How to model simple solution in equil. w/ atmospheric CO2, N2, O2?

Conceptual Models > Equilibrium assumptions

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jheadley:
All I'm trying to do is just establish a baseline to make sure I'm doing things right.

10 mM of KCl should have a conductivity of around 1400 (that I can get), and it should have a pH of around 5.6. That I can also get, if I only put CO2(g) in the equilibrium phase calc.

But as soon as I try to add O2 and N2, my pH goes down to 1, and my pe goes up to 19 or so. It looks like it's saying that the O2 and N2 should react to form about 100 mM of NO3-.

How do I just model a simple beaker of KCl, with all its dissolved gasses?

Ultimately, I want to input actual samples from a surface stream, measured with field measurements and ICP/MS and IC. But I don't understand why things like the atmosphere are seemingly ignored in the provided examples. I want to use my dissolved oxygen measurements (and maybe NO3- concentrations?) to get a semi-quantitative answer to the oxidizing/reducing conditions present in the stream.

--- Code: --- EQUILIBRIUM_PHASES 1
CO2(g) -3.4 10
O2(g) -0.68 10
N2(g) -0.11 10
SOLUTION 1
temp 25
pH 7
pe 4
redox pe
units mmol/l
density 1
Cl 10
K 10
water 1 # kg
END
-------------------------------------------
Beginning of initial solution calculations.
-------------------------------------------

Initial solution 1.

-----------------------------Solution composition------------------------------

Elements Molality Moles

Cl 1.001e-02 1.001e-02
K 1.001e-02 1.001e-02

----------------------------Description of solution----------------------------

pH = 7.000
pe = 4.000
Specific Conductance (?S/cm, 25?C) = 1396
Density (g/cm?) = 0.99752
Volume (L) = 1.00324
Activity of water = 1.000
Ionic strength (mol/kgw) = 1.001e-02
Mass of water (kg) = 1.000e+00
Total alkalinity (eq/kg) = 2.934e-09
Temperature (?C) = 25.00
Electrical balance (eq) = -2.933e-09
Percent error, 100*(Cat-|An|)/(Cat+|An|) = -0.00
Iterations = 3
Total H = 1.110124e+02
Total O = 5.550622e+01

----------------------------Distribution of species----------------------------

Log Log Log mole V
Species Molality Activity Molality Activity Gamma cm?/mol

OH- 1.124e-07 1.012e-07 -6.949 -6.995 -0.046 -4.03
H+ 1.095e-07 1.000e-07 -6.961 -7.000 -0.039 0.00
H2O 5.551e+01 9.997e-01 1.744 -0.000 0.000 18.07
Cl 1.001e-02
Cl- 1.001e-02 9.014e-03 -2.000 -2.045 -0.045 18.14
H(0) 1.413e-25
H2 7.063e-26 7.079e-26 -25.151 -25.150 0.001 28.61
K 1.001e-02
K+ 1.001e-02 9.010e-03 -2.000 -2.045 -0.046 9.07
O(0) 0.000e+00
O2 0.000e+00 0.000e+00 -42.081 -42.080 0.001 30.40

------------------------------Saturation indices-------------------------------

Phase SI** log IAP log K(298 K, 1 atm)

H2(g) -22.05 -25.15 -3.10 H2
H2O(g) -1.50 -0.00 1.50 H2O
O2(g) -39.19 -42.08 -2.89 O2
Sylvite -4.99 -4.09 0.90 KCl

**For a gas, SI = log10(fugacity). Fugacity = pressure * phi / 1 atm.
For ideal gases, phi = 1.

-----------------------------------------
Beginning of batch-reaction calculations.
-----------------------------------------

Reaction step 1.

Using solution 1.
Using pure phase assemblage 1.

-------------------------------Phase assemblage--------------------------------

Moles in assemblage
Phase SI log IAP log K(T, P) Initial Final Delta

CO2(g) -3.40 -4.87 -1.47 1.000e+01 1.000e+01 -1.313e-05
N2(g) -0.11 -3.29 -3.18 1.000e+01 9.938e+00 -6.171e-02
O2(g) -0.68 -3.57 -2.89 1.000e+01 9.847e+00 -1.533e-01

-----------------------------Solution composition------------------------------

Elements Molality Moles

C 1.314e-05 1.313e-05
Cl 1.002e-02 1.001e-02
K 1.002e-02 1.001e-02
N 1.236e-01 1.234e-01

