Question

$H A$ is a weak acid. At $25^{\circ} C$, the molar conductivity of $0.02\, M\, H A$ is $150\, \Omega^{-1} cm ^{2} mol ^{-1}$. If its $\Lambda_{ m }^{ o }$ is $300\, \Omega^{-1} cm ^{2} mol ^{-1}$, then equilibrium constant of $H A$ dissociation is

• A. 0.001
• B. 0.005
• C. 0.01
• D. 0.02

Answer 1

Answer: (C)

Given, molar conductivity
$\left(\Lambda_{m}\right)=300\, \Omega^{-1} cm ^{2} mol ^{-1}$
Limiting molar conductivity
$\left(\Lambda_{m}^{o}\right)=150\, \Omega^{-1} cm ^{2} mol ^{-1}$
Concentration $(C)=0.02\, M$
$\therefore $ Degree of dissociation
$(\alpha)=\frac{\Lambda_{m}}{\Lambda_{m}^{o}}$
$\alpha=\frac{150}{300} =0.5$
Now, by using following reaction, we can determine the value of its dissociation constant
$K_{a}=\frac{C \alpha^{2}}{(1-\alpha)}$
$K_{a}=\frac{0.02 \times(0.5)^{2}}{(1-0.5)}$
$=\frac{0.02 \times 0.5 \times .5}{0.5}=0.01$