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PRECIPITATION REACTIONS Solubility of Salts Section 18.4

PRECIPITATION REACTIONS Solubility of Salts Section 18.4. Analysis of Silver Group. All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions. Analysis of Silver Group.

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PRECIPITATION REACTIONS Solubility of Salts Section 18.4

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  1. PRECIPITATION REACTIONSSolubility of SaltsSection 18.4

  2. Analysis of Silver Group All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions.

  3. Analysis of Silver Group Although all salts formed in this experiment are said to be insoluble, they do dissolve to some SLIGHT extent. AgCl(s) ↔ Ag+(aq) + Cl-(aq) When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED.

  4. Analysis of Silver Group AgCl(s) ↔Ag+(aq) + Cl-(aq) When solution is SATURATED, expt. shows that [Ag+] = 1.67 x 10-5 M. This is equivalent to the SOLUBILITYof AgCl. What is [Cl-]? [Cl-] is equivalent to the AgCl solubility.

  5. Analysis of Silver Group AgCl(s) ↔Ag+(aq) + Cl-(aq) Saturated solution has [Ag+] = [Cl-] = 1.67 x 10-5 M Use this to calculate Kc Kc = [Ag+] [Cl-] = (1.67 x 10-5)(1.67 x 10-5) = 2.79 x 10-10

  6. Analysis of Silver Group AgCl(s) ↔Ag+(aq) + Cl-(aq) Kc = [Ag+] [Cl-] = 2.79 x 10-10 Because this is the product of “solubilities”, we call it Ksp = solubility product constant See Table 18.2 and Appendix J

  7. Lead(II) Chloride PbCl2(s)  Pb2+(aq) + 2 Cl-(aq) Ksp = 1.9 x 10-5

  8. Solubility of Lead(II) Iodide Consider PbI2 dissolving in water PbI2(s) ↔Pb2+(aq) + 2 I-(aq) Calculate Kspif solubility = 0.00130 M Solution 1. Solubility = [Pb2+] = 1.30 x 10-3 M [I-] = ? [I-] = 2 x [Pb2+] = 2.60 x 10-3 M

  9. Solubility of Lead(II) Iodide Consider PbI2 dissolving in water PbI2(s) ↔Pb2+(aq) + 2 I-(aq) Calculate Kspif solubility = 0.00130 M Solution 2. Ksp = [Pb2+] [I-]2 = [Pb2+] {2 • [Pb2+]}2 Ksp = 4 [Pb2+]3 = 4 (solubility)3 Ksp = 4 (1.30 x 10-3)3 = 8.8 x 10-9

  10. Precipitating an Insoluble Salt Hg2Cl2(s) ↔ Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2 If [Hg22+] = 0.010 M, what [Cl-] is req’d to just begin the precipitation of Hg2Cl2? That is, what is the maximum [Cl-] that can be in solution with 0.010 M Hg22+ without forming Hg2Cl2?

  11. Precipitating an Insoluble Salt Hg2Cl2(s) ↔Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2 Recognize that Ksp = product of maximum ion concs. Precip. begins when product of ion concs. EXCEEDS the Ksp.

  12. Precipitating an Insoluble Salt Hg2Cl2(s) ↔ Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2 Solution [Cl-] that can exist when [Hg22+] = 0.010 M, If this conc. of Cl- is just exceeded, Hg2Cl2 begins to precipitate.

  13. Precipitating an Insoluble Salt Hg2Cl2(s) ↔ Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 Now raise [Cl-] to 1.0 M. What is the value of [Hg22+] at this point? Solution [Hg22+] = Ksp / [Cl-]2 = Ksp / (1.0)2 = 1.1 x 10-18 M The concentration of Hg22+ has been reduced by 1016 !

  14. The Common Ion Effect Adding an ion “common” to an equilibrium causes the equilibrium to shift back to reactant.

  15. Common Ion Effect PbCl2(s) ↔Pb2+(aq) + 2 Cl-(aq) Ksp = 1.9 x 10-5

  16. The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s) ↔ Ba2+(aq) + SO42-(aq) Solution Solubility in pure water = [Ba2+] = [SO42-] = x Ksp = [Ba2+] [SO42-] = x2 x = (Ksp)1/2 = 1.1 x 10-5 M Solubility in pure water = 1.1 x 10-5 mol/L

  17. Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s) ↔ Ba2+(aq) + SO42-(aq) The Common Ion Effect Solution Solubility in pure water = 1.1 x 10-5 mol/L. Now dissolve BaSO4 in water already containing 0.010 M Ba2+. Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSO4 be less than or greater than in pure water?___

