PRECIPITATION AND DISSOLUTION

1
PRECIPITATION AND DISSOLUTION
2
EXTENT OF DISSOLUTION

• We can assess the extent of dissolution or
  precipitation using the equilibrium constant,
  e.g.
• SiO2(s) 2H2O(l) ? H4SiO40
• Assuming SiO2(s) is a pure solid and solution is
  dilute so that water is nearly pure
• Ks0 (H4SiO40) ?H4SiO40H4SiO40
• for neutral species, ? ? 1, so
• Ks0 H4SiO40
• If a solution has an actual H4SiO40 gt Ks0, it
  is supersaturated and SiO2(s) should precipitate
  if kinetics allow if H4SiO40 lt Ks0, the
  solution is undersaturated.

3
SOLUBILITY PRODUCTS

• Note that the concentration of H4SiO40 does not
  depend on the total amount of SiO2(s), as long at
  least some solid is present upon attainment of
  saturation.
• A generalized salt dissolves according to
• AmBn(s) ? mAn nB-m
• and has a solubility product given by
• Ks0 AnmB-n
• again, assuming the solid is pure.

4
OTHER REACTIONS ALSO AFFECT SOLUBILITY

• The solubility of FeS(s) depends not only on
• FeS(s) ? Fe2 S2-
• but also on
• Fe2 H2O ? FeOH H
• S2- H2O ? HS- OH-
• HS- H2O ? H2S0 OH-
• Fe2 HS- ? FeHS
• FeS(s) S2- ? FeS22-, etc.
• Solubility ? ?Fe
• Fe2 FeOH FeHS FeS22-

5
SOLUBILITY COMPLICATED BY METASTABILITY

• A metastable compound is one that may be at
  equilibrium, but is not the most stable in the
  system.
• An active, metastable compound may form first
  from a highly supersaturated solution.
• The active form may convert to the stable form,
  or age only very slowly.
• Metastable or active forms have higher
  solubilities than stable forms.
• Many experimental studies deal with active
  form, but in nature the stable form may be more
  important!

6
Figure 5.2 from Stumm Morgan Solubility and
saturation.
The metastable phase may be an amorphous or
fine-grained or strained version of the stable
phase.
7
ION ACTIVITY PRODUCT AND SATURATION INDEX

• CaCO3(s) ? Ca2 CO32-

(Saturation index)
If IAP gt Ks0, i.e., ? gt 1, solution is
oversaturated. If IAP Ks0, i.e., ? 1,
solution is saturated. If IAP lt Ks0, i.e., ? lt 1,
solution is undersaturated. Because of
uncertainty in analyses and thermodynamic data,
log ? 0 0.5 should be considered saturated.
8
SOLUBILITIES OF SIMPLE SALTS

• Consider the mineral anhydrite
• CaSO4(s) ? Ca2 SO42-
• What is the solubility of anhydrite in pure
  water?

Assuming we have pure anhydrite, aCaSO4 1.
Assuming anhydrite is the only source of Ca2 and
SO42- we have Ca2 SO42- x. Finally,
neglecting activity coefficients we obtain the
following expression
9

• Ks0 Ca2SO42- x2 10-4.5
• x 10-2.25
• x 5.62x10-3 mol L-1
• 5.62 x 10-3 mol L-1 136.14 g/mole 0.766 g L-1
• Now consider the salt Al2(SO4)3(s)
• Al2(SO4)3(s) ? 2Al3 3SO42-
• Let x mol L-1 of Al2(SO4)3(s) dissolved, then
  on complete, congruent dissolution we have
• Al3 2x SO42- 3x
• Ks0 Al32SO42-3 (2x)2(3x)3
• (4x2)(27x3) 108x5 69.19

10

• x 0.9148
• Al3 2(0.9148) 1.830 mol L-1
• SO42- 3(0.9148) 2.744 mol L-1
• Solubility of Al2(SO4)3(s)
• 0.9148 mol L-1 x 342.1478 g mol-1 313 g L-1
• Now consider wulfenite (PbMoO4). A solution
  contains 2x10-8 mol L-1 Pb2 and 3x10-7 mol L-1
  MoO42-. Should wulfenite precipitate from this
  solution, and if so, how much wulfenite will
  form?
• IAP Pb2MoO42- (2x10-8)(3x10-7)
  6x10-15 M2
• Ks0 10-16.0 so ? 6x10-15/10-16.0 60

11

• Thus, this solution is supersaturated with
  respect to wulfenite by a factor of 60, and it
  should precipitate. How much?
• Let y mol L-1 of wulfenite that precipitates,
  then
• Pb2eq 2x10-8 - y MoO42-eq 3x10-7 - y
• Ks0 Pb2eqMoO42-eq (2x10-8 - y)(3x10-7 -
  y)
• 6x10-15 - 3.2x10-7y y2 10-16
• 5.90x10-15 - 3.2x10-7y y2 0

12

• y 3.004x10-7 or 1.964x10-8 mol L-1
• But the first root is impossible because it would
  make Pb2eq and MoO42-eq less than zero, so
  the true root is the second one.
• Thus, 1.964x10-8 mol L-1 x 303.264 g mol-1 5.96
  x 10-6 g L-1 5.96 ppb wulfenite would
  precipitate!
• Check of calculations
• Ks0 Pb2eqMoO42-eq (2x10-8 - y)(3x10-7 -
  y)
• Ks0 (2x10-8 - 1.964x10-8)(3x10-7 - 1.964x10-8)
• (3.60x10-10)(2.8036x10-7) 1.0093x10-16
• 10-15.996 ? 10-16.0
• so the calculations check.

