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I'm studying HPLC detection techniques and stuck on the conductometric suppression technique. It is describing how buffer ions (in my case $\ce{Na+}$ and $\ce{HCO3-}$) are being either adsorbed ($\ce{Na+}$) or protonated ($\ce{H2CO3}$) which makes them electrolytically inactive and leaves just water in the column.

Great! I understand that background conductivity is a problem. What I do not get is why was $\ce{NaHCO3}$ added to the water in the first place?

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  • $\begingroup$ Are buffers used to dissolve the sample? $\endgroup$
    – waterlemon
    Feb 14, 2016 at 16:47

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Ion chromatography usually means chromatography where aqueous (and usually inorganic) ions are the analytes. They are separated from each other by the use of ion exchange resins.

If your analytes are anions, then they will be separated on an anion exchange resin. These resins have a crosslinked organic polymer that contains permanent positive charges. The negative charges that neutralize these resin-bound positive charges come from aqueous solution. Different aqueous anions will have slightly different propensities to bind to and temporarily neutralize the solid-phase cations.

If you don't use an anion-containing eluent such as bicarbonate or hydroxide, the analytes that you would like to eventually detect through conductometric detection will permanently stick there! So adding e.g. $\ce{NaHCO3}$ as an eluent will force them off of the column and ensure that your analytes eventually flow to the detector.

A decent diagram of the process is found in this PDF from Thermo (formerly Dionex). $%edit$

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