Why are the results obtained by changing the pH of my CD-MUSIC almost the same?

[yisongxu]

Hello everyone, I'm working on CD-MUSIC for magnetite adsorption of Am and I want to do it with different pH values. Why do I get almost exactly the same results after changing the pH value under the conditions of pH3, 4, 5, 6, 7, 8 and 9? This is my code for pH 3, and the rest of the pH is adjusted at EQUILIBRIUM_PHASES.

Code: [Select]

SOLUTION 1
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mol/l
    density   1
    Cl        0.1
    Na        0.1
    Am        0.0001
    -water    1 # kg

SOLUTION_MASTER_SPECIES
    Am            Am+3             0     243             243

SOLUTION_SPECIES
Am+3 = Am+3
    log_k     0
Am+3 + H2O = AmOH+2 + H+
    log_k     -5.7
Am+3 + 2H2O = Am(OH)2+ + 2H+
    log_k     -11.2
Am+3 + Cl- = AmCl+2
    log_k     0.4

SURFACE_MASTER_SPECIES
    Mag_a         Mag_aOH     
    Mag_b         Mag_bO-     
    Mag_c         Mag_cOH     
    Mag_d         Mag_dOH     

SURFACE_SPECIES
Mag_aOH = Mag_aOH
    log_k     0
Mag_bO- = Mag_bO-
    log_k     0
Mag_cOH = Mag_cOH
    log_k     0
Mag_dOH = Mag_dOH
    log_k     0
Mag_aOH = Mag_aO- + H+
    log_k     -1
H+ + Mag_aOH = Mag_aOH2+
    log_k     1
H+ + Mag_bO- = Mag_bOH
    log_k     7.6
Mag_cOH = Mag_cO- + H+
    log_k     -9.3
H+ + Mag_cOH = Mag_cOH2+
    log_k     5.1
Mag_dOH = Mag_dO- + H+
    log_k     -17.5
H+ + Mag_dOH = Mag_dOH2+
    log_k     3.6
Am+3 + 3Mag_aOH = (Mag_aOH)3Am+3
    log_k     1
Am+3 + 3Mag_aOH = (Mag_aOH)2(Mag_aO)Am+2 + H+
    log_k     1
Am+3 + H2O + 3Mag_aOH = (Mag_aOH)2(Mag_aO)Am(OH)+ + 2H+
    log_k     1
Am+3 + Mag_aOH = Mag_aOHAm+3
    log_k     1
Am+3 + H2O + Mag_aOH = Mag_aOHAmOH+2 + H+
    log_k     1
Am+3 + 2H2O + Mag_aOH = Mag_aOHAm(OH)2+ + 2H+
    log_k     1
Am+3 + Cl- + Mag_aOH = Mag_aOHAmCl+2
    log_k     1
Am+3 + Mag_aOH = Mag_aOAm+2 + H+
    log_k     1
Am+3 + H2O + Mag_aOH = Mag_aOAmOH+ + 2H+
    log_k     1
Am+3 + 2H2O + Mag_aOH = Mag_aOAm(OH)2 + 3H+
    log_k     1
Am+3 + Cl- + Mag_aOH = Mag_aOAmCl+ + H+
    log_k     1
Am+3 + 2Mag_aOH = (Mag_aOH)2Am+3
    log_k     1
Am+3 + 2Mag_aOH = (Mag_aO)2Am+ + 2H+
    log_k     1
Am+3 + 2Mag_aOH = (Mag_aOH)(Mag_aO)Am+2 + H+
    log_k     1
Am+3 + H2O + 2Mag_aOH = (Mag_aOH)2AmOH+2 + H+
    log_k     1
Am+3 + H2O + 2Mag_aOH = (Mag_aO)2AmOH + 3H+
    log_k     1
Am+3 + H2O + 2Mag_aOH = (Mag_aOH)(Mag_aO)AmOH+ + 2H+
    log_k     1
Am+3 + 2H2O + 2Mag_aOH = (Mag_aOH)2Am(OH)2+ + 2H+
    log_k     1
Am+3 + 2H2O + 2Mag_aOH = (Mag_aO)2Am(OH)2- + 4H+
    log_k     1
Am+3 + 2H2O + 2Mag_aOH = (Mag_aOH)(Mag_aO)Am(OH)2 + 3H+
    log_k     1
Am+3 + Cl- + 2Mag_aOH = (Mag_aOH)2AmCl+2
    log_k     1
Am+3 + Cl- + 2Mag_aOH = (Mag_aO)2AmCl + 2H+
    log_k     1
Am+3 + Cl- + 2Mag_aOH = (Mag_aOH)(Mag_aO)AmCl+ + H+
    log_k     1
Am+3 + 3Mag_aOH = (Mag_aO)3Am + 3H+
    log_k     1
Am+3 + 3Mag_aOH = (Mag_aOH)(Mag_aO)2Am+ + 2H+
    log_k     1
Am+3 + H2O + 3Mag_aOH = (Mag_aOH)3AmOH+2 + H+
    log_k     1
Am+3 + H2O + 3Mag_aOH = (Mag_aO)3AmOH- + 4H+
    log_k     1
Am+3 + H2O + 3Mag_aOH = (Mag_aOH)(Mag_aO)2AmOH + 3H+
    log_k     1
Am+3 + 2H2O + 3Mag_aOH = (Mag_aOH)3Am(OH)2+ + 2H+
    log_k     1
Am+3 + 2H2O + 3Mag_aOH = (Mag_aO)3Am(OH)2-2 + 5H+
    log_k     1
Am+3 + 2H2O + 3Mag_aOH = (Mag_aOH)2(Mag_aO)Am(OH)2 + 3H+
    log_k     1
Am+3 + 2H2O + 3Mag_aOH = (Mag_aOH)(Mag_aO)2Am(OH)2- + 4H+
    log_k     1
Am+3 + Cl- + 3Mag_aOH = (Mag_aOH)3AmCl+2
    log_k     1
Am+3 + Cl- + 3Mag_aOH = (Mag_aO)3AmCl- + 3H+
    log_k     1
Am+3 + Cl- + 3Mag_aOH = (Mag_aOH)2(Mag_aO)AmCl+ + H+
    log_k     1
Am+3 + Cl- + 3Mag_aOH = (Mag_aOH)(Mag_aO)2AmCl + 2H+
    log_k     1

