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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 \(\Lambda^{\circ}\) for \(\mathrm{NaBr}\) is 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. \(306 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)

Short Answer

Expert verified
The molar conductivity of NaBr is 128 S cm虏 mol鈦宦 (option c).

Step by step solution

01

Understanding the Concept

The limiting molar conductivity, \(\Lambda^{\circ}\), is the sum of the individual contributions of the ions to the conductivity. In other words, \(\Lambda^{\circ}\) for an electrolyte AB can be calculated using \(\Lambda^{\circ} (\text{A}^+) + \Lambda^{\circ} (\text{B}^-)\). We will use this property to find \(\Lambda^{\circ}(\text{NaBr})\).
02

Setting up the Equations

Given the limiting molar conductivities:- \(\Lambda^{\circ}(\text{NaCl}) = 126\, \text{S}\, \text{cm}^2 \text{mol}^{-1}\)- \(\Lambda^{\circ}(\text{KBr}) = 152\, \text{S}\, \text{cm}^2 \text{mol}^{-1}\)- \(\Lambda^{\circ}(\text{KCl}) = 150\, \text{S}\, \text{cm}^2 \text{mol}^{-1}\)We can write the following equations:1. \(126 = \Lambda^{\circ}(\text{Na}^+) + \Lambda^{\circ}(\text{Cl}^-)\)2. \(152 = \Lambda^{\circ}(\text{K}^+) + \Lambda^{\circ}(\text{Br}^-)\)3. \(150 = \Lambda^{\circ}(\text{K}^+) + \Lambda^{\circ}(\text{Cl}^-)\)
03

Isolating Individual Ion Contributions

Using the equations, we can solve for the individual ions. Subtract equation 3 from equation 2 to find \(\Lambda^{\circ}(\text{Br}^-) - \Lambda^{\circ}(\text{Cl}^-):\)\((\Lambda^{\circ}(\text{K}^+) + \Lambda^{\circ}(\text{Br}^-)) - (\Lambda^{\circ}(\text{K}^+) + \Lambda^{\circ}(\text{Cl}^-)) = 152 - 150 = 2\)Thus, \(\Lambda^{\circ}(\text{Br}^-) = \Lambda^{\circ}(\text{Cl}^-) + 2\)
04

Solving for \(\Lambda^{\circ}(\text{Na}^+)\)

Substitute \(\Lambda^{\circ}(\text{Br}^-)\) in equation 1:- \(\Lambda^{\circ}(\text{Na}^+) + (\Lambda^{\circ}(\text{Cl}^-)) = 126\)- From Step 3: \(\Lambda^{\circ}(\text{Br}^-) = \Lambda^{\circ}(\text{Cl}^-) + 2\)- \(\Lambda^{\circ}(\text{K}^+) = 150 - \Lambda^{\circ}(\text{Cl}^-)\)Using from step 3:- Substitute in equation 2: \(152 = \Lambda^{\circ}(\text{K}^+) + \Lambda^{\circ}(\text{Br}^-)\)- \(152 = (150 - \Lambda^{\circ}(\text{Cl}^-) + (\Lambda^{\circ}(\text{Cl}^-) + 2))\)- \(152 = 150 + 2 = 152\), confirmed.Thus confirming \(\Lambda^{\circ}(\text{Na}^+) = 126 - \Lambda^{\circ}(\text{Cl}^-)\)
05

Calculating \(\Lambda^{\circ}(\text{NaBr})\)

Now, calculate \(\Lambda^{\circ}(\text{NaBr}) = \Lambda^{\circ}(\text{Na}^+) + \Lambda^{\circ}(\text{Br}^-):\)- \(\Lambda^{\circ}(\text{Na}^+) + \Lambda^{\circ}(\text{Br}^-)) = 126 - \Lambda^{\circ}(\text{Cl}^-) + (\Lambda^{\circ}(\text{Cl}^-) + 2)\)- \(\Lambda^{\circ}(\text{NaBr}) = 126 + 2 = 128\)

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Key Concepts

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

Electrolyte Conductivity
Electrolyte conductivity is a crucial concept in understanding how electric charge is transported through an electrolyte. An electrolyte is a substance that, when dissolved in water, shifts into ions and conducts electricity. The measure of this ability to conduct electricity is known as conductivity. Here, we focus on molar conductivity which is the conductivity of a solution divided by its concentration. This provides insight into how well ions are moving.

In practical terms, the greater the mobility of an ion, the higher the conductivity. The conductivity of an electrolyte is directly related to the nature and concentration of ions present in the solution. As concentrations vary, so does the conductivity, making molar conductivity a useful tool for comparing different electrolytes under standardized conditions.

Understanding this concept helps in areas like chemistry and materials science, where the movement of ions is essential. Electrolyte conductivity can be influenced by factors like temperature, solvent composition, and ion size and charge, making it a complex yet fascinating subject.
Ion Contributions
Each individual ion within an electrolyte contributes to its overall conductivity. The molar conductivity of an electrolyte is essentially the sum of the conductivities imparted by each type of ion present. This is important in calculating limiting molar conductivity, represented as \( \Lambda^{\circ} \), which refers to the conductivity measured at infinite dilution when ion interaction is minimal.

Consider that different ions move at different rates depending on factors like their size, the nature of their charge, and the viscosity of the medium. Positive and negative ions contribute differently. To find the specific contribution of an individual ion, you can use the data obtained from the conductivities of various known electrolytes, as was done in the original exercise.

