JEE Main 2020 — Hydrogen Question with Solution

Zeolite removes only cations from hard water, as given below by reaction:

$$\begin{gathered} 2 \mathrm{NaZ} + {\mathrm{M}}^{+2}\left(\mathrm{aq}\right)\to {\mathrm{MZ}}_2 + 2 {\mathrm{Na}}^{+} \\ \mathrm{Here} \mathrm{M}-\mathrm{Ca}, \mathrm{Mg} \end{gathered}$$

While, synthetic resin removes both cations and anions, as given by following reactions:

For cations-

$2 \mathrm{RNa}\left(\mathrm{s}\right) +{\mathrm{M}}^{+2}\left(\mathrm{aq}\right)\to {\mathrm{R}}_2\mathrm{M} \left(\mathrm{s}\right)+ 2{\mathrm{Na}}^{+}\left(\mathrm{aq}\right)$

For anions-

${\mathrm{RNH}}_3^{+}{\mathrm{OH}}^{-}\left(\mathrm{s}\right) + {\mathrm{X}}^{-}\left(\mathrm{aq}\right) \to {\mathrm{RNH}}_3^{+}{\mathrm{X}}^{-} \left(\mathrm{s}\right)+ {\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

So, synthetic resin method is more efficient as it can exchange both cations as well anions.