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Question
We have taken a saturated solution of AgBr. Ksp of AgBr is 12×10−14. If 10−7 mol of AgNO3 are added to 1 litre of this solution then the conductivity of this solution in terms of 10−7Sm−1 units will be: [Given: λoAg+=4×10−3Sm2mol−1, λoBr−=6×10−3Sm2mol−1,λoNO−3=5×10−3Sm2mol−1]
We have taken a saturated solution of AgBr. Ksp of AgBr is 12×10−14. If 10−7 mol of AgNO3 are added to 1 litre of this solution then the conductivity of this solution in terms of 10−7Sm−1 units will be: [Given: λoAg+=4×10−3Sm2mol−1, λoBr−=6×10−3Sm2mol−1,λoNO−3=5×10−3Sm2mol−1]
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Solution
The correct option is A 39
Ksp(AgBr)=[Ag+][Br−]
12×10−14=(S+10−7)S
S=3×10−7
[Br−]=3×10−7[Ag+]=4×10−7
[NO−3]=10−7
K=1061000(λ0Ag+MAg++λ0Br−MBr−+λ0NO−3MNO−3)
K=1061000(4×10−3×4×10−7+6×10−3×3×10−7+5×10−3×10−7)
K=39×10−13×106=39×10−7Sm−1
Ksp(AgBr)=[Ag+][Br−]
12×10−14=(S+10−7)S
S=3×10−7
[Br−]=3×10−7[Ag+]=4×10−7
[NO−3]=10−7
K=1061000(λ0Ag+MAg++λ0Br−MBr−+λ0NO−3MNO−3)
K=1061000(4×10−3×4×10−7+6×10−3×3×10−7+5×10−3×10−7)
K=39×10−13×106=39×10−7Sm−1
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