Question:
From the following molar conductivities at infinite dilution, \( \Lambda_m^0 \), for \( \mathrm{NH_4OH} \) is
\( \Lambda_m^0 \) for \( \mathrm{Ba(OH)_2} = 446.8 \, \Omega^{-1} \, \mathrm{cm^2 \, mol^{-1}} \)
\( \Lambda_m^0 \) for \( \mathrm{BaCl_2} = 241.6 \, \Omega^{-1} \, \mathrm{cm^2 \, mol^{-1}} \)
\( \Lambda_m^0 \) for \( \mathrm{NH_4Cl} = 130 \, \Omega^{-1} \, \mathrm{cm^2 \, mol^{-1}} \)
From the following molar conductivities at infinite dilution, \( \Lambda_m^0 \), for \( \mathrm{NH_4OH} \) is
\( \Lambda_m^0 \) for \( \mathrm{Ba(OH)_2} = 446.8 \, \Omega^{-1} \, \mathrm{cm^2 \, mol^{-1}} \)
\( \Lambda_m^0 \) for \( \mathrm{BaCl_2} = 241.6 \, \Omega^{-1} \, \mathrm{cm^2 \, mol^{-1}} \)
\( \Lambda_m^0 \) for \( \mathrm{NH_4Cl} = 130 \, \Omega^{-1} \, \mathrm{cm^2 \, mol^{-1}} \)
\( \Lambda_m^0 \) for \( \mathrm{Ba(OH)_2} = 446.8 \, \Omega^{-1} \, \mathrm{cm^2 \, mol^{-1}} \)
\( \Lambda_m^0 \) for \( \mathrm{BaCl_2} = 241.6 \, \Omega^{-1} \, \mathrm{cm^2 \, mol^{-1}} \)
\( \Lambda_m^0 \) for \( \mathrm{NH_4Cl} = 130 \, \Omega^{-1} \, \mathrm{cm^2 \, mol^{-1}} \)
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Use subtraction to eliminate common ions quickly in Kohlrausch law problems.
Updated On: Apr 22, 2026
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Solution and Explanation
Concept:
Using Kohlrausch's law of independent migration of ions:
\[
\Lambda_m^0 = \lambda^+ + \lambda^-
\]
Step 1: Write expressions for given electrolytes.
\[ \Lambda_m^0 [Ba(OH)_2] = \lambda_{Ba^{2+}} + 2\lambda_{OH^-} = 446.8 \] \[ \Lambda_m^0 [BaCl_2] = \lambda_{Ba^{2+}} + 2\lambda_{Cl^-} = 241.6 \] \[ \Lambda_m^0 [NH_4Cl] = \lambda_{NH_4^+} + \lambda_{Cl^-} = 130 \]
Step 2: Eliminate \( \lambda_{Ba^{2+}} \).
\[ (Ba(OH)_2 - BaCl_2) \Rightarrow 2\lambda_{OH^-} - 2\lambda_{Cl^-} = 205.2 \] \[ \Rightarrow \lambda_{OH^-} - \lambda_{Cl^-} = 102.6 \]
Step 3: Find \( \Lambda_m^0 (NH_4OH) \).
\[ \Lambda_m^0 [NH_4OH] = \lambda_{NH_4^+} + \lambda_{OH^-} \] From: \[ \lambda_{NH_4^+} = 130 - \lambda_{Cl^-} \] \[ \lambda_{OH^-} = 102.6 + \lambda_{Cl^-} \] Adding: \[ \Lambda_m^0 [NH_4OH] = (130 - \lambda_{Cl^-}) + (102.6 + \lambda_{Cl^-}) \] \[ = 232.6 \] \[ \therefore \Lambda_m^0 (NH_4OH) = 232.6 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \]
Step 1: Write expressions for given electrolytes.
\[ \Lambda_m^0 [Ba(OH)_2] = \lambda_{Ba^{2+}} + 2\lambda_{OH^-} = 446.8 \] \[ \Lambda_m^0 [BaCl_2] = \lambda_{Ba^{2+}} + 2\lambda_{Cl^-} = 241.6 \] \[ \Lambda_m^0 [NH_4Cl] = \lambda_{NH_4^+} + \lambda_{Cl^-} = 130 \]
Step 2: Eliminate \( \lambda_{Ba^{2+}} \).
\[ (Ba(OH)_2 - BaCl_2) \Rightarrow 2\lambda_{OH^-} - 2\lambda_{Cl^-} = 205.2 \] \[ \Rightarrow \lambda_{OH^-} - \lambda_{Cl^-} = 102.6 \]
Step 3: Find \( \Lambda_m^0 (NH_4OH) \).
\[ \Lambda_m^0 [NH_4OH] = \lambda_{NH_4^+} + \lambda_{OH^-} \] From: \[ \lambda_{NH_4^+} = 130 - \lambda_{Cl^-} \] \[ \lambda_{OH^-} = 102.6 + \lambda_{Cl^-} \] Adding: \[ \Lambda_m^0 [NH_4OH] = (130 - \lambda_{Cl^-}) + (102.6 + \lambda_{Cl^-}) \] \[ = 232.6 \] \[ \therefore \Lambda_m^0 (NH_4OH) = 232.6 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \]
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