13
GEOCHEMICAL DIVIDES

• In the last problem the ratio MoO42-/Pb2
  changed from 3x10-7/2x10-8 15 to
  2.804x10-7/3.60x10-10 779. If wulfenite
  continued to precipitate from this solution, as a
  result of evaporation for example, the ratio
  MoO42-/Pb2 would tend towards infinity!
• This is an example of a geochemical divide. This
  solution is enriched in molybdate and depleted in
  lead because the initial ratio MoO42-/Pb2 gt
  1. Thus, precipitation can only cause this ratio
  to increase. It can never decrease, so the ratio
  MoO42-/Pb2 1 is a divide we can never
  cross! If we started with MoO42-/Pb2 lt 1,
  precipitation would never result in a ratio gt 1!

14
Geochemical divide involving gypsum - CaSO42H2O
Figure 10.1 from Faure (1991) Principles and
Applications of Geochemistry. Initial Ca2
5x10-2 and SO42- 7x10-3 M. Ca2/SO42- is
initially gt 1 and always increases upon
evaporation and precipitation.
15
COMMON-ION EFFECT

• Natural solutions of even simple salts are more
  complex than the examples we have looked at so
  far. Consider simultaneous saturation of a
  solution with both wulfenite (PbMoO4) with pKs0
  16.0 and powellite (CaMoO4) with pKs0 7.94.
• PbMoO4 ? Pb2 MoO42-
• CaMoO4 ? Ca2 MoO42-
• Both reactions contribute MoO42- ions to the
  solution, but the final activity (concentration)
  of MoO42- must be the same for both equilibria.
  The solubility product expressions become

16

• Pb2MoO42- 10-16
• Ca2MoO42- 10-7.94
• and eliminating MoO42- we obtain
• Pb2/Ca2 10-16/10-7.94 10-8.06
• so the concentration of Pb2 will be 8.7x10-9
  times less than the concentration of Ca2. To
  determine actual concentrations, we start with
  the expression
• Pb2 Ca2 MoO42-
• which is valid if wulfenite and powellite are the
  only sources of Pb2, Ca2 and MoO42-. Now,
• Pb2 10-16/MoO42-
• Ca2 10-7.94/MoO42-

17

• But the first term is negligible compared to the
  second term, so we get
• MoO42-2 10-7.94
• MoO42- 10-3.97 1.072x10-4 M
• Pb2 10-16/10-3.97 9.33x10-13 M
• Ca2 10-7.94/10-3.97 10-3.97 1.072x10-4 M
• so, Ca2 is much larger than Pb2. Also note
  that, in the absence of powellite, Pb2
  MoO42- y, so
• y2 Ks0 10-16.0
• y 10-8.0

18

• So Pb2 10-8.0 M, or about 4 orders of
  magnitude more than in the presence of powellite.
  However, in the absence of wulfenite, Ca2
  MoO42- y so
• y2 10-7.94
• y 10-3.97
• So Ca2 10-3.97 M, or the same as in the
  presence of wulfenite!
• The common-ion effect occurs when the presence of
  a more soluble salt depresses the solubility of a
  less soluble salt containing a common ion (can be
  a cation or anion). This effect can be used in
  environmental remediation.

19
REPLACEMENT

• We have established that a solution in
  equilibrium with powellite and wulfenite at 25C
  will have
• Pb2/Ca2 10-8.06
• If a solution for which this ratio is greater
  than the equilibrium value (e.g., 10-4)
  encounters a powellite- bearing rock, powellite
  will dissolve, and wulfenite will precipitate,
  leading to the replacement of powellite by
  wulfenite. This is how some ore deposits form.
  This phenomenon can also be used in remediation.

20
SOLUBILITY OF OXIDES AND HYDROXIDES

• Governing reactions for divalent metals are
• Me(OH)2(s) ? Me2 2OH-
• MeO(s) H2O(l) ? Me2 2OH-
• cKs0 Me2OH-2
• Sometimes it is more appropriate to write
• Me(OH)2(s) 2H ? Me2 2H2O(l)
• MeO(s) 2H ? Me2 H2O(l)

21
TRIVALENT METALS

• For a trivalent metal oxide, e.g., goethite
• FeOOH(s) 3H ? Fe3 2H2O(l)

In general, MeOz/2 zH ? Mez
z/2H2O(l) Me(OH)z zH ? Mez zH2O(l)
Log Mez log cKs0 - z pH
22
Figure 5.3 from Stumm Morgan. Solubility of
oxides and hydroxides. Free metal-ion
concentration in equilibrium with solid oxides or
hydroxides.
23
NEED TO INCLUDE HYDROXIDE COMPLEXES

• Need also to consider the formation hydroxide
  complexes, i.e., hydrolysis.
• For example
• Zn2 H2O(l) ? ZnOH H
• Al(OH)2 H2O(l) ? Al(OH)2 H
• In general, the total solubility of a metal oxide
  or hydroxide in the absence of complexing ligands
  is

24
GRAPHICAL REPRESENTATION OF ZnO SOLUBILITY

• At 25C and 1 bar
• ZnO(s) 2H ? Zn2 H2O(l) log Ks0 11.2
• ZnO(s) H ? ZnOH log Ks1 2.2
• ZnO(s) 2H2O(l) ? Zn(OH)3- H log Ks3 -16.9
• ZnO(s) 3H2O(l) ? Zn(OH)42- 2H log Ks4
  -29.7
• ?ZnT Zn2 ZnOH Zn(OH)3-
  Zn(OH)42-
• ?ZnT Ks0H2 Ks1H Ks3H-1
  Ks4H-2

25

• Another reaction is also possible
• ZnO(s) H2O(l) ? Zn(OH)20
• but Ks2 is poorly known because solubility is a
  minimum here.
• The above reactions demonstrate that ZnO(s) is an
  amphoteric substance.
• To calculate solubility diagram
• Zn2 Ks0H2
• log Zn2 log Ks0 - 2pH
• ZnOH Ks1H
• log ZnOH log Ks1 - pH

26

• Zn(OH)20 Ks2
• log Zn(OH)20 log Ks2
• Zn(OH)3- Ks3H-1
• log Zn(OH)3- log Ks3 pH
• Zn(OH)42- Ks4H-2
• log Zn(OH)42- log Ks4 2pH
• A U-shaped curve results with solubilities high
  at low and high pH, and lower in the middle. This
  is typical of all amphoteric oxides and
  hydroxides.

