$40 g N a O H, 106 g N a_{2} C O_{3}$ and $84 g N a H C O_{3}$ is dissolved in water and the solution is made 1 litre. 20 mL of this stock solution is titrated with $1 N H C l,$ hence which of the following statements are correct?

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The titre reading of $H C l$ will be 80 mL if the methyl orange is used indicator from the very beginning

The titre reading of $H C l$ will be 40 mL if the methyl orange is used indicator after the $1^{s t}$ end point.

The titre reading of $H C l$ will be 60mL. if phenolphthalein is used indicator from the very beginning

The titre reading of $H C l$ will be 40 mL. if phenolphthalein is used indicator from the very beginning

answer is A, C, D.

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Detailed Solution

$40 g N a O H$ in litre $M = 1 N = 1$
$106 g N a_{2} N O_{3}$ in litre $M = 1 N = 1 \times 2 = 2$
$84 g N a H C O_{3}$ in litre $M = 1 N = 1$
A) $H p h = 1 (20) + \frac{1}{2} 2 \times 20 = 40 m l$
C) $M e O H$ after $H p h \frac{1}{2} 2 (20) + 1 \times 20 = 40 m l$
D) $M e O H$ from beginning $1 \times 20 \times 2 (20) + 1 \times 20 = 80 m l$