Phase change reactions precipitation dissolution of inorganic species
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Phase Change Reactions Precipitation-Dissolution of Inorganic Species. Bruce Herbert Geology & Geophysics. Precipitation-Dissolution and Metal-Ligand Properties. Generally, species exhibit similar precipitation-dissolution reactions as complexation reactions in that

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Phase Change Reactions Precipitation-Dissolution of Inorganic Species

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Phase change reactions precipitation dissolution of inorganic species

Phase Change ReactionsPrecipitation-Dissolution of Inorganic Species

Bruce Herbert

Geology & Geophysics


Precipitation dissolution and metal ligand properties

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).


Thermodynamics of precipitation dissolution

Thermodynamics of Precipitation-Dissolution

  • General formula for two component dissolution

    (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.


Thermodynamics of precipitation dissolution1

Thermodynamics of Precipitation-Dissolution

  • The solubility product constant, Kso is defined as

    (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.


Thermodynamics of precipitation dissolution2

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

      (6.3)

    • The relative saturation, , is defined as

      (6.4)


Thermodynamics of precipitation dissolution3

Thermodynamics of Precipitation-Dissolution

  • The relative saturation can be monitored over time to assess the degree of equilibration in a system.

    • If < 1, then the system is undersaturated with respect to the solid phase as defined by the reaction in 6.1.

    • If > 1, then the system is oversaturated with respect to the solid phase as defined by the reaction in 6.1.

    • If = 1, then the system is in equilibration with respect to the solid phase as defined by the reaction in 6.1.


Thermodynamics of precipitation dissolution4

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


Thermodynamics of precipitation dissolution5

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


Example the dissolution of gibbsite

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= Ksoand

    Al(OH)3(s) = Al3+(aq) + 3OH-(aq)(6.6)

Check calcs


Thermodynamics of precipitation dissolution6

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.


Example does cd oh 2 or cdco 3 control cd 2 in solution

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


Example does cd oh 2 or cdco 3 control cd 2 in solution1

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


Reverse experimental procedure

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.


Solubility of oxides and hydroxides

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


Solubility of oxides and hydroxides1

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)


Graphical representation of zno s

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


Graphical representation of zno s1

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:


Graphical representation of zno s2

Graphical representation of ZnO (s)


Phase change reactions precipitation dissolution of inorganic species

Log Activity Zn Species

Concentrations of dissolved Zn species in equilibrium with ZnO as a function of pH.


Phase change reactions precipitation dissolution of inorganic species

Log Activity Al Species

Concentrations of dissolved Al species in equilibrium with gibbsite as a function of pH.


Phase change reactions precipitation dissolution of inorganic species

Log Activity Fe Species

Concentrations of dissolved Fe species in equilibrium with Fe(OH)3 as a function of pH.


Phase change reactions precipitation dissolution of inorganic species

Log activity dissolved Si Species

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.


Case study cotter u mill site

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.


Case study cotter u mill site1

Case Study: Cotter U Mill Site

http://www.epa.gov/region08/superfund/co/lincolnpark/


Case study cotter uranium mill site

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


Case study cotter u mill site2

Case Study: Cotter U Mill Site


Case study cotter u mill site3

Case Study: Cotter U Mill Site


Case study cotter u mill site4

Case Study: Cotter U Mill Site


Case study cotter u mill site5

Case Study: Cotter U Mill Site


Case study cotter u mill site6

Case Study: Cotter U Mill Site


Case study cotter u mill site7

Case Study: Cotter U Mill Site


Appendix

Appendix

  • Understand the principles governing the solubility of quartz.

  • Understand the principles governing the solubility of Al- and Fe-oxyhydroxides.


Silica solubility i

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:

  • At pH < 9, quartz solubility is independent of pH.

  • Quartz is frequently supersaturated in natural waters because quartz precipitation kinetics are slow.


