40mL 0.05M solution of sodium sesquicarbonate dihydrate $(N a_{2} C O_{3} . N a H C O_{3} .2 H_{2} O)$ is titrated against 0.05M HCl solution, xmL of acid is required to reach the phenolphthalein end point while y mL of same acid were required when methyl orange indicator was used in a separate titration. Which of the following is (are) correct statement(s)?

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$y - x = 80 m L$

If the titration is started with phenolphthalein indicator and methyl orange is added after the first end point, 2xmL of HCl would be required further to reach the second end point

$y + x = 160 m L$

If the same volume of same solution is titrated against 0.10M NaOH, x/2mL of base would be required in the presence of phenolphthalein indicator

answer is A, B, C.

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Detailed Solution

The salt has equal moles of $N a_{2} C O_{3}$ and $N a H C O_{3}$ . Also when phenolphthalein indicator is used, only reaction occur till end point is

$\Rightarrow x = 40 m L H e n c e, y - x = 80 a n d y + x = 160$

If methyl orange indicator is added subsequently at phenolphthalein end point, xML of HCl would be required for $N a_{2} C O_{3}$ and xmL for original $N a H C O_{3}$ , i.e, a total of 2xmL HCl would be required further. With NaoH, the amphoteric $N a H C O_{3}$ will be neutralized as