27
Solubility of ZnO as a function of pH and the
speciation of dissolved Zn2
28
CHECK OF CALCULATIONS

• A check of the above calculations can be made by
  calculating the pH of the crossover points.
• Zn2 H2O(l) ? ZnO(s) 2H log Ks0 -11.2
• ZnO(s) H ? ZnOH log Ks1 2.2
• Zn2 H2O(l) ? ZnOH H log Kh1 -9.0

At crossover Zn2 ZnOH so Kh1 H pH
-log Kh1 9.0
29

• ZnOH ? ZnO(s) H log Ks1 -2.2
• ZnO(s) 2H2O(l) ? Zn(OH)3- H log Ks3
  -16.9
• ZnOH 2H2O(l) ? Zn(OH)3- 2H log K -19.1

At crossover ZnOH Zn(OH)3- so K H2 pH
-0.5 log K 9.55
30

• Zn(OH)3- H ? ZnO(s) 2H2O(l) log Ks3
  16.9
• ZnO(s) 3H2O(l) ? Zn(OH)42- 2H log Ks4
  -29.7
• Zn(OH)3- H2O(l) ? Zn(OH)42- H log K -12.8

At crossover Zn(OH)3- Zn(OH)42- so K
H pH -log K 12.8
31
Figure 5.4 from Stumm and Morgan Solubility of
amorphous Fe(OH)3, ZnO and CuO. The possible
occurrence of polynuclear species such as
Fe2(OH)24 and Cu2(OH)22 has been ignored. Such
species do not affect much the solubility
characteristics of the solids shown here.
32
(No Transcript)
33
SOLUBILITY OF CARBONATES

• Closed System
• a) CT is constant. What is the maximum dissolved
  Ca2 concentration as a function of CT and pH?
• CaCO3(s) ? Ca2 CO32-
• Ks0 Ca2CO32-
• Ca2 Ks0/CO32- Ks0/(CT?2)
• ?2 is known as a function of pH, so this equation
  yields the desired result.
• Where pH gt pK2 for H2CO3, ?2 ? 1.0 and CT ?
  CO32-. Solubility is controlled by the above
  equation and is pH-independent.

34

• Where pK1 lt pH lt pK2, ?1 ? 1.0 and CT ? HCO3-
• CaCO3(s) H ? Ca2 HCO3-

log K log Ca2 log CT pH so slope of plot
of log Ca2 vs. pH will be a straight line with
a slope of -1. Where pH lt pK1 lt pK2, ?0 ? 1.0 and
CT ? H2CO3 CaCO3(s) 2H ? Ca2 H2CO3
log K log Ca2 log CT 2pH
35
Figure 5.5 from Stumm Morgan Solubility of
carbonates in a closed system with CT constant
3 x 10-3 M. Diagram gives maximum soluble Me2
concentration as a function of pH. Dashed
portions of curves are metastable.
36
DISSOLUTION OF CALCITE IN PURE WATER

• Species Ca2, H2CO3, HCO3-, CO32-, H, OH-
• Mass action expressions pK1, pK2, pKs0, pKw
• Mass balance (by stoichiometry)
• Ca2 CT H2CO3 HCO3- CO32-
• Electroneutrality
• 2Ca2 H HCO3- 2CO32- OH-
• From the solubility product we have
• Ca2 Ks0/CO32-
• CO32- CT?2 HCO3- CT?1 H2CO3 CT?0

37

• Ca2 Ks0/(?2CT)
• but Ca2 CT so
• Ca2 Ks0/(?2Ca2)
• Ca22 Ks0/?2
• Ca2 CT (Ks0/?2)1/2
• CO32- ?2(Ks0/?2)1/2 HCO3- ?1(Ks0/?2)1/2
  H2CO3 ?0(Ks0/?2)1/2
• now substitute into the charge balance
• 2(Ks0/?2)1/2 H ?1(Ks0/?2)1/2
  2?2(Ks0/?2)1/2 Kw/H
• (Ks0/?2)1/2(2 - ?1- 2?2) H - Kw/H 0
• solving by trial and error we get
• pH 9.9 log Ca2 -3.9 log HCO3- -4.05
• log CO32- -4.40 log Alk -3.62 log OH-
  -4.1

38
GRAPHICAL SOLUTION TO SOLUBILITY OF CALCITE IN
PURE WATER

• Start with charge balance
• 2Ca2 H HCO3- 2CO32- OH-
• Because calcite is a base, OH- gtgt H and
  H2CO3 is negligible so CT Ca2 ? HCO3-
  CO32- and
• 2Ca2 Ca2 CO32- OH-
• Ca2 CO32- OH-
• For calcite, CO32- is almost negligible, but
  for other more insoluble carbonates, it is
  negligible, so
• Ca2 ? OH-

39
Figure 5.6 from Stumm Morgan Solubility of
metal carbonates in a closed system in pure water
(Me2 CT. Dashed portions are metastable.
40
DISSOLUTION OF CALCITE IN A SOLUTION OF STRONG
ACID OR BASE

• All equations are same as in previous example,
  but must modify the charge balance to
• CA - CB (Ks0/?2)1/2(2 - ?1- 2?2) H -
  Kw/H
• solve the above for pH by trial and error and
  then
• Ca2 CT (Ks0/?2)1/2
• CO32- ?2(Ks0/?2)1/2
• HCO3- ?1(Ks0/?2)1/2
• H2CO3 ?0(Ks0/?2)1/2

41
OPEN SYSTEM CaCO3(s)-CO2-H2O

• Because the system is open to the atmosphere,
  Ca2 ? CT, but H2CO3 pCO2KH. We also have
  the charge-balance expression
• 2Ca2 H CT(?1 2?2) OH- and
• Ca2 Ks0/CO32- and
• CT pCO2KH/?0 and
• CO32- CT ?2

