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Published byDarlene Stevison Modified about 1 year ago

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Precipitation Equilibria

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Solubility Product Ionic compounds that we have learned are insoluble in water actually do dissolve a tiny amount. We can quantify the solubility using the equilibrium expression or solubility product.

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Solubility Product Example: What is the solubility of silver chloride in pure water? K sp = 1.8x Write the equilibrium expression: [Ag + ][Cl - ] = K sp =1.8x why are we ignoring the AgCl??

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Solubility Product Use (ICE) Ag + Cl - I00 C+x Exx

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Solubility Product Plug the equilibrium (E) values into the equilibrium expression: [x][x] = x 2 = 1.8x x = 1.3x10 -5 [Ag + ] = 1.3x10 -5 M, and [Cl - ] = 1.3x10 -5 M

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Solubility Product Example: What is the solubility of lead iodide in pure water? K sp = 7.1x10 -9 Write the equilibrium expression: [Pb 2+ ][I - ] 2 = 7.1x10 -9

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Solubility Product Use (ICE) Pb 2+ I-I- I00 C+x+2x Ex2x

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Solubility Product Plug the equilibrium (E) values into the equilibrium expression: [x][2x] 2 =2x 3= 7.1x10 -9 x = 1.2x10 -3 [Pb 2+ ] = 1.2x10 -3 M, then [I - ] = ??

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Solubility Therefore, the solubility of PbI 2 is: 1.2x10 -3 mol/L That is, 1.2x10 -3 moles of PbI 2 will dissolve in 1L of water. Or, multiply by the MW of PbI 2 to find that: (1.2x10 -3 mol/L)(461.0g/mol) = 0.55g/L

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The Common Ion Effect What if there is already an ion dissolved in the water that is common with the ionic compound? For example: What is the solubility of silver chloride in a solution that contains 2.0x10 -3 M Cl - ?

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The Common Ion Effect Write the equilibrium expression: [Ag + ][Cl - ] = 1.8x Use the ‘ICE’ method: Ag + Cl - I02.0x10 -3 C+x Ex2.0x x

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The Common Ion Effect Plug the equilibrium (E) values into the equilibrium expression: [x][2.0x x] = 1.8x Solve: x x10 -3 x – 1.8x = 0 x = 9.0x10 -8 Solubility of AgCl is 9.0x10 -8 mol/L vs. 1.3x10 -5 mol/L when no common ion was present!

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Another Example Add 10.0mL of 0.20M AgNO 3 to 10.0mL of 0.10M NaCl. How much Cl - will remain in solution? First, this is a limiting reagent problem:

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Since we are combining two solutions, find moles: Since we know that AgCl is insoluble, the amount of Ag + remaining in solution is: 2.0x10 -3 – 1.0x10 -3 = 1.0x10 -3 moles [Ag + ] = 0.050M

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To determine [Cl - ], simply use ICE and the equilibrium expression: [0.050+x][x] = 1.8x ignore x in the Ag + term x = 3.6x10 -9 [Cl - ] = 3.6x10 -9 M What is [Ag + ] at equilibrium? Ag + Cl - I C+x E0.050+xx

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So, how do we tell when a ppt will form? We use, P (analagous to Q) If P > Ksp, ppt will form If P< Ksp, no ppt will form If P=Ksp, solution is saturated but no ppt yet Solubility rules we used earlier work only When the concentration is 0.1 mol or greater

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Dissolving ppts Many methods are used to make water-insoluble ionic solids ionize Most commonly H+ is used to react with basic anions a strong acid, often HCl, is used works on virtually all carbonates many sulfides NH 3 or OH - is used to react with metal cations Use K = Ksp x Kf (2 steps)

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Qualitative Analysis Objective is to separate and identify cations present in an “unknown” solution Use ppt reactions to divide the ions into 4 groups Then bring ions into solution, separate and identify

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Groups for Qualitative Analysis Group I Cations that form insoluble chlorides: Ag, Pb, Hg 2 Group II Cations that form insoluble sulfides H 2 S (toxic and stinky) at pH 5 used: Cu, Bi, Hg, Cd, Sn, Sb Group III Cations that from more soluble sulfides Don’t ppt at ph 5 but do at pH 9: Al, Cr, Co, Fe, Mn, Ni, Zn Group IV Soluble chlorides and sulfide Alkaline earth (Mg, Ca, Ba) ppt as carbonates Alkali metal (Man, K) can be identified with flame tests

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