Ionic conductance at infinite dilution of Al^3+ and SO4^2- ions are 60 and 80 ohm^-1 cm^2 eq^-1, res — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand Kohlrausch's Law of Independent Migration of Ions According to Kohlrausch's Law, at infinite dilution, the equivalent conductance of an electrolyte ($\Lambda^\circ_{\text{eq}}$) is equal to the sum of the limiting equivalent ionic conductances of its constituent cations and anions: $$\Lambda^\circ_{\text{eq}} = \lambda^\circ_{\text{eq}}(\text{cation}) + \lambda^\circ_{\text{eq}}(\text{anion})$$ Importantly, when using equivalent conductances, the stoichiometric coefficients of the ions from the salt formula are not multiplied into the terms because the equivalent conductance is already defined per unit of charge. Step 2 - Calculate the Limiting Equivalent Conductance ($\Lambda^\circ_{\text{eq}}$) of \ce{Al2(SO4)3} We are given: * Equivalent ionic conductance of $\ce{Al^3+}$: $\lambda^\circ_{\text{eq}}(\ce{Al^3+}) = 60\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}$ * Equivalent ionic conductance of $\ce{SO4^2-}$: $\lambda^\circ_{\text{eq}}(\ce{SO4^2-}) = 80\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}$ Substituting these values into the equivalent conductance formula: $$\Lambda^\circ_{\text{eq}}(\ce{Al2(SO4)3}) = \lambda^\circ_{\text{eq}}(\ce{Al^3+}) + \lambda^\circ_{\text{eq}}(\ce{SO4^2-})$$ $$\Lambda^\circ_{\text{eq}}(\ce{Al2(SO4)3}) = 60\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1} + 80\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}$$ $$\Lambda^\circ_{\text{eq}}(\ce{Al2(SO4)3}) = \boxed{140\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}}$$ Thus, the equivalent conductance at infinite dilution of $\ce{Al2(SO4)3}$ is $140\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}$, which confirms that **Option (B) is correct**. Step 3 - Determine the Valence Factor ($z$) of \ce{Al2(SO4)3} To relate the molar conductance ($\Lambda^\circ_{\text{m}}$) to the equivalent conductance ($\Lambda^\circ_{\text{eq}}$), we need the valence factor ($z$ or n-factor) of the salt, which represents the total magnitude of positive or negative charge produced per formula unit of the electrolyte upon complete dissociation: $$\ce{Al2(SO4)3(aq) -> 2Al^3+(aq) + 3SO4^2-(aq)}$$ * Total positive charge from the $\ce{Al^3+}$ cations: $2 \times (+3) = +6$ * Total negative charge from the $\ce{SO4^2-}$ anions: $3 \times (-2) = -6$ * Therefore, the valence factor is: $$z = 6$$ Step 4 - Calculate the Limiting Molar Conductance ($\Lambda^\circ_{\text{m}}$) of \ce{Al2(SO4)3} The relationship between molar conductance and equivalent conductance is: $$\Lambda^\circ_{\text{m}} = z \times \Lambda^\circ_{\text{eq}}$$ Substituting the values of $z = 6$ and $\Lambda^\circ_{\text{eq}}(\ce{Al2(SO4)3}) = 140\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}$: $$\Lambda^\circ_{\text{m}}(\ce{Al2(SO4)3}) = 6 \times 140\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}$$ $$\Lambda^\circ_{\text{m}}(\ce{Al2(SO4)3}) = \boxed{840\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}}$$ Thus, the molar conductance at infinite dilution of $\ce{Al2(SO4)3}$ is $840\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$, which confirms that **Option (C) is correct**. Step 5 - Evaluate the Options and Explain the Distractors * **Option (A) is incorrect:** This option states that the molar conductance is $140\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$. This is incorrect because $140\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}$ is the *equivalent* conductance, not the molar conductance. * **Option (B) is correct:** As calculated in Step 2, the equivalent conductance of the salt is $140\ \Omega^{-1}\text{ cm}^2\text{ eq}^{-1}$. * **Option (C) is correct:** As calculated in Step 4, the molar conductance of the salt is $840\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$. * **Option (D) is incorrect:** This option states that the molar conductance is $23.33\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$. This erroneous value is obtained if one incorrectly divides the equivalent conductance by the valence factor ($\frac{140}{6} \approx 23.33$) instead of multiplying it. $$\text{Correct Options: } \boxed{B, C}$$