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

Given `: lambda_(X^-)^(@) = lambda_(Y^-)^(@)`
Adding `lambda_(H^+)^(@)` on both sides
`lambda_(X^-)^(@) + lambda_(H^+)^(@) = lambda_(Y^-)^(@) + lambda_(H^+)^(@)`
or `Lambda_(HX)^(@) = Lambda_(HY)^(@) " " …. (i)`
Degree of dissociation is given as
`alpha_(HX) = (Lambda_(HX)^c)/(Lambda_(HX)^(@)) ` and `alpha_(HY) = (Lambda_(HY)^c)/(Lambda_(HY)^@)`
or `(alpha_(HX))/(alpha_(HY)) = (Lambda_(HX)^(c))/(Lambda_(HY)^(c)) xx (Lambda_(HY)^(@))/(Lambda_(HX)^(@)) = (Lambda_(HX)^(c))/(Lambda_(HY)^(c)) " " ... (ii)`
But `Lambda_(HX)^(c) = (1)/(10) Lambda_(HY)^(c) = 0.1 Lambda_(HY)^(c)`
`therefore alpha_(HX) = 0.1 alpha_(HY)`
Now `K_(a (HX)) = (c alpha^(2) HX)/(1 - alpha_(HX)) = c alpha^(2) HX = 0.01 xx (0.1 alpha_(HY))^(2) " " .. . (iii)`
`K_(a) (HY) = c alpha^(2) HY = 0.1 (alpha_(HY))^(2) " ".... (iv)`
Dividing eq. (iii) by eq. (iv)
`(K_(a (HX)))/(K_(a (HY))) = (0.01 xx 0.01 xx 0.01 alpha^(2) HY)/(0.1 alpha^(2) HY) = 10^(-3)`
Taking log
`log K_(a (HX)) - log K_(a (HY)) = -3`
`-pK_(a) (HX) + pK_(a) (HY) = -3`
`or pK_(a) (HX) - pK_(a) (HY) = 3`