We have taken a saturated solution of `AgBrK_(sp)` of `AgBr` is `12 xx 10^(-14)`. If `10^(-7)` mole of `AgNO_(3)` are added to 1 litre of this solution find conductivity (specific conductance) of this solution in terms of `10^(-7) Sm^(-1) mol^(-1)`.
Given: `lambda_((Ag^(+)))^(@) = 6 xx10^(-3) S m^(2) mol^(-1), lambda_((Br^(-)))^(@) = 8 xx 10^(-3) S m^(2) mol^(-1), lambda_((NO_(3)^(-)))^(@) = 7 xx 10^(-3) S m^(2) mol^(-1)`

Answer

`55SM^(-1)`

Explanation

`AgBr(s)hArrAg^(+)+Br^(-)`
`(S+10^(-7))xxS=K_(SP)=12xx10^(-14)`
`S=3xx10^(-7)M`
`[Ag^(+)]=4xx10^(-7)M,[Br^(-)]=3xx10^(-7)M,[NO_(3)^(-)]=10^(-7)M`
`K_("total")=lamda_((Ag^(+)))^(@)lamda_((Ag^(+)))^(@)+lamda_(Br^(-))^(@)-lamda_(Br^(-))^(@)+lamda_((NO_(3))^(-))^(@)lamda_((NO_(3)^(-)))^(@)`
`lamd_(KCl)^(@)=lamda_(k)^(@)+lamda_(Cl^(-))^(@)`
`K_(KCl)=4xx10^(-4)xx6xx10^(-3)+3xx10^(-4)xx8xx10^(-3)+1xx10^(-4)xx7xx10^(-3)`
`K_(KCl)=24+24+7`
`K_(KCl)=55Scm^(-1)`