\(\Lambda _m^0\) for NaCl, HCl, and NaOAc are 126.4, 425.9, and 91.0 S cm² mol^-1 respectively. Calculate \(\Lambda ^0\) for HOAc.

390.5 S cm² mol^-1

Given,

\(\Lambda _m^0 \text{ NaCl = } \Lambda _m^0 Na^+  +\Lambda _m^0 Cl^- \text{ = 126.4 S cm}^2\text{ mol}^{-1}\) ----- (i)

\(\Lambda _m^0 \text{ HCl = } \Lambda _m^0 H^+  +\Lambda _m^0 Cl^- \text{ = 425.9 S cm}^2\text{ mol}^{-1}\) ----- (ii)

\(\Lambda _m^0 \text{ NaOAc = } \Lambda _m^0 Na^+  +\Lambda _m^0 OAc^- \text{ = 91.0 S cm}^2\text{ mol}^{-1}\) ----- (iii)

Subtracting equation (i) from equation (ii), followed by the addition of (iii), we get

\(\Lambda _m^0 \text{ HCl − }\Lambda _m^0 \text{ NaCl + }\Lambda _m^0 \text{ NaOAc = }\Lambda _m^0 H^+  +\Lambda _m^0 Cl^-\text{ − }\Lambda _m^0 Na^+  \text{ − }\Lambda _m^0 Cl^- \text{ + }\Lambda _m^0 Na^+  +\Lambda _m^0 OAc^-\)

or, \(\Lambda _m^0 H^+  +\Lambda _m^0 OAc^- \text{ = }425.9 − 126.4 + 91.0\)

or. \(\Lambda _m^0 \text{ NaOAc = 390.5 S cm}^2\text{ mol}^{-1} \)