JEE Main 2022 · 28 July · morning · Single correct · +4 / −1
The metal salts formed during softening of hardwater using Clark's method are :
- (a) Ca(OH)₂ and Mg(OH)₂
- (b) CaCO₃ and Mg(OH)₂
- (c) Ca(OH)₂ and MgCO₃
- (d) CaCO₃ and MgCO₃
Answer and solution
Answer (b) CaCO₃ and Mg(OH)₂
Solution
Chemistry ma’am · 4 steps
Lime goes in and carbonate comes out for calcium, but magnesium settles as the less soluble hydroxide.
- What Clark's method does. It removes temporary hardness, which comes from dissolved bicarbonates of calcium and magnesium. A calculated amount of slaked lime, Ca(OH)₂, is added, and the hardness ions leave as insoluble solids.
- Calcium bicarbonate. Lime turns the bicarbonate into carbonate, and calcium carbonate precipitates: Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃ v + 2H₂O.
Mg(HCO₃)₂ + 2Ca(OH)₂ → 2CaCO₃ v + Mg(OH)₂ v + 2H₂O
- Magnesium bicarbonate. MgCO₃ is fairly soluble, but Mg(OH)₂ is much less soluble. So with lime present, magnesium comes out as the hydroxide, and that is why it needs two moles of lime.
- So. The salts precipitated are CaCO₃ and Mg(OH)₂, which is option (b).
The idea to keep. In lime softening, each metal leaves as its least soluble form. For calcium that is CaCO₃, and for magnesium it is Mg(OH)₂, not MgCO₃.
Answer from the NTA final answer key. Solution by PrepWiser.