solubility of corundum

[njordan]

Hi,

I want to obtain the total concentration of all Al species in equilibrium with corundum in 0.1M NaCl.

I used the following script:

SOLUTION 1
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mol/kgw
    Na        0.10515 charge
    Cl        0.10515
    -water    1 # kg

PHASES
fix_pH
    H+ = H+
    log_k     0.0
   
EQUILIBRIUM_PHASES
Corundum    0    1
   
SAVE Solution 1
END

SELECTED_OUTPUT
    -file                 test.txt
    -simulation           true
    -distance             false
    -ionic_strength       true
    -totals               Al Na Cl
    -molalities           Al+3  Al(OH)+2  Al(OH)2+  Al(OH)3  Al(OH)4-
    -equilibrium_phases   Corundum
    -saturation_indices   Corundum

USE solution 1; EQUILIBRIUM_PHASES; Corundum 0 0; fix_pH    -3.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Corundum 0 0; fix_pH    -4.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Corundum 0 0; fix_pH    -5.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Corundum 0 0; fix_pH    -6.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Corundum 0 0; fix_pH    -7.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Corundum 0 0; fix_pH    -8.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Corundum 0 0; fix_pH    -9.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Corundum 0 0; fix_pH    -10.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Corundum 0 0; fix_pH    -11.00 NaOH; end

However, I don´t get the expected behavior with increasing total Al concentration in acidic and basic pH range...

Thanks in advance for the help!

[dlparkhurst]

You need to add corundum in the EQUILIBRIUM_PHASES definitions, say "Corundum 0 1" for each of the pH calculations, as you used in your first calculation.

[njordan]

thank you very much!

Using the same principle, I need to do the same for amorphous Eu(OH)3 in 1 M NaCl, for that I use the SIT database.

SOLUTION 1
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mol/kgw
    Na        1.0215 charge
    Cl        1.0215
    -water    1 # kg

PHASES
fix_pH
    H+ = H+
    log_k     0.0

EQUILIBRIUM_PHASES
Eu(OH)3(am)    0    1

   
SAVE Solution 1
END


SELECTED_OUTPUT
    -file                 test.txt
    -simulation           true
    -distance             false
    -ionic_strength       true
    -totals               Eu Na Cl
    -molalities           Eu+3  Eu(OH)+2 Eu(OH)2+ Eu(OH)3 Eu(OH)4- EuCl+2 EuCl2+
    -equilibrium_phases   Eu(OH)3(am)

USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -3.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -4.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -5.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -6.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -7.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -8.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -9.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -10.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -11.00 NaOH; end

But here, the calculations are not converging at all.... Whats' wrong?

Thanks in advance for the help!

[dlparkhurst]

For some of your solution you need to add acid to reach the desired pH and for some you need to add base. The ones needing addition of acid fail with your setup. There is a trick described in the footnote of Example 8 to automatically allow the needed reactant to be added. Here HCl or NaOH is added to produce the desired pH values.

Code: [Select]

SOLUTION 1
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mol/kgw
    Na        1.0215 charge
    Cl        1.0215
    -water    1 # kg
PHASES
fix_pH
    H+ = H+
    log_k     0.0
NaCl
    NaCl = Na+ + Cl-
    log_k -20
END
SELECTED_OUTPUT
    -file                 test.txt
    -simulation           true
    -distance             false
    -ionic_strength       true
    -totals               Eu Na Cl
    -molalities           Eu+3  Eu(OH)+2 Eu(OH)2+ Eu(OH)3 Eu(OH)4- EuCl+2 EuCl2+
    -equilibrium_phases   Eu(OH)3(am)
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; NaCl 0 10; fix_pH    -3.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; NaCl 0 10; fix_pH    -4.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; NaCl 0 10; fix_pH    -5.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; NaCl 0 10; fix_pH    -6.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; NaCl 0 10; fix_pH    -7.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; NaCl 0 10; fix_pH    -8.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; NaCl 0 10; fix_pH    -9.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; NaCl 0 10; fix_pH    -10.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; NaCl 0 10; fix_pH    -11.00 NaOH; end


[njordan]

Thank you.

But by using your script, the ionic strength is not constant in the whole pH range, it is 4.5 mol/kg at pH 3.5, 5 mmol/kg at pH 7 and 1 mmol/kg at pH 11.  And the Na+ concentration defined as ~1mol/kg in the solution 1 (background electrolyte) is completely vanished at acidic pH .

[dlparkhurst]

You can either switch to use something like KBr so that Na and Cl concentrations are not changed, or you can simply replace NaOH in the fix pH with HCl for those cases that need acid.

[njordan]

Thank you.

