Transcript Aqueous Complexes (or speciation)

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Dissolved Inorganic Carbon (DIC)
• Initially, DIC in groundwater comes from CO2
– CO2(g) + H2O ↔ H2CO3°
– PCO2: partial pressure (in atm)
– PCO2 of soil gas can be 10-100 times the PCO2 of
atmosphere
• In groundwater, CO2 usually increases along a
flow path due to biodegradation in a closed
system
– CH2O + O2  CO2 + H2O
• CH2O = “generic” organic matter
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Dissolved Inorganic Carbon (DIC)
• Equilibrium expression with a gas is known as
Henry’s Law:
• CO2 + H2O  H2CO3; KCO2 = 10-1.47
• H2CO3  HCO3- + H+; Ka1 = 10-6.35
• HCO3-  CO32- + H+; Ka2 = 10-10.33
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Total DIC = 10-1 M
0
-
CO2 (aq)
--
HCO3
CO3
pH = 10.33
pH = 6.35
–4
–6
-
Species with HCO3 (log molal)
–2
–8
–10
Common pH range
in natural waters
–12
–14
–16
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3
4
5
6
7
8
9
10
11
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pH
Walt Tue Feb 21 2006
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Alkalinity
• Alkalinity = acid neutralizing capability (ANC)
of water
– Total effect of all bases in solution
– Typically assumed to be directly correlated to
HCO3- concentration in groundwater
– HCO3- = alkalinity
0.82
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Salts (Electrolytes)
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Salts
• When you mix an acid + base, H+ and OH- form H2O
• The remaining anion and cation can form a salt
– e.g., mix H2SO4 + CaOH, make CaSO4
– mix HCl + NaOH, make NaCl
• Salts are named after the acid they come from
– e.g., chlorides, carbonates, sulfates, etc.
• All minerals are salts except oxides, hydroxides, and
native elements
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Solubility of Salts
• Remember: A saturated solution of a salt is in a
state of equilibrium
• Al2(SO4)3(s)  2Al3+ + 3SO42– Can write our familiar equilibrium expression with
an equilibrium constant
– Ksp = ([Al3+]2)([SO42-]3)
• Ksp = solubility product constant
• Activities of solids = 1 by definition
• Ksp values can be calculated (or looked up)
– Ksp for Al2(SO4)3(s) = 69.19 (at 25°C)
• Very large Ksp which means the salt is very soluble
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Solubility
• Al2(SO4)3(s)  2Al3+ + 3SO42-
– What is the solubility of Al2(SO4)3?
• What are the activities of Al3+ and SO42- in a saturated
solution of Al2(SO4)3?
–
– Thus in a saturated solution of Al2(SO4)3
• [Al3+] = 2x = 1.829 mol/L
• [SO42-] = 3x = 2.744 mol/L
– Solubility of Al2(SO4)3 = 0.915 x the molecular
weight (342.148 g/mol) = 3.13 x 102 g/L
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Solubility
• We often want to know whether a solution is
saturated with respect to a mineral
• e.g., Is a solution with 5 x 10-2 mol/L Ca2+ and
7 x 10-3 mol/L SO42- saturated with respect to
gypsum (CaSO42H2O)?
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Gypsum Solubility
• CaSO4 2H2O  Ca2+ + SO42-
– Ksp = [Ca2+] [SO42-] = 10-4.6
– If the solution is saturated, the product of the
activities would = 10-4.6
– (Note that [Ca2+] and [SO42-] don’t have to be equal)
– (Ca2+)(SO42-) = (5x10-2)(7x10-3) = IAP = 10-3.45
– SI = -3.45 – (-4.6) = 1.15
– Because SI > 0, gypsum predicted to precipitate
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How much gypsum would precipitate
to reach equilibrium (saturation)?
• CaSO42H2O  Ca2+ + SO42- + 2H2O
– Ksp = [Ca2+] [SO42-] = 10-4.6
– As gypsum precipitates (reverse reaction), the IAP
will decrease because [Ca2+] and [SO42-] are being
used up
– Once the IAP = Ksp, the solution will be in
equilibrium with respect to gypsum
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How much gypsum would precipitate
to reach equilibrium (saturation)?
