Question

The conductivity of $0.001M$ acetic acid is $5\times {10}^{-2}Sc{m}^{-1}$ and ${\wedge }^{\infty }$ is $390.5Sc{m}^{2}mo{l}^{-1}$ then the calculated value of dissociation constant of acetic acid would be:

• A. $81.78\times {10}^{-4}$
• B. $81.78\times {10}^{-5}$
• C. $16.38\times {10}^{-6}$
• D. $18.78\times {10}^{-5}$

Answer 1

Answer: (C)

$16.38\times {10}^{-6}$

$M=0.001$ $mol/L=\frac{0.001}{{10}^{-3}}$ $mol/c{m}^3$

$\therefore \lambda =5\times {10}^{-5}\times \frac{1000}{{10}^{-3}}=5\times {10}^6\times {10}^{-5}$

$⟹\lambda =50$

Degree of dissociation, $\alpha =\frac{50}{390.5}=0.128$

$C{H}_{3}COOH\leftrightharpoons C{H}_{3}CO{O}^{-}+H^{+}$

$c-c\alpha$ $c\alpha$ $c\alpha$

$K_a=\frac{c^2{\alpha }^2}{c-c\alpha }=c{\alpha }^2[c-c\alpha \approx c]$

$\therefore K_a={10}^{-3}\times (0.128{)}^2$

$⟹K_a=16.38\times {10}^{-6}$