Question

The limiting molar conductivities of $HCl,$ $C{H}_{3}COONa$ and $NaCl$ are respectively 425, 90 and $125mhoc{m}^{2}mo{l}^{-1}$ at $25{}^{\circ }C$ . The molar conductivity of $0.1MC{H}_{3}OOH$ solution is $7.8mhoc{m}^{2}mo{l}^{-1}$ at the same temperature. The degree of dissociation of $0.1M$ 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 $NaCl={\lambda }_{N{a}^{+}}^o+{\lambda }_{C{l}^{-}}^o$ $=125mhoc{m}^{2}mo{l}^{-1}$ ... (i)
$\wedge {}^{\circ }$ for $HCl={\lambda }_{H^{+}}^o+{\lambda }_{C{l}^{-}}^o$ $=425mhoc{m}^{2}mo{l}^{-1}$ ...(ii)
$\wedge {}^{\circ }$ for $C{H}_{3}COONa={\lambda }_{C{H}_{3}CO{O}^{-}}^o+{\lambda }_{N{a}^{+}}^o$
$=90mhoc{m}^{2}mo{l}^{-1}$ ...(iii)
Adding Eqs. (ii) and (iii) and subtracting (i), we get
${\lambda }_{H^{+}}^o+{\lambda }_{C{l}^{-}}^{{}^{\circ }}+{\lambda }_{C{H}_{3}CO{O}^{-}}^o+{\lambda }_{N{a}^{+}}^o-{\lambda }_{Na}^o-{\lambda }_{C{l}^{-}}^o$
$=425+90-125$
$=390mhoc{m}^{2}mo{l}^{-1}$
or ${\wedge }_{C{H}_{3}COOH}^o={\lambda }_{C{H}_{3}CO{O}^{-}}^o+{\lambda }_{H^{+}}^o$
$=390mhoc{m}^{2}mo{l}^{-1}$
Thus, the molar conductivity of
$C{H}_{3}COOH$ at infinite dilution $\wedge {}^{\circ }=390mhoc{m}^{2}mo{l}^{-1}$
The molar conductivity of $0.1MC{H}_{3}COOH$ solution $({\wedge }_m^c)$
$=7.8mhoc{m}^{2}mo{l}^{-1}$
Degree of dissociation $(\alpha )=\frac{{\wedge }_m^c}{{\wedge }_m^o}=\frac{7.8}{390}=0.02$