Calcium and magnesium ion form a $10^{5}$ L sample of hard water was quantitatively precipitated as carbonates and weight of precipitate obtained was found to be 568 g. On strong heating 264 g of weight is lost. The molarity of the $C a^{2 +}$ ions in hard water is $x \times 10^{- 5}$ M. Find x

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Solution

Let the mass of the $C a C O_{3}$ precipitate obtained be 'a' g and that of $M g C O_{3}$ be (568-a) g.
Given, lost mass is 264 g
So, the amount of weight lost = $\frac{molar mass of C O_{2}}{molar mass of C a C O_{3}} \times m a s s o f C a C O_{3} + \frac{molar mass of C O_{2}}{molar mass of M g C O_{3}} \times m a s s o f M g C O_{3}$
$\frac{44}{100} \times a + \frac{44}{84} \times (568 - a)$ = 264 g
On solving, a = 400 g
Moles of $C a C O_{3}$ = $\frac{g i v e n m a s s}{m o l a r m a s s}$ = $\frac{400}{100}$ = 4 mol
Molarity = $\frac{n u m b e r o f m o l e s}{v o l u m e o f t h e s o l u t i o n i n L}$ = $\frac{4}{10^{5}}$ = $4 \times 10^{- 5}$ M

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