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The amount of mineral precipitation
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Topic: The amount of mineral precipitation (Read 2237 times)
Jay Chou
Frequent Contributor
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
Global Moderator
Posts: 4036
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
Frequent Contributor
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
Global Moderator
Posts: 4036
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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The amount of mineral precipitation