Question

The conductivity of $0.001M$ acetic acid is $5\times 10^{-5}S$ $c{m}^{-1}$ and ${\wedge }^{\circ }$ is $390.5Sc{m}^{2}mo{l}^{-1}$. 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)

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