Question

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

Answer 1

The ratio of measure conductivity (specific conductance) of an electrolyte solution to its molar concentration is known as molar conductivity. The molar conductivity of weak electrolytes depends on the concentration of the solution. The more dilute a solution, the greater it's molar conductivity, due to increased ionic dissociation.

Complete answer:
The molar conductivity of each species is proportional to its electrical mobility per electric field.
Weak electrolytes have a lower degree of association like acetic acid at higher concentrations. Such weak electrolytes increase their degree of dissociation while dilution leads to a change in the molar conductivity at 1 mole of electrolyte.
Let $\alpha$ be the degree of dissociation, which is the ratio of molar conductivity at concentration c to limiting molar conductivity,
$\alpha =\dfrac{{{\lambda }_{m}}}{{{\lambda }^{o}}_{m}}$
Given, the molar conductivity of a solution of a weak acid HX (0.10M) is 10 times smaller than the molar conductivity of a solution of weak acid HY (0.10M).
Let, the molar conductivity of weak acid HY is,${{\lambda }_{m}}(HY)=a$
Then the molar conductivity of weak acid HX is, ${{\lambda }_{m}}(HX)=\dfrac{a}{10}$
Consider, the degree of dissociation of weak acids HX and HY are ${{\alpha }_{1}}$ and ${{\alpha }_{2}}$ respectively.
$\dfrac{{{\alpha }_{1}}}{{{\alpha }_{2}}}=\dfrac{\dfrac{{{\lambda }_{m}}(HX)}{{{\lambda }^{o}}_{m}(HX)}}{\dfrac{{{\lambda }_{m}}(HY)}{{{\lambda }^{o}}_{m}(HY)}}=\dfrac{1}{10}$ --- (1) since, $({{\lambda }^{o}}_{{{X}^{-}}}\approx {{\lambda }^{o}}_{{{Y}^{-}}})$
HX is a weak acid, then does not dissociate completely,
$HX\rightleftharpoons$ ${{H}^{+}}$ + ${{X}^{-}}$
0.1M - - (before dissociation)
$0.1(1-{{\alpha }_{1}})$ $0.1{{\alpha }_{1}}$ $0.1{{\alpha }_{1}}$ (After dissociation)
${{K}_{{{a}_{1}}}}=\dfrac{[{{H}^{+}}][{{X}^{-}}]}{[HX]}=\dfrac{0.1{{\alpha }_{1}}X0.1{{\alpha }_{1}}}{0.1(1-{{\alpha }_{1}})}=0.1{{\alpha }_{1}}^{2}$ --- (2)
Consider the degree of ionization of both acids to be <<1
HY is a weak acid, then does not dissociate completely,
$HY\rightleftharpoons$ ${{H}^{+}}$ + ${{Y}^{-}}$
0.1M - - (before dissociation)
$0.1(1-{{\alpha }_{2}})$ $0.1{{\alpha }_{2}}$ $0.1{{\alpha }_{2}}$ (After dissociation)
${{K}_{{{a}_{2}}}}=\dfrac{[{{H}^{+}}][{{Y}^{-}}]}{[HY]}=\dfrac{0.1{{\alpha }_{2}}X0.1{{\alpha }_{2}}}{0.1(1-{{\alpha }_{2}})}=0.1{{\alpha }_{2}}^{2}$ --- (3)
From equation (1), (2), and (3)
$\dfrac{{{K}_{a}}(HX)}{{{K}_{a}}(HY)}=\dfrac{{{\alpha }_{1}}^{2}}{{{\alpha }_{2}}^{2}}=\dfrac{1}{100}$
$\log {{K}_{a}}(HX)-\log {{K}_{a}}(HY)=-2$
Hence, $p{{K}_{a}}(HX)-p{{K}_{a}}(HY)$=2

Note:
It is possible to calculate the molar conductivity for any electrolyte from individual ions of molar conductivity with the help of the Kohlrausch Law of independent migration of ions. Like Acetic acids are weak electrolytes, is possible to determine the value of its dissociation constants, if we know the value of molar conductivity of a solution, its ions, and the concentration of weak acid.