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Question
Conductivity of a saturated solution of a sparingly soluble salt AB (1 : 1 electrolyte) at 298 K is 1.85 × 10−5 S m−1. Solubility product of the salt AB at 298 K `(Λ_"m"^∘)_"AB"` = 14 × 10−3 S m2 mol−1.
Options
5.7 × 10−3
1.32 × 10−12
7.5 × 10−12
1.74 × 10−12
Solution
1.74 × 10−12
Explanation:
κ = 1.85 × 10−5 S m−1
`(Λ_"m"^∘)` = 14 × 10−3 S m2 mol−1
Ksp = ?
Ksp = `((κ xx 10^-3)/Λ^∘)^2`
= `((1.85 xx 10^-5 xx 10^-3)/(14 xx 10^-3))^2`
= (0.1321 × 10−5)2
= 0.01745 × 10−10
= 1.745 × 10−12
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