Question

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

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

Answer 1

Answer: (C)

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

So $\alpha =\frac{{\Lambda }_v}{\Lambda \infty }=\frac{50}{390.5}=0.13$

Now, ${\kappa }_a=\frac{c{\alpha }^2}{1-\alpha }=\frac{10^{-3}\times (0.13{)}^2}{0.87}=1.9\times 10^{-5}\simeq 19\times 10^{-6}$