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Phase Change Reactions
Precipitation-Dissolution of Inorganic Species
Bruce Herbert
Geology & Geophysics
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Precipitation-Dissolution and Metal-Ligand Properties
■ Generally, species exhibit similar precipitation-dissolution reactions as
complexation reactions in that
■ The more stable solid phases of a hard metal will be precipitates with
a hard base (all other factors being equal).
■ The more stable solid phases of a soft metal will be precipitates with
a soft base (all other factors being equal).
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Thermodynamics of Precipitation-Dissolution
■ General formula for two component dissolution
+
lM a L b ( s) = aM m
+
bL
( aq )
( aq )
K dis
(M m + ) a (Ll - ) b
=
(M a L b )
(6.1)
■ where M is a metal, L a ligand, a and b are stoichiometric coefficients, m
and n are the charges of the ions, and Kdis is the equilibrium dissolution
constant.
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Thermodynamics of Precipitation-Dissolution
■ The solubility product constant, Kso is defined as
K so º K dis (Ma L b )
(6.2)
■ If the solid is in its Standard State, as is commonly assumed, then
Kdis=Kso. If the solid is not in its Standard State, the IAP will be a
function of all of the thermodynamic variables that affect the activity of
the solid.
■ Precipitation-dissolution reactions often occur over much longer time
scales than complexation reactions in solution. Species in the solution
phase will come to equilibration among themselves before they reach
equilibration with the solid phase.
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Thermodynamics of Precipitation-Dissolution
■ We can use this fact to define two useful criteria for precipitation-dissolution
reactions:
■ The ion activity product, IAP is defined as
IAP º (M
m+ a
l-
) (L )b
(6.3)
■ The relative saturation, , is defined as
W º IAP
(6.4)
K so
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Thermodynamics of Precipitation-Dissolution
■ The relative saturation can be monitored over time to assess the degree
of equilibration in a system.
■ If W < 1, then the system is undersaturated with respect to the
solid phase as defined by the reaction in 6.1.
■ If W > 1, then the system is oversaturated with respect to the solid
phase as defined by the reaction in 6.1.
■ If W = 1, then the system is in equilibration with respect to the solid
phase as defined by the reaction in 6.1.
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Thermodynamics of Precipitation-Dissolution
■
MINTEQA2 and PHREEQ calculates a saturation index, SI
SI = log [IAP/Kso]
(6.5)
■ If the SI < 0, then the system is undersaturated
■ If the SI > 0, then the system is oversaturated
■ IF the SI ≈ 0, then the system is at equilibrium
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Thermodynamics of Precipitation-Dissolution
■ If SI ≠ 0, then we can make one of three conclusions concerning the
system of interest:
■ The reaction is not in equilibrium
■ No solid phase corresponding to the reaction as written exists in the
system
■ The reaction is at (possibly metastable) equilibrium, but the solid
phase is not in the Standard State assumed in computing Kso
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Example: The Dissolution of Gibbsite
■ The value of Kdis can be calculated with Standard-State chemical potentials.
■ Assuming that gibbsite is in its Standard State, then Kdis = Kso and
Al(OH)3(s) = Al3+(aq) + 3OH-(aq)
Species
Al(OH)3(s)
Al3+(aq)
OH-(aq)
where
DG° = å n j m° j
j
(6.6)
Standard State Chemical Potentials, u°
-1154.8 ± 1.3 kJ mol -1
-489.9 ± 4.0 kJ mol -1
-157.2 kJ mol-1
K = P j a nj
j
DG° = -RTlnK
Check calcs
log Kdis =
[-1154.8 ± 1.3 - 489.9 ± 4.0 - 3(157.2)]
kj
= -33.9 mol
5.708
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Thermodynamics of Precipitation-Dissolution
■ The central problem with precipitation-dissolution reactions, as
environmental geologist, is to predict which solid phase controls aqueous
activities of a metal or ligand.
■ The controlling phase at equilibrium will be the one which results in the
smallest value of the aqueous activity of the ion.
■ The corollary is also true: the chemical potential, m(aq), is smallest whenever
the aqueous activity is at a minimum. At that time the chemical potential of
a species in the solid and aqueous phases will be equal.
m (aq) = m (s)
at equilibrium
(6.8)
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Example: Does Cd(OH)2 or CdCO3 control (Cd2+) in solution?
