Question

Molar conductance of a $0.2$ M solution of a weak acid, $HA$ is $2.8\times {10}^{-2}$ $S{m}^2$ $mo{l}^{-1}$. If the limiting molar conductance of $HA$ is $560$ S $m^2$ $mo{l}^{-1}$, the dissociation constant of the acid is:

• A. $5\times {10}^{-10}$
• B. ${10}^{-10}$
• C. $5\times {10}^{-5}$
• D. $2\times {10}^{-15}$

Answer 1

Answer: (A)

$5\times {10}^{-10}$

Degree of dissociation= $\alpha =\frac{\text{Molar conductance of}0.2Mso{l}^n}{\text{Limiting molar conductvity}}$

$\alpha =\frac{2.8\times {10}^{-2}}{560}=5\times {10}^{-5}$

$\Rightarrow HA\rightleftharpoons H^{+}+A^{-}$

$C(1-\alpha )$ $C\alpha$ $C\alpha$

$K_a=\frac{C^2{\alpha }^2}{C(1-\alpha )}\Rightarrow K_a\frac{C{\alpha }^2}{1-\alpha }$

$\because \alpha <<1\Rightarrow K_a=C{\alpha }^2$

$\therefore K_a=0.2\times (5\times {10}^{-5}{)}^2$

$=0.2\times 25\times {10}^{-10}=5\times {10}^{-10}$