1. Mineral Solubility
Dissolution Reactions
Activity-Ratio Diagrams
Phosphorus Fertilizer Reactions in Calcareous Soil
Gypsum in Acid Soil
Clay Mineral Weathering

2. Dissolution Reactions
Solubility depends on relative strength of bonds in mineral
compared with bonds in solvation complex
CaSO4 • 2H2O, solvation complexes energetically favorable
Little covalent character to bonds
Al(OH)3, solvation complexes not energetically favorable
Much higher extent of covalent character

3. To solubilize latter, must destabilize bonds in lattice as with attack by
H+ or ligand exchange

4. Kinetics of dissolution of easily dissolved solids controlled by film diffusion
Kinetics for clay minerals and hydrous oxides controlled by surface reaction
Zeroth order
d[M] / dt = k
k depending on surface area, temperature, pressure, and
concentrations of H+ and strongly complexing ligands
 d[M] / dt = k* [H+]N

5. However, equilibrium is achieved and solution concentrations of
constituent ions are set by thermodynamic equilibrium constant
Al(OH)3 (s) = Al3+(aq) + (OH-)3(aq)
(Al3+)(OH-)3 / (Al(OH)3) = Kdis
If minerals are pure and crystalline, activities of solid minerals = 1
Otherwise (solid solution or poorly crystalline), activities different from 1

6. Ksp = (Al3+)(OH-)3 = Kdis (Al(OH)3)
Where reaction involves OH-, commonly this species is
formally replaced by H+
Al(OH)3(s) + 3H+(aq) = Al3+(aq) + 3H2O
For which
*Ksp = *Kdis (Al(OH)3) / (H2O)3 = (Al3+) / (H+)3

7. The products
(Al3+)(OH-)3 or (Al3+) / (H+)3 called ion activity products
Generally,
MaLb(s) = aMm+(aq) + bLl-(aq)
IAP = (Mm+)a (Ll-)b

8. If one compares measured to equilibrium IAPs,
can say whether equilibrium exists
Relative saturation
= IAP / Ksp
< 1, undersaturated
= 1, equilibrium
 > 1, supersaturated

9. d[Al(OH)3] / dt = k(Ω – 1)
Do problem 4.

10. Near equilibrium, the rate of precipitation of calcite is proportional
to Ω - 1, where Ω = IAP / Kso. How many times larger is the rate
of precipitation of calcite when the IAP = 10-7 than when it is 10-8?
d[Ca(CO)3] / dt = k(Ω – 1)
(Ω – 1) = 10-7/ = = 29.20
(Ω – 1) = 10-8/ = =
Therefore ~ 15x

11. Activity-Ratio Diagrams
Does a particular solid phase control solution concentrations of certain ions
and, if so, which solid phase?
Guess set of solids and write appropriate dissolution reactions
Express Kdis equation in log form and rearrange to give form
log [ (solid phase) / (ion of interest) ] =
-log Kdis +  log [ (solution activities)]
3. Plot log [ (solid phase) / (ion of interest) ] versus a log [(solution activity)]
Commony, -log(H+) = pH
To construct linear plots, must arbitrarily set other log [(solution activity)]

12. Example calculation
Consider what Ca-mineral may be controlling (Ca2+) in a arid region soil
Compare anhydrite (CaSO4), gypsum (CaSO4 • 2H2O) and calcite (CaCO3)
CaSO4(s) = Ca2+(aq) + SO42-(aq)
log Kdis = (25 C and 1 atm)
CaSO4 • 2H2O(s) = Ca2+(aq) + SO42-(aq) + 2H2O
log Kdis = -4.62
and

13. CaCO3(s) = Ca2+(aq) + CO32-(aq)
log Kdis = 1.93
One could also write calcite dissolution as an acidic hydrolysis reaction
CaCO3(s) + 2H+(aq) = Ca2+(aq) + CO2(g) + H2O
log Kdis = 9.75

14. For anhydrite,
log Kdis = log (Ca2+) + log (SO42-) – log (CaSO4)
From which by rearrangement
log [ (anhydrite) / (Ca2+) ] = - log Kdis + log (SO42-)
Similarly,
log [ (gypsum) / (Ca2+) ] = -log Kdis + log (SO42-) + 2 log (H2O)
log [ (calcite) / (Ca2+) ] = -log Kdis + 2pH + log (CO2) + log (H2O)

15. Various choices for independent variable
pH, (SO42-), (CO2) = partial pressure in atm, or (H2O) = relative humidity
For H2O, activity typically = 1 but may be less under arid conditions
Let’s use pH and set (SO42-) = 0.003, PCO2 = and (H2O) = 1
Substituting,
log [ (anhydrite) / (Ca2+) ] = 1.86
log [ (gypsum) / (Ca2+) ] = 2.10
log [ (calcite) / (Ca2+) ] = pH

16. Interpret this figure

17. According to this approach
The solid phase that controls solubility is the one that
produces the largest activity ratio for the free ionic species in solution
Largest log [ (solid) / (Ca2+) ] at certain pH
pH < 7.8, gypsum controls but pH > 7.8, calcite controls
Anhydrite doesn’t come into play under these conditions

