Hello,
I am investing desorption behaviour of Cr(VI), by the different desorbing agents like HNO3, HCl, H2SO4, KOH and deionized water. HNO3, HCl, H2SO4 had showed more desorption capacity of Cr(VI) (>70%) compared to KOH (50%). What might me the reasons?
The desorption percentage of Cr(VI) ions from organic adsorbent materials is higher for HNO3, HCl, and H2SO4 compared to other eluents like EDTA and KOH because these strong acids have a high affinity for Cr(VI) ions and can effectively remove them from the surface of the adsorbent material. The high ionic strength of these acids enhances the interaction between the ions and the organic adsorbent material, leading to a higher desorption percentage.
On the other hand, EDTA and KOH have a lower ionic strength and are less effective in removing Cr(VI) ions from the adsorbent material. They may also react with the organic adsorbent material, reducing its ability to retain Cr(VI) ions.
In conclusion, the desorption of Cr(VI) ions from organic adsorbent materials is influenced by the strength of the eluent and its ability to interact with the ions. Strong acids like HNO3, HCl, and H2SO4 have a high affinity for Cr(VI) ions and are more effective in removing them from the adsorbent material.
Hi Prof Rodolfo Bolaños Barrera,
Thanks for your nice explanation. Could you please suggest some reference materials in favour of citing this explanation in a paper?
Best regards,
Mahbub
University of Technology Sydney, Australia
Here are some references on the topic: