At , the limiting molar conductivities () of , , and are experimentally established as , , and respe — Electrochemistry Chemistry Question
Question
At $298\text{ K}$, the limiting molar conductivities ($\Lambda_m^\circ$) of $Ba(OH)_2$, $BaCl_2$, and $NH_4Cl$ are experimentally established as $524 \times 10^{-4}$, $280 \times 10^{-4}$, and $130 \times 10^{-4}\text{ S m}^2\text{ mol}^{-1}$ respectively. By applying Kohlrausch's law, calculate the exact limiting molar conductivity of weak $NH_4OH$ in the format $x \times 10^{-4}\text{ S m}^2\text{ mol}^{-1}$. What is $x$?
💡 Solution & Explanation
By Kohlrausch's Law: $\Lambda_m^\circ(NH_4OH) = \lambda^\circ(NH_4^+) + \lambda^\circ(OH^-)$. We formulate this using the strong electrolytes: $\Lambda_m^\circ(NH_4OH) = \Lambda_m^\circ(NH_4Cl) + \frac{1}{2}\Lambda_m^\circ(Ba(OH)_2) - \frac{1}{2}\Lambda_m^\circ(BaCl_2)$. The factor of $1/2$ corrects for the $Ba^{2+}$ ion and the two $OH^-$ or $Cl^-$ ions per formula unit. Substituting values: $130 + \frac{524}{2} - \frac{280}{2} = 130 + 262 - 140 = 392 - 140 = 252$.