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Standard XII

Chemistry

Ostwald Dilution Law

A sample of h...

Question

A sample of hard water contains $96$ ppm of $S O_{4}^{2 -}$ and $183$ ppm of $H C O_{3}^{-}$ with ${C a}^{2 +}$ as the only cation. Few moles of $C a O$ will be required to remove $H C O_{3}^{-}$ from $1000$ kg of this water. If $1000$ kg of this water is treated with the amount of $C a O$ calculated above, the concentration (in ppm) of residual ${C a}^{2 +}$ ions will be :
(Assume $C a C O_{3}$ to be completely insoluble in water. The ${C a}^{2 +}$ ions in one litre of the treated water are completely exchanged with hydrogen ions. One ppm means one part of the substance in one million part of water, mass/mass)

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Solution

Sample of hard water contains $96$ ppm $S O_{4}^{2 -}$ and $40$ ppm ${C a}^{2 +} (C a S O_{4})$ .
Also, it contains $183$ ppm $H C O_{3}^{-}$ and $60$ ppm ${C a}^{2 +} [C a {(H C O_{3})}_{2}]$ .
To remove $C a {(H C O_{3})}_{2}$ from $10^{3}$ kg or $10^{6}$ g sample of hard water which contains $243 g C a {(H C O_{3})}_{2}$ or $\frac{3}{2}$ moles of $C a {(H C O_{3})}_{2}$ , $C a O$ required is $\frac{3}{2}$ moles.
$C a {(H C O_{3})}_{2} + C a O \to C a C O_{3} + H_{2} O + C O_{2}$
Thus, moles of $C a O$ required $= \frac{3}{2}$ or $1.5$
Also, ${C a}^{2 +}$ ions left in solution are of $C a S O_{4}$ $= 40$ ppm

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