIVITY COEFFICIENTS
43
EFFECT OF ACTIVITY COEFFICIENTS ON GYPSUM
SOLUBILITY

• Question what is the solubility of gypsum in
  pure water at 25C and 1 bar? For the
  equilibrium
• CaSO42H2O(s) ? Ca2 SO42- 2H2O(l)
• we can calculate log KSP -4.41. We can also see
  that, if gypsum dissolves in pure water, then the
  stoichiometry of the reaction is such that the
  molarity of calcium should equal the molarity of
  sulfate.
• Thus, solubility of gypsum MCa2.
• So whats the problem? Catch 22! We dont know
  the concentrations, so we cant calculate the
  ionic strength, so we cant calculate the
  activity coefficients, so we cant calculate the
  concentrations!

44
WHAT TO DO?

• We start by making an initial assumption that the
  activity coefficients are equal to 1 and solve
  the problem by iteration.
• We write
• but because we assume activity coefficients are
  equal to 1, we write
• This is what the solubility would be if we
  ignored activity coefficients altogether.

45
THE NEXT STEP

• Now, having the concentration of Ca2 and SO42-,
  we can calculate the ionic strength according to
• Applying the Debye-Hückel formula we get
• which are about half the originally assumed
  values. We calculate a new estimate for the
  molality of Ca2
• This is used to calculate a new ionic strength
  and the whole process is repeated until
  convergence.

46
RESULTS OF ITERATIONS
47
FINAL ANSWER

• The final calculated molarity of Ca2 is
  13.78?10-3.
• This is 2.21 times the calculated molality of
  Ca2 assuming activity coefficients are unity.
• We see that activity coefficient corrections are
  very important for this solution.
• It is customary to express solubility in g L-1 of
  gypsum
• (13.78?10-3 mol L-1)(172.1 g mol-1) 2.37 g L-1

48
Charge Balance

• More than 90 of all the dissolved solids in
  waters can be attributed to eight ions
• Ca2, Mg2, Na, K, Cl-, SO42-, HCO3-, and CO2-
• Direct analyses can be done for the first six.
• HCO3-, and CO2- are determined by titrating with
  an acid to an endpoint of 4.4
• This is reported a total alkalinity
• The proportion of carbonate and bicarbonate are
  calculated using the sampling temperature and pH.
• To check on the analyses a cation-anion balanced
  in performed.

49
Charge Balance

• To perform a charge balance
• Convert all ionic concentrations to equivalents
  per liter
• Sum charges on cations ?zmc
• Sum charges on anions ?zma
• Determine charge balance error
• CBE ?zmc - ?zma / ?zmc ?zma

50
Charge Balance Example

• CBE ?zmc - ?zma / ?zmc ?zma
• ?zmc 2(0.00584) 2(0.0016)
• 0.01168 0.0032
• 0.01488
• ?zma (0.00475) 2(0.00518)
• 0.00475 0.01036
• 0.01511
• CBE ?zmc - ?zma / ?zmc ?zma
• (0.01488 0.01511) / (0.01488
  0.01511)
• 0.00023 / 0.02999
• .0077

51
More on Solubility

• Precipitation of ions as insoluble (low
  solubility) minerals is a major mechanism by
  which metal concentrations are limited in
  groundwater.
• A useful generalization is that highly charged
  cations (2, 3, 4) form insoluble precipitates
  with highly charged anions
• SO42-
• CO32-
• PO43-
• OH-
• Chalcophile elements (Pb, Cd, Hg) form insoluble
  sulfides.

52
Solubility
53
Solubility Example
Dissolution of silver chloride AgCl ? Ag Cl-

• If a solution is saturated with silver chloride,
  the activities of Ag and Cl- obey the equation
• Ksp products / reactants Ag Cl-
• Note the activity of solids are set to 1 by
  convention, so in this case AgCl 1.
• Hence the solubility of AgCl Ag Cl-,
  since one mole of AgCl dissolves in water to give
  one mole of Ag ions and one mole of Cl- ions.
• The solubility product, Ksp, for AgCl at 25C is
  10-9.8, therefore the solubility of
• AgCl is
• Solubility Ag Cl- ?10-9.8 10-4.9
  mol/liter

54
Common Ion Effect

• Note that the solubility of AgCl depends on
  there being no Ag or Cl- ions already in the
  solution when we start dissolving AgCl.
• Lets say there was already some Ag in the
  solution, then once the solution becomes
  saturated,
• Agtotal Aginitial Agadded and
  Cl-total Cl-added
• So, we can say Ksp Aginitial Agadded
  Cl-added
• In this case the solubility of AgCl will be
  equal to the solubility of the chloride ion, i.e.
  less than the solubility of AgCl in pure water.
  This known as the common-ion effect, which often
  results in precipitation of compounds near the
  confluence of two streams of different
  compositions.

