Problem 186
Question
The limiting molar conductivities \(\Lambda^{\circ}\) for \(\mathrm{NaCl}, \mathrm{KBr}\) and \(\mathrm{KCl}\) are 126,152 and \(150 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) respectively. The \(\Lambda^{0}\) for \(\mathrm{NaBr}\) is \(\quad[2004]\) (a) \(278 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) (b) \(178 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) (c) \(128 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) (d) \(306 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
Step-by-Step Solution
Verified Answer
The limiting molar conductivity of NaBr is 128 S cm²/mol (option c).
1Step 1: Understanding Limiting Molar Conductivity
Limiting molar conductivity (0) is the conductivity of an electrolyte at infinite dilution, where each ion moves independently. To find the value for a salt like NaBr, we can use the known 0 values for similar ions from other known salts.
2Step 2: Use Kohlrausch's Law of Independent Migration
According to Kohlrausch's law, 0 of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions. That means:\[4(NaBr) = 4(Na^+) + 4(Br^-)\]
3Step 3: Identify Relevant Ion Conductivities
We know 0 values for NaCl, KBr, and KCl. Using these, we'll isolate conductivities for the specific ions:
- For NaCl: 0(Na^+) + 0(Cl^-) = 126
- For KBr: 0(K^+) + 0(Br^-) = 152
- For KCl: 0(K^+) + 0(Cl^-) = 150.
4Step 4: Calculate Ion Conductivities
From KCl and KBr equations, calculate separately 0(Br^-) and 0(Cl^-):
1. Subtract KCl from KBr:
- 0(Br^-) - 0(Cl^-) = 152 - 150 = 2.
2. With this, solve:
- 0(Br^-) = 0(Cl^-) + 2.
5Step 5: Solve for NaBr
With ion values known or expressed, and knowing 0(Na^+), we find:\[4(NaBr) = 4(Na^+) + (4(Cl^-) + 2)\]Solve for actual numbers:- 4(Na^+) + 4(Cl^-) = 126 - Plug 4(Cl^-) = 4(Na^+) + 2 into the equation:\[4(Na^+) + 4(Br^-) = 126 + 2 = 128\]Thus, 4(NaBr) = 128 S cm²/mol.
Key Concepts
Kohlrausch's LawIon ConductivityElectrolyte
Kohlrausch's Law
Kohlrausch's Law plays a vital role in understanding the behavior of ionic solutions, especially concerning their conductive properties. It states that the limiting molar conductivity of an electrolyte at infinite dilution is equal to the sum of the limiting molar conductivities of its individual ions. This is expressed as \[ \Lambda^{\circ} = \lambda^{\circ}_{+} + \lambda^{\circ}_{-} \]where \( \lambda^{\circ}_{+} \) and \( \lambda^{\circ}_{-} \) are the limiting molar conductivities of the cation and anion, respectively.
This law is particularly useful because it allows scientists to predict the conductivity of a salt not directly measured by using known values from similar salts. By breaking down the conductivity into contributions from individual ions, one can calculate the expected behavior of newly formed compounds. This principle of additive contributions is what makes Kohlrausch's Law a cornerstone in the study of electrolytes and their conductive characteristics.
This law is particularly useful because it allows scientists to predict the conductivity of a salt not directly measured by using known values from similar salts. By breaking down the conductivity into contributions from individual ions, one can calculate the expected behavior of newly formed compounds. This principle of additive contributions is what makes Kohlrausch's Law a cornerstone in the study of electrolytes and their conductive characteristics.
Ion Conductivity
Ion conductivity refers to the ability of ions to conduct electric current through a solution. Each ion contributes uniquely to the overall conductivity of the solution, depending on its mobility and charge. Highly mobile ions like protons or hydroxide ions typically result in higher conductivity values.
Understanding ion conductivity helps in determining the conduction mechanism within a solution. For example:
Understanding ion conductivity helps in determining the conduction mechanism within a solution. For example:
- Cations move towards the cathode, contributing positively to the current.
- Anions move towards the anode, also aiding in the flow of electricity.
Electrolyte
The term 'electrolyte' refers to a substance that produces an electrically conducting solution when dissolved in a polar solvent, such as water. Electrolytes are essential in many processes in both nature and technology.
Electrolytes can be classified into two major types based on their dissociation:
Electrolytes can be classified into two major types based on their dissociation:
- **Strong Electrolytes:** These completely dissociate into ions in solution. Examples include salts like NaCl and acids like HCl.
- **Weak Electrolytes:** These only partially dissociate, resulting in an equilibrium between ions and unionized molecules in solution. Examples are acetic acid and ammonia.
Other exercises in this chapter
Problem 184
Consider the following \(E^{\circ}\) values \(E\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=+0.77 \mathrm{~V}\) \(E\left(\mathrm{Sn}^{2+} / \mathrm{Sn}\righ
View solution Problem 185
3\. In a hydrogen-oxygen fuel cell, combustion of hydrogen occurs to [2004] (a) produce high purity water (b) generate heat (c) remove adsorbed oxygen from elec
View solution Problem 187
In a cell that utilizes the reaction \(\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathr
View solution Problem 188
The \(E\left(\mathrm{M}^{3+} / \mathrm{M}^{2+}\right)\) values for \(\mathrm{Cr}, \mathrm{Mn}, \mathrm{Fe}\) and \(\mathrm{Co}\) are \(-0.41,+1.57,+0.77\) and \
View solution