SURFACE 1
    -sites DENSITY
    Mag_aOH    6.68      71.016    1
        -capacitance 3.22   2.25
    Mag_bO-    2.23
    Mag_cOH    4.45
    Mag_dOH    1.48
    -cd_music

PHASES
Fix_H+
    H+ = H+
    log_k     0

EQUILIBRIUM_PHASES 1
    Fix_H+    -3 HCl       10
SELECTED_OUTPUT 1
    -file                 selected_output_1.sel
    -molalities           Am+3  AmOH+2  Am(OH)2+  AmCl+2
                          (Mag_aO)2Am(OH)2-  (Mag_aO)2Am+  (Mag_aO)2AmCl  (Mag_aO)2AmOH
                          (Mag_aO)3Am  (Mag_aO)3Am(OH)2-2  (Mag_aO)3AmCl-  (Mag_aO)3AmOH-
                          (Mag_aOH)(Mag_aO)2Am(OH)2-  (Mag_aOH)(Mag_aO)2Am+  (Mag_aOH)(Mag_aO)2AmCl  (Mag_aOH)(Mag_aO)2AmOH
                          (Mag_aOH)(Mag_aO)Am(OH)2  (Mag_aOH)(Mag_aO)Am+2  (Mag_aOH)(Mag_aO)AmCl+  (Mag_aOH)(Mag_aO)AmOH+
                          (Mag_aOH)2(Mag_aO)Am(OH)+  (Mag_aOH)2(Mag_aO)Am(OH)2  (Mag_aOH)2(Mag_aO)Am+2  (Mag_aOH)2(Mag_aO)AmCl+
                          (Mag_aOH)2Am(OH)2+  (Mag_aOH)2Am+3  (Mag_aOH)2AmCl+2  (Mag_aOH)2AmOH+2
                          (Mag_aOH)3Am(OH)2+  (Mag_aOH)3Am+3  (Mag_aOH)3AmCl+2  (Mag_aOH)3AmOH+2
                          Mag_aOAm(OH)2  Mag_aOAm+2  Mag_aOAmCl+  Mag_aOAmOH+
                          Mag_aOHAm(OH)2+  Mag_aOHAm+3  Mag_aOHAmCl+2  Mag_aOHAmOH+2

END

[dlparkhurst]

The sorption log Ks put almost all of the Am on the surface. Concentrations in solution are 1e-10 or less.

