Problem with pH range vs USER PUNCH

[lrossi]

Hello,

Following experiments on Zn co-precipitation with Mg carbonates, we are attempting to model Zn speciation in solution. (Experimental conditions: NaCl, NaHCO3 and ZnCl2 + addition of NaOH to vary pH). This varies from 8.6 to 11.5 in the laboratory.
Here's an example of a script concerning our control solutions (i.e. without Mg carbonate). However, we are unable to obtain results over a pH range wider than 9.5 to 10.5.

It's true that we're not completely at equilibrium in our experiments. But is it still possible, under our experimental conditions, to obtain PhreeqC results over a wider pH range (ideally between pH 7 and 12)?
In fact, with this script, modeling stops at pH 10.5.
Do you see a problem in the script or in writing the database (which we've updated with the literature)?
In view of the literature at 25?C, we have not considered Zn-Cl complexes.


Thank you in advance for your time and help.
Best regards
L.R

Code: [Select]

SOLUTION_SPECIES
H2O + Mg+2 = MgOH+ + H+
    log_k     -11.6825
    delta_h   70.1 kJ
    -analytical_expression 0.598231 1.331513e-007 -3660.892581 -0.0003664608 4.047512 0
CO2 + H2O = HCO3- + H+
    log_k     -6.349
    delta_h   9.06 kJ
    -analytical_expression -528.756415 -0.090452 30072.651897 190.677125 -2071812.21964 0
HCO3- = CO3-2 + H+
    log_k     -10.337
    delta_h   14640 kJ
    -analytical_expression 63.649684 0.004110834 -2377.370157 -26.211854 -211481.598893 0
HCO3- + Mg+2 = MgHCO3+
    log_k     1.09
    delta_h   0.6 kJ
    -analytical_expression -8.8935 0.01694 1474.786 0 0 0
CO3-2 + Mg+2 = MgCO3
    log_k     3.01
    delta_h   13.8 kJ
    -analytical_expression 5.5093 -0.000171429 -734.208 0 0 0

Ca+2 + HCO3- = CaHCO3+
    log_k     1.047
    delta_h   1.457 kJ
    -analytical_expression 491.854704 0.08534 -28199.037707 -178.910624 1869510.082793 0
CO3-2 + Ca+2 = CaCO3
    log_k     3.327
    delta_h   15.889 kJ
    -analytical_expression 861.719643 0.148548 -45945.622198 -314.459569 2625013.107826 0




