The molar conductivity of , OH^- and Cl^- at infinite dilution is 150, 200 and 75 Ω^-1 cm^2 mol^-1, — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Definition and Formula for Degree of Dissociation ($\alpha$) The degree of dissociation ($\alpha$) of a weak electrolyte (such as ammonium hydroxide, $\ce{NH4OH}$) at a given concentration is defined as the fraction of the total electrolyte that is ionized in the solution. It is directly related to the molar conductivity at concentration $C$ ($\Lambda_m$) and the limiting molar conductivity at infinite dilution ($\Lambda_m^\circ$) by the formula: $$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$$ We are given the molar conductivity of a $0.01\text{ M}$ solution of $\ce{NH4OH}$: $$\Lambda_m = 22\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ To calculate $\alpha$, we must first determine the limiting molar conductivity ($\Lambda_m^\circ$) of $\ce{NH4OH}$. Step 2 - Apply Kohlrausch's Law to Find the Limiting Molar Conductivity of Ammonium Ions According to Kohlrausch's Law of Independent Migration of Ions, at infinite dilution, the molar conductivity of an electrolyte is the sum of the individual limiting molar conductivities of its constituent cations and anions. For ammonium chloride ($\ce{NH4Cl}$), which is a strong electrolyte: $$\Lambda_m^\circ(\ce{NH4Cl}) = \lambda^\circ(\ce{NH4^+}) + \lambda^\circ(\ce{Cl^-})$$ Given values: * $\Lambda_m^\circ(\ce{NH4Cl}) = 150\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ * $\lambda^\circ(\ce{Cl^-}) = 75\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ Substitute these values into the equation to find the limiting molar conductivity of the ammonium ion ($\lambda^\circ(\ce{NH4^+})$): $$150\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1} = \lambda^\circ(\ce{NH4^+}) + 75\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ $$\lambda^\circ(\ce{NH4^+}) = 150\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1} - 75\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ $$\lambda^\circ(\ce{NH4^+}) = 75\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ Step 3 - Calculate the Limiting Molar Conductivity of Ammonium Hydroxide ($\Lambda_m^\circ(\ce{NH4OH})$) Using Kohlrausch's Law, we can now express the limiting molar conductivity of the weak base $\ce{NH4OH}$ as: $$\Lambda_m^\circ(\ce{NH4OH}) = \lambda^\circ(\ce{NH4^+}) + \lambda^\circ(\ce{OH^-})$$ Given: * $\lambda^\circ(\ce{OH^-}) = 200\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ * $\lambda^\circ(\ce{NH4^+}) = 75\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$ (calculated in Step 2) Substitute these values into the formula: $$\Lambda_m^\circ(\ce{NH4OH}) = 75\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1} + 200\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ $$\Lambda_m^\circ(\ce{NH4OH}) = 275\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$$ Step 4 - Calculate the Degree of Dissociation ($\alpha$) of $\ce{NH4OH}$ Now, substitute the values of the molar conductivity at $0.01\text{ M}$ ($\Lambda_m$) and the limiting molar conductivity ($\Lambda_m^\circ$) into the expression for the degree of dissociation: $$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$$ $$\alpha = \frac{22\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}}{275\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}}$$ $$\alpha = \mathbf{0.080}$$ Thus, the degree of dissociation ($\alpha$) of the $0.01\text{ M}$ ammonium hydroxide solution is exactly $0.080$ (or $8.0\%$). Step 5 - Evaluate and Explain the Options * **Option (A) is incorrect:** This value ($0.146$) is obtained if one incorrectly calculates $\Lambda_m^\circ(\ce{NH4OH})$ as $\Lambda_m^\circ(\ce{NH4Cl}) = 150\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$, leading to $\alpha = \frac{22}{150} \approx 0.146$. This completely neglects Kohlrausch's law and the contributions of $\ce{OH^-}$ and $\ce{Cl^-}$ ions. * **Option (B) is incorrect:** This value ($0.063$) represents a mathematical error during calculation. * **Option (C) is correct:** As mathematically demonstrated, using Kohlrausch's law to find the correct limiting molar conductivity of $\ce{NH4OH}$ ($275\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$) yields a degree of dissociation of $0.080$. * **Option (D) is incorrect:** This value ($0.293$) is obtained if the numerator and denominator are incorrectly paired or if an arithmetic error is made. $$\text{Correct Option: } \boxed{\text{C}}$$