Question

The conductivity of $0.00241M$ acetic acid is $7.896\times {10}^{-5}S{cm}^{-1}$ and ${\Lambda }^{\infty }$ is $390.5S{cm}^2{mol}^{-1}$ than the calculated value of dissociation constant of acetic acid would be :

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

Answer 1

The correct option is C $18.78\times {10}^{-6}$

given that

K = 7.896 $\times {10}^{-5}5m^{-1}$

c = m 0.00241 m

molar conductivity

$\wedge m=\frac{k\times 100}{m}$

$\wedge m=\frac{7.8986\times {10}^{-}5\times 1000}{0.00241}$

= 32.6 $c{m}^{2}mol{e}^{-}1$

degree of dissociation :-

$\alpha =\frac{\wedge m}{\wedge m}$

$\alpha =\frac{32.6}{390.5}=0.084$

dissociation constant

k = $\frac{c\alpha }{1-\alpha }=\frac{0.00241\times 0.084}{1-0.084}$

=18.6 $\times {10}^{-}6mol/L$