We have taken a saturated solution of `AgBr,` whose `K_(sp)` is `12xx10^(-14`. If `10^(-7)M" of "AgNO_(3)` are added to 1 L of this solution, find the conductivity of the solution in terms of `10^(-7)"S.m"^(-1)` units. Given that `lambda^(@)` values of the `Ag^(+), Br^(-) and NO_(3)^(-)` ions are `6xx10^(-3), 8xx10^(-3) and 7xx10^(-3)"S.m"^(2)."mol"^(-1)` respectively.

Explanation

`AgBr(s)hArrAg^(+)(aQ)+Br^(-)(aq)" …[1]"`
`AgNO_(aq)rarrAg^(+)(aq)+NO_(3)^(-)" …[2]"`
Suppose the solubility of `AgBr` in `10^(-7)"M "AgNO_(3)` is `"s mol.:"^(-1)`. Subtituting in equation (1) and (2), we get,
`therefore" Total "[Ag^(+)]=(s+10^(-7))M, K_(sp)(AgBr)=[Ag^(+)][Br^(-)]`
`"or, "12xx10^(-14)=(s+10^(-7))s" or, "s^(2)+10^(-7)s=12xx10^(-14)`
`"or, "s^(2)+10^(-7)s-12xx10^(-14)=0" or, "s=3xx10^(-7)M`
`therefore" "[Br^(-)]=3xx10^(-7)M=3xx10^(-7)xx10^(3)m^(-3)=3xx10^(-4)"mol.m"^(-3)`
`[Ag^(+)]=3xx10^(-7)+10^(-7)=4xx10^(-7)M`
`=4xx10^(-7)xx10^(3)m^(-3)=4xx10^(-4)"mol.m"^(-3)`
`[NO_(3)^(-)]=10^(-7)M=10^(-7)xx10^(3)m^(-3)=1xx10^(-4)"mol.m"^(-3)`
We know, `lambda=(kappa)/(C)" or, "kappa =lambda xxC`.
`therefore" "kappa_(Br^(-))=3xx10^(-4)xx8xx10^(-3)"S.m"^(-1)=24xx10^(-7)"S.m"^(-1)`
`therefore" "kappa_(Ag^(+))=4xx10^(-4)xx6xx10^(-3)"s.m"^(-1)=24xx10^(-7)"S.m"^(-1)`
`therefore" "kappa_(NO_(3)^(-))=1xx10^(-4)xx7xx10^(-3)"S.m"^(-1)=7xx10^(-7)"S.m"^(-1)`
`therefore" "kappa_("total")=kappa_(Br^(-)+kappa_(Ag^(+))+kappa_(NO_(3)^(-))`
`24xx10^(-7)+24xx10^(-7)+7+10^(-7)=55xx10^(-7)"S.m"^(-1)`