Problem 271 - A Complete Resource Book in Chemistry for JEE Main

Question

The molar conductivities \(\Lambda^{\circ} \mathrm{NaOA} \mathrm{c}\) and \(\Lambda^{\circ} \mathrm{HCl}\) at infinite dilution in water at \(25^{\circ} \mathrm{C}\) are \(91.0\) and \(426.2 \mathrm{~S}\) \(\mathrm{cm}^{2} / \mathrm{mol}\) respectively. To calculate \(\Lambda^{\circ} \mathrm{HOAc}\), the additional value required is (a) \(\Lambda^{\circ} \mathrm{H}_{2} \mathrm{O}\) (b) \(\Lambda^{\circ} \mathrm{KCl}\) (c) \(\Lambda^{\circ} \mathrm{NaOH}\) (d) \(\Lambda^{\circ} \mathrm{NaCl}\)

Step-by-Step Solution

Answer

(c) \(\Lambda^{\circ} \mathrm{NaOH}\) is needed to determine \(\Lambda^{\circ} \mathrm{HOAc}\).

Step 1: Identify Known Values

We are given the molar conductivities \(\Lambda^{\circ} \mathrm{NaOAc} = 91.0 \, \mathrm{S} \cdot \mathrm{cm}^2 / \mathrm{mol}\) and \(\Lambda^{\circ} \mathrm{HCl} = 426.2 \, \mathrm{S} \cdot \mathrm{cm}^2 / \mathrm{mol}\). We need to calculate \(\Lambda^{\circ} \mathrm{HOAc}\).

Step 2: Understand Required Concept

To calculate the molar conductivity of acetic acid (\(\Lambda^{\circ} \mathrm{HOAc}\)), we need the conductivities of its ions, \(\Lambda^{\circ} \mathrm{H^+}\) and \(\Lambda^{\circ} \mathrm{OAc^-}\).

Step 3: Utilize Key Equations

Knowing \(\Lambda^{\circ} \mathrm{NaOAc}\) and \(\Lambda^{\circ} \mathrm{HCl}\) provides us with partial information. We can obtain \(\Lambda^{\circ} \mathrm{Na^+}\), \(\Lambda^{\circ} \mathrm{Cl^-}\), and \(\Lambda^{\circ} \mathrm{H^+}\) through given assets.

Step 4: Determine Missing Data for Completion

We need \(\Lambda^{\circ} \mathrm{NaOH}\) because it allows us to calculate \(\Lambda^{\circ} \mathrm{OH^-}\), which completes the pieces required to compute \(\Lambda^{\circ} \mathrm{HOAc} = \Lambda^{\circ} \mathrm{H^+} + \Lambda^{\circ} \mathrm{OAc^-}\).

Key Concepts

Molar Conductivity at Infinite Dilution

Molar conductivity at infinite dilution refers to the molar conductivity of a solution when the concentration approaches zero. At this point, the conductivity is only dependent on the types of ions present and not on how closely they are packed together. This means that the ions move without any interaction from each other, which makes these calculations exceptionally useful.

With infinite dilution, the molar conductivity \( \Lambda^{\circ} \) can relate closely to studying electrochemical cells and ion behavior in solutions. Understanding molar conductivity at infinite dilution helps enhance the explanation of the conductivity contributions from individual ions. This is important because it allows us to deduce other conductivity values using relationships such as \( \Lambda^{\circ}_{\text{total}} = \Lambda^{\circ}_{\text{cation}} + \Lambda^{\circ}_{\text{anion}} \).

For the given exercise, the infinite dilutions of \( \mathrm{NaOAc} \) and \( \mathrm{HCl} \) are key components in the process of calculating the molar conductivity of acetic acid. Without reaching infinite dilution, these precise calculations could not happen.

Ionic Conductivity

Ionic conductivity is a measure of an ion's ability to conduct electricity in a solution. It is expressed in units of Siemens per centimeter per mole (S cm²/mol). The value is based on the movement of ions like sodium (Na⁺), chloride (Cl⁻), and acetate (OAc⁻) in their aqueous environment.

In our given problem, we use the ionic conductivities of \( \mathrm{NaOAc} \) and \( \mathrm{HCl} \) to derive the required conductivity information for the ions of acetic acid (\( \mathrm{HOAc} \)). For each ion, such as sodium or hydrogen, its contribution to overall conductivity is calculated separately before summing them together for the total ionic conductivity.

• Na⁺ and Cl⁻ contribute to the conductivity measures in their respective solutions, like NaCl or HCl.
• H⁺ and OAc⁻ form the basis of what is needed to compute acetic acid’s molar conductivity.

Understanding ionic conductivity is essential in predicting the behavior of ions in electrochemical reactions.

Acetic Acid Conductivity Calculation

To calculate the conductivity of acetic acid (\( \Lambda^{\circ} \mathrm{HOAc} \)), we first determine which ionic conductivities are necessary. Since we know \( \Lambda^{\circ} \mathrm{NaOAc} \) and \( \Lambda^{\circ} \mathrm{HCl} \), we can use relationships from known ions.

For this, we require the ionic conductivities of the ions sodium (Na⁺), acetate (OAc⁻), hydrogen (H⁺), and chloride (Cl⁻). With these, the calculation goes as follows:

• Acquire Na⁺ and NaOAc⁻ from \( \Lambda^{\circ} \mathrm{NaOAc} \).
• Use H⁺ from \( \Lambda^{\circ} \mathrm{HCl} \) to calculate the ionic contribution of OAc⁻.
• With \( \Lambda^{\circ} \mathrm{NaOH} \), find the missing \( \mathrm{OH}^- \).

As each ion adds to the overall molar conductivity of acetic acid, putting the pieces together forms the complete conductivity equation, \( \Lambda^{\circ} \mathrm{HOAc} = \Lambda^{\circ} \mathrm{H^+} + \Lambda^{\circ} \mathrm{OAc^-} \). This highlights how utilizing known infinity dilution values helps solve such electronic behavior questions.

Ionic Equations in Electrochemistry

In electrochemistry, ionic equations play a crucial role in understanding reaction mechanisms and conductivity. These equations illustrate how ions react in solution and help explain conductivity in electrochemical cells.

Each ionic equation represents the dissociation process of ionic compounds in water, showing ions as separate entities. For example, HCl dissociates to \( \mathrm{H^+} \) and \( \mathrm{Cl^-} \). This process allows for the differentiation between the cations and anions contributing to the conductivity.

• NaOAc dissociates and presents ions \( \mathrm{Na^+} \) and \( \mathrm{OAc^-} \).
• In solutions like NaOH, additional possible reactions can aid in determining unknown values.

Ionic equations are fundamental for solving unknown conductivities, forecasting electrochemical reactions, and understanding the electrical properties of ionic solutions. They provide the framework to develop the required calculations for determining the conductivity of various acids, bases, and salts.