I used the following script where I replaced NaOH with HCl for the acidic pH

SOLUTION 1
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mol/kgw
    Na        1.0215 charge
    Cl        1.0215
    -water    1 # kg

PHASES
fix_pH
    H+ = H+
    log_k     0.0

EQUILIBRIUM_PHASES
Eu(OH)3(am)    0    1

SAVE Solution 1
END

SELECTED_OUTPUT
    -file                 test.txt
    -simulation           true
    -distance             false
    -ionic_strength       true
    -totals               Eu Na Cl
    -molalities           Eu+3  Eu(OH)+2 Eu(OH)2+ Eu(OH)3 Eu(OH)4- Eu(CO3)+ Eu(CO3)2- Eu(CO3)3-3 Eu(HCO3)+2 EuCl+2 EuCl2+
    -equilibrium_phases   Eu(OH)3(am)

USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -3.00 HCl; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -4.00 HCl; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -5.00 HCl; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -6.00 HCl; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -7.00 HCl; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -8.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -9.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -10.00 NaOH; end
USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 1; fix_pH    -11.00 NaOH; end

Now the ionic strength is constant from pH 8 to 11, but is almost 5 mol/kg between pH 3 and 6.

[dlparkhurst]

Seems OK to me. Given the solubility constant for Eu(OH)3, you dissolved all of your Eu(OH)3 (1 mol), which requires 3 equivalents of acid (plus some) to get to the pH values of 6 and less. So yes, the ionic strength will be 3 from the acid added, plus contributions from the initial solution and the dissolved Eu, or approximately 5.

Other programs will let you fixed the ionic strength, but I think PHREEQC's approach is more realistic.

[njordan]

Thank you very much.

Just a short last question: if I want to see how much solid would precipitate (if oversaturation is reached) at a contant total Eu concentration, I need to use this script, correct? (0 as target SI and 0 moles of solid present at the beginning)

SOLUTION 1
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mol/kgw
    Na        1.0215 charge
    Cl        1.0215
    Eu         0.001
    -water    1 # kg

PHASES
fix_pH
    H+ = H+
    log_k     0.0

SAVE Solution 1
END

USE solution 1; EQUILIBRIUM_PHASES; Eu(OH)3(am) 0 0; fix_pH    -3.00 HCl; end

[dlparkhurst]

Fine, except that Eu(OH)3 won't precipitate at pH 3. You will need a much higher pH for the solubility of Eu(OH)3 to be less than 0.001.

[njordan]

Thank you. Yes true, I intend to due it at high pH as well, I just wanted to check the structure of the script.

Something else please.

When I write in the selected output:
 -equilibrium_phases   Eu(OH)3(am)

In the output file, I get two columns: Eu(OH)3(am) and   d_Eu(OH)3(am)

Eu(OH)3(am) is the number of moles of the solid while d_Eu(OH)3(am) is the number of moles of dissolved solid, I guess.

How is d_Eu(OH)3(am) calculated? I thought it would be the sum of all aquatic Eu species, but that is not the case....

[dlparkhurst]

Have you looked at the description of SELECTED_OUTPUT in the manual?

[njordan]

Yes, but I obviously did not get it from the first time, sorry for that.

d_Eu(OH)3(am) is the mole transfer of the reactant that is necessary to produce equilibrium with the pure phase, so in my case: the moles of Eu(OH)3(am) after the reaction - 1 (initial amount), correct?

[dlparkhurst]

Yes.

[njordan]

Thank you.

I want to calculate the speciation in NaClO4 as well, I used the following:

SOLUTION_MASTER_SPECIES
    Cl            Cl-              0     Cl              35.4527
    Cl(-1)        Cl-              0     Cl             
    Cl(7)         ClO4-            0     Cl

SOLUTION_SPECIES
1.000Cl-     + 2.000O2     = ClO4-
-log_k 15

SOLUTION 1
Cl(7)  1
END
USE solution 1
REACTION
END

At acidic pH, the concentration of EuCl++ equals that of EuOH++.

To decrease the concentration of Eu-chloro complexes which are not existing in NaClO4, can I simply change the log K of 1.000Cl-     + 2.000O2     = ClO4-
from 15 to 30, or should I rather decrease the log K of the Eu-chloro complexes in the database?

[dlparkhurst]

I think you want to increase the log K for the ClO4- reaction until ClO4- species are the predominant Cl species in solution.

[njordan]

yes exactly, so I guess this is the right way, correct?

[dlparkhurst]

Yes, assuming you want to limit Cl to one redox state (Cl(7)).

[njordan]

Yes, this is exactly my intention. Thank you!

[njordan]

I have a follow up question please.

I would like to calculate with Phreeqc a pH dependent solubility diagram of Eu(OH)3(am), that is log [Eu]total (10-10M to 10-3M) vs. pH.

Is it possible to use Phreeqc for that? 

[dlparkhurst]

I am not sure what you want. You are already calculating the solubility as a function of pH. If you are talking about a predominant species diagram, you could look at PhreePlot.