Question

The conductivity of $0.001 \,M$ acetic acid is $5 \times 10^{-5} S$ $cm^{-1}$ and $\wedge^{\circ}$ is $390.5 \,S\, cm^{2}\, mol^{-1}$. 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

Answer: (C)

$\wedge^{c}_{m}$ of $0.001M$ solution $= K = \frac{1000}{\text{Molarity}}$
$= 5 \times 10^{-5} \times \frac{1000}{0.001} = 50$
Degree of dissociation $\left(\alpha\right) = \frac{\wedge^{c}_{m}}{\wedge^{\circ}_{m}} = \frac{50}{390.5} =0.128$
For weak acids dissociation constant $\left(K\right) = \frac{C\alpha^{2}}{\left(1 -\alpha\right)}$
$K = \frac{0.001 \times \left(0.128\right)^{2}}{\left(1 -0.128\right)} = 18.78 \times10^{-6}$