The equivalent conductances of two strong electrolytes at infinite dilution in $$H_2O$$ (where ions move freely through a solution) at $$25^\circ C$$ are given below: $$\wedge^\circ CH_3COONa = 91.0$$ S $$cm^2$$/equiv $$\wedge^\circ_{HCl} = 426.2$$ S $$cm^2$$/equiv. What additional information/quantity one needs to calculate $$\wedge^\circ$$ of an aqueous solution of acetic acid?

Kohlrausch’s law of independent ionic migration states that the limiting equivalent conductance of an electrolyte is the sum of the limiting conductances of its constituent ions:

$$\Lambda^{\circ} (AB) = \lambda^{\circ}_{A^{z+}} + \lambda^{\circ}_{B^{z-}}$$

To find $$\Lambda^{\circ}$$ for acetic acid $$CH_3COOH$$ we must know the two ionic contributions $$\lambda^{\circ}_{H^+}$$ and $$\lambda^{\circ}_{CH_3COO^-}$$ and then add them:

$$\Lambda^{\circ}_{CH_3COOH} = \lambda^{\circ}_{H^+} + \lambda^{\circ}_{CH_3COO^-}$$

The data supplied in the problem are

• $$\Lambda^{\circ}_{CH_3COONa} = 91.0\;{\rm S\,cm^2\;equiv^{-1}}$$
• $$\Lambda^{\circ}_{HCl} = 426.2\;{\rm S\,cm^2\;equiv^{-1}}$$

Express these in terms of ionic conductances:

$$\Lambda^{\circ}_{CH_3COONa} = \lambda^{\circ}_{Na^+} + \lambda^{\circ}_{CH_3COO^-}\quad -(1)$$
$$\Lambda^{\circ}_{HCl} = \lambda^{\circ}_{H^+} + \lambda^{\circ}_{Cl^-}\quad -(2)$$

Equations (1) and (2) contain four unknown ionic terms. One more independent equation is required to solve for the four ionic conductances. The most convenient additional electrolyte is sodium chloride:

$$\Lambda^{\circ}_{NaCl} = \lambda^{\circ}_{Na^+} + \lambda^{\circ}_{Cl^-}\quad -(3)$$

With equations (1), (2) and (3) we can algebraically eliminate $$\lambda^{\circ}_{Na^+}$$ and $$\lambda^{\circ}_{Cl^-}$$ to obtain $$\lambda^{\circ}_{CH_3COO^-}$$ and $$\lambda^{\circ}_{H^+}$$ and hence calculate $$\Lambda^{\circ}_{CH_3COOH}$$.

Therefore, the additional information needed is the limiting equivalent conductance of NaCl.

Option A which is: $$\Lambda^{\circ}_{NaCl}$$