Questions regarding the use of the CD-MUSIC model

[yisongxu]

Hello, dear teachers. I recently just used the pheeqc software to fit the surface complexation model and have a few questions to ask. I was working on the adsorption of americium Am by magnetite. In the "SELECTED_OUTPUT" section, select and output the concentrations of several forms of americium in the liquid phase, as well as the concentrations of several species combined with magnetite and americium. The initial concentration of americium added in the reaction was 0.0001mol/L, and I calculated eight pH conditions. Shouldn't the result obtained be that the remaining concentration in the liquid phase plus the complex species is equal to the initial 0.0001mol/L? Why is it that in my calculation results, the total sum of some in the liquid phase is higher than 0.0001, and the total sum of some complex species is higher than 0.0001? Their total sum is 0.000258mol/L at 8 pH values. Why is that? Is there a problem with where I input it?

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

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     -16.4
H+ + Mag_aOH = Mag_aOH2+
    log_k     2.5
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     6.9
Am+3 + 3Mag_aOH = (Mag_aOH)2(Mag_aO)Am+2 + H+
    log_k     0.2
Am+3 + H2O + 3Mag_aOH = (Mag_aOH)2(Mag_aO)Am(OH)+ + 2H+
    log_k     -7.5

SURFACE 1
    -equilibrate with solution 1
    -sites DENSITY
    Mag_a      6.68      71.016    1
        -capacitance 3.22   2.25
    Mag_b      2.23
    Mag_c      4.45
    Mag_d      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+  (Mag_aOH)3Am+3
                          (Mag_aOH)2(Mag_aO)Am+2  (Mag_aOH)2(Mag_aO)Am(OH)+

END

[dlparkhurst]

You have defined the initial condition or your surface to be in equilibrium with SOLUTION 1. So, the initial surface contains Am in addition to the Am that is in solution. The total Am in your reaction system is the sum of the Am in the initial solution and the Am in the initial surface definition.

[yisongxu]

I got the result I wanted. Thank you for your answer