1

2

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

3

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- CO3

2- + H+; Ka2 = 10-10.33

4Walt Tue Feb 21 2006

2 3 4 5 6 7 8 9 10 11 12–16

–14

–12

–10

–8

–6

–4

–2

0

pH

Sp

eci

es

with

HC

O3- (

log

mo

lal)

CO2(aq) CO

3--

HCO3-

Total DIC = 10-1 M

pH = 6.35 pH = 10.33

Common pH rangein natural waters

5

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

6

Salts (Electrolytes)

7

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

8

Solubility of Salts• Remember: A saturated solution of a salt is in a

state of equilibrium• Al2(SO4)3(s) 2Al3+ + 3SO4

2-

– 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

9

Solubility• Al2(SO4)3(s) 2Al3+ + 3SO4

2-

– What is the solubility of Al2(SO4)3?• What are the activities of Al3+ and SO4

2- in a saturated solution of Al2(SO4)3?

– – Thus in a saturated solution of Al2(SO4)3

• [Al3+] = 2x = 1.829 mol/L• [SO4

2-] = 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

10

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 SO4

2- saturated with respect to gypsum (CaSO42H2O)?

11

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

12

How much gypsum would precipitate to reach equilibrium (saturation)?

• CaSO42H2O Ca2+ + SO42- + 2H2O

– Ksp = [Ca2+] [SO42-] = 10-4.6

– As gypsum precipitates (reverse reaction), the IAP will decrease because [Ca2+] and [SO4

2-] are being used up

– Once the IAP = Ksp, the solution will be in equilibrium with respect to gypsum

13

How much gypsum would precipitate to reach equilibrium (saturation)?

• CaSO42H2O Ca2+ + SO42- + 2H2O

– Ksp = [Ca2+] [SO42-] = 10-4.6

– The solution initially has 5x10-2 mol/L Ca2+ and 7x10-3 mol/L SO4

2- – 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

14

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

15

Ca2+

SO42-

Ca2+

SO42-

= 7.1

= 79.2

(Equilibrium reached)

16

Geochemical Divide

• The initial ratio of species can affect which minerals precipitate

• GEOCHEMICAL DIVIDE– If [Ca2+] / [SO4

2-] had been < 1 instead of > 1, then [SO4

2-] 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

17

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– CaSO42H2O Ca2+ + SO4

2- + 2H2O• Ksp (gypsum) = [Ca2+] [SO4

2-] = 10-4.6

– BaSO4 Ba2+ + SO42-

• Ksp (barite) = [Ba2+] [SO42-] = 10-10.0

– Barite is much less soluble than gypsum

18

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

19

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 [SO4

2-] using simultaneous equations– [SO4

2-]2 = 10-4.6 + 10-10.0

– [SO42-] = 10-2.3 mol/L

– Note that barite only contributes a negligible amount of [SO4

2-]

20

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

• CaSO42H2O Ca2+ + SO42- + 2H2O

– The increase in [SO42-] can cause the solution to

become supersaturated with respect to barite, which is less soluble than gypsum

21

Gypsum and Barite equilibrium• CaSO42H2O Ca2+ + SO4

2- + 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

22

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)

23

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)

24

Calcite Solubility

• CaCO3 Ca2+ + CO32-; Ksp = 10-8.35 (1)

• HCO3- CO3

2- + 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)

25

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(CO3

2-) + (HCO3-) + (OH-)

• Now have 6 equations and 6 unknowns

26

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

27

0 2 4 6 8 10 12 14–10

–8

–6

–4

–2

0

pH

log

a C

a++

HCO3-

CO2(aq)

CO3--

Calcite

25°C

Walt Tue Feb 21 2006

Dia

gram

HC

O3- ,

T

=

25 °

C ,

P

=

1.01

3 ba

rs,

a [m

ain]

=

10

–3,

a [H

2O

] =

1;

Sup

pres

sed:

CaH

CO

3+

H2CO3

28

Calcite Solubility• The reaction we just used for calcite

dissolution generally doesn’t occur in nature– CaCO3 Ca2+ + CO3

2-

• Dissolution of calcite done primarily by acid– In natural systems, primary acid is– CaCO3 + H2CO3 Ca2+ + 2HCO3

-

CO2

29

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

30

Calcite Solubility• CO2(g) CO2(aq)

• CO2(aq) + H2O H2CO3

• CaCO3 + H2CO3 Ca2+ + 2HCO3-

• Increase 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

31

CaCO3 + H2CO3 Ca2+ + 2HCO3-

At Saturation

~25 mg/L calcite could be precipitated

●

●

What if we increase PCO2 from atmospheric by 10x?

32

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?

33

0 2 4 6 8 10 12 14–10

–8

–6

–4

–2

0

pH

log

a C

a++

HCO3-

CO2(aq)

CO3--

Calcite

25°C

Walt Tue Feb 21 2006

Dia

gram

HC

O3- ,

T

=

25 °

C ,

P

=

1.01

3 ba

rs,

a [m

ain]

=

10

–3,

a [H

2O

] =

1;

Sup

pres

sed:

CaH

CO

3+

H2CO3

34

0 2 4 6 8 10 12 14–10

–8

–6

–4

–2

0

pH

log

a C

a++

HCO3-

CO2(aq)

CO3--

Calcite

25°C

Walt Tue Feb 21 2006

Dia

gram

HC

O3- ,

T

=

25 °

C ,

P

=

1.01

3 ba

rs,

a [m

ain]

=

10

–3,

a [H

2O

] =

1;

Sup

pres

sed:

CaH

CO

3+

H2CO3

35

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

36

Falling SpringsSt. Clair County

37

Stalactite

Stalagmite

38

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

39

0 2 4 6 8 10 12 14–10

–8

–6

–4

–2

0

pH

log

a C

a++

HCO3-

CO2(aq)

CO3--

Calcite

25°C

Walt Tue Feb 21 2006

Dia

gram

HC

O3- ,

T

=

25 °

C ,

P

=

1.01

3 ba

rs,

a [m

ain]

=

10

–3,

a [H

2O

] =

1;

Sup

pres

sed:

CaH

CO

3+

No calcite at pH < ~5.5

H2CO3

40

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

41

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”

42

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

43

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

44

Geochemical Cycles and Kinetics (reaction rates)

45

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

46

Hydrologic (Water) Cycle

47

Carbon Cycle

48

Rock Cycle

SedimentMetamorphic

Rocks

IgneousRocks

SedimentaryRocks

Magma formation

Intrusion

Crystalliza

tion

WeatheringTransport

Deposition

Burial

Diagenesis

Lithification

Deformation

Recrystallization

Segregation

Weathering, etc.

SubductionWeathering, etc.

Subduction

49

Transfers between reservoirs

Reservoir 2

Reservoir 1

Reservoir 3

d 4,1n

dtd

1,2 ndt

d3,4 n

dt d 2,3n

dt

Reservoir 4

dn = rate of transfer of a componentdt from one reservoir to another = Flux

n = component concentrationt = time

50

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

51

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

52

Steady State

Reservoir 2

Reservoir 1

Reservoir 3

d1,2 n

dt

d 2,3n

dt

d1,2ndt

d2,3ndt

=

d2ndt

= 0

53

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) 2n

d1,2n

dt

T =

54

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?

50 L5 L/min 5 L/min

55

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

50 L5 L/min 5 L/min

56

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

57

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