• CaSO42H2O  Ca2+ + SO42- + 2H2O
– Ksp = [Ca2+] [SO42-] = 10-4.6
– The solution initially has 5x10-2 mol/L Ca2+ and 7x10-3 mol/L
SO42– To reach equilibrium, x moles precipitate:
• [Ca2+] = 5x10-2 - x; [SO42-] = 7x10-3 - x;
• Substitute into eq. above: [5x10-2 - x] [7x10-3 - x] = 10-4.6
• Eventually get x = 6.45 x 10-3
– Amount of gypsum that will precipitate in this solution is
6.45 x 10-3 x 172.17 (mc. wt.) = 1.11 g/L
• At this point, IAP = Ksp and the solution is saturated with respect to
gypsum, and no more will precipitate
• Equilibrium has been reached
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Changing solution composition due to
precipitation of gypsum
• As gypsum precipitates, the [Ca2+] / [SO42-]
ratio increases from 7.1 to 79.2
– The precipitation of a salt reduces the
concentrations of ions and changes the chemical
composition of remaining solution
– In our example, if precipitation continues, [SO42-]
will be used up, and none will remain in solution
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(Equilibrium reached)
Ca2+ = 79.2
SO42-
Ca2+ = 7.1
SO42-
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Geochemical Divide
• The initial ratio of species can affect which
minerals precipitate
• GEOCHEMICAL DIVIDE
– If [Ca2+] / [SO42-] had been < 1 instead of > 1, then
[SO42-] would have become concentrated relative
to [Ca2+]
– End up with a different final solution
– May lead to precipitation of different minerals
– This is important during the evolution of brines by
evaporative concentration
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Precipitation of Salts in Natural Waters
• Natural waters are complex, may have more
than 1 salt precipitating
• Let’s consider 2 sulfate minerals, gypsum and
barite
– CaSO42H2O  Ca2+ + SO42- + 2H2O
• Ksp (gypsum) = [Ca2+] [SO42-] = 10-4.6
– BaSO4  Ba2+ + SO42• Ksp (barite) = [Ba2+] [SO42-] = 10-10.0
– Barite is much less soluble than gypsum
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Precipitation of Salts in Natural Waters
• [SO42-] has the same value in both equilibria:
– [Ba2+] 10-4.6 / [Ca2+] = 10-10.0
– [Ba2+] / [Ca2+] = 10-5.4
– [Ca2+] is 250,000 x [Ba2+] when the solution is
saturated with respect to both minerals
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Gypsum and Barite equilibrium
• [SO42-] has the same value in both equilibria:
– [Ba2+] 10-4.6 / [Ca2+] = 10-10.0
– [Ba2+] / [Ca2+] = 10-5.4
– [Ca2+] is 250,000 x [Ba2+] when the solution is
saturated with respect to both minerals
• Solve for [SO42-] using simultaneous equations
– [SO42-]2 = 10-4.6 + 10-10.0
– [SO42-] = 10-2.3 mol/L
– Note that barite only contributes a negligible
amount of [SO42-]
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Gypsum and Barite equilibrium
• Suppose a saturated solution of barite comes
into contact with gypsum
– It is likely that the solution is undersaturated with
respect to gypsum, which is much more soluble
than barite
– If gypsum dissolves, [SO42-] will increase
• CaSO42H2O  Ca2+ + SO42- + 2H2O
– The increase in [SO42-] can cause the solution to
become supersaturated with respect to barite,
which is less soluble than gypsum
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Gypsum and Barite equilibrium
• CaSO42H2O  Ca2+ + SO42- + 2H2O
• Ba2+ + SO42-  BaSO4
– Barite precipitates as gypsum dissolves until [Ca2+]
/ [Ba2+] approaches 250,000
• Then replacement of gypsum by barite stops because
solution is saturated with respect to both minerals
– This is called the common ion effect
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Precipitation of Salts in Natural Waters
• Replacement of 1 mineral by another is
common in geology
– Introduction of a common ion causes solution to
become supersaturated with respect to the less
soluble compound
– Thus the more soluble compound is always