42

• By trial and error we get pH 8.4 and
• log Ca2 -3.3
• log HCO3- -3.0
• log CO32- -5.0
• log Alk -3.0
• log OH- -5.6

43
A SIMPLIFICATION AND A GRAPHICAL SOLUTION

• Starting with the charge-balance expression
• 2Ca2 H HCO3- 2CO32- OH-
• The terms H, OH- and CO32- are all
  negligible, leaving
• 2Ca2 ? HCO3-
• So the solution can be found by substituting to
  get

Or by using the graph on the following page.
44
Figure 5.7 from Stumm Morgan. Solubility of
MeCO3(s) as a function of pH at constant pCO2
10-3.5 atm. If no excess acid or base are added,
the equilibrium composition of the solution is
given by the vertical bars. The inset gives -log
Me2 for pure MeCO3(s) in suspensions in
equilibrium with the above pCO2 as a function of
Ks0.
45
OPEN SYSTEM CaCO3(s)-CO2-H2O WITH STRONG ACID OR
BASE

• Again we modify the charge-balance
• CB 2Ca2 H CT(?1 2?2) OH- CA
• and we obtain by substitution

Which is solved by trial and error.
46
CALCITE SOLUBILITY IN SEAWATER

• Is seawater saturated with respect to calcite?
• In seawater at 25?C, the solubility of calcite is
  given by
• cKs0 CaTCO32-T 5.94 x 10-7
• where
• CaT total concentration of soluble Ca(II)
• Ca2 Ca-complexes
• CO32-T total concentration of soluble
  carbonate
• CO32- CO32- - complexes
• also pcKH 1.53 pcK1 6.00 pcK2 9.11

47

• Again starting with the charge-balance
• 2Ca2 H CT(?1 2?2) OH-
• and making the substitutions

The resulting equation is solved to obtain pH
8.36, CaT 1.5 x 103 M, CO32-T 3.95 x 10-4
M, HCO3-T 3.13 x 10-3 M and Alk 3.92 x
10-3 M. This problem would have been much more
difficult if we dealt explicitly with the
infinite dilution activity scale and complexes.
48
SO, AFTER ALL THAT, IS SEAWATER SATURATED WITH
RESPECT TO CALCITE?

• The actual concentrations of CaT and CO3,T can be
  calculated from pH 8.2 and Alk 2.4 x 10-3
  eq/L for seawater.
• CO32-T CT?2 (Alk?2)/(?1 2?2)
  3.87x10-4 M
• CaT 1.06x10-2 M
• CaTactCO32-Tact (1.06x10-2)(3.87x10-4)
  4.1x10-6 M2
• ? IAP/cKs0 4.1x10-6 M2/ 5.94 x 10-7 M2 6.9
• Thus, surface seawater is supersaturated with
  respect to calcite by a factor of 7 and should
  precipitate.

49
EFFECT OF PRESSURE AND TEMPERATURE ON THE
SOLUBILITY OF CALCITE IN SEAWATER

• How do pH, CO32-T, Ca2T and ? vary when
  seawater is cooled to 5?C and subjected to PT
  1000 atm?
• Assumptions
• 1) CaCO3(s) does not precipitate or dissolve
• 2) Water initially calcite supersaturated
• 3) Borate does not affect calculations
• Ca2T, CT and Alk are conservative properties
  and remain independent of pressure and
  temperature.

50

• At any P and T
• Alk ? CT(?1 2?2)
• ?1 and ?2 can be calculated with cK1 and cK2
  valid for seawater at the appropriate pressure
  and temperature.
• The pressure correction is

With the values of cKi corrected for pressure, we
now determine pH from the charge-balance
expression as before
51

• With the value of pH obtained we can now
  calculate
• CO32-T CT?2
• ?? IAP/cpKs0
• and we find that, Ca2T, CO32-T and,
  therefore, IAP, do not change much with P and T.
  However, pH and cpKs0 change considerably, so ?
  changes. Calcite solubility increases
  significantly with decreasing T and increasing P.
  Calcite displays retrograde solubility, i.e.,
  solubility decreases with increasing temperature.

52
Figure 5.8 Stumm Morgan. Effect of T and P on
composition and extent of calcite supersaturation
of an enclosed seawater sample. The initial
composition of the seawater at 25C and 1 atm pH
8.2, CT 2.18x10-3 M, Carb-Alk 2.4x10-4 eq
L-1. Ca2T 1.06x10-2 M.
53
REACTION PATHS FOR CALCITE DISSOLUTION IN
GROUNDWATER

• How does groundwater composition change as
  rainwater reacts with calcite?
• Two endmember cases
• 1) Water in contact with a large reservoir at
  constant pCO2
• 2) Water becomes isolated
• We need to calculate the reaction progress, i.e.,
  the compositional changes as a function of CaCO3
  dissolution.
• At T 10C and I 4x10-3 M we have pcKH
  1.27 pcK1 6.43 pcK2 10.38 pcKs0 7.95.

54

• 1) Reservoir with constant pCO2
• Same equations as used previously
• HCO3- (?1/?0)KHpCO2
• which provides a linear relationship between log
  HCO3- and pH.
• 2) Closed system
• When the water becomes separated, dissolved CO2
  is consumed according to
• H2CO3 CaCO3(s) ? Ca2 2HCO3-
• when the system is closed, acidity does not
  change with the extent of CaCO3 dissolution.

55

• Acy 2H2CO3 HCO3- H - OH-
• Acy ? CT(2?0 ?1) constant
• For the initial conditions we can calculate Acy
  using
• CT KHpCO2/?0
• Knowing Acy, we can calculate CT and HCO3-
  CT?1 at selected pH values. This gives the curve
  in the following figure.

