1. S.I. = log Q / K FROM A WATER ANALYSIS: AT EQUILIBRIUM:
Calculate DG0R from tabulated thermo data
DG0R = -RT ln Keq
Calculate a value for Keq  that number defines the ratio:
FROM A WATER ANALYSIS:
Determine ai for each ion (consider speciation, calc I, find gi for each iteratively)
Calculate Q for any given reaction via:
S.I. = log Q / K

2. Goldschmidt’s rules of Substitution
The ions of one element can extensively replace those of another in ionic crystals if their radii differ by less than about 15%
Ions whose charges differ by one may substitute readily if electrical neutrality is maintained – if charge differs by more than one, substitution is minimal

3. Goldschmidt’s rules of Substitution
When 2 ions can occupy a particular position in a lattice, the ion with the higher ionic potential forms a stronger bond with the anions surrounding the site
Substitution may be limited when the electronegativities of competing ions are different, forming bonds of different ionic character

4. Goldschmidt’s rules updated…
After Wood (2003): Equilibrium partitioning depends primarily on 2 energies of substitution into the crystal (a) the energy of elastic strain generated by inserting an ion which is either too large or too small for the site. (b) the electrostatic work done in inserting an ion which is either more or less highly charged than the major ion normally occupying the site.
The theory requires modifications to Goldschmidt’s rules (2) and (3).
Rule (2) should now be: The site has a preferred radius of ion (rO) which enters most easily. For ions of the same charge, those which are closest in radius to rO enter most easily. Ions which are smaller or larger are discriminated against.
Rule (3): The site has a preferred charge ZO . For ions of similar size, but different charge the one whose charge is closest to ZO enters most easily.

5. FeS2 What ions would substitute nicely into pyrite?? S- radius=219 pm
Fe2+ radius=70 pm

6. Coupled Substitution
When the ion of a major element in a mineral is replaced with something having a different charge, the charge imbalance created must be neutralized by addition of a counter ion
Example  addition of Al3+ in a silicate structure (replacing Si4+) requires addition of a Na+ or K+ (Key to understanding feldspar chemistry…). When 2 Al3+ are added for Si4+, this then can be balanced by adding a Ca2+ ion

8. Goldschmidt’s Classifications
Principally for how trace elements/ transition metals are distributed:
Atmophile elements are generally extremely volatile (i.e., they form gases or liquids at the surface of the Earth) and are concentrated in the atmosphere and hydrosphere.
Lithophile elements are those showing an affinity for silicate phases and are concentrated in the silicate portion (crust and mantle) of the earth.
Siderophile elements have an affinity for a metallic liquid phase. They are depleted in the silicate portion of the earth and presumably concentrated in the core.
Chalcophile elements have an affinity for a sulfide liquid phase. They are also depleted in the silicate earth and may be concentrated in the core. They are also often associated with ore deposits formed via aqueous processes.

9. Atmophile elements are generally extremely volatile
Lithophile elements are those showing an affinity for silicate phases
Siderophile elements have an affinity for a metallic liquid phase.
Chalcophile elements have an affinity for a sulfide liquid phase.

10. Acid-Base Geochemistry
Arrhenius’ definition:
Acid  any compound that releases a H+ when dissolved in water
Base  any compound that releases an OH- when dissolved in water
Bronstead-Lowry’s definition:
Acid  donates a proton
Base  receive/accept a proton
Lewis’ definition:
Acid  electron pair donor acceptor
Base  electron pair donor

11. Conjugate Acid-Base pairs
Generalized acid-base reaction:
HA + B  A + HB
A is the conjugate base of HA, and HB is the conjugate acid of B.
More simply, HA  A- + H+ HA is the conjugate acid, A- is the conjugate base
H2CO3  HCO3- + H+

12. Hydrolysis Mz + H2O  M(OH)z-1 + H+
Reaction of a cation, which generates a H+ from water is a hydrolysis reaction
Described by the equilibrium constant Ka
Hydrolysis also describes an organic reaction in which the molecule is cleaved by reaction with water…

