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InterviewSolution
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The solubility of AgBr in water is 1.28 × 10-5 mol/dm3 at 298 K. Calculate the solubility product of AgBr at the same temperature. |
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Answer» Given : S = 1.28 × 10-5 mol dm-3; Ksp = ? AgBr dissociates as, AgBr ⇋ \(Ag^+_{(aq)}\)+ \(Br^-_{(aq)}\) Ksp = [Ag+] [Br-] As the solubility of AgBr in water is 1.28 × 10-5 moles/dm3. [Ag+] = [Br-] = 1.28 x 10-5 mol dm3 ∴ Ksp = [1.28 × 10-5] [1.28 × 10-5] = 1.638 × 10-10 ∴ Solubility product of AgBr = 1.638 × 10-10 |
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