We have taken a saturated solution of `AgBr`, whose `K_(sp)` is `12xx10^(-14).` If `10^(-7)M` of `AgNO_(3)` are added to `1L` of this solutino, find the conductivity `(` specific conductance `)` of the solution in terms of `10^(-7)S m^(-1)` units.
Given `:`
`lambda^(@)._((Ag^(o+)))=6xx10^(-3)S m^(2) mol^(-1)`
`lambda^(@)._((Br^(c-)))=8xx10^(-3)S m^(2)mol^(-1)`
`lambda^(@)._((NO_(3)^(C-)))=7XX10^(-3)S m^(2) mol^(-1)`

Answer

`k=55xx10^(-7)Sm^(-1)`
`=55Scm^(-1)(` in terms of `10^(-7) S m^(-1))`

Explanation

Suppose the solubility of `AgBr `in `10^(-7) AgNO_(3)` in `S mol L^(-1)`

Total `[Ag^(o+)]=(S+10^(-7))M`
`K_(sp)` of `AgBr=[Ag^(o+)][Br^(c-)]`
`12xx10^(-14)=(S+10^(-7))(S)=S^(2)+10^(-7)S`
`S^(2)+10^(-7)S-12xx10^(-14)=0`
On solving, `S=3xx10^(-7)M`
Hence, `[Br^(c-)] =3xx10^(-7)M=3xx10^(-7)xx10^(3)m^(3)`
`=3xx10^(-4)m^(3)`
`[Ag^(o+)]=(3xx10^(-7)+10^(-7))M=4xx10^(-7)xx10^(3)m^(3)`
`=4xx10^(-4)m^(3)`
`[NO_(3)^(c-)]=10^(-7)M=10^(-7)xx10^(+3)=1xx10^(-4)m^(3)`
`:. lambda=(k)/(c)or k=lambdaxxc`
`:. k_(Br^(c-))=3xx10^(-4)xx8xx10^(-3)S m^(-1)`
`=24xx10^(-7)S m^(-1)`
`k_(Ag^(o+))=4xx10^(-4)xx6xx10^(-3)Sm^(-1)`
`=24xx10^(-7)S m^(-1)`
`k_(NO_(3)^(c-))=1xx10^(-4)xx7xx10^(-3)=7xx10^(-7)Sm^(-1)`
`k_(Tot a l)=k_(Br^(c-))+k_(Ag^(o+))+k_(NO_(3)^(c-))`
`(` Specific conductivity of solutino `)`
`=(24xx10^(-7)+24xx10^(-7)+7xx10^(-7))Sm^(-1)`
`=55xx10^(-7)Sm^(-1)`
`=55 S m^(-1)(` in terms of `10^(-7) S m^(-1))`