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 $y$ is

• A. 0.880
• B. 0.88
• C. 0.8800

Answer 1

Answer: (A), (B), (C)

${\alpha }_1=\frac{{\wedge }_m^c}{{\wedge }_m^0}=\frac{y\times {10}^2}{4\times {10}^2}=\frac{y}{4}=\alpha$

On dilution conductivity increases three times
${\alpha }_1=\frac{3y\times {10}^2}{4\times {10}^2}=3{\alpha }_1=3\alpha$
$\begin{array}{ccccc}HA& \rightleftharpoons & H^{+}& +& A^{-}\\ C& & O& & O\\ C(1-\alpha )& & C\alpha & & C\alpha \\ & & & & \\ & & & & \end{array}$
$K_a=\frac{C{\alpha }^2}{1-\alpha }$
Since temperature is constant $K_a$will be constant
$\frac{C_1{\alpha }_1^2}{1-{\alpha }_1}=\frac{C_2{\alpha }_2^2}{1-{\alpha }_2}$

$\Rightarrow \frac{C\times {\alpha }^2}{1-\alpha }=\frac{\left(\frac{C}{20}\right(3\alpha {)}^2)}{(1-3\alpha )}$
$\Rightarrow \frac{1}{1-\alpha }=\frac{9}{20}\frac{1}{(1-3\alpha )}$
$\Rightarrow 20-60\alpha =9-9\alpha ;$
$\Rightarrow \alpha =\frac{11}{51}=0.2156$
$\Rightarrow \alpha =0.22$

$\alpha =\frac{y}{4};y=4\alpha =4\times 0.22=0.88$

From above question we found the value of $\alpha$.