• More Aqueous Geochemistry
• First a look back at calcite solubility as a function of pH and Temperature

a
H
3



K
1
K
CO
2 p
CO
2

2 K cal

2
HCO
3

K
2

Ca
2

M
3
Ca
2


K
1
4
K
K
2 cal

K
Ca
2

CO
2
 p
CO
2
2
HCO
3

2

The pH of pure water in equilibrium with calcite at 25°C as a function of the partial pressure of CO
2
. Note that pH decreases linearly with increasing CO
2 partial pressure.
8.5
8.0
7.5
7.0
6.5
-4.0
-3.5
-3.0
-2.5
log p
CO
2
-2.0
-1.5
-1.0
3

Changes to Ocean pH as a function of PCO2

Why do we care about mineral precipitation in CO2 sequestration?

• CaCO3(s) <==> Ca 2+ (aq) + CO
3
-2 (aq)
• In writing the equilibrium constant expression for a heterogeneous equilibria, we ignore the concentrations of pure liquids and solids. So the equilibrium constant expression for the equilibria above is:
• K sp
= [Ca 2+ ][CO
3
-2 ]
• This equilibrium constant is called a solubility-product constant. Even though [CaCO3] is excluded from the equilibrium constant expression, some undissolved CaCO3(s) must be present in order for the system to be at equilibrium. In general, the solubility product constant (K sp
) is the equilibrium constant for the equilibrium that exists between a solid ionic solute and its ions in a saturated aqueous solution. The rules for writing the solubility-product expression are the same as those for writing any other equilibrium constant expression: The solubility product is equal to the product of the concentrations of the ions involved in the equilibrium, each raised to the power of its coefficient in the equilibrium equation.

•
•
• The solubility of a substance is affected not only by temperature but also by the presence of other solutes. We now consider the case in which the other solute has an ion in common with the substance involved in the equilibrium, in other words we are going to look at the common-ion effect in solubility. Just as in the case of the common-ion effect in acid-base equilibria, the common-ion effect in solubility is just a special case of LeChattelier's principle. Consider the equilibrium of CaCO3(s) with its saturated solution, if we add a solute containing
Ca 2+ or a solute containing CO3 -2 to the saturated solution, the equilibrium will shift to the left making CaCO3 less soluble.
• CaCO3(s) <==> Ca 2+ (aq) + CO
3
-2 (aq)
<----Addition of Ca 2+ or CO3 -2 shifts equilibrium left decreasing solubility--------<<
• Example: What if a solution undersaturated with respect to calcite passes through an evaporite bed with anhydrite (Ca SO
4
–
4
). Anhydrite will dissolve putting CA ++ and SO into solution. The sulfate will not affect the solubility of calcite but the added Calcium might raise the Ca ++ concentration enough to exceed the Solubility Product of Calcite.

Carbonates ( such as calcite) are one of the mineral groups that may form to sequester CO2. But remember that in the last class we say that adding CO2 to the aqueous solution caused a pH decrease. How does that affect the precipitation of calcite?
http://www.ldeo.columbia.edu/~sanpisa/Ocea nSed%20project/class%20project/caco3phsol.J
PG

These are very common reactions in weathering and hydrothermal alteration
CO
H
2
2
+ H
CO
HCO
3
-
3
2
O = H
2
= H + + HCO
= H +
CO
+ CO
3
3
3
-
-2

Feldspar compositional variation
• Or - Orthoclase KAlSi
3
O
8
• Ab - Albite NaAlSi
3
O
8
• An - Anorthite CaAl
2
Si
2
O
8
Feldspars exhibit two solid solutions
One from Or to Ab and the other from
Ab to An.
Labradorite: An50 to An70
Feldspars are the most common mineral in the earths crust !!

Solid solution is not seen just in Feldspar.
One other example is Carbonate which is important for this class

Feldspar compositional variation
• Or - Orthoclase KAlSi
3
O
8
• Ab - Albite NaAlSi
3
O
8
• An - Anorthite CaAl
2
Si
2
O
8
Feldspars exhibit two solid solutions
One from Or to Ab and the other from
Ab to An.
Labradorite: An50 to An70
Feldspars are the most common mineral in the earths crust !!

• Feldspar rich sandstone (Arkose)-- Reservoir
• Shales—Caprock
• Basalts

I. Gaus et al. / Chemical Geology 217 (2005) 319–337

I. Gaus et al. / Chemical Geology 217 (2005) 319–337

If you have kinetic rate constants you can calculate how long these reactions will take.

• Feldspar is not the only mineral you need to be concerned with in a sandstone reservoir.
• The amount of initial carbonate (probably as cement) is also important

Ca-rich plagioclase feldspar is an essential mineral in basalt.
In felsic rocks both plagioclase and
Alkali feldspars are important

Reaction of Ca-rich feldspar to calcite
Volcanic glass may be important too. It is easily reacted.

Another important mineral in basalt is OLIVINE

http://www.esc.ethz.ch/research/efficientconv ersion/cotwo/mineralization.jpg