Problem 50 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A \(2.0 \mathrm{~g}\) sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of \(\mathrm{CO}_{2}\) ceases. The volume of \(\mathrm{CO}_{2}\) at \(750 \mathrm{~mm}\) Hg pressure and at \(298 \mathrm{~K}\) is measured to be \(123.9 \mathrm{~mL}\). A \(1.5 \mathrm{~g}\) of the same sample requires 150 mL. of \((\mathrm{M} / 10) \mathrm{HCl}\) for complete neutralisation. Calculate the \(\%\) composition of the components of the mixture. $$ \text { Page } 36,1,773-Q+\quad[1992-5 \text { Marks }] $$

Step-by-Step Solution

Answer

The exact percentage composition requires solving equations iteratively. Calculate and use stoichiometry relations.

Step 1: Convert Pressure to Standard Units

Convert the given pressure from mm Hg to atm since standard conditions use atm. Given pressure is 750 mm Hg. Use the conversion: \(1 \text{ atm} = 760 \text{ mm Hg}\), so:\[P = \frac{750 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 0.9868 \text{ atm}\]

Step 2: Use Ideal Gas Law

Use the Ideal Gas Law to find the moles of \( \mathrm{CO}_{2} \). The Ideal Gas Law is \( PV = nRT \). Here:- \(P = 0.9868 \text{ atm}\)- \(V = 123.9 \text{ mL} = 0.1239 \text{ L}\)- \(R = 0.0821 \text{ L atm/mol K}\)- \(T = 298 \text{ K}\)Plug in these values to solve for \(n\):\[n = \frac{PV}{RT} = \frac{0.9868 \times 0.1239}{0.0821 \times 298} = 0.00505 \text{ mol}\]

Step 3: Calculate Mass of Carbonate in 2.0 g Sample

The CO2 evolved comes from sodium carbonate and sodium bicarbonate. The molar mass of \(\mathrm{Na}_2\mathrm{CO}_3\) is 106 g/mol and \(\mathrm{NaHCO}_3\) is 84 g/mol. Moles of CO2: 0.00505 mol. Each mole of \(\mathrm{Na}_2\mathrm{CO}_3\) and \(\mathrm{NaHCO}_3\) evolves one mole of CO2.Set up the equation for mass: \[ x + y = 2.0 \quad (1)\]Where x is the mass of \(\mathrm{Na}_2\mathrm{CO}_3\) and y is the mass of \(\mathrm{NaHCO}_3\).

Step 4: Calculate Equivalent Acid Required for Titration

Determine the acidity (amount of HCl required for the bicarbonate neutralization) which corresponds to the given 1.5 g sample. \(150 \text{ mL} \text{ of } M/10 \text{ HCl} = 0.015 \text{ mol/L}\).\[0.015 \times 0.150 = 0.00225 \text{ mol of HCl}\]This is equal to the amount of bicarbonate because:- 1 mole \(\text{NaHCO}_3\) reacts with 1 mole \(\text{ HCl}\) to give \(\text{NaCl}\), \(\text{H}_2\text{O}\), and \(\text{CO}_2\).

Step 5: Solve the System of Equations

Using the stoichiometry of reactions:- \(1 \text{ mol } \text{Na}_2\text{CO}_3 \equiv 2 \text{ mol }\text{HCl}\)- \(1 \text{ mol } \text{NaHCO}_3 \equiv 1 \text{ mol }\text{HCl}\)From the given, the proportion of acid required indicates the presence of \(\text{NaHCO}_3\) and \(\text{Na}_2\text{CO}_3\). Solve the equations based on these masses: \(x = 2z + y \text{ for masses.}\)Iterate to quantify each compound.

Step 6: Determine Percent Composition

Calculate the percentage of each component using their mass values:- Mass of \(\mathrm{Na}_2\mathrm{CO}_3\), \(\mathrm{NaHCO}_3\), and \(\mathrm{Na}_2\mathrm{SO}_4\).- The percentage of each equals \( \left(\frac{\text{mass of the component}}{2.0}\right) \times 100 \).

Key Concepts

Stoichiometry

Stoichiometry is a key concept in chemistry that involves calculating the amounts of reactants and products in chemical reactions. It is based on the balanced chemical equations that represent how molecules and atoms interact. Using stoichiometry, we can predict the amount of substances consumed and produced in a reaction. It involves using mole ratios derived from balanced equations to determine unknown quantities.
For example, when sodium carbonate (\(\text{Na}_2\text{CO}_3\)) and sodium bicarbonate (\(\text{NaHCO}_3\)) react with acids, they release carbon dioxide gas (\(\text{CO}_2\)). By understanding the stoichiometry of these reactions, we can calculate the amount of \(\text{CO}_2\) generated and use that data to find the mass of the original compounds.

• This assumes that each mole of carbonate, bicarbonate yields one mole of \(\text{CO}_2\).
• In the problem, stoichiometry helps set equations that allow solving for the unknown masses of sodium carbonate and bicarbonate based on their roles in the reaction.

This process helps in unraveling the composition of complex mixtures by observing how they react.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the four key properties of gases: pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and moles (\(n\)). The equation is represented as \(PV = nRT\), where \(R\) is the ideal gas constant. This relationship assumes the gas behaves ideally, meaning the particles do not interact except for elastic collisions.
In practical terms, when you have a known volume of gas collected at certain temperature and pressure conditions, you can determine the amount in moles using this law. This was applied in the exercise to find the moles of \(\text{CO}_2\) produced by the reaction:

• First, convert given measurements into appropriate units (e.g., pressure from mm Hg to atm, volume from \(\text{mL}\) to \(\text{L}\)).
• Insert these values into the equation to solve for \(n\), revealing how much gas is present.

This powerful tool bridges the measurable properties of gases with the amounts of substances involved in chemical reactions.

Percent Composition

Percent composition refers to the percentage by mass of each component in a chemical mixture. It is a helpful tool for analyzing mixtures quantitatively. In chemistry, it allows us to understand the proportion of elements or compounds in a sample.
To find the percent composition, you take the mass of each component and divide it by the total mass of the mixture, then multiply by 100 to get a percentage.

• The problem illustrates this by calculating the percentage of sodium carbonate, sodium bicarbonate, and sodium sulfate in a 2.0 g sample.
• By first determining the mass of each component using chemical reactions and stoichiometry, those values are then expressed as percentages of the total sample.

This approach gives insight into mixture composition, essential for both laboratory work and industrial applications where precise formulation is crucial.

Neutralization Titration

Neutralization titration is a technique used to determine the concentration of an unknown acid or base solution. It involves adding a titrant of known concentration gradually until the solution neutralizes, indicated by a visible change (like a color shift in an indicator).
In the context of the exercise, this method quantifies the bicarbonate content in the mixture by using hydrochloric acid (HCl) to neutralize it. Here's how it works:

• As \(\text{NaHCO}_3\) reacts with \(\text{HCl}\), it releases \(\text{CO}_2\) and forms \(\text{NaCl}\) and \(\text{H}_2\text{O}\).
• The amount of \(\text{HCl}\) used directly corresponds to the \(\text{NaHCO}_3\) present.
• Measurements and calculations based on the \(\text{HCl}\) volume help find the bicarbonate's mass and, consequently, its percentage in the original mixture.

This titration technique is widely employed in chemistry to analyze substances and is integral in controlling pH levels in various products.