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# Calculate the weight of ${\text{CaO}}$ required to remove hardness of $1000000{\text{L}}$ of water containing $1.62{\text{g}}$ of ${\text{Ca}}{\left( {{\text{H}}{{\text{CO}}_3}} \right)_2}$ per litre.

Last updated date: 13th Jun 2024
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Hint: Hardness can be defined as a property of water to form insoluble particles with soap instead of lather. Temporary hardness can be removed by the following methods:
Boiling
Base exchange process

Hardness is the soap destroying property due to the presence of bicarbonates, sulfates and chlorides of calcium and magnesium. There are two types of hardness-temporary and permanent hardness. Temporary hardness is due to the presence of bicarbonates of calcium and magnesium. Permanent hardness is due to the presence of sulfates, nitrates and chlorides of calcium and magnesium.
Permanent hardness can be removed by the following methods:
Base exchange process
It is given that in $1{\text{L}}$ of water, amount of ${\text{Ca}}{\left( {{\text{H}}{{\text{CO}}_3}} \right)_2}$, ${{\text{m}}_{{\text{Ca}}{{\left( {{\text{H}}{{\text{CO}}_3}} \right)}_2}}} = 1.62{\text{g}}$
Thus in $1000000{\text{L}}$ of water, amount of ${\text{Ca}}{\left( {{\text{H}}{{\text{CO}}_3}} \right)_2}$, ${{\text{m}}_{{\text{Ca}}{{\left( {{\text{H}}{{\text{CO}}_3}} \right)}_2}}} = 1.62{\text{g}} \times {\text{1000000L = 1620kg}}$
The reaction of calcium bicarbonates in water is treated with calcium oxide to remove hardness, which gives calcium carbonate and water. The reaction is given below:
${\text{Ca}}{\left( {{\text{H}}{{\text{CO}}_3}} \right)_2} + {\text{CaO}} \to 2{\text{Ca}}{{\text{CO}}_3} + {{\text{H}}_2}{\text{O}}$
$1{\text{mol}}$ of ${\text{Ca}}{\left( {{\text{H}}{{\text{CO}}_3}} \right)_2}$ reacts with $1{\text{mol}}$ of ${\text{CaO}}$.
$\left[ {40 + 2\left( {1 + 12 + \left( {3 \times 16} \right)} \right)} \right] = 162{\text{g}}$ of ${\text{Ca}}{\left( {{\text{H}}{{\text{CO}}_3}} \right)_2}$ reacts with $40 + 16 = 56{\text{g}}$ of ${\text{CaO}}$
i.e. $162{\text{g}}$ of ${\text{Ca}}{\left( {{\text{H}}{{\text{CO}}_3}} \right)_2}$ reacts with $56{\text{g}}$ of ${\text{CaO}}$
Similarly, $1{\text{g}}$ of ${\text{Ca}}{\left( {{\text{H}}{{\text{CO}}_3}} \right)_2}$ reacts with $\dfrac{{56}}{{162}} = 0.345{\text{g}}$ of ${\text{CaO}}$
So $1620{\text{kg}}$ of ${\text{Ca}}{\left( {{\text{H}}{{\text{CO}}_3}} \right)_2}$ reacts with $0.35{\text{g}} \times {\text{1620}} \times {\text{1}}{{\text{0}}^3}{\text{g = 560kg}}$
Hence mass of required to remove hardness of $1000000{\text{L}}$ of water containing $1.62{\text{g}}$ of ${\text{Ca}}{\left( {{\text{H}}{{\text{CO}}_3}} \right)_2}$ per litre is ${\text{560kg}}$

Note:
Water hardness can be measured by two methods:
By titration with standard soap solution.
In this method, total or permanent hardness can be measured.
By titration with hydrochloric acid
In this method, temporary hardness can be measured.
Moreover, permanent hardness is the difference between total hardness and temporary hardness.