JEE Main 2021 · 26 August · morning · Single correct · +4 / −1
Which one of the following methods is most suitable for preparing deionized water?
- (a) Synthetic resin method
- (b) Clark's method
- (c) Calgon's method
- (d) Permutit method
Answer and solution
Answer (a) Synthetic resin method
Solution
Chemistry ma’am · 3 steps
Deionized means every ion is gone, cations and anions both, and only the two-resin method removes all of them.
- What deionized means. Deionized water has no dissolved ions left, neither cations like Ca²⁺, Mg²⁺, Na⁺ nor anions like Cl⁻, SO₄²⁻, HCO₃⁻. Softening only removes the hardness-causing Ca²⁺ and Mg²⁺, which is not enough.
Method What it does Ions left in water Synthetic resin Cation resin swaps every cation for H⁺, anion resin swaps every anion for OH⁻ none, since H⁺ and OH⁻ form water Clark's Adds Ca(OH)₂ to precipitate bicarbonates as CaCO₃ Removes only temporary hardness, other ions stay Calgon's Sodium hexametaphosphate ties up Ca²⁺ and Mg²⁺ in a soluble complex The complex, Na⁺ and all anions Permutit Sodium zeolite swaps Ca²⁺ and Mg²⁺ for Na⁺ Na⁺ and all anions - How resins work. In the cation exchanger, 2RSO₃H + Ca²⁺ → (RSO₃)₂Ca + 2H⁺. In the anion exchanger, RNH₃⁺OH⁻ + Cl⁻ → RNH₃⁺Cl⁻ + OH⁻. The released ions then combine as H⁺ + OH⁻ → H₂O, so nothing ionic remains.
- So. Only the synthetic resin (ion exchange) method removes both cations and anions completely, so it is the method for deionized water. The answer is (a).
The idea to keep. Softening removes only Ca²⁺ and Mg²⁺. Deionizing needs a cation resin giving H⁺ and an anion resin giving OH⁻, which together turn every ion into plain water.
Answer from the NTA final answer key. Solution by PrepWiser.