JEE Main 2020 · 7 January · morning · Single correct · +4 / −1
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) more efficient as it can exchange only cations.
- (c) more efficient as it can exchange both cations as well as anions.
- (d) less efficient as it exchanges only anions.
Answer and solution
Answer (c) more efficient as it can exchange both cations as well as anions.
Solution
Chemistry ma’am · 3 steps
Ask what each method leaves behind: zeolite swaps only cations, resins remove cations and anions both.
- Zeolite process. Sodium zeolite, Na₂Z, swaps its Na⁺ for Ca²⁺ and Mg²⁺. The anions such as Cl⁻ and SO₄²⁻ are not touched, so the water still carries dissolved sodium salts.
- Synthetic resins. Water passes through two resins. The cation exchange resin swaps H⁺ for metal ions: 2RSO₃H + Ca²⁺ → (RSO₃)₂Ca + 2H⁺. The anion exchange resin swaps OH⁻ for anions: RNH₃OH + Cl⁻ → RNH₃Cl + OH⁻.
H⁺ + OH⁻ → H₂O
Method Ions exchanged Water you get Regenerated by Zeolite cations only soft, but still has Na⁺ salts brine (NaCl solution) Synthetic resins cations and anions deionised (demineralised) dilute acid and dilute alkali - So. The released H⁺ and OH⁻ combine to water, so almost no ions are left. The resin method is more efficient because it exchanges both cations and anions, which is option (c).
The idea to keep. Zeolite only softens water: the anions stay and pair up with Na⁺. Ion exchange resins remove both kinds of ion and give deionised water, and both methods can be regenerated.
Answer from the NTA final answer key. Solution by PrepWiser.