In the mixture of $N a H C O_{4}$ and $N a_{2} C O_{3}$ , volume of a given $H C l$ required is $x m L$ with phenolphthalein indicator and $y m L$ with methyl orange indicator in same titration. Hence, volume of $H C l$ for complete reaction of $N a_{2} C O_{3}$ present in the original mixture is

2 x

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y

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x / 2

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(y - x)

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Solution

The correct option is D $2 x$
In presence of phenolphthalein, $50$ % $N a_{2} C O_{3}$ is neutralised whereas $N a H C O_{3}$ remains unaffected. In presence of methyl orange, both $N a_{2} C O_{3}$ and $N a H C O_{3}$ will be $100$ % neutralised.
Let volume of $H C l$ for complete reaction of $N a_{2} C O_{3} = V_{1} m L$ and volume of $H C l$ for complete reaction of $N a H C O_{3} = V_{2} m L$ . With phenolphthalein, $50$ % $N a_{2} C O_{3}$ will be neutralized.
$∴ \frac{V_{1}}{2} = x, V_{1} = 2 x$ .

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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 :

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