Problem 113 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Equivalence conductance at infinite dilution of \(\mathrm{NH}_{4} \mathrm{Cl}, \mathrm{NaOH}\) and \(\mathrm{NaCl}\) are 129.8, \(217.4\) and \(108.9 \Omega^{-1} \mathrm{~cm}^{2}\) \(\mathrm{mol}^{-1}\), respectively. If the equivalent conductance of \(0.01 \mathrm{~N}\) solution of \(\mathrm{NH}_{4} \mathrm{OH}\) is \(9.532 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\), then the degree of dissociation of \(\mathrm{NH}_{4} \mathrm{OH}\) at this temperature is (a) \(0.04 \%\) (b) \(2.1 \%\) (c) \(4.0 \%\) (d) \(44.7 \%\)

Step-by-Step Solution

Answer

The degree of dissociation of NH4OH at this temperature is 4.0%.

Step 1 - Calculate Equivalent Conductance at Infinite Dilution for NH4OH

Use Kohlrausch's law of independent migration of ions which states that the equivalent conductance of a electrolyte at infinite dilution can be represented as the sum of the equivalent conductance of the individual ions. For NH4OH, this can be represented as: \(\Lambda_{NH4OH}^\circ = \Lambda_{NH4^+}^\circ + \Lambda_{OH^-}^\circ\). Equivalent conductance at infinite dilution for NH4OH can be calculated using the equivalent conductance at infinite dilution for NH4Cl and NaOH, subtracting the value for NaCl to eliminate the effect of the Na+ ion: \(\Lambda_{NH4OH}^\circ = (\Lambda_{NH4Cl}^\circ + \Lambda_{NaOH}^\circ) - \Lambda_{NaCl}^\circ\)

Step 2 - Substitute Given Values to Find Lambda Infinity for NH4OH

Substitute the given values into the equation from Step 1: \(\Lambda_{NH4OH}^\circ = (129.8 + 217.4) - 108.9 \Omega^{-1} \mathrm{cm}^{2} \mathrm{mol}^{-1}\)

Step 3 - Calculate the Degree of Dissociation (Alpha)

The degree of dissociation \(\alpha\) is given by the ratio of the equivalent conductance at a given concentration (\(\Lambda_c\)) to the equivalent conductance at infinite dilution (\(\Lambda^\circ\)). Therefore, \(\alpha = \frac{\Lambda_c}{\Lambda^\circ}\). Substitute the value of \(\Lambda_{NH4OH}^\circ\) obtained in Step 2 and the given \(\Lambda_c\) for the 0.01 N solution of NH4OH to find \(\alpha\)

Step 4 - Substitute Values and Calculate Alpha

Substitute the calculated value for \(\Lambda_{NH4OH}^\circ\) and the given value for \(\Lambda_c\): \(\alpha = \frac{9.532}{\Lambda_{NH4OH}^\circ} \)

Step 5 - Convert Alpha to Percentage

To express the degree of dissociation as a percentage, multiply the value of \(\alpha\) by 100: \(\alpha (\%) = \alpha \times 100\). This gives the degree of dissociation in percentage form.

Key Concepts

Kohlrausch's Law

Kohlrausch's Law is a fundamental principle in physical chemistry that provides a deep insight into the conductive behavior of electrolytes in solutions. This law posits that at infinite dilution, every ion contributes to the total conductivity of the electrolyte independently of the presence of other ions. In essence, when an electrolyte is sufficiently diluted, interactions between ions are minimized, and each ion's contribution, known as its ionic conductance, becomes constant.

The law is often expressed as: \[\begin{equation}\begin{aligned}\text{Total equivalent conductance} (\Lambda^\circ) &= \text{Sum of ionic conductances of the cation (}\Lambda_{\text{cation}}^\circ\text{)} \ &+ \text{the anion (}\Lambda_{\text{anion}}^\circ\text{)}.\end{aligned}\end{equation}\]In practice, Kohlrausch's Law enables the calculation of the equivalent conductance of an electrolyte at infinite dilution (\[\begin{equation}\Lambda^\circ\end{equation}\]) by adding the equivalent conductance of its constituent ions. This is particularly useful when the electrolyte in question cannot be measured directly, as it was demonstrated in the textbook exercise where the equivalent conductance for NH4OH at infinite dilution was calculated by using values from NH4Cl, NaOH, and NaCl, adjusting for the common ion.

• The concept is critical for understanding ionic mobility in solutions.
• It allows the calculation of unknown conductance values.
• Kohlrausch's Law provides a foundation for further studies in electrochemistry.

Degree of Dissociation

The degree of dissociation \[\begin{equation}\alpha\end{equation}\] is a pivotal parameter in understanding the nature of electrolytes in solution. It describes the fraction of molecules that have dissociated into ions in a solution. The degree of dissociation is influenced by factors like the nature of the solute, the solvent, temperature, and the concentration of the solution.

It can be mathematically represented by:\[\begin{equation}\alpha = \frac{\Lambda_{c}}{\Lambda^\circ}\end{equation}\]Where \[\begin{equation}\Lambda_{c}\end{equation}\] is the measured equivalent conductance at a certain concentration, and \[\begin{equation}\Lambda^\circ\end{equation}\] is the equivalent conductance at infinite dilution. The degree of dissociation is unitless and is usually expressed as a percentage. In the exercise provided, the value for \[\begin{equation}\alpha\end{equation}\] was determined using given conductance values and signified how much NH4OH was dissociated at a specific concentration. This concept is crucial for:

• Understanding reaction mechanisms in solutions.
• Predicting the behavior of weak and strong electrolytes.
• Assessing the strength of acids and bases.

The exercise highlighted illustrates the practical application of this concept by calculating the degree of dissociation, linking the theoretical knowledge with real-world chemical behavior.

Physical Chemistry for Competitive Examinations

Physical Chemistry is a significant branch of chemistry that melds the principles of physics and chemistry to study the physical properties of molecules, the forces that act upon them, and their reactions. For students preparing for competitive examinations, having a sound understanding of physical chemistry concepts such as Kohlrausch's Law and the degree of dissociation can be a game-changer.

These concepts form the basis for numerous analytical and theoretical problems that test a student's ability to apply fundamental principles to solve complex problems. Competitive exams often assess a student's proficiency with topics like electrolysis, electrochemical cells, and thermodynamics, which are rooted in physical chemistry.

When it comes to competitive exams, the clarity of concepts and the ability to rapidly apply them in different scenarios is crucial. For instance, through understanding Kohlrausch's Law, one can predict the conductivity of electrolytes, and knowing the degree of dissociation allows for insights into the strength of acids and bases. Problem-solving techniques, as exemplified in the textbook solution, equip students with the skills needed to tackle a wide array of chemical problems.

• Conceptual understanding is critical for tackling challenging questions.
• Applied knowledge differentiates top scorers in competitive exams.
• Focused study on physical chemistry can improve overall chemistry scores.

For optimal preparation, students must practice textbook exercises, grasp the key concepts, and hone their problem-solving skills to succeed in the competitive realm of chemistry examinations.