Processes > Surface Complexation
How to change the pH of Surface problem with mixing by steps from5 to 11
(1/1)
MACAHarding:
Hello,
I also need to graph Co totals as a function of pH change. The following code gives me 5 and 11 for pH.
--- Code: ---TITLE Problem#1: Modeling Competitive Adsorption. Arsenic Adsorption Problem
SOULTION_SPECIES
CrO4-2 + 6H+ + 3e- = Cr(OH)2+ + 2H2O
log_k 100. #67.376
Cu+2 + e- = Cu+
log_k -100. #2.69
END
PHASES
Ferrihydrite
Fe(OH)3 + 3H+ = Fe+3 + 3H2O
log_k 3.191
delta_h -73.374 kJ
Fix_H+
H+ = H+
log_k 0
END
SOLUTION 1 #Backgound water
temp 25
pH 7.04
pe 4 Ferrihydrite 0
redox pe
units mg/kgw
density 1
As 0.000967
B 0.0468
Ba 0.037
C(4) 142
Ca 31.7
Cd 5.41e-005
Cl 15.8
Co 0.000138
Cr 0.0007797
Cu 0.00284
F 0.194
Fe 0.0792
K 2.51
Li 7.59e-3
Mg 8.9
Mn 0.012
Mo 0.000753
Na 25
Ni 0.0007778
Pb 0.000207
S 18.6
Sb 7.47e-005
Se 0.000557
Sr 0.219
V 0.0012
Zn 0.00761
-water 1 # kg
EQUILIBRIUM_PHASES 1
Ferrihydrite 0 0.01
SURFACE 1
-equilibrate with solution 1
Hfo_wOH Ferrihydrite equilibrium_phase 0.2 600
Hfo_sOH Ferrihydrite equilibrium_phase 0.0025
SAVE Solution 1
SAVE Equilibrium_phases 1
SAVE Surface 1
END
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -5 Ca(OH)2; USE Surface 1;
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -6 Ca(OH)2; USE Surface 1;
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -7 Ca(OH)2; USE Surface 1;
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -8 Ca(OH)2; USE Surface 1;
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -9 Ca(OH)2; USE Surface 1;
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -10 Ca(OH)2; USE Surface 1;
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -11 Ca(OH)2; USE Surface 1;
SOLUTION 2 #leachate
temp 25
pH 5
pe 4 Ferrihydrite 0
redox pe
units mg/kgw
density 1
As 0.0723
B 1.73
Ba 0.0631
C(4) 19.3
Ca 117
Cd 0.00209
Cl 16.1
Co 0.00161
Cr 0.00144
Cu 0.002654
F 0.178
Fe 0.457
K 15.1
Li 0.17
Mg 2.92
Mn 0.109
Mo 0.35
Na 18.1
Ni 0.0134
Pb 0.000934
S 333
Sb 0.00453
Se 0.00784
Sr 1.02
V 0.0305
Zn 0.018
-water 1 # kg
MIX 1
1 0
2 1
SELECTED_OUTPUT 1
-file P1.AsAdsorptionProblem.csv
-reset false
-reaction false
-pH true
-totals Co
END
--- End code ---
dlparkhurst:
--- Code: ---TITLE Problem#1: Modeling Competitive Adsorption. Arsenic Adsorption Problem
SOULTION_SPECIES
CrO4-2 + 6H+ + 3e- = Cr(OH)2+ + 2H2O
log_k 100. #67.376
Cu+2 + e- = Cu+
log_k -100. #2.69
END
PHASES
Ferrihydrite
Fe(OH)3 + 3H+ = Fe+3 + 3H2O
log_k 3.191
delta_h -73.374 kJ
Fix_H+
H+ = H+
log_k 0
END
SOLUTION 1 #Backgound water
temp 25
pH 7.04
pe 4 Ferrihydrite 0
redox pe
units mg/kgw
density 1
As 0.000967
B 0.0468
Ba 0.037
C(4) 142
Ca 31.7
Cd 5.41e-005
Cl 15.8
Co 0.000138
Cr 0.0007797
Cu 0.00284
F 0.194
Fe 0.0792
K 2.51
Li 7.59e-3
Mg 8.9
Mn 0.012
Mo 0.000753
Na 25
Ni 0.0007778
Pb 0.000207
S 18.6
Sb 7.47e-005
Se 0.000557
Sr 0.219
V 0.0012
Zn 0.00761
-water 1 # kg
EQUILIBRIUM_PHASES 1
Ferrihydrite 0 0.01
SURFACE 1
-equilibrate with solution 1
Hfo_wOH Ferrihydrite equilibrium_phase 0.2 600
Hfo_sOH Ferrihydrite equilibrium_phase 0.0025
SAVE Solution 1
SAVE Equilibrium_phases 1
SAVE Surface 1
END
USER_GRAPH
10 GRAPH_X -LA("H+")
20 GRAPH_Y TOT("Co")
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -5 Ca(OH)2; USE Surface 1; END
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -6 Ca(OH)2; USE Surface 1; END
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -7 Ca(OH)2; USE Surface 1; END
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -8 Ca(OH)2; USE Surface 1; END
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -9 Ca(OH)2; USE Surface 1; END
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -10 Ca(OH)2; USE Surface 1; END
USE Solution 1; Equilibrium_phases 1; Ferrihydrite 0 0.1; Fix_H+ -11 Ca(OH)2; USE Surface 1; END
END
--- End code ---
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