Question

A solution contains \[N{{a}_{2}}C{{O}_{3}}\] and \[NaHC{{O}_{3}}\]. 10 mL of the solution required 2.5mL of 0.1M \[{{H}_{2}}S{{O}_{4}}\] for neutralization using phenolphthalein as an indicator. Methyl orange is then added when a further 2.5mL of 0.2 M \[{{H}_{2}}S{{O}_{4}}\] was required. Calculate the amount of \[N{{a}_{2}}C{{O}_{3}}\] and \[NaHC{{O}_{3}}\] is one litre of the solution.

Answer 1

Hint: To solve this question, we have to keep following things in mind:
\[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] where N represents normality of the solution and V represents the volume of the solution.
Phenolphthalein is used as an indicator. It turns pink at a pH level of 8.2 and continues to a bright magenta at pH 10 and above. So, for this to work as an indicator at the neutralisation point the pH of solution must be basic. So, this will give the amount of \[N{{a}_{2}}C{{O}_{3}}\]used.
The pH range between 3.1 (red) and 4.4 (yellow) is the colour-change interval of methyl orange and the solution must have pH in this range at the neutralisation point. So, this will give amount of \[NaHC{{O}_{3}}\] used



Complete answer: or Complete step by step answer:
As we know Normality = \[{{n}_{f}}\times \]Molarity
\[{{n}_{f}}\]= n factor for \[{{H}_{2}}S{{O}_{4}}\] is 2 hence,
2.5 mL of 0.1 M \[{{H}_{2}}S{{O}_{4}}\]​= 2.5 mL of 0.2 N ​
=\[\dfrac{1}{2}\]​\[N{{a}_{2}}C{{O}_{3}}\]​ present in 10 mL of mixture
So,
5 mL of 0.2 N \[{{H}_{2}}S{{O}_{4}}\]​ solution = \[N{{a}_{2}}C{{O}_{3}}\]present in 10 mL of mixture
= 5 mL of 0.2 N \[N{{a}_{2}}C{{O}_{3}}\]​
=\[\dfrac{0.2\times 53}{1000}\times 5\]= 0.053 g
Thus, the amount of \[N{{a}_{2}}C{{O}_{3}}\]​ present in the solution = \[\dfrac{0.053}{10}\times 1000\]= 5.3 g/L of mixture
Now between first and second end points,
= 2.5 mL of 0.2 M \[{{H}_{2}}S{{O}_{4}}\]​ is used
= 2.5 mL of 0.4 N \[{{H}_{2}}S{{O}_{4}}\]​ is used
= 5 mL of 0.2 N \[{{H}_{2}}S{{O}_{4}}\]​ is used
=\[\dfrac{1}{2}\]\[N{{a}_{2}}C{{O}_{3}}\]​+\[NaHC{{O}_{3}}\]​ present in 10 mL of mixture
(5−2.5)mL 0.2 N \[{{H}_{2}}S{{O}_{4}}\]​
=\[NaHC{{O}_{3}}\]​ present in 10 mL of mixture
= 2.5 mL of 0.2 N \[NaHC{{O}_{3}}\]​
= \[\dfrac{0.2\times 84}{1000}\times 2.5\]= 0.042 g
Hence, amount of \[NaHC{{O}_{3}}\]​in the given solution is = \[\dfrac{0.042}{10}\times 1000\]= 4.20g/L of mixture.
The amount of \[N{{a}_{2}}C{{O}_{3}}\] and \[NaHC{{O}_{3}}\] is one litre of the solution is 0.053g and 0.042 g respectively.

Note:
Neutralization can be defined as a chemical reaction in which acid and a base react quantitatively with each other. In a reaction in water, neutralization results in there being no excess of hydrogen or hydroxide ions present in the solution if both acid and base are present in equivalent amounts. If we consider a neutralization reaction between a weak acid and base, then some of the energy liberated is used in dissociating the weak acid and base. So, the final amount of energy liberated is a little less.