By using these values, one can also deduce the contributions when substituting one ion for another in different compounds, providing a method to predict and verify molar conductivities for unknown substances, such as determining the conductivity for NaBr through other known values like those for NaCl, KBr, and KCl.
Molar Conductivities Calculation
Calculating molar conductivities involves understanding the basic relation that the limiting molar conductivity, \( \Lambda^{\circ} \), is the sum of the individual contributions of the cation and the anion. This is expressed as \( \Lambda^{\circ}(A^+B^-) = \Lambda^{\circ}(A^+) + \Lambda^{\circ}(B^-) \).

To solve for the conductivity values of specific ions, systems of equations can be set up from known conductivities of related compounds. For example, knowing the values for NaCl, KBr, and KCl allowed the calculation of the conductivity for NaBr. By isolating ion contributions and following algebraic substitution techniques, we solve for unknown values.

In practice, these calculations are pivotal in fields where precise knowledge of ionic behavior is required. From industrial applications to laboratory experiments, accurate molar conductivities enable better design and understanding of chemical processes. Thus, mastering these calculations gives insight into the efficient design of electrolytic systems.

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Most popular questions from this chapter

\(2 \mathrm{Hg} \rightarrow \mathrm{Hg}_{2}^{2+}, \mathrm{E}^{\circ}=+0.855 \mathrm{~V}\) \(\mathrm{Hg} \rightarrow \mathrm{Hg}^{2+}, \mathrm{E}^{\circ}=+0.799 \mathrm{~V}\) Equilibrium constant for the reaction \(\mathrm{Hg}+\mathrm{Hg}^{2+} \rightarrow \mathrm{Hg}_{2}{ }^{2+}\) at \(27^{\circ} \mathrm{C}\) is a. 79 b. 89 c. 69 d. \(82.9\)

Use the following data at \(25^{\circ} \mathrm{C}\) for the questions given below \(\mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}(\mathrm{s}) \quad \mathrm{E}^{\circ}=-0.28 \mathrm{~V}\) \(\mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Mg}(\mathrm{s}) \quad \mathrm{E}^{\circ}=-2.37 \mathrm{~V}\) What is \(\mathrm{K}\) for the equilibrium at \(25^{\circ} \mathrm{C}\) ? a. \(4 \times 10^{-70}\) b. \(4 \times 10^{70}\) c. \(2 \times 10^{-35}\) d. \(2 \times 10^{70}\)

In the following question two statements (Assertion) A and Reason (R) are given. Mark a. if \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\); b. if \(A\) and \(R\) both are correct but \(R\) is not the correct explanation of A; c. \(\mathrm{A}\) is true but \(\mathrm{R}\) is false; d. A is false but \(\mathrm{R}\) is true, e. A and \(R\) both are false. (A): If standard reduction potential for the reaction \(\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag}\) is \(0.80\) volts then for the reaction \(3 \mathrm{Ag}^{+}+3 \mathrm{e}^{-} \rightarrow 3 \mathrm{Ag}\) it will be \(2.4\) volts \((\mathbf{R})\) : If concentration is increased, reduction electrode potential is increased.

For the following cell with hydrogen electrodes at two different pressures \(\mathrm{p}_{1}\) and \(\mathrm{p}_{2}\) \(\operatorname{Pt}\left(\mathrm{H}_{2}\right) \mid \mathrm{H}^{+}\)(aq) \(\mid \operatorname{Pt}\left(\mathrm{H}_{2}\right)\) \(\mathrm{p}_{1}\) \(1 \mathrm{M} \quad \mathrm{p}_{2}\) EMF is given by a. \(\mathrm{RT} / \mathrm{F} \log _{e} \mathrm{p}_{1} / \mathrm{p}_{2}\) b. \(\mathrm{RT} / 2 \mathrm{~F} \log _{\mathrm{c}} \mathrm{p}_{1} / \mathrm{p}_{2}\) c. \(\mathrm{RT} / \mathrm{F} \log _{\mathrm{e}} \mathrm{p}_{2} / \mathrm{p}_{1}\) d. \(\mathrm{RT} / 2 \mathrm{~F} \log _{\mathrm{e}} \mathrm{p}_{2} / \mathrm{p}_{1}\)

Consider the following standard reduction potentials, \(\mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}(\mathrm{s}) \quad \mathrm{E}^{\circ}=-1.66 \mathrm{~V}\) \(\mathrm{I}_{2}(\mathrm{~s})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-}(\mathrm{aq}) \quad \mathrm{E}^{\circ}=+0.54 \mathrm{~V}\) Under standard conditions, a. \(\mathrm{I}^{-}(\mathrm{aq})\) is a stronger oxidizing agent than \(\mathrm{Al}(\mathrm{s})\) and \(\mathrm{I}_{2}(\mathrm{~s})\) is a stronger reducing agent than \(\mathrm{Al}^{3+}(\mathrm{aq})\) b. Al (s) is a stronger oxidizing agent than \(\mathrm{I}^{-}\)(aq), and \(\mathrm{Al}^{3+}\) (aq) is a stronger reducing agent than \(\mathrm{I}_{2}(\mathrm{~s})\). c. \(\mathrm{I}_{2}\) (s) is a stronger oxidizing agent than \(\mathrm{Al}^{3+}\) (aq), and \(\mathrm{Al}(\mathrm{s})\) is a stronger reducing agent than \(\mathrm{I}^{-}(\mathrm{aq})\). d. \(\mathrm{Al}^{3+}(\mathrm{aq})\) is a stronger oxidizing agent than \(\mathrm{I}_{2}\) (s), and I- (aq) is a stronger reducing agent than \(\mathrm{Al}\) (s).

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