Mineral Solubility

Dissolution Reactions

Activity-Ratio Diagrams

Phosphorus Fertilizer Reactions in Calcareous Soil

Gypsum in Acid Soil

Clay Mineral Weathering

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

To solubilize latter, must destabilize bonds in lattice as with attack byH+ or ligand exchange

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

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

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

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

If one compares measured to equilibrium IAPs, can say whether equilibrium exists

Relative saturation

= IAP / Ksp

< 1, undersaturated

= 1, equilibrium

> 1, supersaturated

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

4. Near equilibrium, the rate of precipitation of calcite is proportionalto Ω - 1, where Ω = IAP / Kso. How many times larger is the rateof precipitation of calcite when the IAP = 10-7 than when it is 10-8?

d[Ca(CO)3] / dt = k(Ω – 1)

(Ω – 1) = 10-7/10-8.48 -1 = 101.48 -1 = 29.20

(Ω – 1) = 10-8/10-8.48 -1 = 100.48 -1 = 2.02

Therefore ~ 15x

Activity-Ratio Diagrams

Does a particular solid phase control solution concentrations of certain ions and, if so, which solid phase?

1. Guess set of solids and write appropriate dissolution reactions

2. 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)]

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 = -4.38 (25 C and 1 atm)

CaSO4 • 2H2O(s) = Ca2+(aq) + SO42-(aq) + 2H2O

log Kdis = -4.62

and

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

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)

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 = 0.0003 and (H2O) = 1

Substituting,

log [ (anhydrite) / (Ca2+) ] = 1.86

log [ (gypsum) / (Ca2+) ] = 2.10

log [ (calcite) / (Ca2+) ] = -13.27 + 2pH

Interpret this figure

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

Note that if PCO2 > 0.00032 atm (say, 0.003)

log [ (calcite) / (Ca2+) ] = -11.27 + 2pH

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)

Other Approaches

Phosphate Fertilizer Reactions in Calcareous Soil

CaHPO4 • 2H2O = Ca2+ + HPO42- + 2H2O logKdis = -6.67 DCPDH

CaHPO4 = Ca2+ + HPO42- logKdis = -6.90 DCP

1/6 Ca8H2(PO4)6 • 5H2O + 2/3 H+ =

4/3Ca2+ + HPO42- + 5/6H2O logKdis = -3.28 OCP

1/6 Ca10(OH)2(PO4)6 + 4/3H+ =

5/3 Ca2+ + HPO42- + 1/3H2O logKdis = -2.28 HA

CaCO3 + 2H+ = Ca2+ + CO2 + H2O logKdis = 9.75 calcite

Problem

Assume dissolution of calcite controls calcium concentration

Develop activity / ratio diagrams for

DCPDHDCPOCPHA

-6.57 = log(Ca2+) + log(HPO42-) – log(DCPDH)

= 9.75 - 2pH – log(CO2) – log[(DCPDH) / (HPO42-)]

log[(DCPDH) / (HPO42-)] = 16.32 – log PCO2 – 2pH

log[(DCP) / (HPO42-)] = 16.65 – log PCO2 – 2pH

log[(OCP) / (HPO42-)] = 17.59 – log PCO2 – 2pH

log[(HA) / (HPO42-)] = 21.24 – log PCO2 – 2pH

Some questions before proceeding. How does one arrive at

1/6 Ca8H2(PO4)6 • 5H2O + 2/3 H+ = 4/3Ca2+ + HPO42- + 5/6H2O

from

Ca8H2(PO4)6 • 5H2O = 8Ca2+ + 2HPO42- + 4PO4

3- + 5H2O?

Ca8H2(PO4)6 • 5H2O = 8Ca2+ + 2HPO42- + 4PO4

3- + 5H2O

4PO43- + 4H+ = 4HPO4

2-

1/6 Ca8H2(PO4)6 • 5H2O + 2/3 H+ = 4/3Ca2+ + HPO42- + 5/6H2O

for which log Kdis = -3.28

From this, how does one arrive at

log[(OCP) / (HPO42-)] = 17.59 – log PCO2 – 2pH

using (Ca2+) = Kdis(CaCO3)(H+)2/PCO2(H2O) or

log(Ca2+) = 9.75 – log PCO2 – 2pH

1/6 Ca8H2(PO4)6 • 5H2O + 2/3 H+ = 4/3Ca2+ + HPO42- + 5/6H2O

log[1/6(OCP)/(HPO42-)] - 2/3pH = 3.28 + 4/3(Ca2+)

Substituting log(Ca2+) = 9.75 – log PCO2 – 2pH

log[1/6(OCP)/(HPO42-)] - 2/3pH = 3.28 + 13.00 – 4/3log PCO2 – 8/3pH

log[1/6(OCP)/(HPO42-)] = 16.28 - log PCO2 – 2pH – 1/3 log PCO2

What about hydroxyapatite?

1/6 Ca10(OH)2(PO4)6 + 4/3H+ = 5/3 Ca2+ + HPO42- + 1/3H2O

log[(HA) / (HPO42-)] = 18.63 – log PCO2 – 2pH – 2/3log PCO2

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.

If the initial state of a soil is such that several solid phases can formpotentially with a given ion, the solid phase that forms first will be theone for which the activity ratio is nearest above the initial value in the soil. Thereafter, the remaining accessible solid phases will form in orderof increasing activity ratio, with the rate of formation of a solid phase insequence 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

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+ + SO42- + 6H2O

log[(AlOHSO4 $ 5H2O) / (Al3+)] = 3.8 – pH + log(SO42-) + 6log(H2O)

Set pH = 4.5, (H2O) = 1 and (K+) = 0.0001

log[(jurbanite) / (Al3+) = 8.30 + log(SO42-)

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

= -8.77 + 3pH + 3log(H2O) = 4.73

Clay Mineral Weathering

Inferences on stability in different weathering environments

GibbsiteKaoliniteSmectite