Question

Conductivity of $0.00241 \,M$ acetic acid is $7.896 \times 10^{-5}\, S \,cm^{-1}$. If $\Lambda °_m$ for acetic acid is $390.5 \,S \,cm^2\, mol^{-1}$, what is its dissociation constant?

• A. $2.52 \times 10^5$
• B. $3 \times 10^{-7}$
• C. $1.86 \times 10^{-5}$
• D. $7 \times 10^{-2}$

Answer 1

Answer: (C)

$\Lambda^c_m = \frac{k\times 1000}{M}$
$ = \frac{(7.896 \times 10^{-5})\times 1000}{0.00241}$
$ = 32.76\,S\,cm^2\,mol^{-1}$
$\alpha = \frac{\Lambda^c_m}{\lambda^{\circ}_m} = \frac{32.76}{390.5} $
$ = 8.4 \times 10^{-2}$
$K_a = \frac{c\alpha^2}{1 - \alpha} $
$= \frac{0.00241 \times (8.4 \times 10^{-2})^2}{1-0.084} $
$= 1.86\times 10^{-5}$