Question

Which of the following is/are correct for the titration of a mixture of $NaOH$, $N{a}_{2}C{O}_3$ and inert substances against $HCl$?

• A. Phenolphthalein is used to detect the end point when half equivalent of $N{a}_{2}C{O}_3$ and full equivalent of $NaOH$ is consumed.
• B. Phenolphthalein is used to detect the second end point.
• C. Methyl orange is used to detect the final end point.
• D. Methyl orange is used to detect the first end point.

Answer 1

Answer: (A), (C)

The correct options are
A Phenolphthalein is used to detect the end point when half equivalent of $N{a}_{2}C{O}_3$ and full equivalent of $NaOH$ is consumed.
C Methyl orange is used to detect the final end point.

Consider a mixture of $NaO{H}_{\left(aq\right)}$ and $N{a}_{2}C{O}_{3\left(aq\right)}$.
Reaction between $HC{l}_{\left(aq\right)}$ and $N{a}_{2}C{O}_{3\left(aq\right)}$ takes place in two stages:
$HC{l}_{\left(aq\right)}+N{a}_{2}C{O}_{3\left(aq\right)}\to NaHC{O}_{3\left(aq\right)}+H_2{O}_{\left(l\right)}............\left(1\right)$
$HC{l}_{\left(aq\right)}+NaHC{O}_{3\left(aq\right)}\to NaC{l}_{\left(aq\right)}+C{O}_{2\left(g\right)}+H_2{O}_{\left(l\right)}............\left(2\right)$
While that between $HC{l}_{\left(aq\right)}$ and $NaO{H}_{\left(aq\right)}$ completes in only one step:
$HC{l}_{\left(aq\right)}+NaO{H}_{\left(aq\right)}\to NaC{l}_{\left(aq\right)}+H_2{O}_{\left(l\right)}............\left(3\right)$
Solution mixture of reaction $\left(1\right)$ at the equivalence point is alkaline, that of reaction $\left(2\right)$ is acidic and of reaction $\left(3\right)$ is neutral.

Thus, the whole titration should have three breaks in the pH curve, corresponding to the above three stages. Reactions $\left(1\right)$ and $\left(3\right)$ can be indicated by phenolphthalein and that of reaction $\left(2\right)$ can be indicated by methyl orange.