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

Hint: 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.