\(\Lambda^0_{m(\text{NH}_4\text{OH})}\) is equal to ____

According to Kohlrausch's law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of the cation and the anion.

\(\Lambda^0_{m(\text{NH}_4\text{OH})} = \lambda^0_{m(\text{NH}_4^+)} + \lambda^0_{m(\text{OH}^-)}\)

Using the limiting molar conductivities of the strong electrolytes \(\text{NH}_4\text{Cl}\), \(\text{NaOH}\), and \(\text{NaCl}\):

\(\Lambda^0_{m(\text{NH}_4\text{Cl})} = \lambda^0_{m(\text{NH}_4^+)} + \lambda^0_{m(\text{Cl}^-)}\)

\(\Lambda^0_{m(\text{NaOH})} = \lambda^0_{m(\text{Na}^+)} + \lambda^0_{m(\text{OH}^-)}\)

\(\Lambda^0_{m(\text{NaCl})} = \lambda^0_{m(\text{Na}^+)} + \lambda^0_{m(\text{Cl}^-)}\)

Adding the equations for \(\text{NH}_4\text{Cl}\) and \(\text{NaOH}\), and subtracting the equation for \(\text{NaCl}\):

\(\Lambda^0_{m(\text{NH}_4\text{Cl})} + \Lambda^0_{m(\text{NaOH})} - \Lambda^0_{m(\text{NaCl})} = (\lambda^0_{m(\text{NH}_4^+)} + \lambda^0_{m(\text{Cl}^-)}) + (\lambda^0_{m(\text{Na}^+)} + \lambda^0_{m(\text{OH}^-)}) - (\lambda^0_{m(\text{Na}^+)} + \lambda^0_{m(\text{Cl}^-)})\)

\(\Lambda^0_{m(\text{NH}_4\text{Cl})} + \Lambda^0_{m(\text{NaOH})} - \Lambda^0_{m(\text{NaCl})} = \lambda^0_{m(\text{NH}_4^+)} + \lambda^0_{m(\text{OH}^-)} = \Lambda^0_{m(\text{NH}_4\text{OH})}\)

Answer: \(\Lambda^0_{m(\text{NH}_4\text{Cl})} + \Lambda^0_{m(\text{NaOH})} - \Lambda^0_{m(\text{NaCl})}\)