Question

The limiting molar conductivities of $ HCl, $ $ C{{H}_{3}}COONa $ and $ NaCl $ are respectively 425, 90 and $ 125\text{ }mho\text{ }c{{m}^{2}}\text{ }mo{{l}^{-1}} $ at $ 25{}^\circ C $ . The molar conductivity of $ 0.1M\,C{{H}_{3}}OOH $ solution is $7.8\, mho \,c{{m}^{2}}mo{{l}^{-1}} $ at the same temperature. The degree of dissociation of $0.1\, M$ acetic acid solution at the same temperature is

• A. $0.10$
• B. $0.02$
• C. $0.15$
• D. $0.03$
• E. $0.20$

Answer 1

Answer: (B)

According to Kohlrauschs law,
$ \wedge {}^\circ $ for $ C{{H}_{3}}COOH=\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{o}+\lambda _{{{H}^{+}}}^{o} $
$ \wedge {}^\circ $ for $ NaCl=\lambda _{N{{a}^{+}}}^{o}+\lambda _{C{{l}^{-}}}^{o} $ $ =125\text{ }mho\text{ }c{{m}^{2}}\text{ }mo{{l}^{-1}} $ ... (i)
$ \wedge {}^\circ $ for $ HCl=\lambda _{{{H}^{+}}}^{o}+\lambda _{C{{l}^{-}}}^{o} $ $ =425\text{ }mho\text{ }c{{m}^{2}}\text{ }mo{{l}^{-1}} $ ...(ii)
$ \wedge {}^\circ $ for $ C{{H}_{3}}COONa=\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{o}+\lambda _{N{{a}^{+}}}^{o} $
$ =90\,mho\,c{{m}^{2}}mo{{l}^{-1}} $ ...(iii)
Adding Eqs. (ii) and (iii) and subtracting (i), we get
$ \lambda _{{{H}^{+}}}^{o}+\lambda _{C{{l}^{-}}}^{{}^\circ }+\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{o}+\lambda _{N{{a}^{+}}}^{o}-\lambda _{Na}^{o}-\lambda _{C{{l}^{-}}}^{o} $
$ =425+90-125 $
$ =390\text{ }mho\text{ }c{{m}^{2}}\text{ }mo{{l}^{-1}} $
or $ \wedge _{C{{H}_{3}}COOH}^{o}=\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{o}+\lambda _{{{H}^{+}}}^{o} $
$ =390\text{ }mho\text{ }c{{m}^{2}}\text{ }mo{{l}^{-1}} $
Thus, the molar conductivity of
$ C{{H}_{3}}COOH $ at infinite dilution $ \wedge {}^\circ =390\text{ }mho\text{ }c{{m}^{2}}\text{ }mo{{l}^{-1}} $
The molar conductivity of $ 0.1\text{ }M\text{ }C{{H}_{3}}COOH $ solution $ (\wedge _{m}^{c}) $
$ =7.8\text{ }mho\text{ }c{{m}^{2}}\text{ }mo{{l}^{-1}} $
Degree of dissociation $ (\alpha )=\frac{\wedge _{m}^{c}}{\wedge _{m}^{o}}=\frac{7.8}{390}=0.02 $