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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 $\ll 1$ )
$1$
$2$
$3$
$4$
Solution
$HX \rightleftharpoons H ^{+}+ X ^{-}$
$Ka =\frac{\left[ H ^{+}\right]\left[ X ^{-}\right]}{[ HX ]}$
$HY \rightleftharpoons H ^{+}+ Y ^{-}$
$Ka =\frac{\left[ H ^{+}\right]\left[ Y ^{-}\right]}{[ HY ]}$
$\Lambda_{ m } \text { for } HX =\Lambda_{ m _2}$
$\Lambda_{ m } \text { for } HY =\Lambda_{ m _2}$
$\Lambda_{ m _1}=\frac{1}{10} \Lambda_{ m _2}$
$Ka _2= C ^2$
$Ka _1= C _1 \times\left(\frac{\Lambda_{ m _1}}{\Lambda_{ m _1}^0}\right)^2$
$Ka _2= C _2 \times\left(\frac{\Lambda_{ m _2}}{\Lambda_{ m _2}^0}\right)^2$
$\frac{ Ka _1}{ Ka _2}=\frac{ C _1}{ C _2} \times\left(\frac{\Lambda_{ m _1}}{\Lambda_{ m _2}}\right)^2=\frac{0.01}{0.1} \times\left(\frac{1}{10}\right)^2=0.001$
$pKa _1- pKa _2=3$
