Question

Specific conductance of $0.1\text{}\text{M }\text{H}\text{A}\text{}$ is $3.75\times 10^{- 4 \, }\text{}\text{o}\text{h}\text{m}^{- 1}\text{}\text{c}\text{m}^{- 1}$ . If $\text{\lambda }^{\text{\infty }}$ of $\text{H}\text{A}$ is $250 \, \text{}\text{o}\text{h}\text{m}^{- 1}\text{c}\text{m}^{2}\text{m}\text{o}\text{l}^{- 1},$ then dissociation constant $\text{K}_{\text{a}}$ of $\text{H}\text{A}$ is

• A. $1\times 10^{- 5}$
• B. $2.25\times 10^{- 4}$
• C. $2.25\times 10^{- 5}$
• D. $2.25\times 10^{- 13}$

Answer 1

Answer: (C)

$\lambda _{m}=\frac{1000 x k}{m}=3.75$
$\alpha =\frac{\lambda _{m}}{\lambda _{m}^{\in fty}}=\frac{3 . 75}{250}=0.015=1.5\times 10^{- 2}$
$\therefore Ka=C\left(\alpha \right)^{2}=0.1\left(0 . 015\right)^{2}=2.25\times 10^{- 5}$