Question

The conductivity of $1\times 10^{- 3}$ M acetic acid is $5\times 10^{- 5}Scm^{- 1}$ and $\Lambda^\circ $ is 390.5 S $cm^{2}mol^{- 1}$ . 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)

$\Lambda_{v}=\kappa_{v}\times \frac{1000}{M}=\frac{5 \times 1 0^{- 5} \times 1000}{1 \times 1 0^{- 3}}=50 \, Scm^{2}mol^{- 1}$

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

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