13. AMPHOTERIC SUBSTANCE Now consider the acid-base reaction:
NH3 + H2O  NH4+ + OH-
In this case, water acts as an acid, with OH- its conjugate base. Substances that can act as either acids or bases are called amphoteric.
Bicarbonate (HCO3-) is also an amphoteric substance:
Acid: HCO3- + H2O  H3O+ + CO32-
Base: HCO3- + H3O+  H2O + H2CO30
Some substances can either donate or accept a proton, depending on the pH of the solution. Such substances are termed amphoteric. If pH is low (i.e., the activity of H+ is high), an amphoteric substance will act as a base and accept a proton. However, if pH is high (i.e., H+ ions are scarce), an amphoteric substance will act as an acid and donate a proton. Examples of amphoteric substances include H2O and HCO3- as shown above, as well as HSO4-, H2PO4-, HPO42-, etc.
Acid: HSO4-  SO42- + H+
Base: HSO4- + H+  H2SO40
Acid: H2PO4-  HPO42- + H+
Base: H2PO4- + H+  H3PO40
Acid: HPO42-  PO43- + H+
Base: HPO42- + H+  H2PO4-

14. Strong Acids/ Bases
Strong Acids more readily release H+ into water, they more fully dissociate
H2SO4  2 H+ + SO42-
Strong Bases more readily release OH- into water, they more fully dissociate
NaOH  Na+ + OH-
Strength DOES NOT EQUAL Concentration!

15. Acid-base Dissociation
For any acid, describe it’s reaction in water:
HxA + H2O  x H+ + A- + H2O
Describe this as an equilibrium expression, K (often denotes KA or KB for acids or bases…)
Strength of an acid or base is then related to the dissociation constant  Big K, strong acid/base!
pK = -log K  as before, lower pK=stronger acid/base!

16. Geochemical Relevance?
LOTS of reactions are acid-base rxns in the environment!!
HUGE effect on solubility due to this, most other processes

17. Dissociation of H2O H2O  H+ + OH- Keq = [H+][OH-]
log Keq = -14 = log Kw
pH = - log [H+]
pOH = - log [OH-]
pK = pOH + pH = 14
If pH =3, pOH = 11  [H+]=10-3, [OH-]=10-11
Definition of pH

18. pH
Commonly represented as a range between 0 and 14, and most natural waters are between pH 4 and 9
Remember that pH = - log [H+]
Can pH be negative?
Of course!  pH -3  [H+]=103 = 1000 molal?
But what’s gH+?? Turns out to be quite small  or so…

19. pKx? Why were there more than one pK for those acids and bases??
H3PO4  H+ + H2PO4- pK1
H2PO4-  H+ + HPO42- pK2
HPO41-  H+ + PO43- pK3

20. Methods of solving equations that are ‘linked’
Sequential (stepwise) or simultaneous methods
Sequential – assume rxns reach equilibrium in sequence:
0.1 moles H3PO4 in water:
H3PO4 = H+ + H2PO42- pK=2.1
[H3PO4]=0.1-x , [H+]=[HPO42-]=x
Apply mass action: K=10-2.1=[H+][HPO42-] / [H3PO4]
Substitute x  x2 / (0.1 – x) =  x x = 0, solve via quadratic equation
x=0.024  pH would be 1.61
Next solve for H2PO42-=H+ + HPO4-…

21. BUFFERING
When the pH is held ‘steady’ because of the presence of a conjugate acid/base pair, the system is said to be buffered
In the environment, we must think about more than just one conjugate acid/base pairings in solution
Many different acid/base pairs in solution, minerals, gases, can act as buffers…

22. Henderson-Hasselbach Equation:
When acid or base added to buffered system with a pH near pK (remember that when pH=pK HA and A- are equal), the pH will not change much
When the pH is further from the pK, additions of acid or base will change the pH a lot

