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Hardness of Water and Treatment

One of the fo...

Question

One of the following is an incorrect statement, point it out.

Permanent hardness can be removed by boiling water

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Hardness of water effects soap consumption

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Temporary hardnes is due to bicarbonates of Ca and Mg

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Permanent hardness is due to the soluble

S O_{4}^{2 -} C l^{-}, N O_{3}^{-}

of Ca and Mg

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Solution

The correct option is A Permanent hardness can be removed by boiling water
Hard water, water that contains salts of calcium and magnesium principally as bicarbonates, chlorides, and sulfates. Water hardness that is caused by calcium bicarbonate is known as temporary, because boiling converts the bicarbonate to the insoluble carbonate; hardness from the other salts is called permanent. Calcium and magnesium ions in hard water react with the higher fatty acids of soap to form an insoluble gelatinous curd, thereby causing a waste of the soap.

Theory :
Softening of water:
Temporary hardness:
By boiling:
$C a (H C O_{3})_{2} soluble Heat \to C a C O_{3} ↓ Insoluble + C O_{2} + 2 H_{2} O$
$M g (H C O_{3})_{2} soluble Heat \to M g C O_{3} ↓ Insoluble + C O_{2} + 2 H_{2} O$
Clark’s method:
$C a (H C O_{3})_{2} Soluble + C a (O H)_{2} Lime \to 2 C a C O_{3} Insoluble ↓ + 2 H_{2} O$
$M g (H C O_{3})_{2} Soluble + C a (O H)_{2} Lime \to M g C O_{3} Insoluble ↓ + C a C O_{3} ↓ + 2 H_{2} O$
Permanent hardness:
By washing soda method:
$C a C l_{2} + N a_{2} C O_{3} \to C a C O_{3} precipitate ↓ + 2 N a C l$
$M g S O_{4} + N a_{2} C O_{3} \to M g C O_{3} precipitate ↓ + N a_{2} S O_{4}$
Permutit method:
$N a_{2} A l_{2} S i_{2} O - 8. x H_{2} O$ -These complex salts are also known as zeolites.
$N a_{2} Z Sodium Zeolite + C a^{2 +} From hard water \to C a Z Calcium Zeolite + 2 N a^{+}$
$N a_{2} Z Sodium Zeolite + M g^{2 +} From hard water \to M g Z Magnesium Zeolite + 2 N a^{+}$
Where Z= $A l_{2} S i_{2} O_{8} . x H_{2} O$
The permutit is loosely packed in a big tank over a layer of coarse sand.
Hard water is introduced into the tank from the top.
Water reaches the bottom of the tank and then slowly rises through the permutit layer in the tank.
The cations present in hard water are exchanged for sodium ions.
Calgon's Method:
$N a_{6} P_{6} O_{18} \to 2 N a^{+} + N a_{4} P_{6} O_{18}^{2 -}$
$M^{2 +} + N a_{4} P_{6} O_{18}^{2 -} \to [N a_{2} M P_{6} O_{18}]^{2 -} + 2 N a^{+}$
Where M= Mg, Ca
The complex anion keeps the Mg2+ and Ca2+ ions in solution.
Synthetic resins method:
The resin exchanges $N a^{+}$ ions with $C a^{2 +}$ and $M g^{2 +}$ ions present in hard water to make the water soft. Here R is resin anion.
$2 R N a (s) + M^{2 +} (a q) \to R_{2} M (s) + 2 N a^{+} (a q)$

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Similar questions

H_{2} O

M g^{2 +}

C a^{2 +}

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

2 g C a C O_{3}

A

B

X

X

Y

A, B, X, and Y,

M = M g, C a

Permanent hardness is due to presence of soluble salts of Mg and Ca in the form of chlorides and sulphates in $H_{2} O$ . It can be removed by

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