Ionic conductance at infinite dilution of Al3+ and SO4^2- are 60 and 80 ohm-1 cm2 eq-1, respectively — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand Kohlrausch's Law for Equivalent Conductance According to Kohlrausch's Law of independent migration of ions, at infinite dilution, each ion makes a definite contribution toward the total conductance of an electrolyte, irrespective of the nature of the other ion with which it is associated. In terms of equivalent conductance at infinite dilution ($\Lambda_{\text{eq}}^\circ$), the law states that the limiting equivalent conductance of an electrolyte is equal to the sum of the limiting equivalent conductances of its constituent cations and anions: $$\Lambda_{\text{eq}}^\circ(\text{electrolyte}) = \lambda_{\text{eq}}^\circ(\text{cation}) + \lambda_{\text{eq}}^\circ(\text{anion})$$ Where: * $\lambda_{\text{eq}}^\circ(\text{cation})$ is the limiting equivalent conductance of the cation. * $\lambda_{\text{eq}}^\circ(\text{anion})$ is the limiting equivalent conductance of the anion. Step 2 - Calculate the Equivalent Conductance of \ce{Al2(SO4)3} We are given the following limiting equivalent ionic conductances at infinite dilution: * For $\ce{Al^{3+}}$: $\lambda_{\text{eq}}^\circ(\ce{Al^{3+}}) = 60\text{ \Omega}^{-1}\text{ cm}^2\text{ eq}^{-1}$ * For $\ce{SO4^{2-}}$: $\lambda_{\text{eq}}^\circ(\ce{SO4^{2-}}) = 80\text{ \Omega}^{-1}\text{ cm}^2\text{ eq}^{-1}$ Substituting these values into the Kohlrausch formula for equivalent conductance: $$\Lambda_{\text{eq}}^\circ(\ce{Al2(SO4)3}) = \lambda_{\text{eq}}^\circ(\ce{Al^{3+}}) + \lambda_{\text{eq}}^\circ(\ce{SO4^{2-}})$$ $$\Lambda_{\text{eq}}^\circ(\ce{Al2(SO4)3}) = 60\text{ \Omega}^{-1}\text{ cm}^2\text{ eq}^{-1} + 80\text{ \Omega}^{-1}\text{ cm}^2\text{ eq}^{-1}$$ $$\Lambda_{\text{eq}}^\circ(\ce{Al2(SO4)3}) = \boxed{140\text{ \Omega}^{-1}\text{ cm}^2\text{ eq}^{-1}}$$ Thus, the limiting equivalent conductance of aluminum sulfate is $140\text{ \Omega}^{-1}\text{ cm}^2\text{ eq}^{-1}$. This shows that **Option (B)** is correct. Step 3 - Relate Equivalent Conductance to Molar Conductance The relationship between the standard limiting molar conductance ($\Lambda_{\text{m}}^\circ$) and the standard limiting equivalent conductance ($\Lambda_{\text{eq}}^\circ$) of an electrolyte is given by: $$\Lambda_{\text{m}}^\circ = z \times \Lambda_{\text{eq}}^\circ$$ Where $z$ is the valence factor (or $n$-factor) of the salt, representing the total positive or negative charge produced by the complete dissociation of one formula unit of the electrolyte. For aluminum sulfate ($\ce{Al2(SO4)3}$), the dissociation is: $$\ce{Al2(SO4)3 -> 2Al^{3+} + 3SO4^{2-}}$$ The total positive charge of the cations is: $$z = 2 \times (+3) = +6$$ The total negative charge of the anions is: $$z = 3 \times (-2) = -6$$ Thus, the valence factor of $\ce{Al2(SO4)3}$ is $z = 6$. Substituting this valence factor and the equivalent conductance into the relationship: $$\Lambda_{\text{m}}^\circ(\ce{Al2(SO4)3}) = 6 \times 140\text{ \Omega}^{-1}\text{ cm}^2\text{ eq}^{-1}$$ $$\Lambda_{\text{m}}^\circ(\ce{Al2(SO4)3}) = \boxed{840\text{ \Omega}^{-1}\text{ cm}^2\text{ mol}^{-1}}$$ Thus, the limiting molar conductance of aluminum sulfate is $840\text{ \Omega}^{-1}\text{ cm}^2\text{ mol}^{-1}$. This shows that **Option (C)** is correct. Step 4 - Alternative Verification using Molar Ionic Conductivities We can also calculate the molar conductance using the limiting molar ionic conductances ($\lambda_{\text{m}}^\circ$). The relationship between equivalent and molar conductivities for individual ions is: $$\lambda_{\text{m}}^\circ(\text{ion}) = \text{charge of the ion} \times \lambda_{\text{eq}}^\circ(\text{ion})$$ * For $\ce{Al^{3+}}$: $$\lambda_{\text{m}}^\circ(\ce{Al^{3+}}) = 3 \times 60\text{ \Omega}^{-1}\text{ cm}^2\text{ eq}^{-1} = 180\text{ \Omega}^{-1}\text{ cm}^2\text{ mol}^{-1}$$ * For $\ce{SO4^{2-}}$: $$\lambda_{\text{m}}^\circ(\ce{SO4^{2-}}) = 2 \times 80\text{ \Omega}^{-1}\text{ cm}^2\text{ eq}^{-1} = 160\text{ \Omega}^{-1}\text{ cm}^2\text{ mol}^{-1}$$ According to Kohlrausch's Law in terms of molar conductance: $$\Lambda_{\text{m}}^\circ(\ce{Al2(SO4)3}) = 2\lambda_{\text{m}}^\circ(\ce{Al^{3+}}) + 3\lambda_{\text{m}}^\circ(\ce{SO4^{2-}})$$ $$\Lambda_{\text{m}}^\circ(\ce{Al2(SO4)3}) = 2(180\text{ \Omega}^{-1}\text{ cm}^2\text{ mol}^{-1}) + 3(160\text{ \Omega}^{-1}\text{ cm}^2\text{ mol}^{-1})$$ $$\Lambda_{\text{m}}^\circ(\ce{Al2(SO4)3}) = 360 + 480 = \boxed{840\text{ \Omega}^{-1}\text{ cm}^2\text{ mol}^{-1}}$$ Both methods yield identical results, confirming the accuracy of the calculations. Step 5 - Evaluate and Explain the Options * **Option (A) is incorrect:** This option states that the molar conductance is $140\text{ \Omega}^{-1}\text{ cm}^2\text{ mol}^{-1}$. This is incorrect because $140$ is the equivalent conductance, which must be multiplied by the valence factor $z=6$ to obtain the molar conductance. * **Option (B) is correct:** As calculated in Step 2, the limiting equivalent conductance of $\ce{Al2(SO4)3}$ is indeed $140\text{ \Omega}^{-1}\text{ cm}^2\text{ eq}^{-1}$. * **Option (C) is correct:** As calculated in Step 3 and Step 4, the limiting molar conductance of $\ce{Al2(SO4)3}$ is indeed $840\text{ \Omega}^{-1}\text{ cm}^2\text{ mol}^{-1}$. * **Option (D) is incorrect:** This value ($23.33\text{ \Omega}^{-1}\text{ cm}^2\text{ mol}^{-1}$) is obtained if the equivalent conductance is divided by $6$ instead of multiplied, representing an incorrect application of the relationship between molar and equivalent conductance. $$\text{Correct Options: } \boxed{\text{B, C}}$$