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

Explanation

`wedge^(@)._(m(AgBr))=lambda^(@)._(Ag^(o+))+lambda^(@)._(Br^(c-))`
`=62+78=140.0Scm^(2) mol^(-1)`
`:. wedge^(@)._(m(AgBr))=wedge_(m(AgBr))=(kxx1000)/(S)` ltBRgt `140.0S cm^(2)mol^(-1)=(8.5xx10^(-7)xx10^(3))/(S)`
`:. S=8.5xx10^(-7)xx10^(3)//140.0`
`=0.06xx10^(-4)mol L ^(-1)`
`=0.06xx10^(-4)xxMw_((AgBr))`
`=0.06xx10^(-4)xx188g L^(-1)` ltbr. `1.128xx10^(-3)gL^(-1)`
`:. K _(sp)` of `AgBr=S^(2)M^(2)=(0.06xx10^(-4))^(2)`
`=3.6 xx 10^(-11)mol^(2)L^(-2)`