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$RH_2$(ion exchange resin)can replace $Ca^{2+}$ in hand water as,$RH_2+Ca^{2+}\rightarrow RCa+ 2H^+$.1 litre of hard water passing through $RH_2$ has PH is 2.Hence hardness in PPM of $Ca^{2+}$ is:

$(a)\;200\qquad(b)\;100\qquad(c)\;50\qquad(d)\;125$

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$[H^+]=10^{-2}$
$\therefore Ca^{2+}=\large\frac{10^{-2}}{2}$
$\Rightarrow \large\frac{40\times 10^{-2}}{2}$
$\Rightarrow 0.2g/l$
$\therefore$Hardness=$\large\frac{0.2\times 10^6}{10^3}$
$\Rightarrow$ 200PPM
Hence (a) is the correct answer.
answered Jan 9, 2014 by sreemathi.v
 

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