Question

A mixture of $25{\text{ }}mL$ $NaOH$ and $N{a_2}C{O_3}$ when titrated with $N/10$ $HCl$ using phenolphthalein indicator required 25{\text{ }}mL$$$$HCl . The same volume of mixture when titrated with $N/10$ $HCl$ using methyl orange indicator required $30{\text{ }}mL$ of $HCl$ . Calculate the amount of $N{a_2}C{O_3}$ and $NaOH$ in one liter of this mixture.

Answer 1

Acid-base titration is one of the famous titrations for determining the concentration of acid/ base by neutralizing with a known concentration of standard solution of base/acid respectively. This is a quantitative analysis. The pH indicator is added in a very small amount of solution so that the pH of solution can be determined visually.

Complete step by step solution:
There are different indicators for acid-base titration reaction. Majorly phenolphthalein and methyl orange indicators are used.
Phenolphthalein is an acid-base titration indicator. The main application is, it turns colorless when the solution becomes acidic and it turns pink when the solution turns into basic.
Methyl orange is a pH indicator which is used in acid- base titration indicator. Its main application is it turns red when the solution becomes acidic and it turns yellow when the solution becomes basic.
The phenolphthalein is an indicator, where the sodium hydroxide $NaOH$ gets neutralised and sodium carbonate $N{a_2}C{O_3}$ converted into sodium bicarbonate $NaHC{O_3}$ .
$NaOH\, + \,HCl\, \to \,NaCl\, + \,{H_2}O$
$N{a_2}C{O_3}\, + \,HCl\, \to \,NaHC{O_3}\, + \,NaCl$
Since, 25{\text{ }}mL$$$$N/10$$$$HCl$$$$ \equiv $$$$NaOH$$$$ + {\text{ }}\dfrac{1}{2}\,N{a_2}C{O_3} present in $25{\text{ }}mL$ of mixture.
In another titration, when Methyl orange indicator is used, the sodium hydroxide $NaOH$ gets neutralised and sodium carbonate $N{a_2}C{O_3}$ converted into carbonic acid (${H_2}C{O_3}$) .
$N{a_2}C{O_2} + {\text{ }}2HCl\, \to \,\,2NaCl{\text{ }} + {\text{ }}{H_2}C{O_3}$
$30\,mL\,\dfrac{N}{{10}}\,HCl\, \equiv \,\dfrac{1}{2}\,N{a_2}C{O_3}$ present in $25{\text{ }}mL$ of mixture.
So,
$(30 - 25)\,mL\,\dfrac{N}{{10}}\,HCl\, \equiv \,\dfrac{1}{2}\,N{a_2}C{O_3}$ present in $25{\text{ }}mL$ of mixture.
Hence,
$10\,mL\,\dfrac{N}{{10}}\,HCl\, \equiv \,\dfrac{1}{2}\,N{a_2}C{O_3}$ present in $25{\text{ }}mL$ of mixture.
$\equiv \,10\,mL\,\,\dfrac{N}{2}\,N{a_2}C{O_3}$ solution
The amount of sodium carbonate $N{a_2}C{O_3}$,
$= \,\dfrac{{53\, \times \,10}}{{10\, \times \,1000}}\, = \,0.053\,g$
The amount of sodium carbonate $N{a_2}C{O_3}$ is present in $25{\text{ }}mL$ of mixture,
The amount which is present in one litre of mixture,
$= \dfrac{{0.053}}{{25}}\, \times \,1000\, = 2.12\,g$
$(30 - 10)\,mL\,\dfrac{N}{{10}}\,HCl\, \equiv \,NaOH$ present in $25{\text{ }}mL$ of mixture.
$\equiv \,20mL\,\dfrac{N}{{10}}NaOH$
The amount of sodium hydroxide $NaOH$ in $25{\text{ }}mL$ of mixture,
$= \,\dfrac{{40\, \times \,20}}{{10\, \times \,1000\,}}\, = \,0.08\,g$
The amount which is present in one litre of mixture,

$$= \,\dfrac{{0.08}}{{25}}\, \times \,1000\, = \,3.20\,g \\\ \\\$$

Note: Phenolphthalein is used as an indicator when strong base and weak acid titration are performed. In the same manner methyl orange is used as an indicator when strong acid and weak base titration are performed. These are used in the lab to identify the quantity by the colour change.