How does the treatment of resin R-SO3H with sodium chloride lead to formation of RNa? I am studying the methods of removal of permanent hardness of water, and came across the &quot;synthetic resins method&quot;. From my textbook,

Cation exchange resins contain large organic molecules with $-\text{SO}_3\text{H}$ groups and are water insoluble. Ion exchange resin $\text{RSO}_3\text{H}$ is changed to $\text{RNa}$ by treating with $\text{NaCl}$. The resin then exchanges $\text{Na}^+$ ions with $\text{Ca}^{2+}$ and $\text{Mg}^{2+}$ ions present in hard water to make it soft.

I don't understand how the treatment of $\text{RSO}_3\text{H}$ with $\text{NaCl}$ to form $\text{RNa}$ works. For one thing, in the former compound, since $-\text{SO}_3\text{H}$ is an anion, $\text{R}$ is a cation, but in $\text{RNa}$, $\text{Na}^+$ being a cation, $\text{R}$ is an anion. What is the redox reaction that is responsible for the conversion here?
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.