Washing soda $(N a_{2} C O_{3} .10 H_{2} O)$ is widely used in softening of hard water. If 1L of hard water requires 0.0143 g of washing soda, what is hardness of water in terms of ppm of $C a C O_{3}$ ?

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Solution

Molecular weight of washing soda=286 g/mol
Molecular weight of $C a C O_{3}$ =100 g/mol
$C a S O_{4} + N a_{2} C O_{3} \to C a C O_{3} + N a_{2} S O_{4}$
0.0143 g of $N a_{2} C O_{3} .10 H_{2} O = \frac{0.0143 \times 10^{3}}{286 / 2} = 0.1 m E q$
0.1mEq of $N a_{2} C O_{3} = \frac{0.1}{2} m m o l o f C a C O_{3}$ in 1L
$= 0.05 \times 10^{- 3} \times 100 g i n 10^{3}$ mL
$= 0.005 g i n 10^{3} m L = 5 p p m o f C a C O_{3}$

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