In the mixture of $(N a H C O_{3} + N a_{2} C O_{3})$ volume of HCl required is x mL with phenolphthalein indicator and y mL with methyl orange indicator in the same titration. Hence, volume for complete reaction of $N a_{2} C O_{3}$ is :

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x/2

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(y-x)

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Solution

The correct option is A 2x
$a$ moles of ${N a H C O}_{3}$
$b$ moles of ${N a}_{2} {C O}_{3}$
with $p h p$ , ${N a}_{2} {C O}_{3} ⟶ {N a H C O}_{3}$ . $x$ ml
So, ${N a H C O}_{3} ⟶ H_{2} {C O}_{3}$ also requires $x$ ml
Total volume for ${N a}_{2} {C O}_{3}$ is $2 x$ ml.

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Similar questions

Q. In the mixture of

N a H C O_{3} + N a_{2} C O_{3}

, volume of

H C l

required is

x

mL with phenolphthalein indicator and then

y

mL with methyl orange indicator in same titration. Hence, volume of

H C l

for complete reaction of

N a_{2} C O_{3}

is :

In the mixture of $N a H C O_{3}$ and $N a_{2} C O_{3}$ volume of a given HCl required is 'x' ml with phenolphthalein indicator and further 'y' ml required with methyl orange indicator. Hence volume of $H C l$ for complete reaction of $N a H C O_{3}$ is