Evaporites deposits Evaporites are formed in closed or semi-closed basins where evaporation exceeds precipitation (+ runoff).
Rogers Dry Lake, Mojave Desert, California
The Salar de Uyuni Lake in Bolivia (more than 9,000 km2) is the largest salt playa in the world. The salar (salt pan) has about a meter of briny water (blue). In addition to sodium chloride, NaCl (rock salt), and calcium sulfate, CaSO4, the lake also contains lithium chloride, LiCl, making this the biggest source of lithium in the world. http://eps.mcgill.ca/~courses/c542/ 1/42 Chemical Fractionation and the Chemical Divide As a result of evaporation, chemical fractionation takes place between seawater and the remaining concentrated brines. The fractionation can be accounted for by a variety of mechanisms: 1) Mineral precipitation 2) Selective dissolution of efflorescent crusts and sediment coatings 3) Exchange and sorption on active surfaces 4) Degassing 5) Redox reactions
Mineral precipitation is the most important and the one that can most easily be modeled. The basic assumption of the model is that minerals will precipitate as the solution becomes saturated with respect to a solid phase. In other words, precipitation occurs when the ion activity product of the solution becomes equal to the solubility constant of the mineral and remains constant upon further evaporation. The fate of seawater constituents upon mineral precipitation rests on the concept of the chemical divide. 2/42 Chemical Divides: Branching points along the path of solution evolution
3/42 Brine evolution and chemical divides 2+ 2- If we monitor the evaporation of a dilute Ca and SO4 solution, their concentrations will increase until we reach gypsum saturation. From this point on, 2+ 2- the evolution of the remaining solution will depend on the [Ca ]:[SO4 ] ratio of the starting solution. Two cases are possible: 2+ 2- 1) [Ca ] = [SO4 ], the ratio is equal to the stoichiometry of the solid, the ratio in solution will not change as evaporation progresses because they are both taken out of solution in equal amounts.
2+ 2- 2+ 2- 1/2 -3 [Ca ] , [SO4 ] [Ca ] = [SO4 ] = (Ksp) = 4.9 x 10 m
2+ 2- [Ca ] [SO4 ] = Ksp
degree of evaporation
2+ 2- 2) [Ca ] ≠ [SO4 ] ?? 4/42 Brine evolution and chemical divides
2+ 2- If we start with a solution in which [Ca ] = 2 [SO4 ], which upon evaporation has reached equilibrium with respect to gypsum. If we neglect activity coefficients and assume that the solubility product for gypsum is 10-4.61.
2+ 2- -4.61 2- 2- 2- 2 [Ca ] [SO4 ] = 10 = 2 [SO4 ] [SO4 ] = 2 [SO4 ] , then 2- -3 2+ -3 [SO4 ] = 3.5 x 10 m and [Ca ] = 7.0 x 10 m If we evaporate this solution until the mass of water remaining is a fraction 1/n of the saturated solution (i.e., a concentration factor of n) which results in the precipitation of y moles of gypsum per kilogram of original water. 2+ -3 2- -3 [Ca ]n = n (7.0 x 10 - y) and [SO4 ]n = n (3.5 x 10 - y) if the solution remains in equilibrium with gypsum
-4.61 2 -3 -3 Ksp = 10 = n (7.0 x 10 - y)(3.5 x 10 - y) = n2 (y2 – 1.05 x 10-2 y + 24.5 x 10-3) for any value of n, the quadratic equation can be solved for y and the concentration of 2+ 2- Ca and SO4 calculated from above. 5/42 Brine evolution and chemical divides
6/42 Brine evolution and chemical divides The general principal being that “whenever a binary salt is precipitated during evaporation, and the effective ratio of the two ions in the salt is different from the ratio of the concentrations of the ions in solution, further evaporation will result in an increase in the concentration of the ion present in greater relative concentration in solution, and a decrease in the concentration of the ion present in lower relative concentration.” …
Mono Lake
Dead Sea
7/42 Real solutions Given that the solubility product of gypsum is approximately 2.4 x 10-5 , one can 2+ -2 2- -2 determine if a solution containing [Ca ] = 1 x 10 m and [SO4 ] = 3 x 10 m is saturated with respect to the salt, if we know that the total ion activity coefficients for the two species of interest in this solution are, respectively, 0.225 m-1 and 0.0843 m-1.
2+ 2- IAP = ion activity product = a(Ca ) a(SO4 ) 2+ 2- 2+ 2- = [Ca ] [SO4 ] γ(Ca ) γ(SO4 ) = (1 x 10-2) (3 x 10-2) (0.225) (0.0843) = 5.7 x 10-6 -6 -5 Ω = saturation state = IAP/Kºsp = 5.7 x 10 /2.4 x 10 = 0.24 since Ω < 1 the solution is undersaturated with respect to gypsum in this solution, … Saturation indices are also often used to report the state of a solution with respect to a given mineral.
