Different surfaces gives the same zeta potential!!!

[Tamirian]

Hi all,

I have created the kaolinite surface in my first and second models below. However, in my second model I have considered adsorption of Ca2+ on the edges as well.  How come do I get the same results after running both models? I mean I expected to have different zeta potential after each simulation but they were all the same!!! I wonder if you could help me and tell me what I am doing wrong here.

My first try

SURFACE_MASTER_SPECIES
    Kaolin_al     Kaolin_alOH 
    Kaolin_si     Kaolin_siOH 
SURFACE_SPECIES
Kaolin_alOH = Kaolin_alOH
    log_k     0
Kaolin_siOH = Kaolin_siOH
    log_k     0
H+ + Kaolin_alOH = Kaolin_alOH2+
    log_k     -3
Kaolin_alOH = Kaolin_alO- + H+
    log_k     -3.8
    delta_h   32 kJ
Kaolin_siOH = Kaolin_siO- + H+
    log_k     -7
    delta_h   32 kJ

SURFACE 1
    -sites DENSITY
    Kaolin_alOH    1.15      10        20
    Kaolin_siOH    1.15

SOLUTION 1
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mol/l
    density   1
    Ca        0.2
    Cl        0.2
    Na        0.2
    -water    1 # kg
SAVE solution 1
SAVE surface 1
END

SOLUTION 2 LSW
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mol/l
    density   1
    -water    1 # kg

END

MIX 1
    1    0.7
    2    0.3

SAVE solution 3

END

USE solution 3
USE surface 1


My second model

SURFACE_MASTER_SPECIES
    Kaolin_al     Kaolin_alOH 
    Kaolin_si     Kaolin_siOH 
SURFACE_SPECIES
Kaolin_alOH = Kaolin_alOH
    log_k     0
Kaolin_siOH = Kaolin_siOH
    log_k     0
H+ + Kaolin_alOH = Kaolin_alOH2+
    log_k     -3
Kaolin_alOH = Kaolin_alO- + H+
    log_k     -3.8
    delta_h   32 kJ
Kaolin_siOH = Kaolin_siO- + H+
    log_k     -7
    delta_h   32 kJ
Ca+2 + Kaolin_alOH = Kaolin_alOCa+ + H+
    log_k     -9.7
    delta_h   45 kJ
Ca+2 + Kaolin_siOH = Kaolin_siOCa+ + H+
    log_k     -9.7
    delta_h   45 kJ
CaOH+ + Kaolin_alOH = Kaolin_alOCaOH + H+
    log_k     -4.5
    delta_h   45 kJ
CaOH+ + Kaolin_siOH = Kaolin_siOCaOH + H+
    log_k     -4.5
    delta_h   45 kJ

SURFACE 1
    -sites DENSITY
    Kaolin_alOH    1.15      10        20
    Kaolin_siOH    1.15

SOLUTION 1
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mol/l
    density   1
    Ca        0.2
    Cl        0.2
    Na        0.2
    -water    1 # kg
SAVE solution 1
SAVE surface 1
END

SOLUTION 2 LSW
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mol/l
    density   1
    -water    1 # kg

END

MIX 1
    1    0.7
    2    0.3

SAVE solution 3

END
USE solution 3
USE surface 1

[dlparkhurst]

With your definitions, the sorption of Ca is ~1e-10, so it has a negligible effect on the surface charge.