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`):

Answer

3

Explanation

`(Lambda_(HX))/(Lambda_(HY)) = (1)/(10) = (alpha_(HX)//Lambda_(HX)^(@))/(alpha_(HY)//Lambda_(HY)^(@)) because alpha = (Lambda_(c))/(Lambda^(@))`
As `overset(@)(lambda)_(x^(-)) = overset(@)(lambda)_(y^(-)), overset(@)(Lambda)_(HX) therefore (alpha_(HX))/(alpha_(HY)) = (1)/(10)`
`((K_(a))_(HX))/((K_(a))_(HY)) = (0.01 alpha_(HX)^(2))/(0.1 alpha_(HX)^(2))`