■ Background Data: pH = 7.6; (HCO3-) = 10-3
■ For the Hydroxide Phase:
■ *Kso is the dissolution equilibrium constant for cadmium hydroxide by adding
the ionization of water.
■ Assume both Cd(OH)2(s) and H2O(l) are in their Standard States. Then:
■ log (Cd2+) = log *Kso + 2 log (H+) = log *Kso -2 pH
■ since *Kso = (Cd2+)/(H+)2
■ Then log (Cd2+) = -1.59 if Cd(OH)2(s) is the controlling phase
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Example: Does Cd(OH)2 or CdCO3 control (Cd2+) in solution?
■ Background Data: pH = 7.6; (HCO3-) = 10-3
■ For the Carbonate Phase:
■ Assume CdCO3(s) is in its Standard State. Then:
■ log (Cd2+(aq)) = log Kso - pH - log (HCO3-)
■ then log (Cd2+) = -5.47
■ therefore CdCO3(s) is the controlling phase
CdCO3(s) = Cd2+(aq) + CO32-(aq)
H+(aq) + CO32-(aq) = HCO3-(aq)
log Kso = -11.20
CdCO3(s) + H+(aq) = Cd2+(aq) + HCO3-(aq)
log Kso = -0.87
log Kw = 10.33
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Reverse experimental procedure
■ Determine (Cd2+) in solution using a Cd-sensitive electrode
■ Determine (CO32-) and calculate IAP
■ Compare IAP to published values of Kso.
■ If the two are not equal then either:
■ Equilibrium with the solid phase does not exist
■ Solid phase controlling ion activity is not the one suspected
■ Solid is not in the Standard State.
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Solubility of Oxides and Hydroxides
■ Oxides and hydroxides are often the most common precipitates of
trace metals. Their precipitation-dissolution is strongly affected by pH.
■ Solubility of oxides and hydroxides can be expressed as:
■ M(OH)2(s) = M2+(aq) + 2OH-(aq) Kso = (M2+)(OH-)2
■ MO(s) + H20(l) = M2+(aq) + 2OH-(aq) Kso = (M2+)(OH-)2
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Solubility of Oxides and Hydroxides
■ We can rewrite the reaction in order to include protons
M(OH)2(s) + 2H+(aq) = M2+(aq) + 2H2O(l)
*Kso = (M2+)/(H+)2 = Kso/Kw2
(6.15)
MO(s) + 2H+(aq) = M2+(aq) + H2O(l)
*Kso = (M2+)/(H+)2 = Kso/Kw2
(6.16)
■ where Kw is the hydrolysis constant for water
H2O(l) = H+(aq) + OH-(aq)
Kso = (H+)(OH-)/(H2O)
(6.17)
■ This gives
log[Mz+] = log Kso + z pKw - z pH
(6.18)
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Graphical representation of ZnO (s)
■
The dissolution of zinc oxide as a function of pH is governed by the following reactions
■
Reaction, Logarithmic Form, log *Kso
ZnO(s) + 2H+(aq) = Zn2+(aq) + 2H2O(l)
log (Zn2+) = log *Kso - 2pH, log *Kso = 11.2
ZnO(s) + H+(aq) = ZnOH+(aq)
log (ZnOH+) = log *Kso - pH, log *Kso = 2.2
ZnO(s) + 2H2O(l) = Zn(OH)3-(aq)+ H+(aq)
log (Zn(OH)3-) = log *Kso + pH, log *Kso = -16.9
ZnO(s) + 3H2O(l) = Zn(OH)42-(aq)+ 2H+(aq)
log (Zn(OH)42-) = log *Kso + 2pH, log *Kso = -29.7
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Graphical representation of ZnO (s)
■ The logarithmic equations are equations of straight lines and
can be plotted (using Excel) where pH forms the independent
variable:
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Graphical representation of ZnO (s)
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2
0
Log
Activity Zn
Species
ZnOH
-2
+
ZnO(s)
-4
-6
log concentration
Zn(OH)4
-8
-
Zn(OH)3
Zn
2-
2+
-10
4
6
8
10
12
14
pH
Concentrations of dissolved Zn species in equilibrium with
ZnO as a function of pH.
