Which one of the following methods is most suitable for preparing deionized water?

We begin by recalling what “de-ionized water” means. De-ionized (or demineralised) water is water from which all the dissolved ions, both cations like $$\text{Na}^+, \text{K}^+, \text{Ca}^{2+}$$ and anions like $$\text{Cl}^-, \text{SO}_4^{2-}, \text{HCO}_3^-$$, have been removed. So, to obtain de-ionized water we need a process that can simultaneously take out positive as well as negative ions.

Now we analyse each method named in the options.

Option A tells us about the Synthetic resin method. In this method we employ two kinds of ion-exchange resins:

• A cation-exchange resin, often represented as $$\text{R-H}$$. The functional group releases $$\text{H}^+$$ ions and picks up any metal cation present in water according to the general exchange reaction

$$\text{R-H} + \text{M}^{n+} \; \longrightarrow \; \text{R-M}^{(n-)} + n\,\text{H}^+$$

• An anion-exchange resin, commonly written as $$\text{R'-OH}$$. This resin releases an $$\text{OH}^-$$ ion and absorbs any anion present:

$$\text{R'-OH} + \text{A}^- \; \longrightarrow \; \text{R'-A} + \text{OH}^-$$

After the water has passed successively through both resins, the liberated $$\text{H}^+$$ and $$\text{OH}^-$$ ions combine:

$$\text{H}^+ + \text{OH}^- \; \longrightarrow \; \text{H}_2\text{O}$$

Since every cation and every anion has been exchanged out of the water, the effluent is practically free from all ions; that is exactly the definition of de-ionized water. Thus, the synthetic resin method fulfils the requirement completely.

Option B refers to Calgon’s method. Here we add sodium hexametaphosphate $$\text{(NaPO}_3\text{)}_6$$ (popularly called Calgon) to water. The compound forms soluble complexes specifically with hardness-producing calcium and magnesium ions, preventing them from forming scale. Only $$\text{Ca}^{2+}$$ and $$\text{Mg}^{2+}$$ are affected; other ions remain. Therefore, this method softens water but does not totally de-ionize it.

Option C is Clark’s method. We add calculated amounts of lime $$\text{Ca(OH)}_2$$ to precipitate carbonate hardness as $$\text{CaCO}_3$$ and $$\text{Mg(OH)}_2$$. Again, only temporary hardness is removed; many ions stay dissolved. So Clark’s method cannot yield de-ionized water.

Option D mentions the Permutit (zeolite) method. A natural or synthetic zeolite $$\text{Na}_2\text{Z}$$ exchanges its $$\text{Na}^+$$ ions with $$\text{Ca}^{2+}$$ and $$\text{Mg}^{2+}$$ in hard water as per

$$\text{Na}_2\text{Z} + \text{Ca}^{2+} \; \longrightarrow \; \text{CaZ} + 2\,\text{Na}^+$$

This removes only the calcium and magnesium ions responsible for hardness; all other cations and every anion remain. Therefore, the permutit method merely softens water; it does not create de-ionized water.

Comparing the four methods, only the synthetic resin (ion-exchange) method eliminates both cations and anions completely, giving pure de-ionized water. Thus it is the most suitable technique for the desired purpose.

Hence, the correct answer is Option A.