For the complex ion [AgBr)' K,-1.7 x 107 and Ksp for AgBr= 5.0 x 10". Use...

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For the complex ion [AgBr) K,-1.7 x 107 and Ksp for AgBr= 5.0 x 10. Use this information to calculate Kc for the reaction AgBr Br AgBr2 5.

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Answer 1

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Solubility equilibrium of AgBr AgBr(s)g (a)+ Br(aq) Solubility product, KAg Br5.0x105 Formation constant of[ AgBeT.Kf1.7×10, Form ati on constant, Ki - [Ag.][Br] - 1.7x107 Desiredequation for AgBr in Br: AgBr(s) Br(a)AgB LAges] AgBr Br Equilibrium constant, K. - AgBr 1.7x10)5.0x10-13) = 8.5x10-6

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