negative load of organic matter in mEq/l

[m.gamrani]

how to introduce a negative load of organic matter in mEq/l (milliequivalent per litre)

[dlparkhurst]

Do you want to add organic matter with a negative charge? or remove organic matter?

A negatively charged organic matter depends on the definitions of SOLUTION_MASTER_SPECIES and SOLUTION_SPECIES. Depending on the definitions, when you add organic matter to a solution, it will exist in a negatively charged state.

If you want to remove organic matter from a chemical system, use REACTION or KINETICS.

[m.gamrani]

I want to add negatively charged organic matter since I have negative dissolved organic carbon COD-
I added this organic matter as humic acid in

SOLUTION_MASTER_SPECIES
Ha            HaOH             0     Ha              2000

SOLUTION_SPECIES
HaOH = HaOH
log_k 0

HaOH = HaO- + H+
log_k -4

but the problem is that the concentration is in mEq/l is the solution I use it is in mmol/l
SOLUTION 2
Mmol/L units
pH 9.64
C(4) 67.6
F 0.42
Cl 66.6
Br 0.11
S(6) 0.48
Ca 0.81
Mg 0.16
Na 140.6
K 24.2
If 107.9
Al 0.154
Fe 0.166
-water 1 # kg
Ha 30 is the design in mEq/l of COD-

here’s my script

Code: [Select]

TITLE REVERSE MODELLING l'obtention de l'eau saline basique à partir d'une solution de rivière,"évaporation"


SOLUTION_MASTER_SPECIES
Si SiO2 0 SiO2 28.0843


Ha            HaOH             0     Ha              2000






SOLUTION_SPECIES
SiO2 + 2H2O = H3SiO4- + H+
      -log_k  -9.83; -delta_h 6.12 kcal
      -analytic   -302.3724   -0.050698   15669.69    108.18466   -1119669.0
      -Vm  7.94  1.0881  5.3224  -2.8240  1.4767 # supcrt + H2O in a1


SiO2 + 2H2O  = H2SiO4-2 + 2 H+
      -log_k  -23.0;  -delta_h 17.6 kcal
      -analytic   -294.0184   -0.072650   11204.49    108.18466   -1119669.0


H3SiO4- = HSiO3- + H2O
     -log_k -50




HaOH = HaOH
log_k 0


HaOH = HaO- + H+
log_k -4


       
Ag+ + HaO- = HaOAg
        log_k   0.73


Na+ + HaO- = HaONa
        log_k   -0.18


Mn+2 + HaO- = HaOMn+
        log_k   1.4


Fe+2 + HaO- = HaOFe+
        log_k   1.4


Fe+3 +  HaO- = HaOFe+2
        log_k   3.21




Cu+2 + HaO- = HaOCu+
        log_k   2.21


Ca+2 + HaO- = HaOCa+
        log_k   1.18


Al+3 +  HaO- = HaOAl+2
        log_k   4.3






PRINT; -reset true; inverse true


SOLUTION 1 #eau de rivière non salée
Units mmol/L
      pH 6.4
      C(4) 0.02
      F 0.01
      Cl 0.026
      Br 0.001
      S(6) 0.039
      Ca 0.065
      Mg 0.026
      Na 0.052
      K 0.058
      Si 0.204
      Al 0.0011
      Fe 0.0005
     -water 1 # kg




EQUILIBRIUM_PHASES
CO2(g) -3.5


SAVE Solution 1
END


SOLUTION 2 #saline
Units mmol/L
      pH 9.64
      C(4) 67.6
      F 0.42
      Cl 66.6
      Br 0.11
      S(6) 0.48
      Ca 0.81
      Mg 0.16
      Na 140.6
      K 24.2
      Si 107.9
      Al 0.154
      Fe 0.166
     -water 1 # kg
      Ha 30



EQUILIBRIUM_PHASES
      CO2(g) -3.5


SAVE Solution 2
END


INVERSE_MODELING 1
-solution 1 2    #on cherche la solution 2 a partir du solution 1
-phases
       SiO2(am)
       Calcite
       Dolomite     #Mg-Calcite
       Nahcolite
       stevensite   #Smectite trioctaédrique
       Fe-Kaolinite
       Goethite
       water


-balances
      Cl 0.5 #50% d'incertitude
      Na 0.5 #50% d'incertitude
      S(6) 1 #100% d'incertitude
      Br 1 #100% d'incertitude
      Alkalinity 1 #100% d'incertitude
      F 1
      K 0.5
      Ha 1


-uncertainties 0.93
-range true
-tolerance 1e-10
-mineral_water     true
#-minimal #modèle minimal


PHASES
water
   H2O = H2O
   log_K 0


Stevensite
Ca0.15Na0.33Mg2.485Fe0.2Si4O10 (OH) 2 + 6H + = \
0.15Ca + 2 + 0.33Na + + 2.485Mg + 2 + 0.2Fe + 2 + 4SiO2 + 4H2O
log_k 25.45


Fe-Kaolinite
Al1.9Fe0.1Si2O5(OH)4 + 6H+ = 1.9Al+3 + 0.1Fe+3 + 2SiO2 + 5H2O
     log_k 6.471


END

[dlparkhurst]

I strongly urge you not to eliminate the output with -reset false. If you look at the output, you find this for the distribution of species of Ha. 30 mmol/L corresponds to 32.5 mmol/kgw in this solution. HaO- is by far the predominant species at pH 9.6. So, there are 31.2 meq/kgw of negatively charged Ha. The percent ionized will decrease as you decrease the pH, so eventually, HaOH will be the predominant species, and you will have much less charged Ha species.

Code: [Select]

Ha            3.251e-02
   HaO-            3.122e-02   2.307e-02    -1.506    -1.637    -0.131     (0) 
   HaONa           1.270e-03   1.344e-03    -2.896    -2.871     0.025     (0) 
   HaOCa+          2.284e-05   1.688e-05    -4.641    -4.773    -0.131     (0) 
   HaOH            4.993e-08   5.284e-08    -7.302    -7.277     0.025     (0) 
   HaOFe+          7.172e-13   5.299e-13   -12.144   -12.276    -0.131     (0) 
   HaOAl+2         3.937e-17   1.173e-17   -16.405   -16.931    -0.526     (0) 
   HaOFe+2         1.101e-19   3.280e-20   -18.958   -19.484    -0.526     (0)