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 :

2 x

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y

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\frac{x}{2}

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y - x

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Solution

The correct option is A $2 x$
i. With phenolphthalein indicator, $N a H C O_{3}$ does not react with $H C l$ whereas with $N a_{2} C O_{3}$ is $50 %$ reaction.
$V_{H C l} = x m L$ (Half titre value of $N a_{2} C O_{3}$ )
ii. With methyl orange indicator, $N a H C O_{3}$ reacts completely with $H C l$ and with $N a_{2} C O_{3}$ is $100 %$ reaction.
But $y$ $m L$ of $H C l$ is added after $N a_{2} C O_{3}$ has reacted upto $N a H C O_{3}$ i.e., half titre value of $N a_{2} C O_{3}$ .
$V_{H C l}$ =Full litre value of $N a H C O_{3}$ + Half titre value of $N a_{2} C O_{3}$
$y m L = x^{'} + x m L$
$∴$ Volume of $H C l$ required for complete reaction of $N a_{2} C O_{3} = x m L + x m L = 2 x m L$

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Q. 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 :

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