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0
Al(OH)3(s)
-2
Log
Activity Al
Species
-4
-
Al(OH)4
3+
Al
+
Al(OH)2
2+
log concentration
-8
Al O
H
-6
-10
0
2
4
6
8
10
12
pH
Concentrations of dissolved Al species in equilibrium with
gibbsite as a function of pH.
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2
0
-2
Log
Activity Fe
Species
Fe(OH)3(s)
-4
-6
-
log concentration
Fe(OH)4
-8
-10
Fe
3+
FeOH
2+
+
Fe(OH)2
-12
0
2
4
6
8
10
12
14
pH
Concentrations of dissolved Fe species in equilibrium with
Fe(OH)3 as a function of pH.
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0
-1
Log
activity
dissolved
Si Species
-2
Amorphous silica
-3
2-
Total
H2SiO4
Quartz
-4
H4SiO40
-
H3SiO4
-5
log activity of diss
2
4
6
8
10
12
14
pH
Activities of dissolved silica species in equilibrium with quartz and
amorphous silica at 25°C. Note that silica solubility is pH-independent at
pH < 9, but increases dramatically with increasing pH at pH >9.
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Case Study: Cotter U Mill Site
The Cotter/Lincoln Park site consists of a uranium processing mill located adjacent to the
unincorporated community of Lincoln Park. The mill operated continuously from 1958 until
1979, and intermittently since that time. Mill operations released radioactive materials and
metals into the environment. These releases contaminated soil and groundwater around the
mill and the Lincoln Park area.
For more info: http://www.antenna.nl/wise/uranium/umopcc.html
Davis, A., and D.D. Runnells. 1987. Geochemical Interactions between acidic tailings fluid and
bedrock: Use of the computer model MINTEQ. Applied Geochemistry 2: 231-241.
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Case Study: Cotter U Mill Site
http://www.epa.gov/region08/superfund/co/lincolnpark/
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Case Study: Cotter Uranium
Mill Site
The contaminants of most
concern at the site are
molybdenum and
uranium. The primary
exposure pathways would
be drinking contaminated
water and inhaling
contaminated dust.
Radon, a decay product in
the uranium chain, is also
of potential concern.
Major cleanup activities performed since 1988 include:
Connecting Lincoln Park residents to city water;
Constructing a ground-water barrier at the Soil Conservation Service (SCS) dam
to minimize migration of contaminated ground water into Lincoln Park;
Moving tailings and contaminated soils into a lined impoundment to eliminate them as a
source of contamination; and
Excavating contaminated stream sediments
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Case Study: Cotter U
Mill Site
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Case Study: Cotter U Mill Site
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Case Study: Cotter U Mill Site
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Case Study: Cotter U Mill Site
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Case Study: Cotter U Mill
Site
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Case Study: Cotter U
Mill Site
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Appendix
■ Understand the principles governing the solubility of quartz.
■ Understand the principles governing the solubility of Al- and Feoxyhydroxides.
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SILICA SOLUBILITY - I
■ In the absence of organic ligands or fluoride, quartz solubility is relatively
low in natural waters.
■ Below pH 9, the dissolution reaction is:
SiO2(quartz) + 2H2O(l)  H4SiO40
for which the equilibrium constant at 25°C is:
K = aH
0
4 SiO 4
= 10-4
■ At pH < 9, quartz solubility is independent of pH.
■ Quartz is frequently supersaturated in natural waters because quartz
precipitation kinetics are slow.
