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Question
Calculate the conductivity in (1×10−7Sm−2unit) of a solution prepared by dissolving 10−7 mol of AgNO3 in litre of saturated aqueous solution of AgBr. Given Ksp of AgBr=12×10−14 and λoAg+,λoBr− and λoNO−3 are 6×10−3,8×10−3 and 7×10−3Sm2mol−1 respectively. Neglect the contribution of solvent
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Solution
Solubililty of AgBr(s) in presence of 10−7 mole of AgNO3 solution can be obtained by
Ksp=[Ag+][Br−]
12×10−14=[S+10−7][S]
S2+10−7S−12×10−14=0
∴S=3×10−7M
or [Ag+]=3×10−7+10−7=4×10−7M
[Br−]=3×10−7M
[NO−3]=1×10−7M
Now, k=Ao×M1000(Ao in Sm2mol−1=A×104Scm2mol−1)
kAg+=6×10−3×104×4×10−71000=24×10−9Scm−1
kBr−=8×10−3×104×3×10−71000=24×10−9Scm−1
kNO−3=7×10−3×104×10−71000=7×10−9Scm−1
∴ktotal=kAg++kBr−+kNO−3
=24×10−9+24×10−9+7×10−9
=55×10−9Scm−1=55×10−7Sm−1
=55 (in the given unit)
Hence the conductivity of the solution is 55×10−7Sm−2.
Ksp=[Ag+][Br−]
12×10−14=[S+10−7][S]
S2+10−7S−12×10−14=0
∴S=3×10−7M
or [Ag+]=3×10−7+10−7=4×10−7M
[Br−]=3×10−7M
[NO−3]=1×10−7M
Now, k=Ao×M1000(Ao in Sm2mol−1=A×104Scm2mol−1)
kAg+=6×10−3×104×4×10−71000=24×10−9Scm−1
kBr−=8×10−3×104×3×10−71000=24×10−9Scm−1
kNO−3=7×10−3×104×10−71000=7×10−9Scm−1
∴ktotal=kAg++kBr−+kNO−3
=24×10−9+24×10−9+7×10−9
=55×10−9Scm−1=55×10−7Sm−1
=55 (in the given unit)
Hence the conductivity of the solution is 55×10−7Sm−2.
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