InterviewSolution
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InterviewSolution
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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)`] |
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Answer» 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)` |
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