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The equivalent conductances of two strong electrolytes at infinite dilution in \(\mathrm{H}_{2} \mathrm{O}\) (where ions move freely through a solution) at \(25^{\circ} \mathrm{C}\) are given below: [2007] \(\Lambda^{\circ}\left(\mathrm{CH}_{3} \mathrm{COONa}\right)=91.0 \mathrm{~S} \mathrm{~cm}^{2} /\) equiv. \(\Lambda^{\circ}(\mathrm{HCl})=426.2 \mathrm{~S} \mathrm{~cm}^{2} /\) equiv. What additional information/quantity one needs to calculate \(\Lambda^{\circ}\) of an aqueous solution of acetic acid? (a) \(\Lambda^{\circ}\) of \(\mathrm{CH}_{3} \mathrm{COOK}\) (b) The limiting equivalent conductance of \(\mathrm{H}^{+}\left(\lambda^{\circ}\right)\) (c) \(\Lambda^{\circ}\) of chloroacetic acid \(\left(\mathrm{ClCH}_{2} \mathrm{COOH}\right)\) (d) \(\Lambda^{\circ}\) of \(\mathrm{NaCl}\)

Short Answer

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(b) The limiting equivalent conductance of \(\mathrm{H}^+\left(\lambda^{\circ}\right)\).

Step by step solution

01

Understand the Concept

We are tasked with finding the equivalent conductance of acetic acid at infinite dilution using provided information. The equivalent conductance at infinite dilution for a compound is calculated using the contributions of its constituent ions.
02

Analyze the Available Information

We have \[ \Lambda^{\circ}(\mathrm{CH}_{3}\mathrm{COONa}) = 91.0 \mathrm{~S} \mathrm{~cm}^{2} / \text{equiv} \]\[ \Lambda^{\circ}(\mathrm{HCl}) = 426.2 \mathrm{~S} \mathrm{~cm}^{2} / \text{equiv} \] We need additional information to solve for \[ \Lambda^{\circ}(\mathrm{CH}_3\mathrm{COOH}) \].
03

Determine Required Information

To calculate the \[ \Lambda^{\circ}(\mathrm{CH}_3\mathrm{COOH}) \], we apply Kohlrausch's Law of independent migration of ions which states: \[ \Lambda^{\circ} = \lambda^{\circ}_{+} + \lambda^{\circ}_{-} \]. This means we need the limiting equivalent conductances of ions involved.
04

Select the Correct Quantity

The dissociation of acetic acid (\[ \mathrm{CH}_3\mathrm{COOH} \]) involves \[ \mathrm{CH}_3\mathrm{COO}^- \] and \[ \mathrm{H}^+ \]. We already know the \[ \Lambda^{\circ}_{\mathrm{CH}_3\mathrm{COONa}} \], which gives us the conductance of \[ \mathrm{CH}_3\mathrm{COO}^- \]. Therefore, we need the limiting equivalent conductance for\[ \mathrm{H}^+ \left(\lambda^{\circ}_{\mathrm{H}^+}\right) \]. Therefore, option (b) is the correct answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equivalent Conductance
Equivalent conductance is an essential concept in electrochemistry that helps us understand how well an electrolyte conducts electricity when dissolved in water. It is defined as the conductivity of an electrolyte solution divided by the normality of the solution. This measure allows us to compare the conductivity of different electrolytes on a common basis.

For better visualization, think of equivalent conductance as a measure of how efficiently each equivalent of an ion contributes to the overall conductance. It is denoted as \( \Lambda \) and is expressed in units of \( \mathrm{S} \cdot \mathrm{cm}^2 / \text{equivalent} \).

Equivalent conductance increases with dilution. This happens because, as the solution is diluted, the ions move more freely due to less crowding, which enhances the conductance.
  • Formula: \( \Lambda = \frac{1000 \times \text{conductivity} \ (\kappa)}{\text{concentration (c in eq/L)}} \).
  • Dependence: It's affected by temperature, concentration, and the nature of the ions.
Understanding this helps us explore how different ions in solution contribute to the overall conductivity, which is crucial in predicting the behavior of strong and weak electrolytes.
Strong Electrolytes
Strong electrolytes are substances that completely dissociate into ions in solution. This complete dissociation means that they have high equivalent conductance because more ions are available to conduct electricity.

