lecture 3: mineral solubility

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Lecture 3: Mineral Solubility

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Page 1: Lecture 3: Mineral Solubility

Lecture 3: Mineral Solubility

Page 2: Lecture 3: Mineral Solubility

Solubility Controls Biomineralization

• Organisms produce hard parts by exceeding the solubility of the mineral component

• Increased CO2 in the oceans increases carbonate mineral solubility, making biomineralizationmore difficult

De Yoreo and Dove (2004) Science 306, 1301

Page 3: Lecture 3: Mineral Solubility

Solubility Determines Deep-Sea Sediment Types

from: Marine Chemistry by Schulz and Zabel

Page 4: Lecture 3: Mineral Solubility

Solubility Controls Contaminant Fate

McKinley et al. (2007) Vadose Zone J. Stubbs et al. (2009) Geochim. Cosmochim. Acta

Page 5: Lecture 3: Mineral Solubility

Solubility Affects “Soil” Development on Mars

Layered ferric sulfate, calcium sulfate, and iron oxide, Columbia Hills, Mars

Subsurface perchlorate salts, a sign of soil water transport,

Northern Plains, Mars

Page 6: Lecture 3: Mineral Solubility

Solubility Thermodynamics

• Equilibrium constant for mineral solubility is called the solubility product, Ksp– Convention is to write as a dissolution reaction

• Example: GypsumCaSO4·2H2O(s) = Ca2+ + SO4

2- + 2H2O Ksp = [Ca2+][SO4

2-] log Ksp = -4.58– [CaSO4·2H2O] and [H2O] assumed to equal 1

• We’ll address when these assumption do not hold later today and in the next lecture

Page 7: Lecture 3: Mineral Solubility

Evaluating the Saturation State of Natural Waters

• For minerals we use the saturation index (SI) to evaluate saturation state

SI = log (Q/K)– SI = 0 when solution is saturated

• Mineral in equilibrium with solution– SI < 0 when solution is undersaturated

• Mineral, if present, should dissolve– SI > 0 when solution is supersaturated

• Mineral should precipitate

• Sometimes see SI = Q/K, or Ω = Q/K– “Ion Activity Product (IAP)” = Q– We don’t use these!!!

Barite Q/K in the Central Pacific OCean

Page 8: Lecture 3: Mineral Solubility

“The Dolomite Problem”

Ordered Dolomite: Q/K = 1950, SI = 3.29Disordered Dolomite: Q/K = 550, SI = 2.74

Page 9: Lecture 3: Mineral Solubility

Ocean Acidification and Carbonate Mineral Saturation State

Hoegh-Guldberg et al. (2007) Science

Page 10: Lecture 3: Mineral Solubility

Silica Saturation States of Natural Waters

Type of Water Range (ppm) SIQuartz SIAm. Silica

Rivers/Lakes 5-25 -0.1 to 0.6 -1.4 to -0.7Seawater 0.01-7 -2.8 to 0.05 -4.1 to -1.2Soil Porewater 1-117 -0.8 to 1.3 -2.1 to 0.0Groundwater 5-85 -0.1 to 1.1 -1.4 to -0.1Oil Field Brine 5-60 -0.1 to 1.0 -1.4 to -0.3Hot Springs* 100-600 1.2 to 2.0 -0.1 to 0.7

* SI values calculated for 25°C. Hot spring temperatures range as high as 100°C at the surface, and SIquartz ~ 0 at spring water temperature for many hot springs. When water cools, silica precipitates, forming sinter.

SiO2(s) + 2H2O = H4SiO4(aq)Ksp = [H4SiO4]

log Ksp,Quartz = -4.0 log Ksp,Am. Silica = -2.7

Page 11: Lecture 3: Mineral Solubility

Mineral Solubility and Stability• Thermodynamics predicts that the lowest energy

state should occur– This state is said to be thermodynamically stable

• A solution that is supersaturated is not stable– The saturated mineral phase should precipitate

• However, sometimes mineral precipitation is kinetically-inhibited– Metastable phases often form instead– Solutions must be supersaturated with respect to a

metastable phase for it to precipitate• Metastable phases are always more soluble

than stable phases!!!!