Code: [Select]

SOLUTION_MASTER_SPECIES
    Am            Am+3             0     243             243

SOLUTION_SPECIES
Am+3 = Am+3
    log_k     0
Am+3 + H2O = AmOH+2 + H+
    log_k     -5.7
Am+3 + 2H2O = Am(OH)2+ + 2H+
    log_k     -11.2
Am+3 + Cl- = AmCl+2
    log_k     0.4

SURFACE_MASTER_SPECIES
    Mag_a         Mag_aOH     
    Mag_b         Mag_bO-     
    Mag_c         Mag_cOH     
    Mag_d         Mag_dOH     

SURFACE_SPECIES
Mag_aOH = Mag_aOH
    log_k     0
Mag_bO- = Mag_bO-
    log_k     0
Mag_cOH = Mag_cOH
    log_k     0
Mag_dOH = Mag_dOH
    log_k     0
Mag_aOH = Mag_aO- + H+
    log_k     -1
H+ + Mag_aOH = Mag_aOH2+
    log_k     1
H+ + Mag_bO- = Mag_bOH
    log_k     7.6
Mag_cOH = Mag_cO- + H+
    log_k     -9.3
H+ + Mag_cOH = Mag_cOH2+
    log_k     5.1
Mag_dOH = Mag_dO- + H+
    log_k     -17.5
H+ + Mag_dOH = Mag_dOH2+
    log_k     3.6
Am+3 + 3Mag_aOH = (Mag_aOH)3Am+3
    log_k     1
Am+3 + 3Mag_aOH = (Mag_aOH)2(Mag_aO)Am+2 + H+
    log_k     1
Am+3 + H2O + 3Mag_aOH = (Mag_aOH)2(Mag_aO)Am(OH)+ + 2H+
    log_k     1
Am+3 + Mag_aOH = Mag_aOHAm+3
    log_k     1
Am+3 + H2O + Mag_aOH = Mag_aOHAmOH+2 + H+
    log_k     1
Am+3 + 2H2O + Mag_aOH = Mag_aOHAm(OH)2+ + 2H+
    log_k     1
Am+3 + Cl- + Mag_aOH = Mag_aOHAmCl+2
    log_k     1
Am+3 + Mag_aOH = Mag_aOAm+2 + H+
    log_k     1
Am+3 + H2O + Mag_aOH = Mag_aOAmOH+ + 2H+
    log_k     1
Am+3 + 2H2O + Mag_aOH = Mag_aOAm(OH)2 + 3H+
    log_k     1
Am+3 + Cl- + Mag_aOH = Mag_aOAmCl+ + H+
    log_k     1
Am+3 + 2Mag_aOH = (Mag_aOH)2Am+3
    log_k     1
Am+3 + 2Mag_aOH = (Mag_aO)2Am+ + 2H+
    log_k     1
Am+3 + 2Mag_aOH = (Mag_aOH)(Mag_aO)Am+2 + H+
    log_k     1
Am+3 + H2O + 2Mag_aOH = (Mag_aOH)2AmOH+2 + H+
    log_k     1
Am+3 + H2O + 2Mag_aOH = (Mag_aO)2AmOH + 3H+
    log_k     1
Am+3 + H2O + 2Mag_aOH = (Mag_aOH)(Mag_aO)AmOH+ + 2H+
    log_k     1
Am+3 + 2H2O + 2Mag_aOH = (Mag_aOH)2Am(OH)2+ + 2H+
    log_k     1
Am+3 + 2H2O + 2Mag_aOH = (Mag_aO)2Am(OH)2- + 4H+
    log_k     1
Am+3 + 2H2O + 2Mag_aOH = (Mag_aOH)(Mag_aO)Am(OH)2 + 3H+
    log_k     1
Am+3 + Cl- + 2Mag_aOH = (Mag_aOH)2AmCl+2
    log_k     1
Am+3 + Cl- + 2Mag_aOH = (Mag_aO)2AmCl + 2H+
    log_k     1
Am+3 + Cl- + 2Mag_aOH = (Mag_aOH)(Mag_aO)AmCl+ + H+
    log_k     1
Am+3 + 3Mag_aOH = (Mag_aO)3Am + 3H+
    log_k     1
Am+3 + 3Mag_aOH = (Mag_aOH)(Mag_aO)2Am+ + 2H+
    log_k     1
Am+3 + H2O + 3Mag_aOH = (Mag_aOH)3AmOH+2 + H+
    log_k     1
Am+3 + H2O + 3Mag_aOH = (Mag_aO)3AmOH- + 4H+
    log_k     1
Am+3 + H2O + 3Mag_aOH = (Mag_aOH)(Mag_aO)2AmOH + 3H+
    log_k     1
Am+3 + 2H2O + 3Mag_aOH = (Mag_aOH)3Am(OH)2+ + 2H+
    log_k     1
Am+3 + 2H2O + 3Mag_aOH = (Mag_aO)3Am(OH)2-2 + 5H+
    log_k     1
Am+3 + 2H2O + 3Mag_aOH = (Mag_aOH)2(Mag_aO)Am(OH)2 + 3H+
