JEE Main 2022 · 28 July · evening · Single correct · +4 / −1
The products obtained during treatment of hard water using Clark's method are :
- (a) CaCO₃ and MgCO₃
- (b) Ca(OH)₂ and Mg(OH)₂
- (c) CaCO₃ and Mg(OH)₂
- (d) Ca(OH)₂ and MgCO₃
Answer and solution
Answer (c) CaCO₃ and Mg(OH)₂
Solution
Chemistry ma’am · 3 steps
Lime pulls calcium out as the carbonate, but magnesium leaves as the hydroxide, the less soluble one.
- What Clark's method does. It removes temporary hardness, caused by Ca(HCO₃)₂ and Mg(HCO₃)₂. A calculated amount of slaked lime, Ca(OH)₂, is added to the hard water.
Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃ v + 2H₂O ; Mg(HCO₃)₂ + 2Ca(OH)₂ → 2CaCO₃ v + Mg(OH)₂ v + 2H₂O
- Why magnesium differs. MgCO₃ is fairly soluble in water, but Mg(OH)₂ is much less soluble. So magnesium does not stay as the carbonate. It precipitates as the hydroxide, and that is why magnesium needs twice as much lime.
- So. The precipitates are CaCO₃ and Mg(OH)₂, which is option (c).
The idea to keep. In lime softening, each ion leaves as its least soluble form: calcium as CaCO₃, magnesium as Mg(OH)₂. Lime is the reagent, never the product.
Answer from the NTA final answer key. Solution by PrepWiser.