56
Figure 5.9 Stumm Morgan. Dissolution paths of
calcite. Two idealized cases are shown. The
straight lines represent the case where the water
maintains contact with a large reservoir with
constant CO2 partial pressure. The curves
represent the case where the water becomes
isolated from the CO2 reservoir.
57
MIXING OF GROUNDWATERS

• Suppose we have two groundwaters, both exposed to
  and in equilibrium with pCO2 10-2 atm.
  Groundwater I equilibrates with a formation
  containing siderite (FeCO3) and no calcite.
  Groundwater II equilibrates with a formation
  containing only calcite and no siderite.
  Groundwaters I and II are then mixed in equal
  proportions.
• 1) Compute Me2, HCO3- and H for each
  groundwater.
• 2) Compute the composition of the mixture.
• 3) Is the mixture stable with respect to
  precipitation of FeCO3?

58
1) Compute Me2, HCO3- and H for each
groundwater.

• Derive Kps0 for the reaction
• MeCO3(s) CO2(g) H2O(l) ? Me2 2HCO3-
• We need to add the following reactions
• MeCO3(s) ? Me2 CO32- Ks0
• H2CO3 ? H HCO3- K1
• CO2(g) H2O(l) ? H2CO3 KH
• CO32- H ? HCO3- 1/K2
• MeCO3(s) CO2(g) H2O(l) ? Me2 2HCO3-
• Kps0 Ks0K1KH/K2

59

• Electroneutrality is approximated by
• 2Me2 ? HCO3-

Groundwater I (siderite) log Kps0 log Ks0
log K1 log KH - log K2 log Kps0 -10.24 - 6.3
- 1.5 - (-10.3) -7.74 Fe2
0.63(10-7.74)1/3(10-2)1/3 10-3.45 M HCO3- ?
2Fe2 2(10-3.45) 10-3.15 M
60

• H (10-6.3)(10-1.5)(10-2)/10-3.15 10-6.65 M
• pH 6.65
• Groundwater II (calcite)
• log Kps0 -8.42 - 6.3 - 1.5 - (-10.3) -5.92
• Ca2 10-2.84 M HCO3- 10-2.54 M pH 7.26

61
2) Compute the composition of the mixture.

• Ca2, Fe2 and HCO3- of the mixture can be
  obtained according to the relations
• Ca2mix (Ca2I Ca2II)/2 (10-2.84
  0)/2 10-2.54 M
• Fe2mix (Fe2I Fe2II)/2 (0
  10-3.45)/2 10-3.75 M
• HCO3-mix (HCO3-I HCO3-II)/2
• (10-2.54 10-3.15)/2 10-2.75 M

pH 7.04
62
SUMMARY OF RESULTS
63
3) Is the mixture stable with respect to
precipitation of FeCO3?

• FeCO3(s) ? Fe2 CO32- log Ks0 -10.24
• H CO32- ? HCO3- -log K2 10.3
• FeCO3(s) H ? Fe2 HCO3- log Ks1 0.06
• IAP Fe2HCO3-/H
• (10-3.75)(10-2.75)/(10-7.04) 100.54
• ? 100.54/100.06 3.02
• So siderite should precipitate!

64
BUFFER INTENSITY IN PRESENCE OF CaCO3

• 1) Constant pCO2 case. Presence of calcite (or
  any carbonate) can greatly enhance the buffer
  capacity!

Into the charge balance Na HCO3-
2CO32- OH- - H -2Ca2 we can
substitute the expressions for each of these
species in terms of Ks, H and pCO2.
65
Converting back to species concentrations we
obtain
In the absence of CaCO3 the buffer intensity is
(Butler, p. 67-69)
66

• Thus, the presence of calcite in excess of that
  required for saturation greatly increases the
  buffer intensity of a natural water.
• An extremely narrow range of pH is allowed as
  long as calcite is present.
• 2) Closed system with CaCO3(s) but no gas phase.
  Under these conditions, the variable
• D 2(CT - Ca2)
• is constant when calcite dissolves or
  precipitates. If calcite is present but is not in
  equilibrium with a constant CO2 reservoir, buffer
  intensity not as high and pH range not as narrow
  as in case (1).

67
Figure 4.10 from Butler. Buffer index at constant
partial pressure of CO2 (pCO2 10-3.5 atm) with
solid CaCO3 present. For comparison, ?p is also
shown.
0
?p,s
-1
4Ca2
Ca2
-2
?p
4CO32-
H
-3
CO32-
OH-
log of quantity
pCO2
HCO3-
-4
CO2
-5
CaCO30
-6
CaHCO3
-7
CaOH
-8
-9
1
2
3
4
5
7
8
6
12
13
14
9
11
10
pH
68
Figure 4.11 from Butler. Buffer index of a closed
system in equilibrium with CaCO3 but without a
gas phase. Calculations were made for D 10-3.
For comparison, the buffer index (?C) for a
system without CaCO3 is also given.
69
RELATIVE STABILITIES OF HYDROXIDES VS. CARBONATES

• We may need to answer the question Which phase
  controls solubility? General rule is that the
  least soluble phase is the most stable one, and
  should control solubility.
• Example Fe(II) at 1 atm, 25C and I 6x10-3 M
• Ks0(Fe(OH)2) 10-14.7 mol3 L-3
• Ks0(FeCO3) 10-10.4 mol2 L-2
• We cannot compare these values directly. First,
  the units are different. Second, the relative
  solubility depends on pH, pCO2, etc. So Fe(OH)2
  is not necessarily more soluble than FeCO3.