18. Note that if PCO2 > 0.00032 atm (say, 0.003)
log [ (calcite) / (Ca2+) ] = pH

19. Note if (H2O) < 1 (say, 0.60)
log [ (anydrite) / (Ca2+) ] = 2.38
log [ (gypsum) / (Ca2+) ] = 2.18
Clearly, predictions depend on accurate Kdis values and
accuracy of assumed conditions (e.g., PCO2)

20. Other Approaches

21. Phosphate Fertilizer Reactions in Calcareous Soil
CaHPO4 • 2H2O = Ca2+ + HPO H2O logKdis = DCPDH
CaHPO4 = Ca2+ + HPO42- logKdis = DCP
1/6 Ca8H2(PO4)6 • 5H2O + 2/3 H+ =
4/3Ca2+ + HPO /6H2O logKdis = OCP
1/6 Ca10(OH)2(PO4)6 + 4/3H+ =
5/3 Ca2+ + HPO /3H2O logKdis = HA
CaCO3 + 2H+ = Ca2+ + CO2 + H2O logKdis = 9.75 calcite

22. Problem
Assume dissolution of calcite controls calcium concentration
Develop activity / ratio diagrams for
DCPDH
DCP
OCP HA
-6.57 = log(Ca2+) + log(HPO42-) – log(DCPDH)
= pH – log(CO2) – log[(DCPDH) / (HPO42-)]
log[(DCPDH) / (HPO42-)] = – log PCO2 – 2pH

23. log[(DCP) / (HPO42-)] = 16.65 – log PCO2 – 2pH
log[(OCP) / (HPO42-)] = – log PCO2 – 2pH
log[(HA) / (HPO42-)] = – log PCO2 – 2pH
Some questions before proceeding. How does one arrive at
1/6 Ca8H2(PO4)6 • 5H2O + 2/3 H+ = 4/3Ca2+ + HPO /6H2O
from
Ca8H2(PO4)6 • 5H2O = 8Ca2+ + 2HPO PO H2O?

24. Ca8H2(PO4)6 • 5H2O = 8Ca2+ + 2HPO42- + 4PO43- + 5H2O
4PO H+ = 4HPO42-
1/6 Ca8H2(PO4)6 • 5H2O + 2/3 H+ = 4/3Ca2+ + HPO /6H2O
for which log Kdis = -3.28
From this, how does one arrive at
log[(OCP) / (HPO42-)] = – log PCO2 – 2pH
using (Ca2+) = Kdis(CaCO3)(H+)2/PCO2(H2O) or
log(Ca2+) = 9.75 – log PCO2 – 2pH

25. 1/6 Ca8H2(PO4)6 • 5H2O + 2/3 H+ = 4/3Ca2+ + HPO42- + 5/6H2O
log[1/6(OCP)/(HPO42-)] - 2/3pH = /3(Ca2+)
Substituting log(Ca2+) = 9.75 – log PCO2 – 2pH
log[1/6(OCP)/(HPO42-)] - 2/3pH = – 4/3log PCO2 – 8/3pH
log[1/6(OCP)/(HPO42-)] = log PCO2 – 2pH – 1/3 log PCO2
What about hydroxyapatite?
1/6 Ca10(OH)2(PO4)6 + 4/3H+ = 5/3 Ca2+ + HPO /3H2O
log[(HA) / (HPO42-)] = – log PCO2 – 2pH – 2/3log PCO2

26. Since HA gives the smallest (HPO42-), it should control phosphate
solubility. However, all solids can coexist and there (experimentally)
is a stepwise transformation from DCPDH to HA.

27. If the initial state of a soil is such that several solid phases can form
potentially with a given ion, the solid phase that forms first will be the
one for which the activity ratio is nearest above the initial value in the
soil. Thereafter, the remaining accessible solid phases will form in order
of increasing activity ratio, with the rate of formation of a solid phase in
sequence decreasing as its activity ratio increases. In an open system,
any one of the solid phases may be maintained indefinitely.
Gay-Lussac-Ostwald (GLO) Step Rule
If d[solid] / dt = k(Ω – 1) near equilibrium, apparently
d[solid] / dt << k(Ω – 1) if Ω >> 1

28. Gypsum in Acid Soils
AlOHSO4 $ 5H2O jurbanite logKdis = -3.8
Al4(OH)10SO4 $ 5H2O basaluminite logKdis = 5.63
KAl3(OH)6(SO4)2 alunite logKdis = 0.2
For acidic dissolution,
AlOHSO4 $ 5H2O + H+ = Al3+ + SO H2O
log[(AlOHSO4 $ 5H2O) / (Al3+)] = 3.8 – pH + log(SO42-) + 6log(H2O)
Set pH = 4.5, (H2O) = 1 and (K+) =
log[(jurbanite) / (Al3+) = log(SO42-)

29. log[(gibbsite) / (Al3+)] = -8.11 + 3pH + 3log(H2O) = 5.39

30. Clay Mineral Weathering
Inferences on stability in different weathering environments
Gibbsite
Kaolinite
Smectite