- $\wedge^{{\circ}}_{\text{m}(\text{HCl})}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaCl})}$
- $\wedge^{{\circ}}_{\text{(HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}$
Explanation:
$\wedge^{{\circ}}_{\text{m}(\text{H}_2\text{O)}}=\wedge^{{\circ}}_{\text{m}(\text{HCl})}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaCl})}$
$\wedge^{{\circ}}_{\text{m}(\text{H}^+)}+\wedge^{{\circ}}_{\text{m}(\text{OH}^-)}=\wedge^{{\circ}}_{\text{m}(\text{H}^+)}+\wedge^{{\circ}}_{\text{m}(\text{cl}^-)}+\wedge^{{\circ}}_{\text{m}(\text{Na}^+)}+\wedge^{{\circ}}_{\text{m}(\text{OH}^-)}-\wedge^{{\circ}}_{\text{m}(\text{Na}^+)}-\wedge^{{\circ}}_{\text{m}(\text{cl}^-)}$
$\wedge^{{\circ}}_{\text{m}(\text{HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}=\wedge^{{\circ}}_{\text{m}(\text{H}_2\text{O})}$
$\wedge^{{\circ}}_{\text{m}(\text{H}^+)}+\wedge^{{\circ}}_{\text{m}(\text{NO}^-_3)}+\wedge^{{\circ}}_{\text{m}(\text{Na}^+)}+\wedge^{{\circ}}_{\text{m}(0\text{H}^-)}-\wedge^{{\circ}}_{\text{m}(\text{Na}^+)}-\wedge^{{\circ}}_{\text{m}(\text{NO}^-_3)}=\wedge^{{\circ}}_{\text{m}(\text{H}^+)}+\wedge^{{\circ}}_{\text{m}(\text{OH}^-)}$
$\wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{OH})}+\wedge^{{\circ}}_{\text{m}(\text{HCl})}-\wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{Cl})}=\wedge^{{\circ}}_{\text{m}(\text{H}_2\text{0})}$
However, the sum of molar conductivities of constituent ions gives the molar conductivity of water but here NH4OH is a weak electrolyte of which complete decomposition is not possible.