Answer
414.6k+ views
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
Addition of lime
Addition of sodium carbonates
Base exchange process
Complete step by step answer:
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:
Addition of sodium carbonate
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.
Boiling
Addition of lime
Addition of sodium carbonates
Base exchange process
Complete step by step answer:
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:
Addition of sodium carbonate
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)