Question

For the mixture of ${\text{NaHC}}{{\text{O}}_{\text{3}}}\,{\text{ + }}\,{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ volume, of ${\text{HCl}}$ required is x mL with phenolphthalein indicator and then y mL with methyl orange indicator in same titration. Hence volume of ${\text{HCl}}$ for complete reaction of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is:
A.$2x$
B. $y$
C.$\dfrac{x}{2}$
D. $y - x$

Answer 1

Hint: When ${\text{NaHC}}{{\text{O}}_{\text{3}}}\,{\text{ + }}\,{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is titrated with ${\text{HCl}}$ in the presence of phenolphthalein indicator, the phenolphthalein indicator gives endpoint only at neutralization of $50\% $ amount of the ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ whereas methyl orange indicator gives endpoint at complete neutralization of both of the compounds.

Complete step by step answer:
The mixture of ${\text{NaHC}}{{\text{O}}_{\text{3}}}\,{\text{ + }}\,{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is titrated with ${\text{HCl}}$, the neutralization of compounds depend upon the indicator.
In presence of phenolphthalein indicator, sodium bicarbonate ${\text{NaHC}}{{\text{O}}_{\text{3}}}$ does not neutralize. Only the $50\% $ amount of the sodium carbonate ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ neutralizes.
So, x mL is used in the neutralization of $50\% $ of sodium carbonate ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$. So,
$\dfrac{1}{2}{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\, = \,x\,{\text{mL}}$
In presence of methyl orange indicator, sodium bicarbonate ${\text{NaHC}}{{\text{O}}_{\text{3}}}$ neutralizes completely and sodium carbonate ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ also neutralizes completely.
As only $50\% $ amount of the sodium carbonate ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is left because $50\% $ amount of the sodium carbonate ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is already neutralised in presence of phenolphthalein indicator. So, x mL out of y mL is again will be used to neutralise the left amount of the sodium carbonate${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$.
So, the total amount of hydrochloric acid ${\text{HCl}}$ to neutralise the complete sodium carbonate ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is,
$\,x\,{\text{mL}}\,{\text{ + }}x\,{\text{mL}}\, = {\text{2x }}\,{\text{mL}}$
So, the total amount of hydrochloric acid ${\text{HCl}}$ to neutralise the complete sodium carbonate ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is ${\text{2x }}\,{\text{mL}}$.

Therefore the option (A) ${\text{2x}}$, is correct.

Note:
Out of the y mL, x mL is used to neutralize the sodium carbonate ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ so, the amount of hydrochloric acid ${\text{HCl}}$ used in neutralization of sodium bicarbonate ${\text{NaHC}}{{\text{O}}_{\text{3}}}$ is $y\,{\text{mL}}\, - x\,{\text{mL}}\,$.