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SILICA SOLUBILITY - II
■ Thus, quartz saturation does not usually control the concentration of silica in
low-temperature natural waters. Amorphous silica can control dissolved Si:
SiO2(am) + 2H2O(l)  H4SiO40
for which the equilibrium constant at 25°C is:
K = aH
0
4 SiO 4
= 2 ´10-3
■ Quartz is formed diagenetically through the following sequence of reactions:
opal-A (siliceous biogenic ooze)  opal-A’ (nonbiogenic amorphous silica)  opal-CT
 chalcedony  microcrystalline quartz
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SILICA SOLUBILITY - III
At pH > 9, H4SiO40 dissociates according to:
H4SiO40

H3SiO4-
+
K1 =
H+
aH SiO - aH +
3
4
aH SiO0
4
H3SiO4-  H2SiO42- + H+
K2 =
= 10-9.9
4
aH SiO2 - aH +
2
4
aH SiO 3
= 10-11.7
4
The total solubility of quartz (or amorphous silica) is:
(M ) = M
SiO2 T
H 4 SiO40
+ M H SiO- + M H SiO23
4
2
4
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SILICA SOLUBILITY - IV
The equations for the dissociation constants of silicic acid can be rearranged
(assuming a = M ) to get:
K1M H SiO 0
M H SiO - =
3
M H SiO 2- =
2
4
4
4
aH +
4
K2 M H SiO 3
aH +
4
=
K2 K1M H SiO 0
4
4
aH2 +
We can now write:
(M )
SiO2 T
æ
ö
K
K
K
= M H SiO0 ç1 + 1 + 12 2 ÷
÷
4
4 ç
a
a
+
+
H
H
è
ø
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0
-1
Log
activity
dissolved
Si Species
-2
Amorphous silica
-3
2-
Total
H2SiO4
Quartz
-4
H4SiO40
-
H3SiO4
-5
log activity of diss
2
4
6
8
10
12
14
pH
Activities of dissolved silica species in equilibrium with quartz and
amorphous silica at 25°C. Note that silica solubility is pH-independent at
pH < 9, but increases dramatically with increasing pH at pH >9.
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SILICA SOLUBILITY - V
An alternate way to understand quartz solubility is to start with: SiO2(quartz) + 2H2O(l) 
H4SiO40
Kqtz = M H SiO0 = 10-4
4
4
log Kqtz = log M H SiO0 = -4
4
4
Now adding the two reactions:
SiO2(quartz) + 2H2O(l)  H4SiO40 Kqtz
H4SiO40  H3SiO4- + H+ K1
SiO2(quartz) + 2H2O(l)  H3SiO4- + H+ K
K = Kqtz K1 = (10
-4
)(10 )= 10
-9.9
-13.9
= M H SiO- aH +
3
4
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SILICA SOLUBILITY - VI
Taking the log of both sides and rearranging we get:
log M H SiO - = -13.9 + pH
3
4
Finally adding the three reactions:
SiO2(quartz) + 2H2O(l)  H4SiO40 Kqtz
H4SiO40  H3SiO4- + H+ K1
H3SiO4-  H2SiO42- + H+ K2
SiO2(quartz) + 2H2O(l)  H2SiO42- + 2H+ K
K = Kqtz K1K2 = (10-4 )(10-9.9 )(10-11.7 )= 10-25.6 = M H SiO2- aH2 +
2
4
log M H SiO2- = -25.6 + 2 pH
2
4
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SILICA SOLUBILITY - VII
■
■
■
■
SUMMARY
Silica solubility is relatively low and independent of pH at pH < 9 where
H4SiO40 is the dominant species.
Silica solubility increases with increasing pH above 9, where H3SiO4- and
H2SiO42- are dominant.
Fluoride, and possibly organic compounds, may increase the solubility of
silica.
Saturation with quartz does not control silica concentrations in lowtemperature natural waters; saturation with amorphous silica may.
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Appendix
■ Understand the principles governing the solubility of quartz.
■ Understand the principles governing the solubility of Al- and Feoxyhydroxides.
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SOLUBILITY OF GIBBSITE - I
■ We will use gibbsite to illustrate principles of the solubility of Al-bearing
minerals; the solubility of such minerals is highly pH-dependent.
■ The solubility product for gibbsite is given by:
Al(OH)3(gibbsite)  Al3+ + 3OH-
■ We can also write this in the alternate form:
Al(OH)3(gibbsite) + 3H+  Al3+ + 3H2O(l)
K SP = a Al 3+ a
3
OH -
= 10
-32.64
K gibbsite =
a Al 3+
aH3 +
= 109.36
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SOLUBILITY OF GIBBSITE - II
■ Use of the latter equation shows that the concentration of Al3+ will be
very low in the pH range of most natural waters.
■ For example, at pH = 7, we calculate the concentration of Al3+ to be
2.2910-12 mol L-1!