  18. Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s)↔Ba2+(aq) + SO42-(aq) The Common Ion Effect Solution [Ba2+] [SO42-] initial change equilib. 0 0.010 + y + y 0.010 + y y

  19. The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s) ↔Ba2+(aq) + SO42-(aq) Solution Ksp = [Ba2+] [SO42-] = (0.010 + y) (y) Because y < 1.1 x 10-5 M (= x, the solubility in pure water), this means 0.010 + y is about equal to 0.010. Therefore, Ksp = 1.1 x 10-10 = (0.010)(y) y = 1.1 x 10-8 M = solubility in presence of added Ba2+ ion.

  20. The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s) ↔Ba2+(aq) + SO42-(aq) SUMMARY Solubility in pure water = x = 1.1 x 10-5 M Solubility in presence of added Ba2+ = 1.1 x 10-8 M Le Chatelier’s Principle is followed!

  21. Separating Metal Ions Cu2+, Ag+, Pb2+ Ksp Values AgCl 1.8 x 10-10 PbCl2 1.7 x 10-5 PbCrO4 1.8 x 10-14

  22. Separating Salts by Differences in Ksp A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first? Ksp for Ag2CrO4 = 9.0 x 10-12 Ksp for PbCrO4 = 1.8 x 10-14 Solution The substance whose Ksp is first exceeded precipitates first. The ion requiring the lesser amount of CrO42- ppts. first.

  23. Separating Salts by Differences in Ksp A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first? Ksp for Ag2CrO4 = 9.0 x 10-12 Ksp for PbCrO4 = 1.8 x 10-14 Solution Calculate [CrO42-] required by each ion. [CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+] = 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M [CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2 = 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M PbCrO4 precipitates first

  24. Separating Salts by Differences in Ksp A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. PbCrO4 ppts. first. Ksp (Ag2CrO4)= 9.0 x 10-12 Ksp (PbCrO4) = 1.8 x 10-14 How much Pb2+ remains in solution when Ag+ begins to precipitate?

  25. Separating Salts by Differences in Ksp A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. How much Pb2+ remains in solution when Ag+ begins to precipitate? Solution We know that [CrO42-] = 2.3 x 10-8 M to begin to ppt. Ag2CrO4. What is the Pb2+ conc. at this point? [Pb2+] = Ksp / [CrO42-] = 1.8 x 10-14 / 2.3 x 10-8 M = 7.8 x 10-7 M Lead ion has dropped from 0.020 M to < 10-6 M

  26. Separating Salts by Differences in Ksp • Add CrO42- to solid PbCl2. The less soluble salt, PbCrO4, precipitates • PbCl2(s) + CrO42-PbCrO4 + 2 Cl- • Salt Ksp PbCl2 1.7 x 10-5 PbCrO4 1.8 x 10-14

  27. Separating Salts by Differences in Ksp • PbCl2(s) + CrO42-↔PbCrO4 + 2 Cl- Salt Ksp PbCl2 1.7 x 10-5 PbCrO4 1.8 x 10-14 PbCl2(s) ↔ Pb2+ + 2 Cl- K1 = Ksp Pb2+ + CrO42-↔ PbCrO4 K2 = 1/Ksp Knet = K1 • K2 = 9.4 x 108 Net reaction is product-favored

  28. Exit Problems in Pairs • At what pH will a solution 0.150 M Cu2+ begin to precipitate as Cu(OH)2? The Ksp for Cu(OH)2 is 1.6  10-19.

  29. 2. The Ksp for PbSO4 is 1.8  10-8. What is the concentration of SO42- in a saturated solution of PbSO4?

  30. Solubility and Complex Ions Section 18.6 The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 ------> COMPLEX IONS

  31. Reaction of NH3 with Cu2+(aq)

  32. Dissolving Precipitates by forming Complex Ions Formation of complex ions explains why you can dissolve a ppt. by forming a complex ion. AgCl(s) + 2 NH3e Ag(NH3)2+ + Cl-

  33. Dissolving Precipitates by forming Complex Ions Examine the solubility of AgCl in ammonia. AgCl(s) Ag+ + Cl- Ksp = 1.8 x 10-10 Ag+ + 2 NH3 --> Ag(NH3)2+ Kform = 1.6 x 107 ------------------------------------- AgCl(s) + 2 NH3Ag(NH3)2+ + Cl- Knet = Ksp • Kform = 2.9 x 10-3 By adding excess NH3, the equilibrium shifts to the right.

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