K-struvite precipitation

[Cami]

Hi everyone,
can please someone help me to solve this problem?  I'd like to study the precipitation of others phases beyond k-struvite by looking at the SI.

I tried to add the phases not present in the database in the phases block, but these errors still appear 
"Elements in species have not been tabulated MgHPO4:3H2O"
"Reaction for species has not been defined MgHPO4:3H2O"
even if I added MgHPO4:3H2O in the master species block

The same error occurred If I add k_Struvite in the input file...(MgKPO4:6H2O)
How should I modify my input data? Do I need to add dehydration reactions?

Thank you a lot. I would appreciate any help or suggestions you may have.

SOLUTION 1 #pure solution of P,K and Mg
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mmol/kgw
    density   1
    P         130
    K         130
    Mg        130
    -water    1 # kg


SOLUTION_SPECIES

HPO4-2 + Mg+2 = MgHPO4
    log_k     2.87
    delta_h   3.3 kcal
HPO4-2 + K+ = KHPO4-
    log_k     0.29
HPO4-2 + Na+ = NaHPO4-
    log_k     0.29
MgHPO4:3H2O = Mg+2 + H+ + 3H2O + PO4-3
    log_k     -18.75

PHASES
pH_fix
    H+ = H+
    log_k     0

K_struvite
    Mg+2 + 6H2O + K+ + PO4-3 = MgKPO4:6H2O
    log_k     -10.62
Newberyite
    Mg+2 + 3H2O + HPO4-2 = MgHPO4:3H2O
    log_k     -5.88
Bobierrite
    Mg3(PO4)2:8H2O = 8H2O + 3Mg+2 + 2PO4-3
    log_k     -25.47
Cattiite
    Mg3(PO4)2:22H2O + 2H+ = 22H2O + 2HPO4-2 + 3Mg+2
    log_k     -23.1
Brucite
    Mg(OH)2 = Mg+2 + 2OH-
    log_k     -11.16
Na_struvite
    Na+ + HPO4-2 = NaHPO4-
    log_k     0.29
MgHPO4
    MgHPO4:3H2O = H+ + 3H2O + Mg+2 + PO4-3
    log_k     -18.75
Phosphorrosslerite
    MgHPO4:7H2O=Mg+2+H++PO4-3+7H2O
    log_k     -17.01

EQUILIBRIUM_PHASES 1
    Bobierrite 0 0
    Brucite   0 0
    Cattiite  0 0
    MgHPO4    0 0
    Na_struvite 0 0
    Phosphorrosslerite 0 0
    pH_fix    -5 NaOH      10
        -force_equality
    K_struvite 0 0

END

[dlparkhurst]

PHASES are defined with dissolution reactions; the phase's formula must be the first item on the left-hand-side of the equation.

I have rewritten Newberyite and reversed the log K, but check the equation and log K. Note also, that Newberyite is not present in EQUILIBRIUM_PHASES.

You need a dissociation reaction and log K for Na_struvite. The current definition is for an aqueous species.

Code: [Select]

SOLUTION 1 #pure solution of P,K and Mg
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mmol/kgw
    density   1
    P         130
    K         130
    Mg        130
    -water    1 # kg

SOLUTION_SPECIES
HPO4-2 + Mg+2 = MgHPO4
    log_k     2.87
    delta_h   3.3 kcal
HPO4-2 + K+ = KHPO4-
    log_k     0.29
HPO4-2 + Na+ = NaHPO4-
    log_k     0.29

PHASES
pH_fix
    H+ = H+
    log_k     0

Newberyite
#    Mg+2 + 3H2O + HPO4-2 = MgHPO4:3H2O
#    log_k     -5.88
    MgHPO4:3H2O = Mg+2 + 3H2O + HPO4-2 
    log_k     5.88
Bobierrite
    Mg3(PO4)2:8H2O = 8H2O + 3Mg+2 + 2PO4-3
    log_k     -25.47
Cattiite
    Mg3(PO4)2:22H2O + 2H+ = 22H2O + 2HPO4-2 + 3Mg+2
    log_k     -23.1
Brucite
    Mg(OH)2 = Mg+2 + 2OH-
    log_k     -11.16
#Na_struvite
#    Na+ + HPO4-2 = NaHPO4-
#    log_k     0.29
Phosphorrosslerite
    MgHPO4:7H2O=Mg+2+H++PO4-3+7H2O
    log_k     -17.01

EQUILIBRIUM_PHASES 1
    Bobierrite 0 0
    Brucite   0 0
    Cattiite  0 0
#    Na_struvite 0 0
    Phosphorrosslerite 0 0
    pH_fix    -5 NaOH      100
        -force_equality
   
END


[Cami]

Thank you very much dlparkhurst for your kind reply and suggestions. I increased the pH till 10 and checked the SI of different phases, but no phases have a SI > 0, even if I expected to have precipitation (because of Mg:P:K molar ratio equal to 1:1:1). I varied the element concentration in solution 1 but nothing changed. Do you know how to manage it?

Thanks in advance.

SOLUTION 1 #pure solution of P,K and Mg
    temp      25
    pH        4.5
    pe        4
    redox     pe
    units     mmol/kgw
    density   1
    P         130
    K         130
    Mg        130
    -water    1 # kg

SOLUTION_SPECIES
HPO4-2 + Mg+2 = MgHPO4
    log_k     2.87
    delta_h   3.3 kcal
HPO4-2 + K+ = KHPO4-
    log_k     0.29
HPO4-2 + Na+ = NaHPO4-
    log_k     0.29

PHASES
pH_fix
    H+ = H+
    log_k     0
Newberyite
    MgHPO4:3H2O = 3H2O + HPO4-2 + Mg+2
    log_k     -5.88
Bobierrite
    Mg3(PO4)2:8H2O = 8H2O + 3Mg+2 + 2PO4-3
    log_k     -25.47
Cattiite
    Mg3(PO4)2:22H2O + 2H+ = 22H2O + 2HPO4-2 + 3Mg+2
    log_k     -23.1
Brucite
    Mg(OH)2 = Mg+2 + 2OH-
    log_k     -11.16
Phosphorrosslerite
    MgHPO4:7H2O = H+ + 7H2O + Mg+2 + PO4-3
    log_k     -17.01
K_struvite
    MgKPO4:6H2O = 6H2O + K+ + Mg+2 + PO4-3
    log_k     -10.62
Na_struvite
    MgNaPO4:7H2O= Mg+2+Na++PO4-3+7H2O
    log_k     -11.6


EQUILIBRIUM_PHASES 1
    Bobierrite 0 0
    Brucite   0 0
    Cattiite  0 0
    K_struvite 0 0
    Newberyite 0 0
    Phosphorrosslerite 0 0
    pH_fix    -10 NaOH      10
        -force_equality
    Na_struvite 0 0
   
END

[dlparkhurst]

The results depend critically on the log Ks for the reactions in the PHASES definition.

Your script produces Cattiite as the most stable phase given the log Ks that you define. All other phases are more soluble and have negative saturation indices after equilibration with Cattiite. If you comment Cattiite  from EQUILIBRIUM_PHASES (place # at the beginning of the line), then Bobierrite and Na_struvite  precipitate.

So, the minerals that precipitate depend on the log Ks, and, in the real world, other factors like kinetics, inhibition, and crystallization processes.     

[Cami]

Thanks a lot for your help :)