SI = saturation index = log IAP/Kºsp if SI = 0, the solution is in equilibrium with the mineral. If SI is positive, the solution is supersaturated with respect to the mineral and, if SI is negative, the solution is undersaturated. 8/42 Seawater evaporation and marine evaporites
Zechstein Sea during the Late Permian period over the topography of present day North Western Europe.
9/42 Seawater evaporation and marine evaporites The sequence of minerals obtained upon the evaporation of seawater depends on whether or not phases that crystallize are immediately removed from interaction with the remaining solution.
Once CaCO3 and gypsum have precipitated, upon further evaporation and halite precipitation, glauberite should precipitate by reaction of gypsum with the remaining brine. + 2- CaSO4●2H2O + 2Na + SO4 Na2Ca(SO4)2 + 2H2O Upon further evaporation, glauberite is replaced progressively, in part by anhydrite and then both are replaced by polyhalite according to:
+ 2- Na2Ca(SO4)2 CaSO4 + 2Na + SO4 + 2+ 2- Na2Ca(SO4)2 + CaSO4 + 2K + Mg + SO4 + 2H2O + K2MgCa(SO4)4 ●2H2O + 2Na After glauberite is consumed, polyhalite forms at the expense of anhydrite: + 2+ 2- 2CaSO4 + 2K + Mg + 2SO4 + 2H2O K2MgCa(SO4)4 ●2H2O
10/42 Equilibrium seawater evaporation
Polyhalite coprecipitates with a sequence of hydrated MgSO4 salts: Epsomite: MgSO4●7H2O Hexahydrite: MgSO4●6H2O Kieserite: MgSO4●H2O
Upon further evaporation, carnalite (KMgCl3●6H2O) joins the assemblage and the composition of the solution remains invariant until polyhalite is consumed by the following reaction:
2+ - K2MgCa(SO4)4 ●2H2O + Mg + 6Cl + 10H2O 2- 2KMgCl3●6H2O + 2CaSO4 + 2SO4
And finally, bischofite (MgCl2●6H2O ) forms as part of the final assemblage composed of: halite/kieserite/carnellite/anhydrite/bischofite/solution
11/42 Fractional seawater evaporation If any phase that crystallizes is immediately removed from interaction with the solution (e.g., separate evaporative pens), in addition to the minerals encountered during the equilibrium evaporation, bloedite (Na2Mg(SO4)2●4H2O) and kainite (KMgClSO4●11/4H2O ) appear between polyhalite and carnallite.
The significance of these two evaporitic paths can be tested in the geological record.
12/42 Fractional seawater evaporation: salt pens
13/42 Equilibrium and fractional seawater evaporation
The Permian (~260 Myr) Zechstein (Germany) deposit is nearly 1 km thick and its mineral assemblage compares favourably with the equilibrium sequence, but it could not have formed by a one stage equilibrium evaporation of seawater because it would have required a seawater column at least 8 km thick to produce the halite layer (~900 m) alone ( ... ) and the interstratified layers of polyhalite and glauberite would not have been preserved as they are. constraints on the depositional processes
From: Harvie et al. (1980) Science 208, 498-500 14/42 Marine evaporite deposits Evaporation proceeded in a stratified brine basin as a batch process, each batch of fresh seawater added to the basin evaporated until its density was the same as the underlying brine …
15/42 Messinian gypsum deposits
16/42 Evolution of seawater composition
2+ 1- 3x [Ca ]present, imposed by the solubility of gypsum.
2+ 1 2- [Ca ]present/30, imposed by the simultaneous saturation of halite and gypsum.
3 3- pH cannot be greater than 9.0, 4 otherwise brucite, Mg(OH)2, would precipitate.
2+ - 2+ 4- [Ca ] > 2[HCO3 ], otherwise Ca would 2 not remain upon CaCO3 precipitation and gypsum would not form.
Marine evaporites show the same depositional sequence since the late Precambrian + 2+ 2+ + - 2- - (~ 1 by). Holland (1972) concluded that Na , Mg , Ca , K , Cl , SO4 and HCO3 were never more than twice or less than half as abundant as in present-day seawater. 17/42 Residual brines trapped in fluid inclusions in halite 50 μm
50 μm
- Prior to NaCl saturation, most of the HCO3 and Ca2+ in seawater, and approximately one third of 2- + + the SO4 are lost to the precipitates while Na , K , Mg2+, and Cl- are essentially conserved. Not surprisingly, the evaporation path of Miocene (5-6 Ma) evaporites is similar to today.
