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Author Topic: The amount of mineral precipitation  (Read 2237 times)

Jay Chou

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  • Posts: 12
The amount of mineral precipitation
« on: 26/03/22 07:15 »
hello everyone
I'm working on mineral evaporation.
I want to ask you a question. I simulate evaporation in the following way. How to calculate the molar concentration of precipitated minerals?

SOLUTION_SPREAD
    -temp     5
    -pH       9.73
    -units    mg/l
 Description    Number    Temp      pH       Na       K       Mg       Ca       Cl     S(6)    Alkalinity        Li         B
       DLZBHS         1       5    8.73    61.72    8.87    16.91    42.30    66.07    43.20        240.87    1.9708    3.6511
EQUILIBRIUM_PHASES 1
    Aragonite 0 0 precipitate_only
    Calcite   0 0 precipitate_only
    Dolomite  0 0 precipitate_only
    Magnesite 0 0 precipitate_only
    Mirabilite 0 0 precipitate_only
    Pirssonite 0 0 precipitate_only
    Halite    0 0 precipitate_only
    Glaserite 0 0 precipitate_only
     CO2(g)    -3.5 100
REACTION 1
    H2O(g)     -1
    53.8447  54.3998  54.9549  55.06592  55.23245  55.34347  55.39898 
    55.426735  55.4311758  55.448939  55.45449  55.482245  55.498898  55.504449 moles
END

For example, I want to know  molar concentration of precipitated such as Halt and calcite precipitated.

Thank you for your guidance!
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dlparkhurst

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Re: The amount of mineral precipitation
« Reply #1 on: 26/03/22 15:31 »
https://phreeqcusers.org/index.php/topic,1890.msg6657.html#msg6657

This response to one of your posts uses the Basic function EQUI to retrieve the amounts of Calcite, dolomite, and gypsum that have precipitated. A related function is EQUI_DELTA. Look at the section on the Basic Interpreter in the manual.

If you really want concentration, rather than moles of precipitate, the functions SOLN_VOL and TOT("water") would be relevant to calculate mol/L or mol/kgw.

Some other comments: You need to be careful with charge balance. The charge imbalance of the initial solution will increase proportionally to the concentration factor, so you probably should charge balance on one of the analytes. You do not allow for precipitation of B or Li, so their concentrations may get very large. You also may have problems with convergence for very large concentration factors.

This thread, https://phreeqcusers.org/index.php/topic,1914.msg6721.html#msg6721, discusses several of these issues and presents a KINETICS option to try to reach large concentration factors.
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Jay Chou

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  • Posts: 12
Re: The amount of mineral precipitation
« Reply #2 on: 28/03/22 09:42 »
Hello, I want to use phases to define a new mineral phase: sodium borolyte(NaCa[B5O6(OH)]·5H2O)

How to write the dissolution reaction equation?

I entered it like this, but the software displays an error:
NaCa[B5O6(OH)]·5H2O=Na++Ca+2+[B5O6(OH)6] -3+5H2O

I know NaCa[B5O6(OH)]·5H2O  log K is -5.7258 in 25 ℃

How can I successfully define this new mineral?
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dlparkhurst

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Re: The amount of mineral precipitation
« Reply #3 on: 28/03/22 15:46 »
In a reaction in the PHASES data block, you cannot use the characters "[]", you must use parentheses. A colon ":" is used to signify waters of hydration. The aqueous species used in the reaction definition must be defined in SOLUTION_SPECIES; in your case, B5O6(OH)6-3, is not defined in the pitzer.dat database.
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