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)`

Options

A. 39
B. 55
C. 15
D. 41

Answer

A

Explanation

`ksp(AgBr)=[Ag^+][Br^(-)]`
`12xx10^(-14)=(S+10^(-7)).S`
`S=3xx10^(-7)`
`[Br^(-)]=3xx10^(-7)[Ag^+]=4xx10^(-7)`
`[NO_3^(-)]=10^(-7)`
`K=10^6/1000(lambda_(Ag^+)^@M_(Ag^+)+lambda_(Br^(-))^@M_(Br^(-))+lambda_(NO_3^(-))^@M_(NO_3^(-)))`
`K=10^6/1000(2xx10^(-3)xx4xx10^(-7)+6xx10^(-3)xx3xx10^(-7)+5xx10^(-3)xx10^(-7))`
`K=39xx10^(-13)xx10^6=39xx10^(-7) sm^(-1)`