A 1kg hard water sample (density = 1 g/cc) contains 192ppm of $S O_{4}^{2 -}$ and 183ppm of $H C O_{3}^{-}$ with $C a^{2 +}$ as only cation. Few drops of conc $H_{2} S O_{4}$ is added to hard water sample which is just sufficient to remove entire $H C O_{3}^{-}$ in the form of $C O_{2}$ . Now the hard water sample is passed through cation exchange resin [replaces cations with appropriate no. of $H^{+}$ ] and volume is made up to 2000ml by adding distilled water .10ml of it is pipetted out and titrated by using $10^{- 3} M$ NaOH solution . The volume of NaOH solution required in ml is__________

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answer is 35.

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Detailed Solution

$n_{S O_{4}^{2 -} = \frac{192}{96} \times 10^{- 3} = 2 \times 10^{- 3} \Rightarrow n_{c a^{2}} + = 2 \times 10^{- 3}} n_{H C O_{3}^{-} = \frac{183}{61} \times 10^{- 3} = 3 \times 10^{- 3} n' C a^{2 +} = 1.5 \times 10^{- 3}}$
$(n C a^{2 +})$ total $= 3.5 \times 10^{3}$
$2 H C O_{3}^{-} + H_{2} S O_{4} \to S O_{4}^{2 -} + 2 H_{2} O + 2 C O_{2} ↑$
No $H^{+}$ is introduced
After passing through cation exchange resin
$n_{H^{+} = 2 n_{c a^{2 +}} = 7 \times 10^{- 3}}$
In 10ml $n_{H^{+}} = 16 \times 10^{- 3} \times \frac{10}{2000}$
$n o n^{- 1} = 10^{- 3} \times V \Rightarrow n_{O H^{+}} = n_{H^{+}} 10^{3} \times V = 7 \times \frac{10}{2000} \times 10^{- 3} V = \frac{7}{200} L = \frac{7}{200} \times 1000 = 35 m l$