Question

The temporary hardness of a water sample is due to compound X. Boiling this sample converts X to compound Y. X and Y, respectively are:
A.$Ca{\left( {HC{O_3}} \right)_2}$ and $CaO$
B.$Mg{\left( {HC{O_3}} \right)_2}$ and $MgC{O_3}$
C.$Mg{\left( {HC{O_3}} \right)_2}$ and $Mg{\left( {OH} \right)_2}$
D.$Ca{\left( {HC{O_3}} \right)_2}$ and $Ca{\left( {OH} \right)_2}$

Answer 1

Hint:The hardness of water is due to the presence of soluble bicarbonates, chlorides, and sulfates of calcium and magnesium in it. Hard water is defined as the water that does not produce lather with the soap. Hard water contains high mineral content. It is formed when rainwater and the flowing water percolate through the deposits of chalk and limestone which are made up of magnesium and calcium carbonates.

Complete step by step answer:
-We know that the temporary hardness is due to the bicarbonates of Ca and Mg. If magnesium bicarbonate $Mg{\left( {HC{O_3}} \right)_2}$ which is present in the water is heated or boiled to get magnesium carbonate $MgC{O_3}$ , carbon dioxide gas, and water. The magnesium carbonate formed is insoluble carbonates that settle down and can be removed. Soft water is decanted off. The above reaction involved is:
$Mg{\left( {HC{O_3}} \right)_2}\xrightarrow{\Delta }MgC{O_3} + C{O_2} + {H_2}O$
-The compound X is magnesium bicarbonate, $Mg{\left( {HC{O_3}} \right)_2}$ , and the compound Y is $MgC{O_3}$ .

Thus, the correct option is (C).

Additional Information:
The hardness of water is of two types:
-Temporary Hardness – It is due to the presence of bicarbonates of calcium, magnesium dissolved in water. It is also known as carbonate hardness
-Permanent hardness- It is due to the presence of chlorides and sulfates of calcium, magnesium. It is also known as non- carbonate hardness. It is named for permanent hardness because such hardness cannot be removed by simply boiling water.

Note:
Another method of removing temporary hardness is Clark’s method. When temporary hard water is treated with a calculated quantity of lime, bicarbonates present in water change to insoluble carbonates that settle down. Soft water is decanted off. The reaction involved is:
 $Ca{\left( {HC{O_3}} \right)_2} + Ca{\left( {OH} \right)_2} \to 2CaC{O_3} \downarrow + 2{H_2}O
  Mg{\left( {HC{O_3}} \right)_2} + Ca{\left( {OH} \right)_2} \to CaC{O_3} \downarrow + MgC{O_3} \downarrow + 2{H_2}O$