Question

$HA$ is a weak acid. At $25^{\circ }C$, the molar conductivity of $0.02MHA$ is $150{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$. If its ${\Lambda }_m^o$ is $300{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$, then equilibrium constant of $HA$ dissociation is

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

Answer 1

Answer: (C)

Given, molar conductivity
$\left({\Lambda }_m\right)=300{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$
Limiting molar conductivity
$\left({\Lambda }_m^o\right)=150{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$
Concentration $(C)=0.02M$
$\therefore$ 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
$K_a=\frac{C{\alpha }^2}{(1-\alpha )}$
$K_a=\frac{0.02\times (0.5{)}^2}{(1-0.5)}$
$=\frac{0.02\times 0.5\times .5}{0.5}=0.01$