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Permanent Hardness

A sample of h...

Question

A sample of hard water has its hardness due to $M g S O_{4}$ only. When this water is passed through anion exchange resin, $S O_{4}^{2 -}$ ions are replaced by $O H^{-} . 25 m l$ of hard water sample so treated require $20 m l$ of $10^{- 3} M H_{2} S O_{4}$ . What is the hardness of water expressed in terms of ppm of $C a C O_{3} ?$

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Solution

The following reaction is represented as-

$H_{2} S O_{4} + 2 O H^{-} \to S O_{4}^{2 -} + 2 H_{2} O$

$∴$ In 25 ml treated water,

$n_{O H^{-}} = 20 \times 10^{- 3} \times 2 m m o l$

$∴ n_{{M g S O}_{4}} = \frac{n_{{O H}^{-}}}{2} = 20 \times 10^{- 3} m m o l$

$∴$ In 1L $n_{M g S O_{4}} = \frac{20 \times 10^{- 3}}{25} \times 1000 m m o l$

$= \frac{4}{5} m m o l$

$∴ m_{C a C O_{3}} = \frac{4}{5} \times 100 m g = 80 m g$

i.e. Hardness of water $= 80 mg/L = 80 ppm$

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