Question:

Which one the following removes temporary hardness of water ?

Updated On: Jan 30, 2025
  • Slaked lime
  • Plaster of Paris
  • Epsom
  • Hydrolith
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The Correct Option is A

Solution and Explanation

Temporary hardness is due to presence of bicarbonates of $Ca$ (II) and $Mg$ (II). Temporary hardness is removed by adding lime water or milk of lime. This method is known as Clark's process. Removal of hardness is called softening of water.
$Ca ( OH )_{2}+ Ca \left( HCO _{3}\right)_{2} \rightarrow 2 CaCO _{3} \downarrow+2 H _{2} O$
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Concepts Used:

Group 2 Elements

The group two or alkaline earth metals are s-block elements with two electrons in their s-orbital. They are alkaline earth metals. They are named so because of the alkaline nature of the hydroxides and oxides.

Alkaline earth metals are characterized by two s-electrons. This group of elements includes:

  • Beryllium (Be)
  • Magnesium (Mg)
  • Calcium (Ca)
  • Strontium (Sr)
  • Barium (Ba)
  • Radium (Ra)

Elements whose atoms have their s-subshell filled with their two valence electrons are called alkaline earth metals. Their general electronic configuration is [Noble gas] ns2. They occupy the second column of the periodic table and so-called as group two metals also.