    log_k     1
Am+3 + 2H2O + 3Mag_aOH = (Mag_aOH)(Mag_aO)2Am(OH)2- + 4H+
    log_k     1
Am+3 + Cl- + 3Mag_aOH = (Mag_aOH)3AmCl+2
    log_k     1
Am+3 + Cl- + 3Mag_aOH = (Mag_aO)3AmCl- + 3H+
    log_k     1
Am+3 + Cl- + 3Mag_aOH = (Mag_aOH)2(Mag_aO)AmCl+ + H+
    log_k     1
Am+3 + Cl- + 3Mag_aOH = (Mag_aOH)(Mag_aO)2AmCl + 2H+
    log_k     1
PHASES
Fix_H+
    H+ = H+
    log_k     0
END
SOLUTION 1
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mol/l
    density   1
    Cl        0.1
    Na        0.1
    Am        0.0001
    -water    1 # kg
END
SURFACE 1
    -sites DENSITY
    Mag_aOH    6.68      71.016    1
        -capacitance 3.22   2.25
    Mag_bOH    2.23
    Mag_cOH    4.45
    Mag_dOH    1.48
    -cd_music
END
USER_GRAPH 1
    -axis_titles            "pH" "Am, molality" ""
    -axis_scale y_axis      auto auto auto auto log
    -initial_solutions      false
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
10 GRAPH_X -LA("H+")
20 GRAPH_Y TOT("Am")
  -end
    -active                 true
USE solution 1
USE surface 1
EQUILIBRIUM_PHASES 1
    Fix_H+    -3 HCl       10
END
USE solution 1
USE surface 1
EQUILIBRIUM_PHASES 1
    Fix_H+    -5 HCl       10
END
USE solution 1
USE surface 1
EQUILIBRIUM_PHASES 1
    Fix_H+    -7 HCl       10
END
USE solution 1
USE surface 1
EQUILIBRIUM_PHASES 1
    Fix_H+    -9 HCl       10
END
USE solution 1
USE surface 1
EQUILIBRIUM_PHASES 1
    Fix_H+    -11 HCl       10
END
USE solution 1
USE surface 1
EQUILIBRIUM_PHASES 1
    Fix_H+    -13 NaOH       10
END


[yisongxu]

Is it to modify the log K of Am reacting with magnetite to form complex species? Based on the species that could be found in the references and through AI, I obtained the logK of other species (the logK of species is no longer 1) and then carried out the calculation. The results can be obtained at pH3, 4, 5, 6, 7, 8 and 9. There are differences among species under different pH conditions, but the total adsorption capacity is still 100%. According to my previous experiments, magnetite has almost no adsorption of Am at low pH. The adsorption capacity gradually increases with the increase of pH and gradually reaches saturation. How do I need to modify the code if I want to achieve this result?

[dlparkhurst]

Then the Ks or the number of sorption sites are too large in your PHREEQC simulations.

You can fit try to fit the parameters to your experiments. The number of sites and log Ks are usually highly, inversely correlated, so it may not be possible to find a unique fit. With hydrous ferric oxides in Dzombak and Morel, they fixed the number of sites per gram and fit the constants.

[yisongxu]

Thank you for your reply