Problem 114 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The resistance of \(1 \mathrm{M}-\mathrm{CH}_{3} \mathrm{COOH}\) solution is \(250 \Omega\), when measured in a cell of cell constant \(125 \mathrm{~m}^{-1}\). The molar conductivity, in \(\Omega^{-1} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\) is (a) \(5.0 \times 10^{-4}\) (b) 500 (c) \(2 \times 10^{-3}\) (d) 200

Step-by-Step Solution

Answer

500 \(\Omega^{-1} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\)

Step 1: Understanding Molar Conductivity

Molar conductivity \(\Lambda_m\) of a solution is defined as the conductivity \(\kappa\) of the solution divided by the concentration \(c\) of the solute, given as \(\Lambda_m = \frac{\kappa}{c}\). Here, the conductivity \(\kappa\) can be calculated from the resistance \(R\) using the cell constant \(l/A\), where \(l\) is the distance between electrodes and \(A\) is the area of the electrodes. The relation is \(\kappa = \frac{l}{AR}\).

Step 2: Calculate the Conductivity

First, calculate the conductivity \(\kappa\) using the resistance \(R\) of the cell and the cell constant \(\alpha = l/A\). The formula to find conductivity is \(\kappa = \frac{1}{R} \times \alpha\). Plugging in the values, we get \(\kappa = \frac{1}{250 \Omega} \times 125 \mathrm{~m}^{-1} = 0.5 \Omega^{-1} \mathrm{~m}^{-1}\).

Step 3: Calculate Molar Conductivity

Now that we have the conductivity \(\kappa\), we can find the molar conductivity \(\Lambda_m\) using the concentration \(c\) of the \(1 M\) solution. Since the molarity is 1, the concentration \(c\) is \(1 \mathrm{~mol} \mathrm{~m}^{-3}\). So, \(\Lambda_m = \frac{\kappa}{c} = \frac{0.5 \Omega^{-1} \mathrm{~m}^{-1}}{1 \mathrm{~mol} \mathrm{~m}^{-3}} = 0.5 \Omega^{-1} \mathrm{~m}^{2} \mathrm{~mol}^{-1} = 500 \Omega^{-1} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\).

Key Concepts

Electrolytic Conductivity

Electrolytic conductivity, often represented by the symbol \( \kappa \), is a measure of an electrolyte's ability to conduct electricity in solution. The concept is fundamental to understanding how electric current passes through solutions. In simplified terms, when an ionic compound dissolves in water, it separates into ions that are charged particles. These ions are responsible for carrying the current through the solution.

As an essential parameter in electrolysis and electrochemical cells, the conductivity of a solution is influenced by several factors, including the concentration of ions, the nature of the ions themselves, the temperature of the solution, and the ionic mobility. The higher the concentration of ions and the greater their mobility, the higher the conductivity. However, at very high concentrations, the conductivity may decrease due to ion pairing - the phenomenon where ions of opposite charges begin to associate and thus contribute less to the current flow.

To put it into context with the exercise, an acetic acid \(\mathrm{CH}_{3}\mathrm{COOH}\) solution conducts electricity due to the presence of \(\mathrm{CH}_{3}\mathrm{COO}^{-}\) and \(\mathrm{H}^{+}\) ions in solution. Measuring this physical property can shed light on the solution's behavior and characteristics.

Cell Constant

The cell constant, symbolized by \( \alpha \) or often \( l/A \) in textbooks, is a crucial aspect in the context of molar conductivity. It relates to the geometry of the electrodes used in the conductivity measurement setup. The constant is named so because it remains the same for a given cell as long as its geometric configuration does not change.

The cell constant is the ratio of the distance between electrodes \( l \) to the area of the electrodes \( A \). It has units of \( \mathrm{m}^{-1} \) and is determined empirically. In the provided exercise, the cell constant is given as \( 125 \mathrm{~m}^{-1} \) which implies a certain geometric relationship between the electrodes. The significance of the cell constant comes into play when converting between resistance and conductivity. A higher cell constant indicates a higher sensitivity in the measurement of conductivity, which is beneficial in solutions with low ionic concentrations. Understanding the cell constant allows for accurate measurement and calculation of electrolytic conductivity.

Conductivity Calculations

Conductivity calculations involve the transformation of measured resistance into conductivity, and then further into molar conductivity. These calculations depend not only on the measured resistance of the electrolyte solution \( R \) but also on the aforementioned cell constant \( \alpha \). The formula to convert resistance to conductivity is \( \kappa = \frac{1}{R} \times \alpha \) (Ohm's law), where \( \kappa \) represents the conductivity of the electrolyte solution.

Once the conductivity is determined, the product of this value and the cell constant yields the molar conductivity \( \Lambda_m \), which is given by \( \Lambda_m = \frac{\kappa}{c} \) where \( c \) is the concentration of the solution in \( \mathrm{mol} \mathrm{m}^{-3} \). This step is crucial, as molar conductivity is a standard measure enabling comparisons between different electrolytes. For the exercise, using the given resistance and cell constant values, the molar conductivity of the acetic acid solution is calculated, leading to an understanding of the solute's intrinsic ability to conduct electricity in a given solvent.

In summary, carrying out precise conductivity calculations is a fundamental skill in chemistry, particularly in fields such as electrochemistry and analytical chemistry, where it helps characterize solutions' conductive properties.