\(\rm \wedge^o_m\) for NaCl, HCl and NaOAc are 126.4, 425.9 and 91.0 S cm2 mol-1 respectively. Calculate ∧° for HOAc

Explanation:

Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations. If salt is dissolved in water, the conductivity of the solution is the sum of the conductances of the anions and cations.

\(\Lambda_{\mathrm{m}(\mathrm{HAc})}^0=\lambda_{\mathrm{H}^{+}}^0+\lambda_{\mathrm{Ac}^{-}}^0\)

Given, \(\rm \wedge^o_m\) for NaCl, HCl and NaOAc are 126.4, 425.9 and 91.0 S cm² mol^-1 

\(=\lambda_{\mathrm{H}^{+}}^0+\lambda_{\mathrm{Cl}^{-}}^0+\lambda_{\mathrm{AC}^{-}}^0+\lambda_{\mathrm{Na}^{+}}^0-\lambda_{\mathrm{Cl}^{-}}^0-\lambda_{\mathrm{Na}^{+}}^0\)

\(=\Lambda_{\mathrm{m}(\mathrm{HCl})}^{\circ}+\Lambda_{\mathrm{m}(\mathrm{NaAc})}^{\circ}-\Lambda_{\mathrm{m}(\mathrm{NaCl})}^{\circ}\)

= (425.9 + 91.0 – 126.4) S cm² mol^–1 

= 390.5 S cm2 mol–1