Question

Specific conductance of 0.1 M $C{H}_{3}COOH$ at 25${}^0$C is $3.9\times {10}^{-4}oh{m}^{-1}c{m}^{-1}$ if ${\lambda }^{\infty }\left(H_3{O}^{+}\right)$ and ${\lambda }^{\infty }\left(C{H}_{3}CO{O}^{-}\right)$ at 25${}^0$C are 349.0 and $41.0oh{m}^{-1}c{m}^{2}mo{l}^{-1}$ respectively, degree of ionisation of $C{H}_{3}COOH$ at the given concentration is __________.

• A. 1.0%
• B. 4.0%
• C. 5.0%
• D. 2.0%

Answer 1

The correct option is A 1.0%

The specific conductance $\left(\kappa \right)$ of 0.1 M acetc acid solution is $3.9\times {10}^{-4}{\Omega }^{-1}{cm}^{-1}$.
Its molar conductivity is ${\Lambda }_m=\frac{1000\times \kappa }{C}=\frac{1000\times {10}^{-4}}{0.1}=3.9{\Omega }^{-1}{cm}^2{mol}^{-1}.$
According to kohlrausch law of independent migration of ions, the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the contributions of the molar conductivites of its ions.
Hence,
${\Lambda }_m^{\infty }\left({CH}_{3}COOH\right)={\Lambda }_m^{\infty }\left({CH}_{3}CO{O}^{-}\right)+{\Lambda }_m^{\infty }\left(H_3{O}^{+}\right).$
Substitute the values in the above equation.
${\Lambda }_m^{\infty }\left({CH}_{3}COOH\right)=41.0+349.0=390{\Omega }^{-1}{cm}^2{mol}^{-1}$
The degree of dissociation is the ratio of the equivalent conductance at given concentration to the equivalent conductance at zero concentration. Hence,
$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\infty }}=\frac{3.9}{390}=0.01=1$ %.