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Temporary Hardness

The reaction ...

Question

The reaction involved in the removal of temporary hardness of water is:

A

Ca( ${H C O}_{3}$ ) $_{2}$ $h e a t - - \to$ ${C a C O}_{3}$ + $H_{2}$ O + $C O_{2}$ $↑$

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B

C a C l_{2}

+

N a_{2}

C O_{3}

\to

C a C O_{3}

↓

+ 2NaCl

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C

M g S O_{4}

+

N a_{2}

A l_{2}

S i_{2}

O_{8}

.x

H_{2}

O

\to

Mg

A l_{2}

S i_{2}

O_{8}

.x

H_{2}

O +

N a_{2}

S O_{4}

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D

2

C a C l_{2}

+ (

{N a P O}_{3}

)

_{6}

\to

N a_{2}

[

C a_{2}

(

P O_{3}

)

_{6}

] + 4NaCl

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Solution

The correct option is A
Ca( ${H C O}_{3}$ ) $_{2}$ $h e a t - - \to$ ${C a C O}_{3}$ + $H_{2}$ O + $C O_{2}$ $↑$

Temporary hardness of water is caused due to the dissolved bicarbonate of calcium and magnesium. It can be removed by the process of boiling. When the hard water containing calcium bicarbonate is boiled, then this bicarbonate salt decomposes into insoluble calcium carbonate alongwith evolution of carbon dioxide gas and water. The reaction involved is:
$Ca ({HCO}_{3}$ ) $_{2}$ $Heat \to$ ${CaCO}_{3} + H_{2} O + {CO}_{2} ↑$

This insoluble carbonate is later filtered out.

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Similar questions

Q. Permanent hardness is due to

S O_{4}^{2 -}

C l^{-}

C a^{2 +}

M g^{2 +}

and is removed by the addition of

N a_{2} C O_{3}

.

C a S O_{4} + N a_{2} C O_{3} ⟶ C a C O_{3} + N a_{2} S O_{4}

C a C l_{2} + N a_{2} C O_{3} ⟶ C a C O_{3} + 2 N a C l

100

C a C O_{3}

N a_{2} C O_{3}

required to soften

10

L of hard water is :

Q. Permanent hardness is due to

C l^{⊝}

S O_{4}^{2 -}

M g^{2 +}

C a^{2 +}

and is removed by adding

N a_{2} C O_{3}

C a S O_{4} + N a_{2} C O_{3} \to C a C O_{3} + N a_{2} S O_{4}

C a C l_{2} + N a_{2} C O_{3} \to C a C O_{3} + 2 N a C l

Which of the following statements is/are correct?

Following reaction is used to remove temporary hardness which is due to dissolved bicarbonates of

C a^{2 +}

M g^{2 +}

C a (H C O_{3})_{2} + C a (O H)_{2} \to 2 C a C O_{3} ↓ + 2 H_{2} O

Actually lime (

C a O

) is added which in

H_{2} O

C a (O H)_{2}

and precipitates

C a C O_{3}

C a O + H_{2} O \to C a (O H)_{2}

10

L of hard water requires

0.56

C a O

, temporary hardness of

C a C O_{3}

x \times 10^{2}

x

Q. A process designed to remove organic sulphur from coal prior to combustion involves the reaction.

X - S - Y + 2 N a O H \to X - O - Y + N a_{2} S + H_{2} O

C a C O_{3} \to C a O + C O_{2}

N a_{2} S + C O_{2} + H_{2} O \to N a_{2} C O_{3} + H_{2} S

C a O + H_{2} O \to C a (O H)_{2}

N a_{2} C O_{3} + C a (O H)_{2} \to C a C O_{3} + 2 N a O H

In the processing of 320 metric tons of a coal having 1.0% sulphur content, how much limestone

(C a C O_{3})

must be edecomposed to provide enough

C a (O H)_{2}

to regenerate the NaOH used in the original leaching step?

Q. Hardness of water is measured in terms of ppm (parts per million) of

C a C O_{3}

. It is the amount (in g) of

C a C O_{3}

10^{6} g H_{2} O

. In a sample of water,

10

0.56

C a O

to the remove temporary hardness of

H C O_{3}^{-}

.

C a (H C O_{3})_{2} + C a O ⟶ 2 C a C O_{3} + H_{2} O

Temporary hardness is :

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