The molar conductivity of a solution of a weak acid `HX(0.01 M)` is 10 times smalller than the molar conductivity of a solution of a weak acid `HY (0.10 M)`. If `lamda_(X^(-))^(@) =lamda_(Y^(-))^(@)`, the difference in their `pK_(a)` values, `pK_(a)(HX) - pK_(a)(HY)`, is (consider degree of ionisation of both acids to be `ltlt 1`):

Options

A. `4`
B. `1`
C. `2`
D. `3`

Answer

D

Explanation

`HX(aq.)hArrH^(+)(aq.)+X^(-)(aq.)`
`K_(a) = (C_(H^(+))C_(Y^(-)))/(C_(HX))`
`HY(aq.)hArrH^(+)(aq.)+Y^(-)(aq.)`
`K_(a) = (C_(H^(+))C_(Y^(-)))/(C_(HY))`

Let's assume that
`Lambda_(m)` for `HX = Lambda_(m_(1))`
`Lambda_(m)` for `HX = Lambda_(m_(2))`
We are given
`Lambda_(m_(1)) = (1)/(10)Lambda_(m_(2))`
Accroding to Ostwald's dilution law
`K_(a) = C alpha^(2)` (as `alpha lt lt 1`)
Thus, `K_(a_(1)) = C_(1)(Lambda_(m_(1))/(Lambda_(m_(2))^(@)))^(2)`
`K_(a_(2)) = C_(2)(Lambda_(m_(1))/(Lambda_(m_(2))^(@)))^(2)`
Since
`Lambda_(m1)^(@) = lamda_(H^(+))^(@)+lamda_(X^(-))^(@)`
`Lambda_(m2)^(@) = lamda_(H^(+))^(@)+lamda_(Y^(-))^(@)`
and `lamda_(X^(-))~~lamda_(Y^(-))^(@)`
`Lambda_(m_(1))^(@) ~~ Lambda_(m_(2))^(@)`
Thus, `(K_(a_1))/ (K_(a_2)) = (C_(1))/(C_(2)) xx (Lambda_(m_(1))/(Lambda_(m_(2))))`
`= (0.01 M)/(0.1 M) xx ((1//10 Lambda_(m_(2)))/(Lambda_(m_(2))^(2)))`
`= 0.001` or `10^(-3)`
or `K_(a_(1)) = 10^(-3) K_(a_(2))`
Taking negative log of both sides, we get
`-log K_(a_(1)) = (-log 10^(-3))+(-log K_(a_(2)))`
`pK_(a_(1)) = 3+pK_(a_(2))`
or `pK_(a_(1))-pK_(a_(2)) = 3`