1. 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

2. 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+

3. 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…

4. 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-

5. 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!

6. 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!

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

8. 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

9. 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…

10. 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

11. 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…

12. 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

13. 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

14. 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

15. 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

16. 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.

17. 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!