We have taken a saturated solution of `AgBr,K_(sp)` of AgBr is `12xx10^(-14)`. If `10^(-7)` "mole" of `AgNO_(3)` are added to 1 litre of this solution then the conductivity of this solution in terms of `10^(-7) Sm^(-1)` units will be: [Given: `lambda_((Ag^(+)))^(@)=4xx10^(-3)Sm^(2)"mol"^(-1),lambda_((Br^(-)))^(@)=6xx10^(-3) S m^(2) "mol"^(-1),lambda_((NO_(3)^(-)))^(@)=5xx10^(-3)Sm^(2) "mol"^(-1)`]

Explanation

Suppose the solubility `AgBr` in
`10^(-7) AgNO_(3)` is `S moL^(-1)`
`{:(therefore" "underset("S mol L"^(-1))(AgBr) underset(larr)rarr Aunderset(S)g^(+) +Bunderset(S)r^(-)),(therefore" "underset(10^(-7)M)(AgNO_(3))underset(larr)rarrunderset(10^(-7)M)(Ag^(+))+underset(10^(-7)M)(NO_(3)^(-))):}`
Taking `[Ag^(+)] = (S+10^(-7))M`
`K_(sp)` of `AgBr = [Ag^(+)[ [Br^(-)]`
`12 xx 10^(-14) = (S +10^(-7)) (S) = S^(2) +10^(-7)S`
`S^(2) + 10^(-7)S - 12 xx 10^(-14) = 0`
On solving, `S = 3 xx 10^(-7)M`
`[Br^(-)] = 3 xx 10^(-7) M = 3 xx 10^(-7) xx 10^(3) mol//m^(3)`
`= 3 xx 10^(-4) mol//m^(3)`
Hence,
`[Ag^(+)] = (3 xx 10^(-7) +10^(-7)) M = 4 xx 10^(-4) mol//m^(3)`
`[NO_(3)^(-)] = 10^(-7)M = 10^(-7) xx 10^(+3) = 1 xx 10^(=4)m^(3)`
`: lambda = (k)/(c)` or `k = lambda xx c`
`:. k_(Br^(-)) = 3 xx 10^(-4) xx 8 xx 10^(-3) Sm^(-1) = 24 xx 10^(-7) S m^(-1)`
`k_(Ag^(+)) = 4 xx 10^(-4) xx 6 xx 10^(-3) Sm^(-1) = 24 xx 10^(-7) S m^(-1)`
`k_(NO_(3)^(-)) = 1 xx 10^(-4) xx 7 xx 10^(-3) = 7 xx 10^(-7) S m^(-1)`
`K_("total") = k_(Br^(-)) +K_(Ag^(+)) +K_(NO_(3)^(-))`
Specific conductivity of solution
`= (24 xx 10^(-7) +24 xx 10^(-7) +7 xx 10^(-7)) Sm^(-1)`
`= 55 xx 10^(-7) Sm^(-1)`