# PhreeqcUsers Discussion Forum

## Conceptual Models => Equilibrium assumptions => Topic started by: Jeonghwan Hwang on September 09, 2021, 03:09:55 AM

Title: How to make the condition of constant pCO2 in solution?
Post by: Jeonghwan Hwang on September 09, 2021, 03:09:55 AM
Hello, this is Jeonghwan Hwang.
Firstly, thank you for reading.

I developed a sorption model that performed under constant pCO2 condition with 10^-3.5 bar.

I wrote the following to put the pCO2 value into the solution.
The value of CO2, -3.5056, is the logarithm taken after converting 10^-3.5 bar to atm.

=========================
SOLUTION 1
-units mol/L
pH 8.0
U(+6) 9E-8
Na 0.1
C(4) 1 CO2(g) -3.5056
Cl(+7) 0.1. charge
=========================

I wonder if this method is correct or if there is an additional setting to fix the pCO2.

For example, when calculating the equilibrium of a solution to fix H+,
I know that a sentence such as "Fix_H+ -9.0 NaOH 10.0" is applied.

My whole PHREEQC code is written as below;
Thank you

Sincery

Jeonghwan Hwang

=========================================================
database c://phreeqc/database/LLNL.dat

Solution_Species

#Aqueous U(VI) hydrolysis reactions
UO2+2 + H2O = UO2(OH)+ + H+
log_K   -5.25

UO2+2 + 2H2O = UO2(OH)2 + 2H+
log_K   -12.15

UO2+2 + 3H2O = UO2(OH)3- + 3H+
log_K   -20.25

UO2+2 + 4H2O = UO2(OH)4-2 + 4H+
log_K   -32.40

#Aqueous U(VI) carbonate reactions
UO2+2 + CO3-2 = UO2(CO3)
log_K   9.94

UO2+2 + 2CO3-2 = UO2(CO3)2-2
log_K   16.61

UO2+2 + 3CO3-2 = UO2(CO3)3-4
log_K   21.84

2 UO2+2 + CO3-2 + 3 H2O = (UO2)2(CO3)(OH)3-1 + 3H+
log_K   -0.86

3 UO2+2 + CO3-2 + 3 H2O = (UO2)3(CO3)(OH)3+1 + 3H+
log_K   0.66

3 UO2+2 + 6 CO3-2 = (UO2)3(CO3)6-6
log_K   54.0

SURFACE_MASTER_SPECIES
Bentonite_s     Bentonite_sOH #s
Bentonitea_w     Bentonitea_wOH #w1
Bentoniteb_w   Bentoniteb_wOH #w2

SURFACE_SPECIES
Bentonite_sOH = Bentonite_sOH
log_k  0.0

Bentonitea_wOH = Bentonitea_wOH
log_k  0.0

Bentoniteb_wOH = Bentoniteb_wOH
log_k  0.0

#bentonite properties with tirtation curve
Bentonite_sOH  + H+ = Bentonite_sOH2+
log_k  4.5

Bentonite_sOH = Bentonite_sO- + H+
log_k  -7.9

Bentonitea_wOH  + H+ = Bentonitea_wOH2+
log_k  4.5

Bentonitea_wOH = Bentonitea_wO- + H+
log_k  -7.9

Bentoniteb_wOH  + H+ = Bentoniteb_wOH2+
log_k  6.0

Bentoniteb_wOH = Bentoniteb_wO- + H+
log_k  -10.5

#Strong site
Bentonite_sOH + UO2+2 = Bentonite_sOUO2+ + H+
log_k  3.1

Bentonite_sOH + UO2+2 + H2O = Bentonite_sOUO2OH + 2H+
log_k  -4.6

Bentonite_sOH + UO2+2 + 2H2O = Bentonite_sOUO2(OH)2- + 3H+
log_k  -12.6

Bentonite_sOH + UO2+2 + 3H2O = Bentonite_sOUO2(OH)3-2 + 4H+
log_k  -20.9

#Weak site
Bentonitea_wOH  + UO2+2 = Bentonitea_wOUO2+ + H+
log_k  0.5

Bentonitea_wOH  + UO2+2 + H2O = Bentonitea_wOUO2OH + 2H+
log_k  -5.7

EXCHANGE_MASTER_SPECIES
Bentonite_ex    Bentonite_ex-

EXCHANGE_SPECIES

Bentonite_ex- = Bentonite_ex-
log_K 0

Na+ + Bentonite_ex- = NaBentonite_ex
log_K 0

2 NaBentonite_ex + UO2+2 = UO2(Bentonite_ex)2 + 2Na+
log_K 0.45

SURFACE 1
-sites_units absolute
Bentonite_sOH           5.00E-06
Bentonitea_wOH        1.00E-04
Bentoniteb_wOH        1.00E-04
-no_edl

Exchange 1
NaBentonite_ex         0.002175

SOLUTION 1
-units  mol/L
pH      8.0
U(+6)   9E-8
Na      0.1
C(4) 1 CO2(g) -3.5056
Cl(+7)    0.1.    charge

PHASES
Fix_H+
H+ = H+
log_k  0.0
END

SELECTED_OUTPUT
-file Fig2a.txt
-reset              false
-pH                 true
-molalities      Bentonite_sOUO2+  Bentonite_sOUO2OH  Bentonite_sOUO2(OH)2-  Bentonite_sOUO2(OH)3-2  Bentonitea_wOUO2+  Bentonitea_wOUO2OH  UO2(Bentonite_ex)2
-totals          Bentonite_s Bentonitea_w Bentoniteb_w Bentonite_ex U(3) U(4) U(5) U(6)

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -2.0   NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -2.25  NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -2.5  NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -2.75  NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -3.0   NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -3.25   NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -3.5  NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -3.75  NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -4.0   NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -4.5  NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -5.0   NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -5.5  NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -6.0   NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -6.5  NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -7.0   NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -7.5  NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -8.0   NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -8.5   NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -9.0   NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -9.5   NaOH    10.0
END

USE solution 1
USE surface 1
USE Exchange 1
EQUILIBRIUM_PHASES 1
Fix_H+   -10.0   NaOH    10.0
END
Title: Re: How to make the condition of constant pCO2 in solution?
Post by: dlparkhurst on September 09, 2021, 03:29:24 AM
The following will cause the reacted solution to maintain equilibrium with atmospheric CO2(g). Note that at high pH, the total carbon in equilibrium with atmospheric CO2 will be quite large and ultimately beyond the limits of a phreeqc calculation.

Code: [Select]
`USE solution 1USE surface 1USE Exchange 1EQUILIBRIUM_PHASES 1        Fix_H+   -2.0   NaOH    10.0        CO2(g)   -3.4   10END`