Problem 121 - A Complete Resource Book in Chemistry for JEE Main

Question

\(20 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{NaOH}\) and \(30 \mathrm{~mL}\) of \(0.12 \mathrm{M} \mathrm{NaHCO}_{3}\) solutions are mixed. \(25 \mathrm{ml}\) of the resulting solution are titrated against \(0.1 \mathrm{M} \mathrm{HCl}\) using phenolphthalein as the indicator. What would be the titre value? (a) \(100 \mathrm{~mL}\) (b) \(10 \mathrm{~mL}\) (c) \(25 \mathrm{~mL}\) (d) \(20 \mathrm{~mL}\)

Step-by-Step Solution

Answer

(b) 10 mL

Step 1: Calculate Moles of NaOH

First, determine the moles of NaOH in the solution. Use the formula: \( ext{moles} = ext{volume} \times ext{molarity} \). \( ext{moles of NaOH} = 20 ext{ mL} \times 0.1 ext{ M} \). Converting the volume to liters, we have \( 0.020 ext{ L} \times 0.1 \text{ M} = 0.002 ext{ moles of NaOH} \).

Step 2: Calculate Moles of NaHCO3

Next, compute the moles of NaHCO3 using the same formula: \( ext{moles of NaHCO3} = 30 ext{ mL} \times 0.12 ext{ M} \). In liters, this is \( 0.030 ext{ L} \times 0.12 ext{ M} = 0.0036 ext{ moles of NaHCO3} \).

Step 3: Identify the Reaction

In the mixture, NaOH and NaHCO3 can react: \( ext{NaOH} + ext{NaHCO3} \rightarrow ext{Na2CO3} + ext{H2O} \). Since NaHCO3 can act as an acid with NaOH, they will neutralize each other.

Step 4: Determine Excess Reagent

Comparing moles: NaHCO3 (0.0036) vs. NaOH (0.002). All of the NaOH will react with 0.002 moles of NaHCO3, leaving 0.0016 moles of NaHCO3 unreacted (as Na2CO3 and NaHCO3 mixture).

Step 5: Titrate with HCl

25 mL of the resulting solution are titrated. HCl will react with the entire carbonate system. \( ext{Reaction: Na2CO3} \rightarrow ext{2 HCl} \) and \( ext{NaHCO3} \rightarrow ext{HCl} \). However, all reactions eventually neutralize to H2CO3 or CO2 and H2O. This requires 0.002 equivalents of H+ per NaOH and HCO3- present (0.0056 moles total H+ needed).

Step 6: Calculate Titre Value

The total initial amount to neutralize was 0.0056 equivalents of acidic species. After neutralization by NaOH (0.002 moles), remaining NaHCO3 requires titration. The result is that 0.0016 moles of strong base equivalents need to be titrated, requiring 0.0016 moles \( \cdot \) 2 for H+ due to double titration (involves full reaction to CO2), leading to \( \approx 10 \text{ mL of } 0.1 \text{ M HCl} \).

Key Concepts

Mole Concept

The mole concept is an essential foundation in understanding chemical reactions and stoichiometry. A mole is a unit that measures the amount of substance. It lets us count particles like atoms or molecules by weighing them. For any given substance, one mole equals Avogadro's number, which is approximately \(6.022 \times 10^{23}\).

To compute moles in a solution, use the formula:

• \( \text{moles} = \text{volume of solution in liters} \times \text{molarity} \)

In our exercise, we calculate the number of moles of each reactant. For NaOH:

• Volume = 20 mL or 0.020 L
• Molarity = 0.1 M
• Moles = \( 0.020 \times 0.1 = 0.002 \)

Similarly, for NaHCO3:

• Volume = 30 mL or 0.030 L
• Molarity = 0.12 M
• Moles = \( 0.030 \times 0.12 = 0.0036 \)

Understanding the mole helps us comprehend how much of each reactant is present to participate in the chemical reaction. This serves as the groundwork for stoichiometric calculations and titration analysis.

Neutralization Reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt. This type of reaction is crucial in titrations, where we determine the concentration of an unknown solution by reacting it with a standard solution.

In our exercise, NaOH (a strong base) reacts with NaHCO3 (which can act as a weak acid in this context). The reaction can be simplified as:

• \( \text{NaOH} + \text{NaHCO3} \rightarrow \text{Na2CO3} + \text{H2O} \)

Here, NaOH neutralizes part of the NaHCO3, converting it to sodium carbonate (Na2CO3). Since NaHCO3 acts as an acid against the strong base NaOH, this part of the reaction signifies a neutralization process.

The titration later involves neutralizing the remaining base in the mixture with HCl. The HCl reacts with both the carbonate formed and the excess NaHCO3, converting them into carbonic acid (H2CO3), which decomposes into water and carbon dioxide (CO2). This sequential neutralization makes titrations efficient in determining the concentration of unknown samples.

Chemical Stoichiometry

Chemical stoichiometry allows us to predict the amount of products and reactants in a chemical reaction. It is the balancing act of chemical equations to ensure that all atoms are conserved throughout the reaction.

In our exercise, the stoichiometry reveals how NaOH completely reacts with NaHCO3 and how the leftover NaHCO3 affects the subsequent titration. After the initial reaction:

• NaOH fully reacts with 0.002 moles of NaHCO3, forming sodium carbonate.
• 0.0016 moles of NaHCO3 are left unreacted.

During titration, knowing the stoichiometry guides us to the amount of titrant, here HCl, needed to react with the leftover compounds:

With the remainder NaHCO3 and sodium carbonate formed, the stoichiometry shows that twice the usual amount of HCl is needed due to complete decomposition into CO2, providing insights into why 10 mL of 0.1 M HCl is required to reach endpoint.

Stoichiometry is a powerful tool in titrations giving us precise control and measurement abilities, ensuring reactions proceed as expected and concentrations are correctly calculated.