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Hardness of Water and Treatment

100 ml of sam...

Question

100 ml of sample of hard water is passed through a column of the cation exchange resin $H_{2} R$ . The water coming out of the column requires 20 ml of 0.025 M $N a O H$ for the titration. The hardness of water expressed as ppm of $C a^{2 +}$ ion is:

100

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Solution

The correct option is A 100
$H_{2} R + C a^{2 +} ⟶ R C a + 2 H^{+}$
$H^{+} + O H^{-} ⟶ H_{2} O$
Thus, in term of equivalent
$[H^{+}] \equiv [O H^{-}] \equiv [C a^{2 +}]$
Water coming out of cation exchange resin is acidic, and neutralised by $O H^{-}$ .
From formula, $N V (H^{+}) = N V (O H^{-})$
$N (H^{+}) = \frac{0.025 \times 20}{100} = 5 \times 10^{- 3} N$
Since,

$g m e q . o f [H^{+}] = g m e q . o f [C a^{2 +}]$
We get,
mass of $C a^{2 +} = 10^{- 2} g m i n 100 m l$
So,
in $10^{6}$ ml it will be 100 gm.
Hence, hardness of water is 100 ppm.

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