Question

Conductivity of $0.00241M$ acetic acid is $7.896\times 10^{-5}Sc{m}^{-1}$. If $\Lambda {°}_m$ for acetic acid is $390.5Sc{m}^{2}mo{l}^{-1}$, 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)

${\Lambda }_m^c=\frac{k\times 1000}{M}$
$=\frac{(7.896\times 10^{-5})\times 1000}{0.00241}$
$=32.76Sc{m}^{2}mo{l}^{-1}$
$\alpha =\frac{{\Lambda }_m^c}{{\lambda }_m^{\circ }}=\frac{32.76}{390.5}$
$=8.4\times 10^{-2}$
$K_a=\frac{c{\alpha }^2}{1-\alpha }$
$=\frac{0.00241\times (8.4\times 10^{-2}{)}^2}{1-0.084}$
$=1.86\times 10^{-5}$