Question

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

Answer 1

$\alpha_{1}=\frac{\Lambda_{ m }^{ c }}{\Lambda_{ m }^{0}}=\frac{ y \times 10^{2}}{4 \times 10^{2}}=\frac{ y }{4}=\alpha$
On dilution conductivity increases three times
$\alpha_{2}=\frac{3 y \times 10^{2}}{4 \times 10^{2}}=3 \alpha_{1}=3 \alpha$

$K _{ a }=\frac{ C \alpha^{2}}{1-\alpha}$
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 $
$\alpha=0.22$