Common examples of strong electrolytes are most salts, strong acids like HCl, and strong bases like NaOH. When such substances are dissolved in water, they break apart into cations and anions instantly, leaving little to no molecules in their original form.
  • Characteristics: They exist almost entirely as ions in solution.
  • Examples: NaCl, HCl, KOH.
  • Behavior in Solution: Since they are fully dissociated, their conductance does not change much with varying concentrations.
For strong electrolytes, the Kohlrausch's Law of independent migration of ions lets us calculate the conductance of the solution by adding the limiting conductance values of the individual ions. These ions are free to move independently and contribute to the total conductance.
Understanding strong electrolytes and their properties can greatly simplify the calculation of conductance and the study of ionic behavior in solutions.
Infinite Dilution
Infinite dilution is a theoretical concept where an electrolyte's concentration approaches zero, meaning that ions in the solution do not interact with each other. This makes it easier to measure their individual conductance contributions, and it's particularly useful in applying Kohlrausch’s Law.

In practical terms, it describes a solution in which each ion can move independently without being hindered by others, resulting in maximum conductance values for each ion. This condition allows us to measure the fundamental properties of ions under ideal conditions.
  • Why It's Important: It provides a clear picture of the ion’s effect on conductance without interference from other ions.
  • Application in Calculations: Using the limiting equivalent conductance \( \Lambda^{\circ} \), we can determine conductance at any dilution.
  • Practical Implications: By studying how conductance changes with dilution, we can infer the nature of ion interaction in real-world solutions.
When studying solutions, considering infinite dilution is crucial for understanding the maximum potential of an ion to conduct electricity. It helps establish a baseline for comparing different electrolytes and making accurate predictions about solution behavior.

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Most popular questions from this chapter

A dilute aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is electrolyzed using platinum electrodes. The product at the anode and cathode are (a) \(\mathrm{O}_{2}, \mathrm{H}_{2}\) (b) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}, \mathrm{Na}\) (c) \(\mathrm{O}_{2}, \mathrm{Na}\) (d) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}, \mathrm{H}_{2}\)

The hydrogen electrode is dipped in a solution of \(\mathrm{pH}=\) \(3.0\) at \(25^{\circ} \mathrm{C}\). The potential of hydrogen electrode would be (a) \(-0.177 \mathrm{~V}\) (b) \(0.177 \mathrm{~V}\) (c) \(1.77 \mathrm{~V}\) (d) \(0.277 \mathrm{~V}\)

The electrical conductivity of the flowing aqueous solutions is highest for [2008] (a) \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) (b) \(0.1 \mathrm{M} \mathrm{CH}_{2} \mathrm{FCOOH}\) (c) \(0.1 \mathrm{M} \mathrm{CHF}_{2} \mathrm{COOH}\) (d) \(0.1 \mathrm{M} \mathrm{CH}_{2} \mathrm{ClCOOH}\)

Resistance of \(0.2 \mathrm{M}\) solution of an electrolyte is \(50 \Omega\) The specific conductance of the solution is \(1.4 \mathrm{~S} \mathrm{~m}^{-1}\). The resistance of \(0.5 \mathrm{M}\) solution of the same electrolyte is \(280 \Omega\) The molar conductivity of \(0.5 \mathrm{M}\) solution of the electrolyte in \(\mathrm{S} \mathrm{mt}^{2} \mathrm{~mol}^{-1}\) is: (a) \(5 \times 10^{3}\) (b) \(5 \times 10^{2}\) (c) \(5 \times 10^{-4}\) (d) \(5 \times 10^{-3}\)

The electrochemical series is the arrangement of various electrode systems in the increasing order of their standard reduction potentials. It has several important features. On moving from the top to the bottom in the series, tendency to gain electrons, i.e., to get reduced increases. The electrode systems having negative values of standard reduction potentials act as anode when connected to a standard hydrogen electrode, while those having positive values act as cathode. HCl cannot be stored in an aluminium vessel because (a) \(\mathrm{Al}\) is a highly reactive metal. (b) \(\mathrm{HCl}\) is an oxidizing acid (c) \(\mathrm{E}_{\mathrm{A}^{3+} / / \mathrm{I} 1}^{0}\) is much smaller than \(E_{\mathrm{H}}^{\mathrm{o}} / \mathrm{H}_{2}\) (d) All of these

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