----------------------------Description of solution----------------------------

pH = 1.001 Charge balance
pe = 19.627 Adjusted to redox equilibrium
Specific Conductance (?S/cm, 25?C) = 47432
Density (g/cm?) = 1.00156
Volume (L) = 1.00581
Activity of water = 0.995
Ionic strength (mol/kgw) = 1.326e-01
Mass of water (kg) = 9.989e-01
Total alkalinity (eq/kg) = -1.225e-01
Total CO2 (mol/kg) = 1.314e-05
Temperature (?C) = 25.00
Electrical balance (eq) = -2.933e-09
Percent error, 100*(Cat-|An|)/(Cat+|An|) = -0.00
Iterations = 19
Total H = 1.110124e+02
Total O = 5.581280e+01

----------------------------Distribution of species----------------------------

Log Log Log mole V
Species Molality Activity Molality Activity Gamma cm?/mol

H+ 1.225e-01 9.974e-02 -0.912 -1.001 -0.089 0.00
OH- 1.366e-13 1.010e-13 -12.865 -12.996 -0.131 -3.65
H2O 5.551e+01 9.955e-01 1.744 -0.002 0.000 18.07
C(-4) 0.000e+00
CH4 0.000e+00 0.000e+00 -145.510 -145.497 0.013 35.46
C(4) 1.314e-05
CO2 1.314e-05 1.355e-05 -4.881 -4.868 0.013 34.43
HCO3- 7.797e-11 6.013e-11 -10.108 -10.221 -0.113 25.30
(CO2)2 3.267e-12 3.368e-12 -11.486 -11.473 0.013 68.87
CO3-2 7.993e-20 2.827e-20 -19.097 -19.549 -0.451 -3.96
Cl 1.002e-02
Cl- 1.002e-02 7.474e-03 -1.999 -2.126 -0.127 18.38
H(0) 0.000e+00
H2 0.000e+00 0.000e+00 -44.419 -44.406 0.013 28.61
K 1.002e-02
K+ 1.002e-02 7.445e-03 -1.999 -2.128 -0.129 9.28
N(-3) 0.000e+00
NH4+ 0.000e+00 0.000e+00 -48.848 -48.991 -0.143 18.20
NH3 0.000e+00 0.000e+00 -57.247 -57.234 0.013 24.42
N(0) 1.004e-03
N2 5.022e-04 5.178e-04 -3.299 -3.286 0.013 29.29
N(3) 2.538e-14
NO2- 2.538e-14 1.853e-14 -13.595 -13.732 -0.137 25.28
N(5) 1.225e-01
NO3- 1.225e-01 8.946e-02 -0.912 -1.048 -0.137 29.81
O(0) 5.192e-04
O2 2.596e-04 2.676e-04 -3.586 -3.572 0.013 30.40

------------------------------Saturation indices-------------------------------

Phase SI** log IAP log K(298 K, 1 atm)

CH4(g) -142.70 -145.50 -2.80 CH4
CO2(g) -3.40 -4.87 -1.47 CO2 Pressure 0.0 atm, phi 1.000
H2(g) -41.30 -44.41 -3.10 H2
H2O(g) -1.50 -0.00 1.50 H2O
N2(g) -0.11 -3.29 -3.18 N2 Pressure 0.8 atm, phi 1.000
NH3(g) -59.03 -57.23 1.80 NH3
O2(g) -0.68 -3.57 -2.89 O2 Pressure 0.2 atm, phi 1.000
Sylvite -5.15 -4.25 0.90 KCl

**For a gas, SI = log10(fugacity). Fugacity = pressure * phi / 1 atm.
For ideal gases, phi = 1.

------------------
End of simulation.
------------------

--- End code ---

dlparkhurst:
You have diagnosed the problem correctly. In the SOLUTION calculation, PHREEQC allows redox disequilibrium, such that O2(aq) and N2(aq) can happily coexist. However, in reaction calculations--when SOLUTION is equilibrated with EQUILIBRIUM_PHASES, redox equilibrium is calculated. The O2 and N2 react to form NO3- and low pH.

Unless you are considering the formation of gas bubbles, you can assume N2(aq) is inert and its dissolved concentration relatively small. That is why it is not considered in most examples.

If you are considering the formation of gas bubbles, then N2 will distribute between solution and gas, and may be an important component leading to bubble formation. In that case, phreeqc.dat, pitzer.dat, and Amm.dat have an "element" named Ntg that does not react with the nitrogen system. It is effectively an inert species that only has a dissolved form, Ntg, and a gas (PHASES) form, Ntg(g). If you do want it to react, you can use REACTION or KINETICS to simulate the reaction of Ntg to N species.

--- Code: --- EQUILIBRIUM_PHASES 1
CO2(g) -3.4 10
O2(g) -0.68 10
Ntg(g) -0.11 10
SOLUTION 1
temp 25
pH 7
pe 4
redox pe
units mmol/l
density 1
Cl 10
K 10
water 1 # kg
END
--- End code ---

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