Problem 93 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A solution contains \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NaHCO}_{3} .10 \mathrm{~mL}\) of solution requires \(2.5 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) for neutralisation using phenolphthalein as an indicator. Methyl orange is then added when a further \(2.5 \mathrm{~mL}\) of \(0.2 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) was required. Calculate the amount of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NaHCO}_{3}\) in one litre of the solution

Step-by-Step Solution

Answer

Amount of Na₂CO₃: 25 mmol, NaHCO₃: 50 mmol in 1 L.

Step 1: Understand the Reaction

The solution contains both Na₂CO₃ and NaHCO₃. During titration, Na₂CO₃ reacts completely with H₂SO₄ using phenolphthalein as the indicator and turns colorless when converted to NaHCO₃. NaHCO₃ then further reacts, changing color with methyl orange.

Step 2: Use Phenolphthalein Endpoint

The first step involves the reaction: \[ \text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2 \] 2.5 mL of 0.1 M H₂SO₄ is required, equaling 0.25 mmol. Therefore, the moles of Na₂CO₃ reacted are 0.25 mmol.

Step 3: Use Methyl Orange Endpoint

At the methyl orange endpoint, the reaction is: \[ \text{NaHCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2 \] Additionally, reaction with the remaining NaHCO₃ in the solution occurs. 2.5 mL of 0.2 M H₂SO₄ is needed, giving 0.5 mmol of NaHCO₃ reaction.

Step 4: Calculate Total Amounts in Solution

In the total 10 mL solution, there are 0.25 mmol of Na₂CO₃ and 0.5 mmol of NaHCO₃. To find these in 1 L of solution (1000 mL), multiply by 100: \[ \begin{align*} \text{Na}_2\text{CO}_3: & \ 0.25 \times 100 = 25 \text{ mmol} \ \text{NaHCO}_3: & \ 0.5 \times 100 = 50 \text{ mmol} \end{align*} \]

Key Concepts

Neutralization Reaction

Neutralization reactions are essential processes in chemistry where an acid and a base react to form water and a salt.
In the context of our titration exercise, we observe how sodium carbonate (\( \text{Na}_2\text{CO}_3 \)) and sodium bicarbonate (\( \text{NaHCO}_3 \)) are neutralized by sulfuric acid (\( \text{H}_2\text{SO}_4 \)).
Specifically, each of these compounds goes through unique neutralization steps during titration.

• The \( \text{Na}_2\text{CO}_3 \) is completely neutralized at the phenolphthalein endpoint, resulting in the formation of sodium sulfate, water, and carbon dioxide.
• Subsequently, the \( \text{NaHCO}_3 \) in the solution is further neutralized, which occurs with the addition of a second portion of \( \text{H}_2\text{SO}_4 \), changing the indicator to methyl orange.

This two-step neutralization illustrates how acids and bases react in stages, depending on their composition and the solution's pH at each stage.

Acid-Base Indicators

Acid-base indicators are substances that visibly change their color as the pH of a solution changes.
They are vital in titration for identifying the endpoint of a reaction. In this exercise, two different indicators were used to determine the endpoint of each reaction stage.

Phenolphthalein

Phenolphthalein is an indicator that changes from pink (in basic conditions) to colorless (in acidic conditions) around a pH of 8.3.
It was employed to identify the completion of the first reaction step, where \( \text{Na}_2\text{CO}_3 \) was neutralized.

Methyl Orange

Methyl orange, on the other hand, changes color from yellow to red between a pH of 3.1 to 4.4.
It was used in the second stage of the titration, indicating when the remaining \( \text{NaHCO}_3 \) was neutralized.Choosing an appropriate indicator ensures that the exact point of complete neutralization is identified, which is crucial for accurate titration results.

Molarity Calculations

Molarity is a measure of the concentration of a solute in a solution and is expressed as moles of solute per liter of solution.
In the context of titration, accurate molarity calculations are important for determining the amount of each reactant in the solution.The initial titration requires 2.5 mL of 0.1 M \( \text{H}_2\text{SO}_4 \), which neutralizes \( \text{Na}_2\text{CO}_3 \). This means:\(\text{Moles of } \text{H}_2\text{SO}_4 = 0.1 \, \text{M} \times 2.5 \, \text{mL} / 1000 \ = 0.00025 \, \text{moles} \)
\( = 0.25 \, \text{mmol}\)For the \( \text{NaHCO}_3 \), 2.5 mL of 0.2 M \( \text{H}_2\text{SO}_4 \) is used, so:\(\text{Moles of } \text{H}_2\text{SO}_4 = 0.2 \, \text{M} \times 2.5 \, \text{mL} / 1000 \ = 0.0005 \, \text{moles} \)
\( = 0.5 \, \text{mmol}\)These calculations show how molarity is crucial for assessing how much of each compound reacts and calculating concentrations in the final solution.