The conductivity `(k)` of a saturated solution of `AgBr`(1M) at `298K` is `8.5xx10^(-7)Scm^(-1)`. If `lambda_(Ag^(+))^(@)` and `lambda_(Br^(-))^(@)` are 62 and `78 S cm^(2)mol^(-1)` , respectively, then calculate the solubility and `K_(sp)` of `AgBr`.

Explanation

`Lambda_(m)^(@) = lambda_(m)^(@) (Ag^(+) ) + lambda_(m)^(@) (Br^(-))`
`62 + 78 = 140.0 S cm^(2) mol^(-1)`
Now for saturated solution
`Lambda_(m) ^(@) (AgBr) = (kappa xx 1000)/(s)`
where s is the solubility
`140.0 = (8.5 xx 10^(-7) xx 1000)/(s)`
`or " " s = (8.5 xx 10^(-7) xx 1000)/(140.0) = 6.07 xx 10^(-6) mol L^(-1)`
Now `AgBr hArr Ag^(+) + Br^(-)`
`K_(sp) = [Ag^(+)] [Br^(-)]`
`= (6.07 xx 10^(-6)) xx (6.07 xx 10^(-6))`
`= 36.8 xx 10^(-12) mol^(2) L^(-2)`