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Hardness of Water and Treatment

A sample of h...

Question

A sample of hardwater contains 1.1 mg $C a C l_{2}$ and 0.95 mg $M g C l_{2}$ per litre. Calculate the hardness of water in terms of $C a C O_{3}$ present in per $10^{6}$ parts of water.

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Solution

Molecular weight of $C a C l_{2}$ , $M g C l_{2}$ , and $C a C O_{3}$ are 111.0 g/mol, 95 g/mol, 100 g/mol respectively.
$C a C l_{2} + N a_{2} C O_{3} \to C a C O_{3} + 2 N a C l$
$M g C l_{2} + N a_{2} C O_{3} \to M g C O_{3} + 2 N a C l$
1 mole of $C a C l_{2}$ gives 1 mole of $C a C O_{3}$
So, 111 g $C a C l_{2}$ = 100 g $C a C O_{3}$
1.11 mg $C a C l_{2}$ $= \frac{111}{111} = 1 m g C a C O_{3}$

95 g $M g C l_{2}$ = 100 g $C a C O_{3}$
0.95 mg $M g C l_{2}$ = $\frac{95}{95} = 1 m g C a C O_{3}$
Hardness of $C a C O_{3}$
$= \frac{(1 + 1) \times 10^{- 3} \times 10^{6}}{10^{3}}$ ppm
$= 2$ ppm

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