Modeling metal hydroxide solubility as a function of ligand concentration

[Hysu]

Hello,

I am attempting to model Nd(III) solubility as a function of ligand concentration. I am able to run my code, but the output has a constant pthalate concentration of 1E-3 M.

Code: [Select]

PHASES
Fix_pH
H+ = H+
log_k 0

Pht
Phthalat-2 = Phthalat-2
log_k 0

Nd(OH)3(am)
1.000Nd(OH)3  = 1.00Nd+3 + 3.00H2O  -3.00H+
log_k    17.2   #Thereda 09NEC/ALT

# 25C, 0.1M NaCl, 0M Pht

SELECTED_OUTPUT
 -reset false
 -file Nd.loop
USER_PUNCH
 -start
1150   punch eol$ + 'SOLUTION 1'
1155   punch eol$ + 'pH 8'
1156   punch eol$ + 'temp 25'
1157   punch eol$ + 'pe 10'
1158   punch eol$ + 'redox pe'
1160   punch eol$ + 'Na 100' 
1170   punch eol$ + 'Cl 100 charge'
1180   punch eol$ + 'Nd 1e-10' # Nd(OH)3(am) 0 10'

1205   punch eol$ + 'SELECTED_OUTPUT'
1206   punch eol$ + '-file 25C_0atmCO2_0.1MNaCl_X M [Pht]_Nd Solubility_pHm 8_V11a.prn'
1207   punch eol$ + '-pH false'
1208   punch eol$ + '-molalities H+ Nd+3 NdOH+2 Nd(OH)2+ Nd(OH)3 Nd(OH)4- Nd(Phthalat)+ Nd(Phthalat)2- H2Phthalat HPhthalat- Phthalat-2 Na+' 
1300 for i = 0 to 0.05 step 0.000001
1310   punch eol$ + 'USE SOLUTION 1'
1320   punch eol$ + 'EQUILIBRIUM_PHASES 1'
1325   punch eol$ + 'fix_pH -8 NaOH 10'
1330   punch eol$ + 'Pht' + str$(i) + ' Phthalat-2 0.001'
1340   punch eol$ + 'Nd(OH)3(am) 0 10'
1350   punch eol$ + 'END'
1360 next i
If I change line 1330 to

Code: [Select]

1330   punch eol$ + 'Fix_Pht' + str$(i)

then I get a different constant value around 5.

Additionally, even though I am fixing the pH, the output is either 7.89 or 8.08, respectively.

So, where am I going wrong? Thanks for your help in advance.

[dlparkhurst]

You have a long way to go before you can make this calculation.

First, I don't think you need to write the file Nd.loop. There are other ways to adjust the Phthalate concentrations.

Second, you need more thermodynamic data.  Adding one species of Phthalate will have almost no effect on the Nd solubility.  You need data for aqueous species of Phthalate, Nd, and Phthalate-Nd binding. If you add Na2Phthalate as I have done below, you also need to consider Na-Phthalate binding.

I have defined added enough in the script below to show what is needed. The Nd binding is defined to be the same as Al binding from minteq.v4.dat, simply because Al is a trivalent ion; you should not take these numbers as meaningful.

I did not find any database that had both Nd and Phthalate, so you will have to decide how you want to merge data, and you will need to find or fit the Nd-Phthalate binding constants. I tested this script with llnl.dat.

Code: [Select]

SOLUTION_MASTER_SPECIES
Phthalate Phthalate-2 1 164.117 164.117
SOLUTION_SPECIES
Phthalate-2 = Phthalate-2
log_k 0
SOLUTION_SPECIES
# Definitions from minteq.v4.dat
H+ + Phthalate-2 = H(Phthalate)-
log_k 5.408
delta_h 2.1757 kJ
-gamma 0 0
#                  Id: 3309971
#        log K source: NIST46.2
#      Delta H source: NIST46.2
#T and ionic strength:
2 H+ + Phthalate-2 = H2(Phthalate)
log_k 8.358
delta_h 4.8534 kJ
-gamma 0 0
#                  Id: 3309972
#        log K source: NIST46.2
#      Delta H source: NIST46.2
#T and ionic strength:
Na+ + Phthalate-2 = Na(Phthalate)-
log_k 0.8
delta_h 4.184 kJ
-gamma 0 0
#                  Id: 5009970
#        log K source: NIST46.2
#      Delta H source: NIST46.2
#T and ionic strength:
#####################
# Simply replaced Al with Nd in the following definitions
# You must find thermodynamic data for Nd/Phthalate binding!!!!
#####################
Nd+3 + Phthalate-2 = Nd(Phthalate)+
log_k 4.56
delta_h 0 kJ
-gamma 0 0
#                  Id: 309971
#        log K source: NIST46.2
#      Delta H source: NIST46.2
#T and ionic strength:
Nd+3 + 2 Phthalate-2 = Nd(Phthalate)2-
log_k 7.2
delta_h 0 kJ
-gamma 0 0
#                  Id: 309972
#        log K source: NIST46.2
#      Delta H source: NIST46.2
#T and ionic strength:
PHASES
Fix_pH
H+ = H+
log_k 0

#Pht
#PhthNdat-2 = PhthNdat-2
#log_k 0

Nd(OH)3(am)
1.000Nd(OH)3  = 1.00Nd+3 + 3.00H2O  -3.00H+
log_k    17.2   #Thereda 09NEC/NdT

# 25C, 0.1M NaCl, 0M Pht
END
USER_GRAPH
10 GRAPH_X TOT("Phthalate")
20 GRAPH_Y TOT("Nd")
SOLUTION 1
pH 8
temp 25
pe 10
redox pe
Na 100
Cl 100 charge
Nd 1e-10 Nd(OH)3(am) 0 10
END
USE solution 1
EQUILIBRIUM_PHASES 1
fix_pH -8 NaOH 10
Nd(OH)3(am) 0 10
REACTION
Na2Phthalate 1
0.05 in 20 steps
END




[Hysu]

Sorry, I should have mentioned I do have the necessary data (ThermoChimie and analogies with EU(III)) and that I have only included the code that I thought I was having a problem with. I wasn't sure if I could share the data I have since I am working for a company and it might be internal.

I am having trouble attaching the results in picture format, but the code you provided results in a graph that has all data points at 0. It's just that other people will be reviewing the results and I will have to apply this to other systems, so I would like to have a solution that makes sense without having to explain it (sorry for being nitpicky). In the end, your code solves my problem, so can I have the REACTION output put into the SELECTED_OUPUT file I had defined in the code I provided? This would make data analysis easier.

Thanks for your help, again.

[dlparkhurst]

Of course you can add SELECTED_OUTPUT; writing results to file normally makes no difference in the results, so I eliminate it to simplify the scripts.