Questions about HFO titration

[waynehwm]

Hi phreeqcusers,

I am setting up a model to fit my titration data with PEST. My model seems to work, but I have two questions about the model calculation, hopefully, someone can provide me your insight on this.

My phreeqc script is:

Code: [Select]


SURFACE_MASTER_SPECIES
Hfo_s  Hfo_sOH
Hfo_w  Hfo_wOH
 
SURFACE_SPECIES
Hfo_wOH  + H+ = Hfo_wOH2+
log_k  8.5000    # = pKa1,int
 
Hfo_wOH = Hfo_wO- + H+
log_k -9.947 # = -pKa2,int
 
Hfo_sOH  + H+ = Hfo_sOH2+
log_k  8.5000    # = pKa1,int
 
Hfo_sOH = Hfo_sO- + H+
log_k -9.947 # = -pKa2,int
 
SOLUTION 1
    temp      20
    pH        4.116
    redox     pe
    units     mol/l
    density   1
    N(5)      0.1
    Na        0.1
    -water    0.1 # kg
 
SURFACE 1
Hfo_wOH 5.8E-5 2.1 10.8
Hfo_sOH 1.0E-6 2.1 10.8
-equilibrate with solution 1
 
SELECTED_OUTPUT; -file case.prn; -reset false; -high_p true
USER_PUNCH
-headings pH
 -start
 10 PUNCH -LA("H+")
 -end
REACTION 1
    H2O        0.0556
    NaOH       0.0001
    0.012 0.06 0.108 0.156 0.204 0.252 0.3
    0.348 0.396 0.444 0.492 0.54 0.588 0.636
    0.684 0.732 0.78 0.828 0.876 moles
 
1. I meant to set the initial pH of solution 1+ surface 1 after equilibrium to be 4.116, then start the titration. But the calculation sequence of this model seems to do an equilibrium between solution 1 and surface 1 first (pH 4.116 change to 4.13 for example), then add reaction step one. Can I set the equilibrium pH of the system to be 4.116 then start titration?

2. I define a Na 0.1 mole/L in solution 1, why does the solution composition calculation show a 1.004e-01 molality? Where is the extra 0.004e-1 from?

Code: [Select]

Elements           Molality       Moles

N                 1.004e-01   1.004e-02
Na                1.004e-01   1.004e-02
 Thank you for your time!

[dlparkhurst]

You specified the surface to be in equilibrium with solution 1. That necessitates the calculation that you note before the reaction calculation begins.

If you add zero of your reaction, you would see that the solution and surface do not change, if nitrogen (nitrate) was not a redox element. Straight up, you will find that there is a little reduction of nitrate to dissolved nitrogen at equilibrium, which has a small pH effect. I have added definitions effectively to remove the possibility of producing N2(aq), with the result that the pH of the addition of zero reaction is exactly 4.116.

The difference in concentration is the numerical difference between mol/L and mol/kgw.

[waynehwm]

Understood. I will give this a try, thank you David!

[dlparkhurst]

Sorry, meant to add the script.


SURFACE_MASTER_SPECIES
Hfo_s  Hfo_sOH
Hfo_w  Hfo_wOH
 
SURFACE_SPECIES
Hfo_wOH  + H+ = Hfo_wOH2+
log_k  8.5000    # = pKa1,int
 
Hfo_wOH = Hfo_wO- + H+
log_k -9.947 # = -pKa2,int
 
Hfo_sOH  + H+ = Hfo_sOH2+
log_k  8.5000    # = pKa1,int
 
Hfo_sOH = Hfo_sO- + H+
log_k -9.947 # = -pKa2,int
 
SOLUTION_SPECIES
2 NO3- + 12 H+ + 10 e- = N2 + 6 H2O
   -log_k   0 # 207.08
END
SOLUTION 1
    temp      20
    pH        4.116
    redox     pe
    units     mol/l
    density   1
    N(5)      0.1
    Na        0.1
    -water    0.1 # kg
 
SURFACE 1
Hfo_wOH 5.8E-5 2.1 10.8
Hfo_sOH 1.0E-6 2.1 10.8
-equilibrate with solution 1
 
SELECTED_OUTPUT; -file case.prn; -reset false; -high_p true
USER_PUNCH
-headings pH
 -start
 10 PUNCH -LA("H+")
 -end
REACTION 1
    H2O        0.0556
    NaOH       0.0001
    0 0.012   0.06   0.108   0.156   0.204   0.252   0.3   
    0.348   0.396   0.444   0.492   0.54   0.588   0.636   
    0.684   0.732   0.78   0.828   0.876 moles

[waynehwm]

Sorry, meant to add the script.


SURFACE_MASTER_SPECIES
Hfo_s  Hfo_sOH
Hfo_w  Hfo_wOH
 
SURFACE_SPECIES
Hfo_wOH  + H+ = Hfo_wOH2+
log_k  8.5000    # = pKa1,int
 
Hfo_wOH = Hfo_wO- + H+
log_k -9.947 # = -pKa2,int
 
Hfo_sOH  + H+ = Hfo_sOH2+
log_k  8.5000    # = pKa1,int
 
Hfo_sOH = Hfo_sO- + H+
log_k -9.947 # = -pKa2,int
 
SOLUTION_SPECIES
2 NO3- + 12 H+ + 10 e- = N2 + 6 H2O
   -log_k   0 # 207.08
END
SOLUTION 1
    temp      20
    pH        4.116
    redox     pe
    units     mol/l
    density   1
    N(5)      0.1
    Na        0.1
    -water    0.1 # kg
 
SURFACE 1
Hfo_wOH 5.8E-5 2.1 10.8
Hfo_sOH 1.0E-6 2.1 10.8
-equilibrate with solution 1
 
SELECTED_OUTPUT; -file case.prn; -reset false; -high_p true
USER_PUNCH
-headings pH
 -start
 10 PUNCH -LA("H+")
 -end
REACTION 1
    H2O        0.0556
    NaOH       0.0001
    0 0.012   0.06   0.108   0.156   0.204   0.252   0.3   
    0.348   0.396   0.444   0.492   0.54   0.588   0.636   
    0.684   0.732   0.78   0.828   0.876 moles

Just ran it, exactly like what you said. You are amazing David, thank you so much!!