replaced by less soluble
– Makes sense: less soluble happier as solid, more
soluble happier dissolved (relatively)
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Supersaturation
•
Solutions in nature become supersaturated
with respect to a mineral by:
–
–
–
–
Introduction of a common ion
Change in pH
Evaporative concentration
Temperature variations
•
In general solubilities increase with increasing T, but
not always (e.g., CaCO3)
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Calcite Solubility
•
•
•
•
CaCO3  Ca2+ + CO32-; Ksp = 10-8.35 (1)
HCO3-  CO32- + H+; Ka2 = 10-10.33 (2)
H2CO3  HCO3- + H+; KH2 = 10-6.35 (3)
CO2(g) + H2O  H2CO3; K = 10-1.47 (4)
– If open to atmosphere
• H2O  H+ + OH- (5)
• 7 ions/molecules, need 2 more equations or to fix
something (make constant)
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Calcite Solubility
• Fix PCO2 at 10-3.5 atm
• (4) [H2CO3] = 10-1.47 x 10-3.5 = 10-4.97
– [H2CO3] = 1.07 x 10-5 mol/L
• (6) charge balance:
– 2(Ca2+) + (H+) = 2(CO32-) + (HCO3-) + (OH-)
• Now have 6 equations and 6 unknowns
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Calcite Solubility
• After some algebra:
• (Ca2+) = 5.01 x 10-4 mol/L (20.1 mg/L)
• Solubility (S) of calcite = 5.01 x 10-4 x 100.0787
(MW) = 5.01 x 10-2 g/L
– For calcite in water in equilibrium with CO2, at 25°C
– pH = 8.30
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–6
-
–10
0
2
4
6
8
pH
–3
CO2(aq)
H2CO3
HCO3
–8
CO3
--
25°C
10
12
+
Diagram HCO 3, T = 25 °C , P = 1.013 bars, a [main] = 10 , a [H2 O] = 1; Suppressed: CaHCO 3
-
++
log a Ca
0
–2
Calcite
–4
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Walt T ue Feb 21 2006
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Calcite Solubility
• The reaction we just used for calcite
dissolution generally doesn’t occur in nature
– CaCO3  Ca2+ + CO32-
• Dissolution of calcite done primarily by acid
– In natural systems, primary acid is CO2
– CaCO3 + H2CO3  Ca2+ + 2HCO3-
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Calcite Solubility
• Let us consider CaCO3 solubility as affected by
variations in PCO2, pH, and T
– CO2(g)  CO2(aq)
– CO2(aq) + H2O  H2CO3
– CaCO3 + H2CO3  Ca2+ + 2HCO3-
• Predict changes in solubility from these
reactions
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Calcite Solubility
•
•
•
•
CO2(g)  CO2(aq)
CO2(aq) + H2O  H2CO3
CaCO3 + H2CO3  Ca2+ + 2HCO3Increase PCO2?
– Increases (H2CO3), which increases amount of CaCO3
dissolved (at constant T)
• Decreasing PCO2?
– Decreases (H2CO3), causes saturated solution to
become supersaturated and precipitate CaCO3 until
equilibrium restored
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CaCO3 + H2CO3  Ca2+ + 2HCO3What if we increase PCO2 from atmospheric by 10x?
●
●
~25 mg/L calcite
could be precipitated
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Calcite Solubility
• Why does the pH decrease as PCO2 increases?
– CO2(g)  CO2(aq)
– CO2(aq) + H2O  H2CO3
– CaCO3 + H2CO3  Ca2+ + 2HCO3-
• Increasing PCO2 increases H2CO3, which
dissociates:
– H2CO3  HCO3- + H+
– Increasing H+ in solution decreases pH
– And what happens to calcite as we decrease pH?
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–6
-
–10
0
2
4
6
8
pH
–3
CO2(aq)
H2CO3
HCO3
–8
CO3
--
25°C
10
12
+
Diagram HCO 3, T = 25 °C , P = 1.013 bars, a [main] = 10 , a [H2 O] = 1; Suppressed: CaHCO 3
-
++
log a Ca
0
–2
Calcite
–4
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Walt T ue Feb 21 2006
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–6
-
–10
0
2
4
6
8
pH
–3
CO2(aq)
H2CO3
HCO3
–8
CO3
--
25°C
10
12
+
Diagram HCO 3, T = 25 °C , P = 1.013 bars, a [main] = 10 , a [H2 O] = 1; Suppressed: CaHCO 3
-
++
log a Ca
0
–2
Calcite
–4
14
Walt T ue Feb 21 2006
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PCO2
• What can affect PCO2?
