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Selective Precipitation Chemistry Problem
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[QUOTE="frozonecom, post: 4544420, member: 307849"] [h2]Homework Statement [/h2] A solution of [itex]AgNO_{3}[/itex] is added to a solution containing 0.100 M [itex]Cl^{-}[/itex] and 0.100 [itex]CrO_{4}^{2-}[/itex]. What will be the concentration of the less soluble compund when the more soluble one begins to precipitate?[h2]Homework Equations[/h2] Ksp AgCl = [itex]1.82 x 10^{-10}[/itex] Ksp Ag2CrO4 = [itex]1.2 x 10^{-12}[/itex] [h2]The Attempt at a Solution[/h2]So, by calculating for their molar solubilities, I would know which one would precipitate first (which one is more soluble or less soluble) (2x)^2 (x) = Ksp Ag2CrO4 x=Molar solubility of Ag2CrO4 = 6.69 x 10^-5 M (x)(x) = Ksp AgCl x= Molar solubility of AgCl = 1.35 x 10^-5 M Thus, Ag2CrO4 is more soluble and AgCl is the less soluble compound. Now, how will I find the concentration of the less soluble compound when the more soluble one begins to precipitate? Please guide me. Here's my attempt for a solution. The more soluble compound, Ag2CrO4 will begin to precipitate at this Ag+ concentration [Ag+]^2 [CrO42-] = Ksp Ag2CrO4 [Ag+] = sqrt( Ksp Ag2CrO4 / [CrO42-] ) = 3.46 x 10^-6 M Now, how would I find the concentration of AgCl in the solution? Again, here is my attempt: I think, I should substitute the Ag+ concentration at the formula [Ag+][Cl-] = Ksp AgCl , But, is the Cl- concentration that I will get equal to the concentration of AgCl in the solution? I'm very confused. :(. Anyway, here's an attempt: [Cl-] = Ksp AgCl / [Ag+] = 5.26 x 10^-5 M I mean, the answer CAN be plausible since the concentration seemingly decreased. Is it correct guys? I think what I am having problem with is that why would the concentration of AgCl be equal to the equilibrium conc of Cl- ? [/QUOTE]
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