Understanding Kohlrausch's Law

This problem utilizes Kohlrausch's Law, which states that the molar conductivity of an electrolyte at infinite dilution ($\Lambda_m^\circ$) is the sum of the contributions of its constituent ions.

Let:

• $\Lambda_m^\circ(\text{NH}_4\text{Cl}) = x$
• $\Lambda_m^\circ(\text{NH}_4\text{NO}_3) = y$
• $\Lambda_m^\circ(\text{AgNO}_3) = z$
• $\Lambda_m^\circ(\text{AgCl}) = \Lambda_{AgCl}^\circ$ (what we need to find)

Applying Kohlrausch's Law

We can express the molar conductances of the given electrolytes in terms of their ions:

1. $\Lambda_m^\circ(\text{NH}_4\text{Cl}) = \Lambda_m^\circ(\text{NH}_4^+) + \Lambda_m^\circ(\text{Cl}^-) = x$
2. $\Lambda_m^\circ(\text{NH}_4\text{NO}_3) = \Lambda_m^\circ(\text{NH}_4^+) + \Lambda_m^\circ(\text{NO}_3^-) = y$
3. $\Lambda_m^\circ(\text{AgNO}_3) = \Lambda_m^\circ(\text{Ag}^+) + \Lambda_m^\circ(\text{NO}_3^-) = z$

We want to find the molar conductance of AgCl:

$\Lambda_{AgCl}^\circ = \Lambda_m^\circ(\text{Ag}^+) + \Lambda_m^\circ(\text{Cl}^-)$

Deriving the Molar Conductance of AgCl

To find $\Lambda_m^\circ(\text{Ag}^+) + \Lambda_m^\circ(\text{Cl}^-)$, we can combine the equations:

Consider the combination: (Equation 1) - (Equation 2) + (Equation 3)

$ ( \Lambda_m^\circ(\text{NH}_4^+) + \Lambda_m^\circ(\text{Cl}^-) ) - ( \Lambda_m^\circ(\text{NH}_4^+) + \Lambda_m^\circ(\text{NO}_3^-) ) + ( \Lambda_m^\circ(\text{Ag}^+) + \Lambda_m^\circ(\text{NO}_3^-) ) $

$ = x - y + z $

Simplifying the left side:

$ \Lambda_m^\circ(\text{NH}_4^+) + \Lambda_m^\circ(\text{Cl}^-) - \Lambda_m^\circ(\text{NH}_4^+) - \Lambda_m^\circ(\text{NO}_3^-) + \Lambda_m^\circ(\text{Ag}^+) + \Lambda_m^\circ(\text{NO}_3^-) $

The terms $\Lambda_m^\circ(\text{NH}_4^+)$ and $\Lambda_m^\circ(\text{NO}_3^-)$ cancel out, leaving:

$ \Lambda_m^\circ(\text{Cl}^-) + \Lambda_m^\circ(\text{Ag}^+) $

This is exactly $\Lambda_{AgCl}^\circ$. Therefore:

$ \Lambda_{AgCl}^\circ = x - y + z $

Which can be written as $x + z - y$.