Question

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

Answer 1

Neutralisation reactions take place between an acid and a base to give salt and water. Phenolphthalein changes color at a different pH than methyl orange. We shall calculate the equivalents HCl required to neutralise $N{a_2}C{O_3}$ and $NaOH$ for both the indicators and use that difference to calculate the weight and thus amount of $N{a_2}C{O_3}$ and $NaOH$ in the mixture.

Complete step by step solution:
When phenolphthalein is the indicator, whole of $NaOH$ has been neutralised,
$NaOH + HCl \to NaCl + {H_2}O$
where sodium hydroxide (base) reacts with an acid, hydrochloric acid to give salt (sodium chloride) and water and carbonate converted into bicarbonate, i.e.,
$N{a_2}C{O_3} + HCl \to NaHC{O_3} + NaCl$ where $NaHC{O_3}$ is sodium bicarbonate that results from the reaction of sodium carbonate with hydrochloric acid.
So, 25mL $\dfrac{N}{{10}}HCl \equiv NaOH + \dfrac{1}{2}N{a_2}C{O_3}$ present in 25 mL of mixture.
In another titration, when methyl orange is the indicator, whole of $NaOH$ has been neutralised and carbonate converted into carbonic acid, i.e.,
$N{a_2}C{O_3} + 2HCl \to 2NaCl + {H_2}C{O_3}$
So, 30mL $\dfrac{N}{{10}}HCl \equiv NaOH + N{a_2}C{O_3}$ present in 25 mL of mixture.
Hence, $\left( {30 - 25} \right){\text{mL}}\dfrac{N}{{10}}HCl \equiv \dfrac{1}{2}N{a_2}C{O_3}$ present in 25 mL of mixture.
Thus, we can say that,
$10{\text{mL}}\dfrac{N}{{10}}HCl \equiv N{a_2}C{O_3}$ , present in 25 mL of mixture.
$10{\text{mL}}\dfrac{N}{{10}}N{a_2}C{O_3}$ solution.
Amount of sodium carbonate in the solution can be calculated from the equation given below.
Amount of $N{a_2}C{O_3} = \dfrac{{53 \times 10}}{{10 \times 1000}} = 0.053{\text{g}}$ .
Therefore, $0.053{\text{g}}$ of $N{a_2}C{O_3}$ is present in 25mL of the mixture.
One litre comprises 1000mL. Thus, the amount of $N{a_2}C{O_3}$ in 1000mL of the solution, can be calculated by the unitary method.
Amount of $N{a_2}C{O_3}$ in 1L=1000mL of the solution, is given by, $\dfrac{{0.053}}{{25}} \times 1000 = 2.12{\text{g}}$ .
Now, $\left( {30 - 10} \right){\text{mL}}\dfrac{N}{{10}}HCl \equiv N{a_2}C{O_3}$ , present in 25 mL of mixture.
$\equiv 20mL\dfrac{N}{{10}}NaOH.$
Amount of $NaOH$ present in 25mL of the mixture, is $= \dfrac{{40 \times 20}}{{10 \times 1000}} = 0.08g$ .
Amount of $NaOH$ in 1L=1000mL of the solution is given by, $\dfrac{{0.08}}{{25}} \times 1000 = 3.20g$ .

Note:
Phenolphthalein, an organic compound of the phthalein family that is widely employed as an acid-base indicator. As an indicator of a solution's pH, phenolphthalein is colourless below pH $pH = 8.5$ and attains a pink to deep red hue above $pH = 9.0$ phenolphthalein.
Phenolphthalein is a weak acid and is colorless in solution although its ion is pink. If hydrogen ions ( ${H^ + }$ , as found in an acid) were added to the pink solution, the equilibrium would switch, and the solution would be colorless. Adding hydroxide ions ( $O{H^ - }$ , as found in bases) will change the phenolphthalein into its ion and turn the solution pink.