How does the treatment of resin R-SO3H with sodium chloride lead to formation of RNa?
It is a typing mistake. Cation exchange process can be written as
$$\ce{R-SO3^-H^+ (s) + NaCl (aq) -&gt; R-SO3^-Na+(s) + H^+(aq) +Cl^-(aq)}$$
You can immediately see why the resin must be in sodium form for water softening. If the acid form of the resin is used for softening water, what would you get? Extremely strong acids!!
For a divalent ion, the sulfonic acid groups should be in close proximity to &quot;catch&quot; ions such as $\ce{Ca^{2+}}$. Each calcium ion would quantitatively release two hydrogen ions from the resin.