55
Common Ion Effect

• Another example
• A groundwater saturated with calcite encounters a
  rock formation containing gypsum
• Gypsum is more soluble than calcite. It dissolves
  according to
• CaSO42H2O ? Ca2 SO42- 2H2O
• To the extent this reaction goes to the right, it
  pushes the following reaction to the left
• CaCO3 ? Ca2 CO32-
• Resulting in the precipitation of calcite

56
Solubility and pH

• Example Because they are strong bases, the
  solubilities of hydroxides, sulfides and
  carbonates will depend on the pH.
• What is the solubility of Pb(OH)2 at pH 2, 7,
  and 9?
• Pb(OH)2 Pb2 2OH- with K PbOH2 2.5
  ? 10-16
• (i) At pH 2,
• OH Kw/(10-2) 10-12 and
• Pb (2.5 ? 10-16)/(10-12 )2
• 2.4 x 108 moles/liter
• (i.e., it is completely soluble).

57
Solubility and pH (Cont.)

• (ii) At pH 7,
• OH Kw/(10-7) 10-7
• Pb (2.5 ? 10-16)/(10-7 )2
• 2.5 x 10-2 moles/liter
• (less soluble, about 5200 ppm).
• (iii) At pH 9,
• OH Kw/(10-9) 10-5
• Pb (2.5 ? 10-16)/(10-5 )2
• 2.5 x 10-6 moles/liter
• (much less soluble, about 5 ppm).

58
THE DISSOCIATION OF WATER AND NEUTRAL pH - I

• In this case, we have the reaction
• H2O(l) ? H OH-
• for which the equilibrium constant is
• This is called the self-dissociation constant for
  water
• but because usually a H2O ? 1,
• ??rG ?f GH ?f GOH- - ?f GH2O
• ??rG 0 (-157.3) - (-237.1) 79.80 kJ mol-1

59
THE DISSOCIATION OF WATER AND NEUTRAL pH - II

• If this reaction were occurring in pure water
  with no other solutes, then it would have to be
  true that at equilibrium
• Thus

60
THE DISSOCIATION OF WATER AND NEUTRAL pH - III

• When the solution is said to be
  neutral. Hence, neutral pH at 25C is pH 7.
• Note that, a pH of 7 is only neutral at 25C and
  1 bar, because Kw is a function of both pressure
  and temperature.
• For example, at 0C neutral pH is 7.47 and at
  50C neutral pH is 6.63. By 300C, neutral pH is
  5.7!

61
DISSOCIATION CONSTANT OF WATER AT VARIOUS
TEMPERATURES (FROM FAURE, 1997)
62
THE DISSOCIATION OF WATER AND NEUTRAL pH - IV

• Finally, is a solution with and
• in equilibrium at 25C?
• To answer this we need to calculate the IAP
• Because IAP gt Kw, the solution is not in
  equilibrium. Some OH- and H need to be consumed
  to bring the reaction back to equilibrium.

63
pH (Cont.)
Range of pH values in the natural environment

• Most natural waters have pH between 4-9.
• The acids are usually weak, including carbonic
  acid and organic acids (e.g. fulvic and humic)
• pH values gt 8.5 are rare, occurring only in
  evaporitic lakes, lakes clogged with
  photosynthetic plants, and springs discharging
  from serpentine or ultramafic rocks.

64
pH (Cont.)

• Most reactions in gas/water/rock systems involve
  or are controlled by pH
• Aqueous acid-base equilibria, including
  hydrolysis and polymerization.
• Adsorption, because protons compete with cations
  and hydroxyl ions compete with anions for
  adsorption sites. Also, the surface charge of
  most minerals is pH dependent.
• The formation of metal ligand complexes, because
  protons compete with metal ions to bond with
  weak-acid ions, and OH- competes with other
  ligands that would form complexes.
• Oxidation-reduction reactions, whether
  abiological or biologically mediated. Oxidation
  usually produced protons, whereas reduction
  consumes them.
• The solubility rate of dissolution of most
  minerals is strongly pH-dependent. Weathering of
  carbonate, silicate, and alumino-silicate
  minerals consumes protons and releases metal
  cations.

65
Acidity

• Capacity of water to give or donate protons
• Contributions are
• Immediate acidity species present in solution
• Strong and weak acids
• Salts of strong acids and weak bases
• Hydrolysis of Fe3 and Al3
• Oxidation and hydrolysis of Fe2 and Mn2
• Long-term acidity
• Due to reactions of water with solids in system

66
Acidity (Cont.)

• Acidity gives water a greater capacity to attack
  geological material and is usually accompanied by
  high total dissolved solids (TDS), including
  hardness (the sum of the concentrations of the
  multivalent cations, such as Ca2 and Mg2).
• Acidity increases the solubility of hazardous
  substances such as heavy metals and it is
  corrosive and toxic to fish and other aquatic
  life.

67
Alkalinity

• Alkalinity is the capacity of water to accept
  protons.
• Due primarily to bicarbonate ion (HCO3-) and to a
  minor extent, the carbonate ion (CO32-).
• Carbonate alkalinity HCO3- 2 CO32-
• The contribution of OH- is important above pH10.
• Other bases that contribute to the total
  alkalinity are
• Ligands of fulvic acid
• Organic anions such as formate, acetate, and
  propionate
• Bisulfides, orthophosphates, ammonia, and
  silicates.
• Usually reported as mg/L CaCO3 or meq/L CaCO3