Question

The molar conductivity of a solution of a weak acid $HX (0.01\, M )$ is $10$ times smaller than the molar conductivity of a solution of a weak acid $HY (0.10\, M ) .$ If $\lambda_{ X ^{-}}^{0} \approx \lambda_{ Y ^{-}}^{0}$, the difference in their $pK_{a}$ values, $pK _{ a }( HX )- pK _{a}( HY )$, is (consider-degree of ionization of both acids to be $< < 1$ )

Answer 1

$\alpha_{ HX }=\frac{\Lambda_{ HX }}{\Lambda_{ HX }^{\circ}}; \alpha_{ HY }=\frac{\Lambda_{ HY }}{\Lambda_{ HY }^{\circ}}$
$\therefore \alpha_{ HX }=\frac{1}{10} \alpha_{ HY }$
$\left\{\because \Lambda_{ HX }=\frac{\Lambda_{ HY }}{10}\right\}$
$K _{ a }=\alpha^{2} C$
$K _{ aHX }=\alpha_{ HY }^{2} \times \frac{1}{100} \times C _{1}$
$=\alpha_{ HY }^{2} \times \frac{1}{100} \times 0.01$
$K _{ aHY }=\alpha_{ HY }^{2} \times C _{2}=\alpha_{ HY }^{2} \times 0.1$
$K _{ aHX }=\frac{1}{1000} K _{ aHY }$
$pK _{ a _{ HY }}=-\log K _{ a _{ HX }}=-\log \frac{1}{1000} K _{ a _{ HY }}=3$
$\log K _{ a _{ HY }}$
$pK _{ a _{ HY }}=-\log K _{ a _{ HY }}=-\log K _{ a _{ HY }}$
$pK _{ ka _{ HY }}- pK _{ a _{ HY }}=3$