q show the process of ion exchange

Question

Q. Show the Process of ion exchange chromatography?

The process of ion exchange chromatography is the most important, effective and rapid method for the separation and purification of the lanthanons. In this process, a solution of lanthanide ions is run down a column of a synthetic ion exchange resin. Ion exchange resins are organic polymers consisting of functional groups such as -COOH, -S0₃H or -OH. In these resins, hydrogen ions are mobile and will be exchanged with other cations. Therefore, the lanthanide ions replace the'^' ions and get bound to the resin:

Ln³⁺ 3R- SO₃H ------------------------> Ln (SO₃R) 3 + 3H⁺

After the H⁺ ions have passed through the column, a solution of a complexing agent such as citric acid, α-hydroxyisobutyric acid or EDTA at the appropriate pH is passed through the column to elute, i.e. to wash off the metal ions in a selective manner:

Ln( O₃SR)₃ + NH₄)₃EDTAH-----------------------------> Ln (EDTAH) + 3NH₄O₃SR

As the EDTA solution flows down the column, the lanthanide ions come off the resin and form a complex with EDTA and then go back on the resin a little lower down the column.  This process is repeated several times as the metal ions gradually travel down the column.  The smaller lanthanide ions like LU³⁺ form stronger complexes with EDTA than the larger ions like La³⁺+. Therefore, the smaller and heavier ions spend more time in solution and less time on the column. Thus, the heavier ions are eluted from the column first and the lighter ones the last. Using suitable conditions, all the individual elements will be separated. The eluates are then treated with an oxalate solution to precipitate lanthanides as oxalates which are then ignited to get the oxides:

2Ln (EDTAH) + 3(NH₄) ₂C₂O₄---------------> Ln (C₂O₄)3+ 2(NH₄) ₃EDTAH

Ln₂ (C₂O₄)₃---------------> Ln₂O₃+ 3CO + 3CO₂

 

Samarium europium and ytterbium are prepared by reduction of the oxides with La high temperatures:

Ln₂O₃+ 2La-----------------> La₂O₃ + 2Ln, Ln = Sm and Eu

Other lanthanides are obtained by the reaction of LnCl₃ or LnF3 with Ca metal at 1300 K.  LnCl₃ or LnF3 are prepared by heating Ln₂Oj with appropriate ammonium halide:

 Ln₂O₃ + 6 NH₄X--------------------> LnX₃ - 6 NH₃ + 3H₂O

2LnX₃ + 3Ca---------------------> 2Ln + 3CaX₂