Question

At ${25}^{o}C$, the specific conductance of $0.01M$ aqueous solution of acetic acid is $1.63\times {10}^{-2}S{m}^{-1}$ and the molar conductance at infinite dilution is $390.7\times {10}^{-4}S{m}^{2}mo{l}^{-1}$. Calculate the dissociation constant $\left(K_a\right)$ of the acid.

• A. $2.5\times {10}^{-3}mold{m}^{-5}$
• B. $1.2\times {10}^{-3}mold{m}^{-5}$
• C. $9.8\times {10}^{-5}mold{m}^{-3}$
• D. $1.82\times {10}^{-5}mold{m}^{-3}$

Answer 1

The correct option is D $1.82\times {10}^{-5}mold{m}^{-3}$

$\text{Specific conductance,}\kappa =1.63\times {10}^{-2}S{m}^{-1}$

$\text{Molar concentration,}C=0.01mold{m}^{-3}=0.01\times {10}^{3}mol{m}^{-3}$

$\text{Molar conductivity,}{\Lambda }_m=\frac{\kappa }{C}=\frac{1.63\times {10}^{-2}S{m}^{-1}}{0.01\times {10}^{3}mol{m}^{-3}}=16.3\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

At infinite dilute, the ionization become 100% and molar conducatnce will be the highest which is called limiting molar conductance$\left({\Lambda }_m^0\right)$.

At lower dilution, the ionization (dissociation into ions) is less than 100% and molar conductance $\left({\Lambda }_m\right)$ becomes lower.

The degree of dissociation is given by

​​​​​​$\alpha =\frac{\text{Molar conductivity at a given concentration}}{\text{Molar conductivity at a infinite dilution}}$

​​​​​​$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^0}$

$=\frac{16.3\times {10}^{-4}S{m}^{2}mo{l}^{-1}}{390.7\times {10}^{-4}S{m}^{2}mo{l}^{-1}}=0.0417$

For acetic acid,
At time, $t=0$
$\underset{C}{C{H}_{3}COOH}\rightleftharpoons \underset{0}{C}{H}_{3}CO{O}^{-}+{\underset{0}{H}}^{+}$

At time, $t=t_m$
$\underset{C-C\alpha }{C{H}_{3}COOH}\rightleftharpoons \underset{C\alpha }{C}{H}_{3}CO{O}^{-}+{\underset{C\alpha }{H}}^{+}$

$K_a=\frac{C\alpha \times C\alpha }{C-C\alpha }$

$K_a=\frac{C{\alpha }^2}{1-\alpha }$


$K_a=\frac{0.01mold{m}^{-3}\times {\left(0.0417\right)}^2}{1-0.0417}=1.82\times {10}^{-5}mold{m}^{-3}$