Silica solubility ii

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:

  • 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


Silica solubility iii

SILICA SOLUBILITY - III

At pH > 9, H4SiO40 dissociates according to:

H4SiO40  H3SiO4- + H+

H3SiO4-  H2SiO42- + H+

The total solubility of quartz (or amorphous silica) is:


Silica solubility iv

SILICA SOLUBILITY - IV

The equations for the dissociation constants of silicic acid can be rearranged (assuming a = M ) to get:

We can now write:


Phase change reactions precipitation dissolution of inorganic species

Log activity dissolved Si Species

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.


Silica solubility v

SILICA SOLUBILITY - V

An alternate way to understand quartz solubility is to start with:SiO2(quartz) + 2H2O(l)  H4SiO40

Now adding the two reactions:

SiO2(quartz) + 2H2O(l)  H4SiO40Kqtz

H4SiO40  H3SiO4- + H+K1

SiO2(quartz) + 2H2O(l)  H3SiO4- + H+K


Silica solubility vi

SILICA SOLUBILITY - VI

Taking the log of both sides and rearranging we get:

Finally adding the three reactions:

SiO2(quartz) + 2H2O(l)  H4SiO40Kqtz

H4SiO40  H3SiO4- + H+K1

H3SiO4-  H2SiO42- + H+K2

SiO2(quartz) + 2H2O(l)  H2SiO42- + 2H+K


Silica solubility vii

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 low-temperature natural waters; saturation with amorphous silica may.


Appendix1

Appendix

  • Understand the principles governing the solubility of quartz.

  • Understand the principles governing the solubility of Al- and Fe-oxyhydroxides.


Solubility of gibbsite i

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)


Solubility of gibbsite ii

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+Kh,1

    Al3+ + 2H2O(l)  Al(OH)2+ + 2H+Kh,2

    Al3+ + 4H2O(l)  Al(OH)4- + 4H+Kh,4


Solubility of gibbsite iii

SOLUBILITY OF GIBBSITE - III

  • The mass action expressions for these reactions may be written:

    The total dissolved aluminum concentration is given by:


Solubility of gibbsite iv

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:

    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.


Solubility of gibbsite v

SOLUBILITY OF GIBBSITE - V

The concentration of Al(OH)2+ can be obtained from:

but the concentration of Al3+ has already been calculated so:

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.


Solubility of gibbsite vi

SOLUBILITY OF GIBBSITE - VI

Similarly for the other two species:


Solubility of gibbsite vii

SOLUBILITY OF GIBBSITE - VII

Now, substituting into the mass-balance expression:

we get

and taking the logarithm of both sides and substituting the K values at 25°C:


Phase change reactions precipitation dissolution of inorganic species

Log Activity Al Species

Concentrations of dissolved Al species in equilibrium with gibbsite as a function of pH.


Solubility of zincite zno i

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)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

The solubility of zincite is given by:


Solubility of zincite zno ii

SOLUBILITY OF ZINCITE (ZnO) - II

We start with the mass-action expressions for each of the previous reactions:

Assuming that activity coefficients can be neglected we can now write the following expressions:


Solubility of zincite zno iii

SOLUBILITY OF ZINCITE (ZnO) - III

And the total concentration can be written:


Phase change reactions precipitation dissolution of inorganic species

Log Activity Zn Species

Concentrations of dissolved Zn species in equilibrium with ZnO as a function of pH.


Solubility of fe oh 3 i

SOLUBILITY OF Fe(OH)3 - I

For this problem we have the following thermodynamic data at 25°C:

Fe(OH)3(s) + 3H+ Fe3+ + 3H2Olog 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:


Solubility of fe oh 3 ii

SOLUBILITY OF Fe(OH)3 - II

To get the concentration of Fe3+, we start with the mass-action expression:

And for FeOH+:


Solubility of fe oh 3 iii

SOLUBILITY OF Fe(OH)3 - III

But we already solved for the concentration of Fe3+, so

Now for Fe(OH)2+:


Solubility of fe oh 3 iv

SOLUBILITY OF Fe(OH)3 - IV

Finally, for Fe(OH)4-:

For the total solubility we have:


Phase change reactions precipitation dissolution of inorganic species

Log Activity Fe Species

Concentrations of dissolved Fe species in equilibrium with Fe(OH)3 as a function of pH.


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