■ However, Al3+ forms a series of hydroxide complexes which increase its
solubility somewhat:
Al3+ + H2O(l)  Al(OH)2+ + H+
Al3+ + 2H2O(l)  Al(OH)2+ + 2H+
Al3+ + 4H2O(l)  Al(OH)4- + 4H+
K h,1
K h,2
K h,4
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SOLUBILITY OF GIBBSITE - III
■ The mass action expressions for these reactions may be written:
K h ,1 =
a Al (OH )2+ aH +
a Al 3+
= 10
Kh,2 =
- 4.99
Kh,4 =
a Al (OH )- aH4 +
4
a Al 3+
a Al (OH )+ aH2 +
2
a Al 3+
= 10-10.13
= 10-22.05
The total dissolved aluminum concentration is given by:
(M Al )T = M Al
3+
+ M Al (OH )2 + + M Al (OH )+ + M Al (OH )2
4
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SOLUBILITY OF GIBBSITE - IV
■ We now assume that activity coefficients are unity, so that activity equals
concentration.
■ Next, we rewrite the solubility product of gibbsite to obtain:
M Al 3+ = K gibbsite a
3
H+
= 10
9.36
a
3
H+
log M Al 3+ = log K gibbsite - 3 pH = 9.36 - 3 pH
We see that the logarithm of the concentration of Al3+ in equilibrium with
gibbsite is a straight line function of pH, with a slope of -3. In other words,
the concentration of Al3+ decreases 3 log units for every unit increase in
pH.
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SOLUBILITY OF GIBBSITE - V
The concentration of Al(OH)2+ can be obtained from:
M Al (OH )2+ = Kh,1 M Al 3+ aH-1+ = 10-4.99 M Al 3+ aH-1+
but the concentration of Al3+ has already been calculated so:
M Al (OH )2+ = Kh,1 K gibbsite aH2 + = (10-4.99 )(109.36 )aH2 +
log M Al (OH )2+ = log(Kh,1 K gibbsite )- 2 pH = 4.37 - 2 pH
We see that the logarithm of the concentration of Al(OH)2+ in equilibrium
with gibbsite is also a straight line function of pH, but with a slope of -2.
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SOLUBILITY OF GIBBSITE - VI
Similarly for the other two species:
M Al (OH )+ = Kh ,2 M Al 3+ aH-2+ = 10-10.13 M Al 3+ aH-2+
2
M Al (OH )+ = Kh ,2 K gibbsite aH + = (10-10.13 )(109.36 )aH +
2
log M Al (OH )+ = log(Kh ,2 K gibbsite )- pH = -0.77 - pH
2
M Al (OH )- = Kh ,4 M Al 3+ aH-4+ = 10-22.05 M Al 3+ aH-4+
4
M Al (OH )- = Kh ,4 K gibbsite aH-1+ = (10-22.05 )(109.36 )aH-1+
4
log M Al (OH )- = log(Kh, 4 K gibbsite )+ pH = -12.69 + pH
4
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SOLUBILITY OF GIBBSITE - VII
Now, substituting into the mass-balance expression:
(M Al )T = M Al
3+
+ M Al (OH )2 + + M Al (OH )+ + M Al (OH )2
4
we get
(M Al )T = K gibbsite (aH3
+
+ Kh ,1aH2 + + Kh ,2aH + + Kh ,4aH-1+ )
and taking the logarithm of both sides and substituting the K values at 25°C:
log(M Al )T = 9.36
+ log(aH3 + + 10-4.99 aH2 + + 10-10.13 aH + + 10-22.05 aH-1+ )
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0
Al(OH)3(s)
-2
Log
Activity Al
Species
-4
-
Al(OH)4
3+
Al
+
Al(OH)2
2+
log concentration
-8
Al O
H
-6
-10
0
2
4
6
8
10
12
pH
Concentrations of dissolved Al species in equilibrium with
gibbsite as a function of pH.
