Question

The molar conductivity of a solution of a weak acid HX (0.10 M) is 10 times smaller than the molar conductivity of a solution of of weak acid HY (0.1 M). If $({\lambda }_{x-}^0\approx {\lambda }_{y-}^0)$, the difference in their $p{K}_a$ values, $p{K}_a\left(HX\right)-p{K}_a\left(HY\right)$, is:

(consider degree of ionization of both acids to be << 1)

Answer 1

${\Lambda }_m\left(HX\right)=\frac{x}{10}$ ${\Lambda }_m\left(HY\right)=m$
$\frac{{\Lambda }_m\left(HX\right)/{\Lambda }_m^o\left(HX\right)}{{\Lambda }_m\left(HY\right)/{\Lambda }_m^o\left(HY\right)}=\frac{\left(x/10\right)/{\Lambda }_m^o\left(HX\right)}{x/{\Lambda }_m^o\left(HY\right)}=\frac{{\alpha }_1}{{\alpha }_2}=\frac{1}{10}$ ---- 1
$HX\rightleftharpoons H^{+}+X^{-}$
0.1 - -
0.1(1-${\alpha }_1$) 0.1${\alpha }_1$ 0.1${\alpha }_1$

$K_{a_1}=0.1{\alpha }_1^2$ ----- 2
$HY\rightleftharpoons H^{+}+Y^{-}$
0.1 - -
0.1(1-${\alpha }_2$) 0.1${\alpha }_2$ 0.1${\alpha }_2$

$K_{a_2}=0.1{\alpha }_2^2$ ------- 3
From 1, 2 and 3 equation

$\frac{K_{a_1}}{K_{a_2}}=\frac{{\alpha }_1^2}{{\alpha }_2^2}=\frac{1}{100}$
$log{K}_{a_1}-log{K}_{a_2}=-2$
$p{K}_{a_1}-p{K}_{a_2}=2$