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LaOH Adsorption


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Hi,

 

I am trying to create a new data set for sorbing surfaces to model the sorption of phosphorus onto LaOH.

 

I have loaded all the necessary equations of LaOH, however when I add the sorbing surface to React I don’t observe an adsorption.

 

I attach the data set that I created.

 

Please help me in finding if there is a mistake in my data set and how could I model it correctly on React.

 

Thank you,

 

Stacey

 

LaOH2.sdat

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Hi Stacey,


I noticed a few problems with the dataset which I’ll try to explain below.


All .sdat surface datasets include in the Header a reference to the thermo dataset that supplied all the aqueous species and minerals involved in the surface complexation reactions. The dataset you supplied refers to thermo.tdat, which does not contain any lanthanum species. If you’ve modified that dataset to include lanthanum species, it’s best practice to save the dataset with a new filename, such as thermo+La.tdat, rather than overwriting the original file. Or if you’re using another dataset that already contains lanthanum species, such as thermo_minteq.tdat, you should make sure that the surface dataset points to that particular file. Otherwise, TEdit will issue some warnings when it reads your surface dataset because it doesn’t know the mole weight of Lanthanum.


You’ve specified La(OH)3 as the sorbing mineral, but you haven’t specified the site density of your sorbing site on that mineral. Even if the mineral is present in React, it won’t have any sorbing sites based on the dataset you supplied, so you won’t see any sorption.


The program expects the sorbing sites (the basis surface species) to be uncharged. In FeOH.sdat, for example, you can see >(s)FeOH and >(w)FeOH, both of which have a charge of 0. Your sorbing site has a charge of +2, however. This causes a problem. You’ll have to designate an uncharged surface species as the sorbing site. This will likely affect the charge of all your complexes.


This isn’t really a problem, but it is a little weird that some of your reactions are written in terms of H+ and some in terms of OH-. It’s probably best for anyone viewing the dataset that you stick with one or the other. My recommendation is H+.


Another tip for readability is to designate surface species with a “>” symbol and to include “+” or “-“ symbols in the names of species so that you can immediately tell the charge of a species just by looking at its name. In the FeOH.sdat dataset we distribute, for example, you can find species such as >(s)FeOH2+, >(s)FeOH, >(s)FeO-, and >(w)FePO4- -. It’s really easy to tell just by looking at them that they’re all surface species and that they have charges, respectively, of +1, 0, -1, and -2.


Hope this helps,


Brian Farrell

Aqueous Solutions LLC
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Hi Brian,

 

Thank you for your help. I have corrected my sorbing surface and corrected my database to inlcude La by using thermo.com.V8.R6+ database, however I am still having problems to view adsorption when I load the surface to React. I have specified my mineral and its site density. I am attaching my new version of the sorbing surface if you have time to review it and comment as to where my mistake could be.

 

Thank you once again for your help.

 

Stacey

LaOH2.sdat

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Hi Stacey,


You’re not seeing any sorption because your sorbing mineral, La(OH)3, does not precipitate at all during the run. I took a look at the species in solution and noticed that the La++ ion is by far the predominant lanthanum species. It is much more stable than the La+++ ion. La(OH)3 has a very low saturation index because of the low La+++ activity. If you don’t expect the La++ to be important in your system, you could simply suppress it. Or, you could remove all La(II) species from consideration by decoupling the La++ from La+++. If you suspect the thermo data is wrong, you could also alter the log K for the reaction between La++ and La+++. Whatever you choose, a higher La+++ activity should cause La(OH)3 to precipitate and its sorbing sites to form.


By the way, I noticed that you are gradually titrating HPO4-- into the system as you increase the pH. You end up with more total phosphorous in the system at pH 10 than at pH 5. Is this what you want? Just wanted to check.


Hope this helps,

Brian
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