70

• Need to calculate which phase is least soluble at
  a given set of P, T, pH, CT, etc. conditions. To
  illustrate, lets assume Alk 10-4 eq L-1 and
  pH 6.8.
• Solubility equilibrium with FeCO3(s) yields
• FeCO3(s) ? Fe2 CO32- log cKs0 -10.4
• H CO32- ? HCO3- -log cK2 10.1
• FeCO3(s) H ? Fe2 HCO3- log cKs0 -0.3
• log Fe2 log cKs0 - pH - log HCO3-
• At this pH Fe2 ? Fe(II)T and HCO3- ? Alk
• log Fe(II)T -0.3 - 6.8 4 -3.10

71

• Solubility equilibrium with Fe(OH)2(s)
• Fe(OH)2(s) ? Fe2 2OH- log cKs0 -14.5
• 2H 2OH- ? 2H2O(l) -2log cKw 27.8
• Fe(OH)2(s) 2H ? Fe2 2H2O(l) log cKs0
  13.3
• log Fe2 log cKs0 - 2pH
• log Fe(II)T 13.3 - 2(6.8) -0.30
• Thus, even though FeCO3(s) has a greater
  solubility product than Fe(OH)2(s), under these
  specific conditions, FeCO3(s) is less soluble
  than Fe(OH)2(s), so FeCO3(s) is the most stable
  phase.

72
PREDOMINANCE DIAGRAMS

• A predominance diagram is any diagram with two
  compositional variables as axes and that shows
  the stable phases and solubilities in a
  mineral-solution system.
• pCO2-pH diagrams
• Plots with log pCO2 vs. pH as variables. An
  assumed ?Fe is employed. We will assume ?Fe
  10-4 mol L-1.
• 1) Fe(OH)2(s)/Fe2 boundary
• Fe(OH)2(s) 2H ? Fe2 2H2O(l)

73

• log Ks0Fe(OH)2(s) -2pH pFe 0
• 12.85 - 2pH 4 0
• pH 8.43
• 2) FeCO3(s)/Fe2 boundary
• FeCO3(s) 2H ? Fe2 CO2(g) H2O(l)
• log pCO2 log Kps0FeCO3(s) - 2pH pFe
• 7.40 - 2pH 4
• 11.90 - 2pH

74

• 3) FeCO3(s)/Fe(OH)2(s) boundary
• FeCO3(s) H2O(l) ? Fe(OH)2(s) CO2(g)

log pCO2 -5.45
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76
STABILITY OF CARBONATES IN THE SYSTEM Mg-CO2-H2O

• Brucite Mg(OH)2(s) ?fGo -200.0 kcal mol-1
• Magnesite MgCO3(s) ?fGo -245.3 kcal mol-1
• Nesquehonite MgCO33H2O ?fGo -411.7 kcal mol-1
• Hydromagnesite Mg4(CO3)3(OH)23H2O
• ?fGo -1100.1 kcal mol-1
• Mg2 ?fGo -109.0 kcal mol-1
• CO32- ?fGo -126.2 kcal mol-1
• OH- ?fGo -37.6 kcal mol-1
• H2O(l) ?fGo -56.69 kcal mol-1

77

• Mg(OH)2(s) ? Mg2 2OH- pKs0 11.6
• MgCO3(s) ? Mg2 CO32- pKs0 7.5
• MgCO33H2O ? Mg2 CO32- 3H2O pKs0 4.7
• Mg4(CO3)3(OH)23H2O ? 4Mg2 3CO32- 2OH-
  3H2O
• pKs0 29.5
• We will now construct a pCO2-pH diagram to depict
  the phase relations in this system. First, note
  that the only difference between brucite and
  nesquehonite is the presence of water, so only
  one of these phases will appear on the phase
  diagram.

78

• MgCO3(s) ? Mg2 CO32- log Ks0 -7.5
• Mg2 CO32- 3H2O ? MgCO33H2O log Ks0 4.7
• MgCO3(s) 3H2O ? MgCO33H2O log K -2.8
• log K -3log aH2O -2.8
• log aH2O 0.933 or aH2O 8.58
• But as aH2O can never be greater than 1,
  nesquehonite can never be the stable phase,
  magnesite is always the stable carbonate.
• Magnesite/hydromagnesite boundary
• 4MgCO3(s) ? 4Mg2 4CO32- log Ks0 -30.0
• 4Mg2 3CO32- 2OH- 3H2O ? Mg4(CO3)3(OH)23H2O
  log Ks0 29.5
• 4MgCO3(s) 2OH- 3H2O ? Mg4(CO3)3(OH)23H2O
  CO32-
• log K -0.5

79

• 4MgCO3(s) 2OH- 3H2O ? Mg4(CO3)3(OH)23H2O
  CO32-
• log K -0.5
• 2H2O ? 2OH- 2H log K -28.0
• CO32- H ? HCO3- log K 10.3
• HCO3- H ? H2CO3 log K 6.3
• H2CO3 ? CO2 H2O log K 1.5
• 4MgCO3(s) 4H2O ? Mg4(CO3)3(OH)23H2O CO2
• log K -10.4
• K pCO2/(aH2O)4
• log pCO2 log K -10.4
• Thus, magnesite is stable relative to
  hydromagnesite down to very low pCO2.

80

• Magnesite/brucite boundary
• MgCO3 ? Mg2 CO32- log K -7.5
• Mg2 2OH- ? Mg(OH)2 log K 11.6
• 2H2O ? 2OH- 2H log K -28.0
• CO32- H ? HCO3- log K 10.3
• HCO3- H ? H2CO3 log K 6.3
• H2CO3 ? CO2 H2O log K 1.5
• MgCO3 H2O ? Mg(OH)2 CO2 log K -5.80
• log pCO2 log K -5.80

81

• Magnesite/Mg2 boundary
• MgCO3 ? Mg2 CO32- log K -7.5
• CO32- 2H ? CO2 H2O log K 18.1
• MgCO3 2H ? Mg2 CO2 H2O log K 10.6
• log K 10.6 log Mg2 log pCO2 2pH
• we choose Mg2 10-3 mol L-1
• 10.6 -3 log pCO2 2pH
• log pCO2 13.6 - 2pH

82

• Brucite/Mg2 boundary
• Mg(OH)2(s) ? Mg2 2OH- log K -11.6
• 2OH- 2H ? 2H2O log K 28.0
• Mg(OH)2(s) 2H ? Mg2 2H2O log K 16.4
• log K 16.4 log Mg2 2pH
• but Mg2 10-3 mol L-1
• 16.4 -3 2pH
• pH 9.70

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84
SOLUBILITY OF DOLOMITE

• Conditions of dolomite precipitation are poorly
  understood.
• Precipitation of dolomite does not control
  compositions of natural waters.
• Dissolution of dolomite may exercise some
  control.
• Published values Ks0 range from 10-16.5 to
  10-19.5.

