\(\Lambda _m^0\) for NaCl, HCl, and NaOAc are 126.4, 425.9, and 91.0 S cm2 mol-1 respectively. Calculate \(\Lambda ^0\) for HOAc.
Answer & explanation
Correct answer: option 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. \(\Lambda _m^0 \text{ NaOAc = 390.5 S cm}^2\text{ mol}^{-1} \)