#### Modifi? avec les donn?es de Pascale et al., 2002 ####

1.0000 H2O + 1.0000 Zn++  =  Zn(OH)+ +1.0000 H+
 
        log_k           -8.766
-delta_H 0      
# Enthalpy of formation: -0 kcal/mol


2.0000 H2O + 1.0000 Zn++  =  Zn(OH)2 +2.0000 H+
       
        log_k           -19.024
-delta_H 0

     
3.0000 H2O + 1.0000 Zn++  =  Zn(OH)3- +3.0000 H+
       
        log_k           -28.121
-delta_H 0      
# Enthalpy of formation: -0 kcal/mol


 1.0000 Zn++ + 1.0000 CO3-2  =  ZnCO3 #### Powell et al., 2013 ####
       
        log_k           -6.42
-delta_H 0      
# Enthalpy of formation: -0 kcal/mol

1.0000 Zn++ + HCO3- =  ZnHCO3+ #### Powell et al., 2013 ####
       
        log_k           1.64
-delta_H 0      
# Enthalpy of formation: -0 kcal/mol







### ZnCl Modifi? avec Powell et al., 2013 (IUPAC) ### NON CONSIDEREES DANS LE MODELE.

#1.0000 Zn++ + 1.0000 Cl-  =  ZnCl+
       
   #     log_k           0.4
# -delta_H 0      
# Enthalpy of formation: -0 kcal/mol

#1.0000 Zn++ + 2.0000 Cl-  =  ZnCl2
       
      #  log_k           0.69
#-delta_H 0      
# Enthalpy of formation: -0 kcal/mol

#1.0000 Zn++ + 3.0000 Cl-  =  ZnCl3-
       
    #    log_k           0.48
#-delta_H 0      
# Enthalpy of formation: -0 kcal/mol

#1.0000 Zn++ + 4.0000 Cl-  =  ZnCl4-2 #### Ruaya and Seward 1986 ####
       
    #    log_k           -0.19
#-delta_H 0      
# Enthalpy of formation: -0 kcal/mol

#1.0000 Zn++ + 1.0000 Cl- + 1.0000 OH- =  Zn(OH)Cl #### Wagman et al, 1982 ####
         
   #     log_k           6.51
#-delta_H 0      
# Enthalpy of formation: -0 kcal/mol

#########








PHASES
Fix_H+
    H+ = H+
    log_k     0
Hydromagnesite
    Mg5(CO3)4(OH)2:4H2O + 6H+ = 6H2O + 4HCO3- + 5Mg+2
    log_k     32.29
    delta_h   -274.581 kJ
    -analytical_expression -151.4683 0.1118532 59031 3.805 -5066679 0
Magnesite
    MgCO3 = CO3-2 + Mg+2
    log_k     -7.798
    delta_h   -44.4968 kJ
    -analytical_expression -39.6997 -0.0413957 918.46 17.1936 -122783 0
Brucite
    Mg(OH)2 + 2H+ = 2H2O + Mg+2
    log_k     17.12
Nesquehonite
    MgCO3:3H2O = CO3-2 + 3H2O + Mg+2
    log_k     -5.27
    delta_h   -14.4 kJ
Dypingite
    Mg5(CO3)4(OH)2:8H2O = 4CO3-2 + 8H2O + 5Mg+2 + 2OH-
    log_k     -34.98
NaCl
    NaCl = Cl- + Na+
    log_k     -20
Zincite
    ZnO + 2H+ = H2O + Zn+2
    log_k     11.17
    -analytical_expression -4.0168 0 4527.66 0 0 0
Smithsonite
ZnCO3 = Zn+2 + CO3-2
log_k   -10.93
Hydrozincite
Zn5(OH)6(CO3)2 + 6H+ = 5Zn+2 + 2CO3-2 + 6H2O
log_k   9.65
Zn(OH)2(beta1)
       Zn(OH)2 +2.0000 H+  =  + 1.0000 Zn++ + 2.0000 H2O
        log_k           11.72
Zn(OH)2(beta2)
       Zn(OH)2 +2.0000 H+  =  + 1.0000 Zn++ + 2.0000 H2O
        log_k           11.76
Zn(OH)2(epsilon)
       Zn(OH)2 +2.0000 H+  =  + 1.0000 Zn++ + 2.0000 H2O
        log_k           11.38
Zn(OH)2(gamma)
       Zn(OH)2 +2.0000 H+  =  + 1.0000 Zn++ + 2.0000 H2O
        log_k           11.7     
Zn(OH)2(delta)
       Zn(OH)2 +2.0000 H+  =  + 1.0000 Zn++ + 2.0000 H2O
        log_k           11.81  




#La concentration fix?e en Zn est de 1ppm. Celle en chlore correspond ? la quantit? ajout?e de ZnCl2 pour Zn = 1ppm.

SOLUTION 1
    temp      25
    pH        7 charge
    pe        4
    redox     pe
    units     mol/l
    density   1
    Cl        0.0750306
    Na        0.125
    Zn        1.52952E-05
    C         0.05
  -water    1# kg


SAVE SOLUTION 1
END


USE SOLUTION 1
EQUILIBRIUM_PHASES 1
CO2(g) -3.5
O2(g) 1


REACTION 1
NaOH 10
1 moles in 1000 steps



USER_PUNCH 1
    -headings
    -start
10 FOR pH = 7 to 12 step 0.25
20 PUNCH "USE solution 1", EOL$
30 PUNCH "USE EQUILIBRIUM_PHASES 1", EOL$
40 PUNCH "NaCl 0 NaOH 10", EOL$
50 PUNCH "Fix_H+ ", -pH, "NaOH 10", EOL$
60 PUNCH "END", EOL$
70 NEXT pH
    -end

SELECTED_OUTPUT 1
    -file                 selected_output_1.sel

USER_GRAPH
    -headings               pH Zn+2 ZnOH+ Zn(OH)2 Zn(OH)3- ZnCO3 ZnHCO3+   
    -axis_titles            "pH" "Zn mol/L"
    -axis_scale x_axis      0 12 auto auto
    -axis_scale y_axis      auto
    -initial_solutions      false
    -connect_simulations    true
    -plot_concentration_vs  x
  -start

10 GRAPH_X -LA("H+")
60 GRAPH_Y MOL("Zn+2")
40 GRAPH_Y MOL("ZnOH+")
20 GRAPH_Y MOL("Zn(OH)2")
50 GRAPH_Y MOL("Zn(OH)3-")
130 GRAPH_Y MOL("ZnCO3")
140 GRAPH_Y MOL("ZnHCO3+")

  -end
END



[dlparkhurst]

I didn't check the script carefully, but I think the problem is trying to maintain atmospheric pCO2 above pH 10.5. A pH of 11 requires about 20 moles of NaOH and 10 moles of carbon (CO3-2) per kilogram of water to produce atmospheric partial pressure of CO2. That is more than the solubility of NaOH and not a feasible solution composition.

If your experiment is done in the open atmosphere, at pH greater than 10.5 I doubt you achieve equilibrium with CO2 gas. If you allow your solution to stand and continue to react, I suspect the pH would be falling due to CO2(g) slowly reacting CO2(g) + 2OH- = CO3-2 + H2O.

The following shows the effect:

Code: [Select]

SOLUTION
pH 11
Na 1 charge
END
USE solution 1
EQUILIBRIUM_PHASES
CO2(g) -3.4 10
END

[lrossi]

Hello Sir,
I understand. The solution was in contact with the atmosphere during our experiments and certainly not at equilibrium. The idea is to calculate the Zn speciation of our experimental solutions. Isn't there any other way of calculating it at our experimental pH (i.e. not at equilibrium)? Maybe by fixing the pH?

Thank you in advance,
Sincerely
L.R

[dlparkhurst]

You can fix the pH, but you cannot also have equilibrium with atmospheric CO2(g) for pH above 10.

Another way to look at is the amount of C(4) in the system is unknown.