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Why is $\ce{R-SO3H}$ first treated with NaCl and then the RNa form i.e., the sodium form of resin is used for treating hard water? Why cannot we directly treat hard water with $\ce{R-SO3H}$? What is the purpose of NaCl? enter image description here

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  • $\begingroup$ It depends if you want to have as counterparts of anions of strong acids Na+ or H+. For bicarbonates, the latter leads to release of CO2. $\endgroup$
    – Poutnik
    Commented Nov 8, 2022 at 14:09
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    $\begingroup$ The short answer is do you want to drink HCl or NaCl rather? $\endgroup$
    – ACR
    Commented Nov 8, 2022 at 14:19
  • $\begingroup$ @AChem can you explain elabrately with reaction please, I'm not getting?? $\endgroup$ Commented Nov 8, 2022 at 15:40
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    $\begingroup$ The ionex in H cycle converts dissolved salts to respective acids. The ionex in Na cycle converts salts to respective sodium salts. $\endgroup$
    – Poutnik
    Commented Nov 8, 2022 at 16:08
  • $\begingroup$ @DAYA'SSIMULATION, Can you please read Wikipedia article on ion-exchange, en.wikipedia.org/wiki/Ion-exchange_resin. Also, can you write ion-exchange equation in the main question? I want to see if you have understood ion-exchange itself or not? $\endgroup$
    – ACR
    Commented Nov 8, 2022 at 16:15

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Hard water is a water containing the ions $\ce{Ca^{2+}, HCO3^-, SO4^{2-}}$. When this water crosses a layer made of beads having the formula $\ce{RSO3H}$, the following reaction happens : $$\ce{Ca^{2+} + 2 RSO3H -> (RSO3)2Ca + 2 H+}$$ $\ce{(RSO3)2Ca}$ remains fixed on the resin beads. But the ion $\ce{H+}$ replaces $\ce{Ca^{2+}}$ in solution. So the solution becomes acidic. If the negative ion remaining in solution is $\ce{HCO3^-}$, this is not too serious, because both ions $\ce{H+}$ and $\ce{HCO3^-}$ react and produce $\ce{H2CO3}$ which is quickly decomposed into $\ce{H2O}$ and $\ce{CO2}$. But if the negative ion is $\ce{SO4^{2-}}$, it produces a solution containing sulfuric acid $\ce{H2SO4}$ which is rather corrosive.

This unfortunate result can be avoided if the resin has been previously treated by a solution of $\ce{NaCl}$. Such a solution contains the ions $\ce{Na+}$ which react with the resin according to the reaction $$\ce{RSO3H + Na+ ->RSO3Na + H+}$$ Here too, the resin is modified, becoming $\ce{RSO3Na}$, and the $\ce{H+}$ ions are released into the solution. But as this treatment has been done in advance, before adding hard water, these acidic ions can be eliminated from the resin beads by washing. With this new resin beads containing now $\ce{RSO3Na}$, hard water may be used to cross the modified resin layer. In this case, the $\ce{Ca^{2+}}$ ions of the hard water will react with the modified resin according to : $$\ce{Ca^{2+} + 2 RSO3Na -> (RSO3)2Ca + 2 Na+}$$ After this treatment, the resin is once more modified, becoming $\ce{(RSO3)2Ca}$ but the water coming out of the resin layer contains $\ce{Na+}$ ions as only positive ions. The negative ions of the solution have not been changed. They are $\ce{HCO3^-}$ and $\ce{SO4^{2-}}$. The solution getting out of this treatment with $\ce{RSO3Na}$ corresponds to a solution made by dissolving $\ce{NaHCO3}$ and $\ce{Na2SO4}$ which are both not corrosive at all.

This is why it is better to use the modified resin $\ce{RSO3Na}$ instead of the original acidic resin $\ce{RSO3H}$, when treating hard water.

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  • $\begingroup$ Better yet treat with a mixed bed column or a subsequent anion exchange resin adding an equivalent amount of OH-. Neutral water! $\endgroup$
    – jimchmst
    Commented Nov 8, 2022 at 22:16

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