$1.03 g$ mixture of sodium carbonate and calcium carbonate require $20 m L N H C l$ for complete neutralisation. Calculate the percentage of sodium carbonate and calcium carbonate in the given mixture.

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Solution

$N a_{2} C O_{3} 106 + 2 H C l 2 \times 36.5 \to 2 N a C l + H_{2} O + C O_{2}$
Eq. mass $53 1 g e q . 36.5 1 g e q .$
$C a C O_{3} 100 + 2 H C l 2 \times 36.5 \to C a C l_{2} + H_{2} O + C O_{2}$
Eq. mass $50 1 g e q . 36.5 1 g e q .$
Let $x g C a C O_{3}$ be present in the mixture.
Mass of $N a_{2} C O_{3}$ in the mixture $= (1.03 - x) g$
No. of $g$ equivalent of $C a C O_{3} = \frac{x}{50}$
No. of $g$ equivalents of $N a_{2} C O_{3} = \frac{(1.03 - x)}{53}$
No. of $g$ equivalents in $20 m L N H C l = \frac{N o r m a l i t y \times V o l .}{1000}$
$= \frac{1 \times 20}{1000} = \frac{1}{50}$
At equivalence point,
No. of $g$ equivalent of $C a C O_{3} +$ No. of $g$ equivalent of $N a_{2} C O_{3} =$ No. of gram equivalent of $H C l$
$\frac{x}{50} + \frac{1.03 - x}{53} = \frac{1}{50}$
or $x = 0.50$
$C a C O_{3} = 0.50 g$ , % $C a C O_{3} = \frac{0.50}{1.03} \times 100 = 48.54$
$N a_{2} C O_{3} = 0.53 g$ , % $N a_{2} C O_{3} = \frac{0.53}{1.03} \times 100 = 51.46$ .

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1.03 g mixture of sodium carbonate and calcium carbonate require 20 mL N HCl for complete neutralisation. Calculate the percentage of sodium carbonate and calcium carbonate in the given mixture.

$1.03 g$ mixture of sodium carbonate and calcium carbonate require $20 m L N H C l$ for complete neutralisation. Calculate the percentage of sodium carbonate and calcium carbonate in the given mixture.