From: Holland (1986) Nature 320, 27-33 18/42 Evolution of seawater composition
Late Cretaceous: 66-98 Ma Early Cretaceous: 98-144 Ma Permian: 245-296 Ma Silurian: 408-438 Ma Cambrian: 505-570 Ma Precambrian: > 570 Ma
From: Lowenstein et al. (2001) Science 294, 1086-1088. 19/42 Manganese nodules
20a/42 Manganese nodules
From: Piper et al. (1985) AAPG. 20b/42 Manganese nodules
21/42 Sediment accumulation rates All sedimentation rates in the deep-ocean are based on one of four methods of absolute dating:
14 1) C (t1/2 = 5700 yrs) produced by cosmic ray interactions in the atmosphere. 230 231 2) Th (t1/2 = 75,000 yrs), Pa (t1/2 = 32,500 yrs), two radioactive isotopes produced from the decay of dissolved uranium in seawater. 10 6 3) Be (t1/2 = 1.7 x 10 yrs) produced by cosmic ray interactions in the atmosphere. 40 40 4) Ar (t1/2 = 1 byrs) produced by the decay of K in volcanic ash.
A number of relative stratigraphic dating methods can also be used. 1) Reversals in the polarity of the Earth’s magnetic field 2) Faunal changes 3) Climate change
22/42 Production and decay of 14C
About 100 14C atoms are generated by cosmic ray collisions over each square centimeter of the Earth’s surface each minute. Over time, the concentration of 14C has built up in the atmosphere to a point that it is nearly constant, i.e., its rate of production is equal to the rate of decay. 23/42 Rate of radioactive decay
Rate = k [A] k = ln 2/t1/2 = 0.693/t1/2 24/42 14C dating
t1/2 = 5730 years
Assuming that the (14C/C) at any place in the sea has always been the same as today: 14 14 -λt ( C/C)today = ( C/C)formation x e 14 14 t = 8200 ln [( C/C)formation / ( C/C)today] 25/42 Accelerator Mass Spectrometry
University of Ottawa
26/42 14C dating of a sediment core
From: Broecker and Peng (1982) Tracers in the Sea 27/42 Uranium and Thorium decay series
From: Broecker and Peng (1982) Tracers in the Sea 28/42 230Th dating Idealized profile
A230Th-excess = A230Th-total – A238U
t1/2 = 75,000 yrs
29/42 230Th dating
The principle is the same for 231 210 Pa (t1/2 = 32,500 yrs), Pb (t1/2 = 22.3 yrs), 234 and Th (t1/2 = 24.1 days) dating.
From: Broecker and Peng (1982) Tracers in the Sea 30/42 Reversals in Earth’s magnetic field polarity Over geological times, at irregular intervals, the polarity of the Earth’s magnetic field reverses.
31/42 Reversals in Earth’s magnetic field polarity
Normal polarity
Reversed polarity
32/42 Recording the Earth’s magnetic field
Thermoremanent magnetization (resulting from thermal cooling) Curie point = ~ 500оC
33/42 Magnetic stratigraphy – dating magnetic reversals (~1965, Berkeley and USGS Menlo Park)
Chrons or epochs: Periods of predominantly normal or reversed polarity.
Subchrons or events: short-term magnetic reversals (1000’s to 200,000 years).
34/42 Recording the Earth’s magnetic field
From: Willie (1976) The Way the Earth Works 35/42 Reversals in Earth’s magnetic field polarity
From: Willie (1976) The Way the Earth Works 36/42 Faunal succession and index fossils Principle of faunal succession- Fossil organisms succeed one another in a definite and determinable order ... If the period of time over which they lived is dated, fossil organisms can be used to establish the amount of time represented by a sedimentary deposit.
Index fossils – fossils that are widespread geographically and are limited to a short span of geological time.
37/42 Climate Change -Glacial cycles This technique is less reliable because climate changes are repetitive at intervals of roughly the same duration. Nevertheless, the method can be quite useful over the last 11-12 climate cycles.
Marine oxygen isotope record
38/42 Glacial – Interglacial cycles
From: Broecker and Peng (1982) Tracers in the Sea 39/42 Recent sedimentation events For the determination of rapid and recent sedimentation events, a number of other methods can be used: 210 234 137 1) Pb (t1/2 = 22.3 yrs), Th (t1/2 = 24.1 days) and Cs (t1/2 = 21 years). 2) Volcanic ash deposits from dated eruptions 3) Episodic sedimentation events (e.g., St. Jean-Vianney landslide of 1971, Saguenay flood of 1996).
La Baie, July 1996 St. Jean-Vianney, May 4, 1971 40/42 Uranium and Thorium decay series
From: Broecker and Peng (1982) Tracers in the Sea 41/42 137Cs dating Cesium-137 is a radioactive isotope formed mainly as a fission product of nuclear reactors and nuclear weapons testing. It has a half-life of about 30.1 years.
1963
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