The molar conductivities $$\Lambda^\circ_{NaOAc}$$ and $$\Lambda^\circ_{HCl}$$ at infinite dilution in water at $$25^\circ C$$ are $$91.0$$ and $$426.2\,S\,cm^2/mol$$ respectively. To calculate $$\Lambda^\circ_{HOAc}$$, the additional value required is

Kohlrausch’s law of independent ionic migration states that the molar conductivity at infinite dilution is the sum of the ionic conductivities of the individual ions:
$$\Lambda^\circ_{AB}= \lambda^\circ_{A^{z+}}+\lambda^\circ_{B^{z-}}.$$

To obtain $$\Lambda^\circ_{HOAc}=\lambda^\circ_{H^+}+\lambda^\circ_{OAc^-},$$ we need the two ionic conductivities $$\lambda^\circ_{H^+}$$ and $$\lambda^\circ_{OAc^-}$$.

We already have:

• $$\Lambda^\circ_{HCl}= \lambda^\circ_{H^+}+\lambda^\circ_{Cl^-}=426.2\,S\,cm^2\!/\!mol$$
• $$\Lambda^\circ_{NaOAc}= \lambda^\circ_{Na^+}+\lambda^\circ_{OAc^-}=91.0\,S\,cm^2\!/\!mol$$

If we also know

• $$\Lambda^\circ_{NaCl}= \lambda^\circ_{Na^+}+\lambda^\circ_{Cl^-},$$

then adding the first two and subtracting the third eliminates the $$\lambda^\circ_{Na^+}$$ and $$\lambda^\circ_{Cl^-}$$ terms:

$$\Lambda^\circ_{HCl}+ \Lambda^\circ_{NaOAc}-\Lambda^\circ_{NaCl}$$ $$=\bigl(\lambda^\circ_{H^+}+\lambda^\circ_{Cl^-}\bigr)+\bigl(\lambda^\circ_{Na^+}+\lambda^\circ_{OAc^-}\bigr)-\bigl(\lambda^\circ_{Na^+}+\lambda^\circ_{Cl^-}\bigr)$$ $$=\lambda^\circ_{H^+}+\lambda^\circ_{OAc^-}=\Lambda^\circ_{HOAc}.$$

Therefore, the extra data needed is $$\Lambda^\circ_{NaCl}.$p>

Option D which is: $$\Lambda^\circ_{NaCl}$$