Question

The conductivity of $1\times {10}^{-3}$ M acetic acid is $5\times {10}^{-5}Sc{m}^{-1}$ and ${\Lambda }^o$ is 390.5 $Sc{m}^{2}mo{l}^{-1}$. The dissociation constant of acetic acid is:

• A. $81.78\times {10}^{-4}$
• B. $81.78\times {10}^{-5}$
• C. $19\times {10}^{-6}$
• D. $18.78\times {10}^{-5}$

Answer 1

The correct option is C $19\times {10}^{-6}$

The conductivity of $1\times {10}^{-3}$ M acetic acid is $5\times {10}^{-5}Sc{m}^{-1}$ and ${\Lambda }^o$ is 390.5 $Sc{m}^{2}mo{l}^{-1}$. The dissociation constant of acetic acid is $18.78\times {10}^{-6}$.

The calculations are as shown below.

${\Lambda }_v=k_v\times \frac{1000}{M}=\frac{5\times {10}^{-5}\times 1000}{1\times {10}^{-3}}=50Sc{m}^{2}mo{l}^{-1}$

The expression for the degree of dissociation is

$\alpha =\frac{{\Lambda }_v}{{\Lambda }_{\infty }^{\prime }}=\frac{50}{390.35}=0.13$

The expression for the dissociation constant is,

$k_a=\frac{c{\alpha }^2}{1-\alpha }=\frac{{10}^{-3}\times (0.13{)}^2}{1-0.13}=1.9\times {10}^{-5}\simeq 19\times {10}^{-6}$