The equivalent conductance of `M//32` solution of a weak monobasic acid is `8.0 ` and at infinite dilution is `400 `. The dissociation constant of this acid is :

Options

A. `1.25 xx 10^(-6)`
B. `6.25 xx 10^(-4)`
C. `1.25 xx 10^(-4)`
D. `1.25 xx 10^(-5)`

Answer

D

Explanation

Let the weak monobasic acid be `HA`:
`HA hArrH^(+)(aq.)+A^(+)(aq.)`
`{:(I.C.,C,0,0),(Eq.C.,C-Calpha,Calpha,Calpha):}`
where `alpha` is the degree pf ionization while `C` is the initial conc. Of he acid.
`K_(a) = ([H^(+)][A^(-)])/([Ha])`
`= ((Calpha)(Calpha))/(C-Calpha) = (Calpha^(2))/(C(1-alpha)`
`(Calpha^(2))/(1-alpha)`
Since `alphaltlt1`, we have
`K_(a) = C alpha^(2)`
If `Lambda_(eq.)^(@)` and `Lambda_(eq.)^(oo)` are equivalent conductances at any given concentration and at infinite dilution then
`alpha = Lambda_(eq.)^(C)/(Lambda_(eq.)^(oo))`
`= (8.0 "mhos cm"^(2))/(400 "mhos cm"^(2))`
`2 xx 10^(-2)`
Substituting the value of concentration and degree of ionization, we gave
`K_(a) = (1)/(32)(2 xx 10^(-2))^(2)`
`= 0.125 xx 10^(-4)`
`= 1.25 xx 10^(-5)`