Question

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

Answer 1

Answer: 3.00

${\alpha }_{HX}=\frac{{\Lambda }_{HX}}{{\Lambda }_{HX}^{\circ }};{\alpha }_{HY}=\frac{{\Lambda }_{HY}}{{\Lambda }_{HY}^{\circ }}$
$\therefore {\alpha }_{HX}=\frac{1}{10}{\alpha }_{HY}$
$\left\{\because {\Lambda }_{HX}=\frac{{\Lambda }_{HY}}{10}\right\}$
$K_a={\alpha }^{2}C$
$K_{aHX}={\alpha }_{HY}^2\times \frac{1}{100}\times C_1$
$={\alpha }_{HY}^2\times \frac{1}{100}\times 0.01$
$K_{aHY}={\alpha }_{HY}^2\times C_2={\alpha }_{HY}^2\times 0.1$
$K_{aHX}=\frac{1}{1000}{K}_{aHY}$
$p{K}_{a_{HY}}=-\log K_{a_{HX}}=-\log \frac{1}{1000}{K}_{a_{HY}}=3$
$\log K_{a_{HY}}$
$p{K}_{a_{HY}}=-\log K_{a_{HY}}=-\log K_{a_{HY}}$
$p{K}_{k{a}_{HY}}-p{K}_{a_{HY}}=3$