Question

At $298K$, the limiting molar conductivity of a weak monobasic acid is $4\times 10^{2}Sc{m}^{2}mo{l}^{-1}.$ At $298K$, for an aqueous solution of the acid the degree of dissociation is $\alpha$ and the molar conductivity is $y\times 10^{2}Sc{m}^{2}mo{l}^{-1}$. At $298K$, upon $20$ times dilution with water, the molar conductivity of the solution becomes $3y\times 10^{2}Sc{m}^{2}mo{l}^{-1}$
The value of $\alpha$ is _______.

Answer 1

Answer: 0.22

${\alpha }_1=\frac{{\Lambda }_m^c}{{\Lambda }_m^0}=\frac{y\times 10^2}{4\times 10^2}=\frac{y}{4}=\alpha$
On dilution conductivity increases three times
${\alpha }_2=\frac{3y\times 10^2}{4\times 10^2}=3{\alpha }_1=3\alpha$

$K_a=\frac{C{\alpha }^2}{1-\alpha }$
Since temperature is constant $K_a$ will be constant
$\frac{C_1{\alpha }_1^2}{1-{\alpha }_1}=\frac{C_2{\alpha }_2^2}{1-{\alpha }_2}$
$\frac{C\times {\alpha }^2}{1-\alpha }=\frac{\left(\frac{C}{20}\right)(3\alpha {)}^2}{1-3\alpha }$
$\frac{1}{1-\alpha }=\frac{9}{20}-\frac{1}{(1-3\alpha )}$
$20-60\alpha =9-9\alpha ;$
$\alpha =\frac{11}{51}=0.2156$
$\alpha =0.22$