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Precipitation Equilibria

Precipitation Equilibria. 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. Solubility Product.

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Precipitation Equilibria

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  1. Precipitation Equilibria

  2. 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.

  3. Solubility Product • Example: What is the solubility of silver chloride in pure water? Ksp = 1.8x10-10 • Write the equilibrium expression: [Ag+][Cl-] = Ksp =1.8x10-10 why are we ignoring the AgCl??

  4. Solubility Product • Use (ICE)

  5. Solubility Product • Plug the equilibrium (E) values into the equilibrium expression: [x][x] = x2= 1.8x10-10 x = 1.3x10-5 [Ag+] = 1.3x10-5M, and [Cl-] = 1.3x10-5M

  6. Solubility Product • Example: What is the solubility of lead iodide in pure water? Ksp = 7.1x10-9 • Write the equilibrium expression: [Pb2+][I-]2 = 7.1x10-9

  7. Solubility Product • Use (ICE)

  8. Solubility Product • Plug the equilibrium (E) values into the equilibrium expression: [x][2x]2 =2x3= 7.1x10-9 x = 1.2x10-3 [Pb2+] = 1.2x10-3M, then [I-] = ??

  9. Solubility • Therefore, the solubility of PbI2 is: 1.2x10-3 mol/L • That is, 1.2x10-3 moles of PbI2 will dissolve in 1L of water. • Or, multiply by the MW of PbI2 to find that: (1.2x10-3mol/L)(461.0g/mol) = 0.55g/L

  10. 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 2.0x10-3M Cl-?

  11. The Common Ion Effect • Write the equilibrium expression: [Ag+][Cl-] = 1.8x10-10 • Use the ‘ICE’ method:

  12. The Common Ion Effect • Plug the equilibrium (E) values into the equilibrium expression: [x][2.0x10-3 +x] = 1.8x10-10 • Solve: x2 + 2.0x10-3x – 1.8x10-10 = 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!

  13. Another Example • Add 10.0mL of 0.20M AgNO3 to 10.0mL of 0.10M NaCl. How much Cl- will remain in solution? • First, this is a limiting reagent problem:

  14. 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 = 1.0x10-3moles [Ag+] = 0.050M

  15. To determine [Cl-], simply use ICE and the equilibrium expression: [0.050+x][x] = 1.8x10-10 ignore x in the Ag+ term x = 3.6x10-9 [Cl-] = 3.6x10-9M What is [Ag+] at equilibrium?

  16. 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

  17. 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 NH3 or OH- is used to react with metal cations Use K = Ksp x Kf (2 steps)

  18. 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

  19. Groups for Qualitative Analysis • Group I • Cations that form insoluble chlorides: Ag, Pb, Hg2 • Group II • Cations that form insoluble sulfides • H2S (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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