Mineral Stability
NaCl(s)  Na  Clsaturation
equilibrium
Na+
disequilibrium
0
Time
Law of Mass Action
c
d
(C ) (D)
( ) = conc. at equilibrium

K
a
b
(A) (B)
aA  bB  cC  dD
Activity is the effective concentration
a = gc
g= activity coefficient
c = concentration
Rules when dealing with equilibrium constants
• when adding equations, their equilibrium constants are multiplied
• when subtracting equations, their equilibrium constants are divided
• when reversing an equation, its equilibrium constant is inverted
Mineral Dissociation
C d(O H)2  C d(O H) (aq)  O H

2
C d(O H) (aq)  C d

 OH
C d(O H)2 (s)  C d(O H)2 (aq)
K1
K2
K aq
Cd(OH)2(s) is amphoteric:
Cd(OH)2  OH-  Cd(OH)-3 (aq)
Ka
Combine (add) equations, multiply K:
2
Cd(O H)2 (s)  Cd

 2O H
KT = K1 x K2
KT
Mineral Dissociation
C d(O H)2  C d(O H) (aq)  O H

2
C d(O H) (aq)  C d

 OH
C d(O H)2 (s)  C d(O H)2 (aq)
K1
K2
K aq
Reverse the reaction, invert K:
Cd(O H)2 (aq)  Cd(O H)2 (s)
1/Kaq
Subtracting equations, divide K:
Cd(O H)2 (aq)  Cd(O H)  O H-
K1/Kaq
Brucite dissociation — Solubility of a sparingly soluble base
Mg(OH)+(aq) + OH- K1
Mg(OH)2(s)
Mg(OH)+(aq)
Mg2+(aq) + OH-
K2
pK1 = 8.6, pK2 = 2.6
1 — Weak acids and
bases do not control
the pH of natural
environments but
rather respond to it
Dominantly Mg+ when pH <11.4
(most geologic environments)
Faure p. 123, Fig. 9.2
2 — The solubility of
sparingly soluble
bases is pH dependent
Weak acids and bases — pH control of dissociation
H2S(aq)
pH < 7.0
HS- + S2-
pH = 7.0-12.9
pH > 12.9
These relationships occur
in all weak acids and bases
because their dissociation
into ions is controlled by
the pH of the environment
Faure p. 125, Fig. 9.3
Silicic acid dissociation — solubility of amorphous silica
SiO2(amorphous) + H2O
H4SiO4
Solubility of amorphous silica rises
rapidly at pH ≥8 because it is based
on the sum of the concentrations of
the Si-bearing ions in solution
Agate
Opal
A decrease of 0.1 pH from 8.5 to 8.4 of a saturated
solution can deposit 1.37 mg of amorphous SiO2
per liter of solution
Faure p. 127, Fig. 9.4
SiO2•nH2O
Dissolution and the solubility of salts
The simplest weathering reaction is dissolution
of soluble salts, for example:
CaSO4(anhydrite)  Ca2+ + SO42Solubility - The total amount of a substance that will
dissolve in a solution at equilibrium.
Solubility Product
The solubility of a mineral is governed by the solubility product,
the equilibrium constant for a reaction such as:
CaSO4(anhydrite)  Ca2+ + SO42The solubility product is given by:
K SP 
aCa 2  aSO 2 
4
aCaSO4
If anhydrite is a pure solid, then aCaSO4 = 1.
and in dilute solutions: aCa2+  (Ca2+) and aSO42-  (SO42-).
K SP 
aCa 2  aSO 2 
4
aCaSO4
KSP  [Ca2+][SO42-] = 10-4.5
What is the solubility of anhydrite in pure water?
If anhydrite dissolution is the only source of both Ca2+ and
SO42-, then:
[Ca2+] = [SO42-] = x
x2 = 10-4.5
x = 10-2.25 = 5.62x10-3 mol/L
MWanhydrite = 136.14 g/mol
Solubility = (5.62x10-3 mol/L)(136.14 g/mol) = 0.765 g/L
A more complicated salt…
Al2(SO4)3(s)  2Al3+ + 3SO42-
K SP 
a
2
Al 3
a
3
SO42 
a Al2 ( SO4 )3 ( s )
Assume aAl2(SO3)3 = 1, aAl3+  [Al3+] and aSO42-  [SO42-]
KSP  [Al3+]2[SO42-]3 = 69.19
Let x = the total number of moles of Al2(SO4)3(s) dissolved.
[Al3+] = 2x
and
[SO42-] = 3x
Al2(SO4)3(s)  2Al3+ + 3SO42-
K SP 
2
3
a Al
a
3
SO 2 
4
a Al2 ( SO4 )3 ( s )
69.19 = (2x)2(3x)3
69.19 = 108x5
x = 0.9147 mol/L
[Al3+] = 2x = 1.829 mol/L
[SO42-] = 3x = 2.744 mol/L
x = (0.9147 mol/L)(342 g/mol)
x = 312.8 g/LAl2(SO4)3(s)
= 69.19
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
index

Ca  SO 

2
act
KSP
2
4 act
IAP

KSP
Suppose a groundwater is analyzed to contain 5x10-2 mol/L Ca2+ and
7x10-3 mol/L SO42-. Is this water saturated with respect to anhydrite (CaSO4(s))?
CaSO4  Ca2+ + SO42KSP = 10-4.5
IAP = (5x10-2)(7x10-3) = 3.5x10-4 = 10-3.45
 = 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
(equilibrium conditions)
If  < 1, i.e., IAP < KSP, the solution would be undersaturated;
the mineral should dissolve
Another example…
Suppose the drainage from a tailings pile contains 5x10-3 mol/L
SO42-, 10-4 mol/L Fe3+, 10-3 mol/L K+, and has pH = 3. Is this
water saturated with respect to the mineral jarosite,
KFe3(SO4)2(OH)6?
KSP = 10-96.5
IAP = [K+][Fe3+]3[SO42-]2[OH-]6
[OH-] = Kw/[H+] = 10-14/10-3 = 10-11 mol/L
IAP = (10-3)(10-4)3(5x10-3)2(10-11)6 = 10-85.60
 = 10-85.60/10-96.5 = 1010.898 = 7.91x1010
So the solution would be highly supersaturated
and jarosite should precipitate