Question

In comparison to the zeolite process for the removal of permanent hardness, the synthetic resins method is :

• A. less efficient as the resins cannot be regenerated
• B. less efficient as it exchanges only anions
• C. more efficient as it can exchange only cations
• D. more efficient as it can exchange both cations as well as anions

Answer 1

The Correct Option is D

(a) Zeolite method removes only cations $\left(Ca^{2+} and\, Mg^{2+} ion\right)$ present in hard water $2NaZ+M^{2+} \left(aq\right) \to MZ_{2} \left(s\right)+2Na^{+}\left(aq\right)\left(M\to Mg, ca\right)$
(b) Synthetic resin method removes cations $\left(Ca^{2+} and\, Mg^{2+} ion\right) and \,anions\left(like\, cl^{-}, Hco^{-}_{3}, so_{4}^{2-} etc.\right)$
$(i) 2 RNa(s)+M^{2+} \left(aq\right)\to R_{2} M\left(s\right)+2Na^{+}\left(aq\right)$
$ ( cation\, exchange \quad\left(M\to Mg, Ca\right) resin)$
(ii) $RNH_{3}^{+} OH^{-}\left(s\right)+X^{-}\left(aq\right)\to RNH_{3}^{+} x^{-}\left(s\right)+OH^{-}\left(aq\right)$
$Anion\, exchange \left(x^{-}=CI^{-},HCO_{3}^{-}, SO_{4}^{2-} \, resin\right)\quad etc)$