Page 12: Lecture 3: Mineral Solubility

Silica Saturation States of Natural Waters

Type of Water Range (ppm) SIQuartz SIAm. Silica

Rivers/Lakes 5-25 -0.1 to 0.6 -1.4 to -0.7Seawater 0.01-7 -2.8 to 0.05 -4.1 to -1.2Soil Porewater 1-117 -0.8 to 1.3 -2.1 to 0.0Groundwater 5-85 -0.1 to 1.1 -1.4 to -0.1Oil Field Brine 5-60 -0.1 to 1.0 -1.4 to -0.3Hot Springs* 100-600 1.2 to 2.0 -0.1 to 0.7

* SI values calculated for 25°C. Hot spring temperatures range as high as 100°C at the surface, and SIquartz ~ 0 at spring water temperature for many hot springs. When water cools, silica precipitates, forming sinter.

SiO2(s) + 2H2O = H4SiO4(aq)Ksp = [H4SiO4]

log Ksp,Quartz = -4.0 log Ksp,Am. Silica = -2.7

Page 13: Lecture 3: Mineral Solubility

Stable Versus Metastable• Carbonates

– STABLE: Calcite [CaCO3], Dolomite [CaMg(CO3)2]– METASTABLE: Aragonite [CaCO3], Mg-calcite*

• Sulfides– STABLE: Pyrite [FeS2]– METASTABLE: Mackinawite [Nanocrystalline FeS]

• Iron oxides– STABLE: Hematite [Fe2O3], Magnetite [Fe3O4]– METASTABLE: Ferrihydrite [Fe(OH)3]– Goethite [FeOOH] is metastable with respect to

hematite at 25° but stable below ~15°C

Page 14: Lecture 3: Mineral Solubility

Stable Versus Metastable• Clays and Zeolites

– Talc, Muscovite stable– Kaolinite*, Illite, Smectite, Clinoptilolite metastable

• Clay stability difficult to assess– Rarely occur as simple phases with definitive

compositions– Difficult to measure thermodynamic properties

• Metastable phases may interconvert:– Smectite to Illite in marine sediments

Page 15: Lecture 3: Mineral Solubility

Solid Solutions

• A solid solution occurs when an element substitutes into a mineral, and the substituting element can occur on its own in an isostructural phase

• Terminology needed:– Isostructures: Two minerals of different

composition but same structure– Polymorphs: Two minerals of the same

composition but different structures

Page 16: Lecture 3: Mineral Solubility

Isostructures that Form Solid Solutions

AragoniteCaCO3

StrontianiteSrCO3

CalciteCaCO3

OtaviteCdCO3

Corundumα-Al2O3

Hematiteα-Fe2O3

Diasporeα-AlOOH

Goethiteα-FeOOH

Page 17: Lecture 3: Mineral Solubility

Solubility of Ideal Solid Solutions

• Consider the CaCO3-CdCO3 solid solutionCaCO3 = Ca2+ + CO3

2- log Ksp = -8.48CdCO3 = Cd2+ + CO3

2- log Ksp = -12.1• For a normal solid, we set concentration of

mineral equal to 1 in the equilibrium equation• For a solid solution, these are set to the mole

fraction:

[ ] Cd

SS

XCaCd

CdCdCOmolmol

mol==

+

3

See Section 4.5 of Textbook for more details and examples

Page 18: Lecture 3: Mineral Solubility

Effect of T on Solubility• The van’t Hoff equation demonstrates how

T affects an equilibrium constant:log K2 – log K1 = (ΔHr

o/2.303R)(1/T1 – 1/T2)– If ΔHr

o < 0, mineral solubility decreases with increasing T

– If ΔHro > 0, mineral solubility increases with

increasing T• Examples: ΔHr

o(Calcite) = -10.6 kJ/mol; ΔHr

o(Quartz) = 25.1The example here refers to mineral solubility, but the van’t Hoff equation applies

to any equilibrium constant. (Hint: This applies to gas solubility on PS1)

Page 19: Lecture 3: Mineral Solubility

Effect of T on Solubility

Page 20: Lecture 3: Mineral Solubility

Common Ion Effect

• Solutes in real systems often have more than one origin

• This leads to the common ion effect– Predicted concentration of a solute in complex

systems differs from in simple systems– This makes the solid less soluble than would

be expected for single-phase system• Example: Effect of Na2SO4 on Gypsum

solubility

Page 21: Lecture 3: Mineral Solubility

Common Ion Effect in Groundwater: Fluorite Solubility