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${RH}_{2} (ion exchange resin) can replace {Ca}^{2 +} in hard water$
${RH}_{2} + {Ca}^{2 +} ⟶ RCa + 2 H^{+}$
I L of hard water after passing through RH₂ has pH 2.
Hence, hardness in ppm of Ca²⁺ is:

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200

100

125

answer is A.

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Detailed Solution

${RH}_{2} + {Ca}^{2 +} ⟶ RCa + 2 H^{+}$

Energy Ca²⁺ ion is replaced by two H⁺ ions

pH = 2

$\begin{array}{ll} ∴ [H^{+}] = 10^{- 2} M \\ ∴ [{Ca}^{2 +}] = \frac{10^{- 2}}{2} M = \frac{10^{- 2} \times 40}{2} g L^{- 1} \\ ∴ ppm of {Ca}^{2 +} = \frac{10^{- 2} \times 40 \times 10^{6}}{2 \times 10^{3}} {Ca}^{2 +} = 200 \end{array}$

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