How does an ion exchange column work?

[pm1]

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I just got a question on a TBR CBT and I am quite confused.

If someone could please explain it for me it would be HIGHLY appreciated!

Thank you!

 

[dsoz]

I just got a question on a TBR CBT and I am quite confused.

If someone could please explain it for me it would be HIGHLY appreciated!

Thank you!

read
http://en.wikipedia.org/wiki/Ion_exchange

One type is basically, the resin exchanges anions and cations. If it exchanges H+ for cations and OH- for anions, then the H+ and OH- make water.

dsoz

 

[osprey099]

I just got a question on a TBR CBT and I am quite confused.

If someone could please explain it for me it would be HIGHLY appreciated!

Thank you!

You are basically pouring a solute (Ex: CaCl2) into a column and there is a cation in your column. Your goal is to precipitate the cation initially in the column by making it come out of the column and replace it with the cation you insert into the column (in this case Ca2+). Therefore, you need a cation that is less soluble with Cl- than Ca2+ is with Cl- because less soluble -> more likely to precipitate. For our case, we can use Na+. This will happen because once you pour the CaCl2 into the column and it dissociates, the Na+ cations and the Cl- anions will form NaCl precipitate and come out of the column and the Ca2+ will remain in the column. The net result is that you replaced the column with a different cation but this experiment shows you that less soluble -> quicker to precipitate.

 

[pm1]

You are basically pouring a solute (Ex: CaCl2) into a column and there is a cation in your column. Your goal is to precipitate the cation initially in the column by making it come out of the column and replace it with the cation you insert into the column (in this case Ca2+). Therefore, you need a cation that is less soluble with Cl- than Ca2+ is with Cl- because less soluble -> more likely to precipitate. For our case, we can use Na+. This will happen because once you pour the CaCl2 into the column and it dissociates, the Na+ cations and the Cl- anions will form NaCl precipitate and come out of the column and the Ca2+ will remain in the column. The net result is that you replaced the column with a different cation but this experiment shows you that less soluble -> quicker to precipitate.

Nice! Thank you! 🙂

 

[mtchin08]

You are basically pouring a solute (Ex: CaCl2) into a column and there is a cation in your column. Your goal is to precipitate the cation initially in the column by making it come out of the column and replace it with the cation you insert into the column (in this case Ca2+). Therefore, you need a cation that is less soluble with Cl- than Ca2+ is with Cl- because less soluble -> more likely to precipitate. For our case, we can use Na+. This will happen because once you pour the CaCl2 into the column and it dissociates, the Na+ cations and the Cl- anions will form NaCl precipitate and come out of the column and the Ca2+ will remain in the column. The net result is that you replaced the column with a different cation but this experiment shows you that less soluble -> quicker to precipitate.

wait... NaCl is more soluble than CaCl2 right? So the cation in the column must be more soluble. Thus, when Ca2+ comes into the column, it is more insoluble compared to the Na+, and Ca2+ will precipitate into a solid and stay in the column while only Na+ cations flow out. The more soluble sodium pass and into the water supply. If you use a more soluble cation, it will just flow through, leaving the Na+ in its place. If I said anything incorrect, my bad. Just trying to make sense of this comment.