85
CAN WE OBTAIN DOLOMITE Ks0 FROM FIELD DATA?

• CaMg(CO3)2 ? Ca2 Mg2 2CO32- log
  Ks0,dolomite
• 2Ca2 2CO32- ? 2CaCO3 -2log Ks0,calcite
• CaMg(CO3)2 Ca2 ? 2CaCO3 Mg2 log Kexchange

For solutions in equilibrium with both calcite
and dolomite, the Mg/Ca ratio should be constant
at a given T and P. Well waters in Florida have
Mg2/Ca2 0.78. Assuming Ks0,calcite
5x10-9, we obtain Ks0,dolomite 2.0x10-17, in
agreement with some laboratory measurements.
86
Fig. 5.13 from Stumm Morgan. Stability
relations in the system Ca2-Mg2-CO2-H2O at 25C
and 1 bar. Based on Ks0 2x10-17 for dolomite.
Seawater should precipitate dolomite but no
convincing evidence has been found for
present-day dolomite precipitation.
87
SOLUBILITY OF SULFIDES

• Estimate the solubility of ?-CdS at pH 4.5 and
  pH2S 1 atm (25C, I 1 mol L-1).
• We have the following thermodynamic data
• log cKpso(CdS) -5.8
• log cKH(H2S) -1.05
• log cK1(H2S) -6.90
• log cK2(HS-) -14.0
• For now we assume that no hydroxide, sulfide or
  carbonate complexes are formed.

88

• The constant cKps0 refers to the reaction
• ?-CdS(s) 2H ? Cd2 H2S(g)

-5.8 log Cd2 log pH2S(g) 2pH -5.8 log
Cd2 0 9 log Cd2 -14.8 Cd2
1.585x10-15 mol L-1 Cd2 1.78x10-8 µg/L (ppb)
89
PREDOMINANCE DIAGRAM FOR SULFIDES

• Construct a predominance diagram in the system
  Cd2-H2S-CO2-H2O at pCO2 10-3.5 atm.
• We need the following additional thermodynamic
  data
• log cKps0(CdCO3) 6.44
• log cK1(H2CO3) -6.04
• log cK2(HCO3-) -9.57
• log cKH(CO2) -1.51
• Our diagram will plot log (pCO2/pH2S) vs. pH.

90

• CdCO3/CdS boundary
• CdCO3(s) H2S(g) ? CdS(s) H2O(l) CO2(g)
• log K log (cKps0(CdCO3)/cKps0(CdS))
• 6.44 - (-5.8) 12.24 log (pCO2/pH2S)
• CdS/Cd2 boundary
• CdS(s) 2H ? Cd2 H2S(g)

log cKps0 -5.8 log Cd2 log pH2S
2pH Assume Cd2 10-4 mol L-1 -1.8 log pH2S
2pH
91

• However, to plot this on the log (pCO2/pH2S) vs.
  pH diagram, we must subtract log pCO2 from both
  sides of the equation to get
• -1.8 - log pCO2 log pH2S - log pCO2 2pH
• but log pCO2 -3.5 so
• 1.7 log (pH2S/pCO2) 2pH
• log (pCO2/pH2S) 2pH - 1.7

92

• CdCO3/Cd2 boundary
• CdCO3(s) 2H ? Cd2 CO2(g) H2O(l)

log cKps0 6.44 log Cd2 log pCO2
2pH Set Cd2 10-4 mol L-1 and pCO2 10-3.5
atm 6.44 -4 -3.5 2pH pH 6.97
93
log pCO2 -3.5
94
PHOSPHATES - RELATIVE STABILITIES OF CALCITE AND
APATITE

• To calculate a predominance diagram we must be
  concerned about the pH-dependent distribution of
  phosphate and carbonate. We write the reactions
  in terms of the predominant species at the pH of
  interest.
• Choose log PT (where PT is the total phosphate
  concentration) and pH as axes for our diagram.
• Assume CT 10-4 mol L-1 and Ca is conserved in
  the solid phases.

95

• pH lt pK1(phosphate) lt pK1(carbonate)
• 10CaCO3(s) 6H3PO40 2H2O ? Ca10(PO4)6(OH)2(s)
  10H2CO30
• log K 39.5 10log H2CO30 - 6log H3PO40
• 39.5 10(-4) - 6log PT
• log PT -13.25
• pK1(phosphate) lt pH lt pK1(carbonate)
• 10CaCO3(s) 6H2PO4- 6H 2H2O ?
  Ca10(PO4)6(OH)2(s) 10H2CO30
• log K 52.1 10log H2CO30 - 6log H2PO4-
  6pH
• 52.1 10(-4) -6log PT 6pH
• log PT pH - 15.35
• pK1(carbonate) lt pH lt pK2(phosphate)
• 10CaCO3(s) 6H2PO4- 2H2O ? Ca10(PO4)6(OH)2(s)
  10HCO3- 4H
• -10.9 10log HCO3- - 6log H2PO4- - 4pH
• log PT -4/6pH - 4.85

