Question

$$ RH_{2} $$ (ion exchange resin ) can replace $$ Ca^{2+} $$ in hard water as :
$$ RH_{2}+Ca^{2+}\rightarrow RCa+2H^{+} $$.

One litre of hard water after passing through $$ RH_{2} $$ has pH = 2. Hence, hardness in ppm of $$ Ca^{2+} $$ is:

Answer 1

Correct option is A. 200
According to the given reaction

$$ RH_{2}+Ca^{2+}\rightarrow RCa+2H^{+} $$

Each mole $$ Ca^{2+} $$ ion replaced by 2 moles $$ H^{+}$$

1 mole $$ H^{+} $$ replaced $$ \Rightarrow \dfrac{1}{2} = 0.5\,mole \,Ca^{2+} $$

Given,
$$ pH = 2 $$

$$ H^{+} = 10^{-2} = 0.01 $$

0.01 mole $$ H^{+} $$ replaced $$ = 0.01\times 05 = 0.005\,moles\,Ca^{2+} $$

Mass $$ Ca^{2+}$$ replaced $$ = 0.005\times 40 = 0.2\,g = 200\,mg $$

Concentration or Hardness of $$ Ca^{2+} = 200\,mg/L $$

$$ = 200\,ppm $$

Hence, the correct option is $$\text{A}$$