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ENVIRONMENTAL GEOCHEMISTRY AT TEXAS A&M UNIVERSITY
SOLUBILITY OF ZINCITE (ZnO) - I
The thermodynamic data for solubility problems can be presented in another way. At 25°C and 1 bar:
ZnO(s) + 2H+  Zn2+ + H2O(l)
ZnO(s) + H+  ZnOH+
ZnO(s) + 2H2O(l)  Zn(OH)3- + H+
ZnO(s) + 3H2O(l)  Zn(OH)42- + 2H+
log Ks0 = 11.2
log Ks1 = 2.2
log Ks3 = -16.9
log Ks4 = -29.7
The solubility of zincite is given by:
(M Zn )T = M Zn
2+
+ M Zn (OH )+ + M Zn (OH )- + M Zn (OH )23
4
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SOLUBILITY OF ZINCITE (ZnO) - II
We start with the mass-action expressions for each of the previous reactions:
Ks0 =
aZn2 +
a
K s1 =
2
H+
K s 3 = aZn (OH )- aH +
3
aZnOH +
aH +
K s 4 = aZn (OH )2 - aH2 +
4
Assuming that activity coefficients can be neglected we can now write the following
expressions:
M Zn2+ = K s 0aH2 +
log M Zn2+ = log Ks 0 - 2 pH
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SOLUBILITY OF ZINCITE (ZnO) - III
log M ZnOH + = log K s1 - pH
M ZnOH + = K s1aH +
M Zn (OH )3
Ks3
=
aH +
log M Zn (OH )- = log K s 3 + pH
3
And the total concentration can be written:
M Zn (OH )2 4
Ks4
= 2
aH +
log(M Zn )T
log M Zn (OH )2- = log K s 4 + 2 pH
4
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ö
K
K
2
s
3
s
4
= logç K s 0aH + + K s1aH + +
+ 2 ÷
ç
÷
a
a
+
+
H
H ø
è
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2
0
Log
Activity Zn
Species
ZnOH
-2
+
ZnO(s)
-4
-6
log concentration
Zn(OH)4
-8
-
Zn(OH)3
Zn
2-
2+
-10
4
6
8
10
12
14
pH
Concentrations of dissolved Zn species in equilibrium with
ZnO as a function of pH.
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SOLUBILITY OF Fe(OH)3 - I
For this problem we have the following thermodynamic data at 25°C:
Fe(OH)3(s) + 3H+  Fe3+ + 3H2O
log Ks0 = 3.96
Fe3+ + H2O  FeOH2+ + H+
log Kh,1 = -3.05
Fe3+ + 2H2O  Fe(OH)2+ + 2H+
log Kh,2 = -6.31
Fe(OH)3(s) + H2O  Fe(OH)4- + H+
log Ks4 = -18.7
These reactions are a mix of two different types of reactions, but the same principles
apply. The total solubility is given by:
(M Fe )T = M Fe
3+
+ M Fe (OH )2+ + M Fe (OH )+ + M Fe (OH )2
4
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SOLUBILITY OF Fe(OH)3 - II
To get the concentration of Fe3+, we start with the mass-action expression:
Ks0 =
aFe3+
a
= 10
3
H+
3.96
M Fe3+ = K s 0aH3 +
And for FeOH+:
log M Fe3+ = log K s 0 - 3 pH
K h ,1 =
aFe (OH )2 + aH +
aFe3+
= 10-3.05
M Fe (OH )2+ =
Kh ,1M Fe3+
aH +
log M Fe (OH )2 + = log Kh ,1 + log M Fe3+ + pH
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ENVIRONMENTAL GEOCHEMISTRY AT TEXAS A&M UNIVERSITY
SOLUBILITY OF Fe(OH)3 - III
But we already solved for the concentration of Fe3+, so
log M Fe (OH )2 + = log K h ,1 + log K s 0 - 3 pH + pH
= log( Kh ,1K s 0 ) - 2 pH
Now for Fe(OH)2+:
Kh,2 =
aFe (OH )+ aH2 +
2
aFe3+
= 10-6.31
M Fe (OH )+ =
Kh , 2 M Fe3+
2
aH2 +
log M Fe (OH )+ = log Kh,2 + log M Fe3+ + 2 pH
2
log M Fe (OH )+ = log Kh , 2 + log K s 0 - 3 pH + 2 pH
2
= log( Kh , 2 K s 0 ) - pH
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SOLUBILITY OF Fe(OH)3 - IV
Finally, for Fe(OH)4-:
Kh,4 = aFe (OH )- aH + = 10
-18.7
4
M Fe (OH )4
Kh,4
=
aH +
log M Fe (OH )- = log Kh , 4 + pH
4
For the total solubility we have:
log( M Fe )T = log(K s0a3H + + K h,1K s0a2H + + K h,2K s0 aH + + K S0 a-1
H+ )
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2
0
-2
Log
Activity Fe
Species
Fe(OH)3(s)
-4
-6
-
log concentration
Fe(OH)4
-8
-10
Fe
3+
FeOH
2+
+
Fe(OH)2
-12
0
2
4
6
8
10
12
14
pH
Concentrations of dissolved Fe species in equilibrium with
Fe(OH)3 as a function of pH.
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