96

• pK2(phosphate) lt pH lt pK2(carbonate)
• 10CaCO3(s) 6HPO42- 2H 2H2O ?
  Ca10(PO4)6(OH)2(s) 10HCO3-
• 32.3 10log HCO3- - 6log HPO42- 2pH
• log PT 1/3pH - 12.05
• pK2(carbonate) lt pH lt pK3(phosphate)
• 10CaCO3(s) 6HPO42- 2H2O ? Ca10(PO4)6(OH)2(s)
  10CO32- 8H
• -70.2 10log CO32 - 6log HPO42- - 8pH
• log PT -4/3pH 5.03
• pK3(phosphate) lt pH
• 10CaCO3(s) 6PO43- 2H2O ? Ca10(PO4)6(OH)2(s)
  10CO32- 2H
• 1.80 10log CO32 - 6log PO43- - 2pH
• log PT -1/3pH - 6.97

97
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98
ACTIVITY OF SOLID PHASEORWHAT TO DO WHEN SOLID
NOT PURE

• Up to now we have assumed that asolids ? 1 in all
  our solubility calculations. This may not always
  be a valid assumption, especially where there is
  solid solution!
• Example Two-component solid-solution between
  AgCl(s) and AgBr(s)
• AgCl(s) Br- ? AgBr(s) Cl-

99

• Thus, the distribution coefficient is simply
  equal to the ratio of the solubility products of
  the two end member phases.
• aAgBr XAgBr?AgBr and aAgCl XAgCl?AgCl

100

• The amount of dissolution of Br- into solid AgCl
  thus depends on
• 1) the solubility product ratio
• 2) solution composition (aBr-/aCl-)
• 3) effects of non-ideal mixing (activity
  coefficients)
• To a first approximation, ?AgBr/ ?AgCl ? 1 and
  ?Br-/?Cl- ? 1 so

101
QUANTITATIVE IMPORTANCE OF SOLID SOLUTION

• Consider AgBr0.1Cl0.9(s) in equilibrium with
  solution.
• Solid XAgBr 0.1 XAgCl 0.9
• Solution Cl- 10-4.9 mol L-1 Br- 10-8.4
  mol L-1
• Ag 10-4.9 mol L-1.
• D 10-9.7/10-12.3 391 ? (10-4.9/10-8.4)(0.1/0.9
  ) 351
• Br- is enriched in the solid phase relative to
  solution, i.e., Br- partitions preferentially
  into solid.

102

• If we did not consider solid solution formation,
  we would erroneously conclude that the solution
  was undersaturated with respect to AgBr(s)
  because
• IAP 10-4.910-8.4 10-13.3 lt Ks0 10-12.3
• However, the solutions is saturated with Br-
  because it is in equilibrium with a solid
  solution.
• IAP 10-4.910-8.4/0.1 10-12.3 Ks0
• On the other hand, for AgCl(s)
• IAP 10-4.9 10-4.9 10-9.8 ? Ks0 10-9.7
• So the solubility of the minor component is
  greatly affected by solid solution, but the major
  component is less affected.

103

• Example
• Let us examine whether Sr2 in the ocean is
  controlled by solid solution in calcite and
  estimate the XSrCO3. The following is known
• pcKs0(CaCO3) 6.1 pcKs0(SrCO3) 6.8
• CO32- 10-3.6 mol L-1 Sr2 ? 10-4 mol L-1

Assuming saturation of seawater with respect to
strontian calcite Ca2 cKs0/CO32-
10-6.1/10-3.6 ? 10-2.5 mol L-1
104

• So XSrCO3 ?? 0.004 and XCaCO3 ? 0.996
• Because D lt 1, Sr is preferentially partitioned
  into the solution phase. A mole fraction of Sr of
  0.004 is reasonable for marine calcite.
• If the solution were in equilibrium with pure
  strontianite (SrCO3), then
• Sr2 ? Ks0(SrCO3)/CO32- 10-6.8/10-3.6
  10-3.2 mol L-1
• Which is much higher. Again, the solubility of
  the minor component is greatly reduced.

105
CALCIUM SELENATE

• CaSO42H2O ? Ca2 SO42- 2H2O
• log Ks0 -4.6
• CaSeO42H2O ? Ca2 SeO42- 2H2O
• log Ks0 -3.09
• Assume we have a water with Ca2 10-1 mol L-1
  and SeO42- 100 ppb 1.27x10-6 mol L-1.
• If we assume aCaSeO4 XCaSeO4 1, then
• IAP 10-1(1.27x10-6) 1.27x10-7 ltlt 10-3.09
• and solution is undersaturated in selenate. If
  XCaSeO4 10-5 (13 ppb), then IAP
  10-1(1.27x10-6)/10-5 1.27x10-2 gt 10-3.09, and
  solution is saturated with selenate.

106
STOICHIOMETRIC SATURATION

• Stoichiometric saturation - refers to a
  metastable equilibrium between an aqueous phase
  and a multi-component solid, where, for kinetic
  reasons, the composition of the solid does not
  change.
• For example, magnesian calcite
• Ca(1-x)MgxCO3 ? (1-x)Ca2 xMg2 CO32-
• If this reaction was in complete equilibrium,
  there should be a differential partitioning of Ca
  and Mg between calcite and solution, e.g.,

107

• A value of D 0.02 indicates that Mg prefers the
  solution to the solid. However, for kinetic
  reasons when a mineral is dissolving, this
  equilibrium may not be attained fully.
• If the solid composition does not change
  throughout the dissolution process, then it can
  be treated thermodynamically as a one-component
  phase with an activity of 1.

The last term is usually less than 1, so the
solubility of magnesian calcite is less than that
of calcite. Stoichiometric saturation is probably
the rule in low-temperature aqueous environments.
108
NON-IDEALITY IN AQUEOUS SOLUTION

• Consider the dissolution of fluorite
• CaF2(s) ? Ca2 2F-

At low to moderate ionic strength, activity
coefficients are less than one, so the presence
of inert salts would increase the solubility of
fluorite (salting in). At sufficiently high
ionic strength, activity coefficients can become
greater than one, and solubility will decrease
(salting out).