1
Question
The conductivity of solution prepared by dissolving
10−7 mole of AgNO3 in 1 litre of saturated aqueous solution of AgBr having
Ksp of AgBr = 12×10−14 and λ0Ag+, λ0Br−, λ0NO−3 as 6×10−3, 8×10−3 and 7×10−3Sm2mol−1 respectively and neglecting contribution of solvent is found to be equal to p×10–6Sm–1
then, the value of p is
10−7 mole of AgNO3 in 1 litre of saturated aqueous solution of AgBr having
Ksp of AgBr = 12×10−14 and λ0Ag+, λ0Br−, λ0NO−3 as 6×10−3, 8×10−3 and 7×10−3Sm2mol−1 respectively and neglecting contribution of solvent is found to be equal to p×10–6Sm–1
then, the value of p is
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Solution
AgBr(s)⇌Ag++Br−
s+10−7 s
Ksp=12×10−14=(s+10−7)s
⇒s=3×10−7
k=6×10−3×4×10−710−3+8×10−3×3×10−710−3+7×10−3×10−710−3=5.5×10−6
s+10−7 s
Ksp=12×10−14=(s+10−7)s
⇒s=3×10−7
k=6×10−3×4×10−710−3+8×10−3×3×10−710−3+7×10−3×10−710−3=5.5×10−6
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