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Hardness of Water and Treatment

Permanent har...

Question

Permanent hardness is removed by boiling water with:

A

C a {(O H)}_{2}

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B

{N a}_{2} C O_{3}

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C

C a C O_{3}

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D

C a O

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Solution

The correct option is B ${N a}_{2} C O_{3}$
Temporary hardness is removed by boiling water with $N a O H$ .

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

Permanent hardness is removed by boiling water with sodium carbonate.

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

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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 :

C a S O_{4}

causes permanent hardness in water. It can be removed by precipitating it as

C a C O_{3}

N a_{2} C O_{3}

. If the amount of

N a_{2} C O_{3}

required to soften 100 L of hard water, which is 200 ppm

C a C O_{3}

x

g, then value of

x

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. 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?

Q. Both temporary and permanent hardness is removed on boiling water with :-

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