23. Buffering example Let’s convince ourselves of what buffering can do…
Take a base-generating reaction:
Albite + 2 H2O = 4 OH- + Na+ + Al SiO2(aq)
What happens to the pH of a solution containing 100 mM HCO3- which starts at pH 5??
pK1 for H2CO3 = 6.35

24. After 12.5 mmoles albite react (50 mmoles OH-):
Think of albite dissolution as titrating OH- into solution – dissolve 0.05 mol albite = 0.2 mol OH-
0.2 mol OH-  pOH = 0.7, pH = ??
What about the buffer??
Write the pH changes via the Henderson-Hasselbach equation
0.1 mol H2CO3(aq), as the pH increases, some of this starts turning into HCO3-
After 12.5 mmoles albite react (50 mmoles OH-):
pH=6.35+log (HCO3-/H2CO3) = 6.35+log(50/50)
After 20 mmoles albite react (80 mmoles OH-):
pH=6.35+log(80/20) = = 6.95

25. Bjerrum Plots 2 D plots of species activity (y axis) and pH (x axis)
Useful to look at how conjugate acid-base pairs for many different species behave as pH changes
At pH=pK the activity of the conjugate acid and base are equal

26. Bjerrum plot showing the activities of reduced sulfur species as a function of pH for a value of total reduced sulfur of 10-3 mol L-1.
In slide 8 we saw that, in the pH range of most natural waters, bicarbonate was the predominant species in the CO2-H2O system. In this slide, we see that the predominant species in the H2S-H2O system over the pH range of most natural waters is H2S0 (pH < 7.0) or HS- (pH > 7.0). This diagram can be constructed in exactly the same way as outlined for the previous diagram. Note that, as expected, the positions of the lines representing the concentrations of H+ and OH- have not changed.

27. In most natural waters, bicarbonate is the dominant carbonate species!
Bjerrum plot showing the activities of inorganic carbon species as a function of pH for a value of total inorganic carbon of 10-3 mol L-1.
Although Bjerrum plots can be constructed rigorously by solving the combined mass-action and mass-balance expressions in the system for the concentrations of each of the species, there is a faster, approximate route to the construction of these diagrams. Once the total carbonate concentration (CT) is chosen and the pK values are known, the first step is to plot points with pH coordinates equal to the pK values, and concentration coordinates equal to log CT At pH = pK, the concentrations of two species are equal, and therefore equal to CT/2, the log of which is log CT For example, at pH = pK1 = 6.35, the concentrations of H2CO3* and HCO3- are equal to one another and to CT/2. Likewise, at pH = pK2 = 10.33, the concentrations of HCO3- and CO32- are equal to one another and to CT/2. The points where species concentrations are equal are called cross-over points. At pH < pK1 = 6.35, H2CO3* accounts for more than 99% of CT, so the concentration of H2CO3* plots as a horizontal line with a Y-intercept of log CT. As pH nears pK1, the line must bend down to intersect the HCO3- line at the first cross-over point. The HCO3- line extends from the first cross-over point towards lower pH with a slope of +1. At pK1 < pH < pK2, HCO3- accounts for the bulk of CT, so its concentration now plots as a horizontal line. In this pH range, the H2CO3* line descends away from the cross-over point towards higher pH with a slope of -1. As pH approaches pK2, the HCO3- line drops down to the second cross-over point. At pH > pK2, CO32- is the predominant species, so its concentration now plots as a horizontal line at log CT, and the HCO3- line descends from the second cross-over point towards higher pH with a slope of -1. In the range pH > pK2, the H2CO3* line now descends towards higher pH with a slope of -2. As the CO32- line passes through the second cross-over point towards lower pH into the region where pK1 < pH < pK2, it descends with a slope of +1. When this same line crosses under the first cross-over point into the region where pH < pK1, its slope changes to +2.
In most natural waters, bicarbonate is the dominant carbonate species!