– May decrease due to photosynthesis of aquatic
plants; may allow algae to precipitate CaCO3
– Degradation of organic matter in soil zones can
increase PCO2
• CH2O + O2 → CO2 + H2O
– Caves, PCO2 exolves in caves forming speleothems
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Falling Springs
St. Clair County
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Stalactite
Stalagmite
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Calcite Solubility and pH
• Solubility increases very significantly with
increasing acidity of solution (lower pH)
– [Ca2+] = 1013.30 [H+]2
– log [Ca2+] = 13.30 – 2 pH
• Solubility changes 100x with 1 pH unit change
– Calcite cannot persist in even mildly acidic waters
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–6
-
–10
0
2
4
6
8
pH
–3
+
–2
CO2(aq)
H2CO3
HCO3
–8
CO3
--
25°C
10
12
Diagram HCO 3, T = 25 °C , P = 1.013 bars, a [main] = 10 , a [H2 O] = 1; Suppressed: CaHCO 3
-
++
log a Ca
0
No calcite at
pH < ~5.5
Calcite
–4
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Walt T ue Feb 21 2006
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Calcite Solubility and T
• Solubility also affected by T, because
equilibrium constants change
– Solubility of calcite decreases with increasing
temperature
– As particles sink in the oceans, the water gets
colder, and CaCO3 dissolves; none reaches the
deep sea bottom
• Calcite compensation depth (CCD)
• 4.2 – 5.0 km deep
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Chemical Weathering
• Calcite dissolution is a form of chemical weathering
• Congruent dissolution: no new solid phases formed
• Incongruent dissolution: new solid formed
– Al silicates usually dissolve incongruently
• Products of chemical weathering
– New minerals (clays, oxides, …)
– Ions/molecules dissolved; help determine water quality
– Unreactive mineral grains (e.g., quartz, garnet, muscovite)
are major source of “sediment”
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Incongruent Dissolution
• KAlSi3O8 + 9H2O + 2H+  Al2Si2O5(OH)4 + 2K+ +
4H4SiO4
• Let’s predict how reaction responds to
changes in environmental parameters
– What if K+ and/or H4SiO4 removed by flowing
groundwater?
– What if there’s an abundance of H2O?
– If these particular conditions persist, achieving
equilibrium (saturation) may not be possible
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Reaction Equilibrium
• Can a chemical reaction achieve equilibrium in
nature?
• Water/rock ratio is a key variable
– The higher the water/rock ratio, the more likely the
reaction goes to completion, not equilibrium
• Products removed
– If the ratio is small, the reactions can control the
environment and equilibrium is possible
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Geochemical Cycles and Kinetics
(reaction rates)
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Geochemical Cycles
• Material is being cycled continuously in the
Earth’s surface system
• We can think of the Earth’s surface as
consisting of several reservoirs connected by
“pipes” through which matter moves
– Crust, hydrosphere, atmosphere
• All chemical elements are cycled
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Hydrologic (Water) Cycle
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Carbon Cycle
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Rock Cycle
Igneous
Rocks
Metamorphic
Rocks
Weathering, etc.
Sediment
Subduction
Sedimentary
Rocks
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Transfers between reservoirs
Reservoir 1
Reservoir 4
Reservoir 2
Reservoir 3
dn = rate of transfer of a component
dt from one reservoir to another
= Flux
n = component concentration
t = time
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Steady State
• At Earth’s creation, there was a finite amount
of each element
• Very little input of material since then
(meteorites, extraterrestrial dust)
• Since these cycles have been going on for a
very long time, we assume they are essentially
at steady state
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Steady State
• Steady state means that the composition of
reservoirs in a cycle does not change over time
– No accumulation or loss of the material of interest
– Input + any production in the reservoir = outflow + any
consumption in the reservoir
– The mass balance = 0; no creation or loss of material
• In the Earth surface environment, especially the
oceans and atmosphere, cycling of materials has
been occurring at near steady-state
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Steady State
Reservoir 1
d1,2n d2,3n
dt = dt
Reservoir 2
d2n
dt = 0
Reservoir 3
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Residence Time
• Residence time is the average time a molecule
spends in a reservoir between the time it
arrives and the time it leaves
• Determined by dividing the amount in the
reservoir by the flux in (or out)
T=
2n
d1,2n
dt
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Residence time (example)
• Suppose we have a 50 L tank of water, with a
flux in = flux out = 5 L/min; what is the
residence time?
5 L/min
5 L/min
50 L
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Residence time (example)
• Because the water flux in = flux out, it’s at steady
state, so dnH2O/dt = 0
• T = 50L = 10 minutes
5L/min
• This is the time required to add or subtract 50 L of
water
5 L/min
5 L/min
50 L
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Residence time (example)
• However, since there is mixing in the reservoir,
not every molecule of H2O is replaced every
10 minutes
• Some molecules will remain in longer, some
exit more quickly
5 L/min
5 L/min
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Cycles and reaction rates
• As materials cycle through the Earth, they are
moved and transformed at various rates (the
“pipes” connecting reservoirs)
• Transport processes and chemical reactions
take time
• Kinetics is the study of reaction rates
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