1 / 53

Chapter 16 Precipitation Equilibria

Chapter 16 Precipitation Equilibria. Outline. 1. Precipitate formation: the solubility product constant (K sp ) 2. Dissolving precipitates. Revisiting Precipitation. In Chapter 4 we learned that there are compounds that do not dissolve in water These were called insoluble

brady-white
Download Presentation

Chapter 16 Precipitation Equilibria

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 16 Precipitation Equilibria

  2. Outline 1. Precipitate formation: the solubility product constant (Ksp) 2. Dissolving precipitates

  3. Revisiting Precipitation • In Chapter 4 we learned that there are compounds that do not dissolve in water • These were called insoluble • A reaction that produces an insoluble precipitate was assumed to go to completion • In reality, even insoluble compounds dissolve to some extent, usually small • An equilibrium is set up between the precipitate and its ions • Precipitates can be dissolved by forming complex ions

  4. Two Types of Equilibria • AgCl (s) ⇌ Ag+ (aq) + Cl- (aq) • Solid exists in equilibrium with the ions formed when a small amount of solid dissolves • AgCl (s) + 2NH3 (aq) ⇌ Ag(NH3)2+ (aq) + Cl- (aq) • Formation of a stable complex ion can cause an otherwise insoluble compound to dissolve • There are multiple equilibria at work in this example, in similar fashion to the equilibria underlying the function of a buffer (Chapter 14)

  5. Precipitate Formation: Solubility Product Constant, Ksp • Consider mixing two solutions: • Sr(NO3)2 (aq) • K2CrO4 (aq) • The following net ionic equation describes the reaction: • Sr2+ (aq) + CrO42- (aq)  SrCrO4 (s)

  6. Figure 16.1 – Precipitation of SrCrO4

  7. Ksp Expression • SrCrO4 (s) ⇌Sr2+ (aq) + CrO42- (aq) • The solid establishes an equilibrium with its ions once it forms • We can write an equilibrium expression, leaving out the term for the solid (recall that its concentration does not change as long as some is present) • Ksp is called the solubility product constant

  8. Interpreting the Solubility Expression and Ksp • Ksp has a fixed value at a given temperature • For strontium chromate, Ksp = 3.6 X 10-5 at 25 °C • The product of the two concentrations at equilibrium must have this value regardless of the direction from which equilibrium is approached

  9. Example 16.1

  10. Ksp and the Equilibrium Concentration of Ions • Ksp SrCrO4 = [Sr2+][CrO42-] = 3.6 X 10-5 • This means that if we know one ion concentration, the other one can easily be calculated • If [Sr2+] = 1.0 X 10-4 M, then • If [CrO42-] = 2.0 X 10-3, then

  11. Example 16.2

  12. Example 16.1, (Cont’d)

  13. Table 16.1

  14. Ksp and Precipitate Formation • Ksp values can be used to predict whether a precipitate will form when two solutions are mixed • Recall the use of Q, the reaction quotient, from Chapter 12 • We can calculate Q at any time and compare it to Ksp • The relative magnitude of Q vs. Ksp will indicate whether or not a precipitate will form

  15. Q and Ksp • If Q > Ksp, a precipitate will form, decreasing the ion concentrations until equilibrium is established • If Q < Ksp, the solution is unsaturated; no precipitate will form • If Q = Ksp, the solution is saturated just to the point of precipitation

  16. Figure 16.2

  17. Example 16.3

  18. Example 16.3, (Cont’d)

  19. Example 16.3, (Cont’d)

  20. Ksp and Water Solubility • One way to establish a solubility equilibrium • Stir a slightly soluble solid with water • An equilibrium is established between the solid and its ions • BaSO4 (s) ⇌ Ba2+ (aq) + SO42- (aq) • If we set the concentration of the ions equal to a variable, s:

  21. Precipitation Visualized

  22. Example 16.4

  23. Example 16.4, (Cont'd)

  24. Example 16.4, (Cont'd)

  25. Calculating Ksp Given Solubility • Instead of calculating solubility from Ksp, it is possible to calculate Ksp from the solubility • Recall that solubility may be given in many different sets of units • Convert the solubility to moles per liter for use in the Ksp expression

  26. Example 16.5

  27. Example 16.5, (Cont'd)

  28. Ksp and the Common Ion Effect • BaSO4 (s) ⇌ Ba2+ (aq) + SO42- (aq) • How would you expect the solubility of barium sulfate in water to compare to its solubility in 0.10 M Na2SO4? • Solubility must be less than it is in pure water • Recall LeChâtelier’s Principle • The presence of the common ion, SO42-, will drive the equilibrium to the left • Common ions reduce solubility

  29. Visualizing the Common Ion Effect

  30. Example 16.6

  31. Example 16.6, (Cont'd)

  32. Selective Precipitation • Consider a solution of two cations • One way to separate the cations is to add an anion that precipitates only one of them • This approach is called selective precipitation • Related approach • Consider a solution of magnesium and barium ions

  33. Selective Precipitation, (Cont'd) • Ksp BaCO3 = 2.6 X 10-9 • Ksp MgCO3 = 6.8 X 10-6 • Carbonate ion is added • Since BaCO3 is less soluble than MgCO3, BaCO3 precipitates first, leaving magnesium ion in solution • Differences in solubility can be used to separate cations

  34. Figure 16.3

  35. Figure 16.4 – Selective Precipitation

  36. Example 16.7

  37. Example 16.7, (Cont'd)

  38. Dissolving Precipitates • Bringing water-insoluble compounds into solution • Adding a strong acid to react with basic anions • Adding an agent that forms a complex ion to react with a metal cation

  39. Strong Acid • Zn(OH)2 (s) + 2H+ (aq)  Zn2+ (aq) + 2H2O • This reaction takes place as two equilibria: • Zn(OH)2 (s) ⇌ Zn2+ (aq) + 2OH- (aq) • 2H+ (aq) + 2OH- (aq) ⇌ 2H2O • Zn(OH)2 (s) + 2H+ (aq) ⇌ Zn2+ (aq) + 2H2O • Because the equilibrium constant for the neutralization is so large, the reaction goes essentially to completion • Note that for the second equilibrium, K = (1/Kw)2 = 1 X 1028

  40. Example 16.8

  41. Example 16.8, (Cont'd)

  42. Example 16.8, (Cont'd)

  43. Insoluble Compounds that Dissolve in Strong Acid • Virtually all carbonates • The product if the reaction is H2CO3, a weak acid that decomposes to carbon dioxide • H2CO3 (aq)  H2O + CO2 (g) • Many sulfides • The product of the reaction is H2S, a gas that is also a weak acid • H2S (aq) ⇌ H+ (aq) + HS- (aq)

  44. Visualizing Selective Dissolving of Precipitates

  45. Example 16.9

  46. Complex Formation • Ammonia and NaOH can dissolve compounds whose metal cations form complexes with NH3 and OH- • As with the addition of a strong acid, multiple equilibria are at work: • Zn(OH)2 (s) ⇌ Zn2+ (aq) + 2OH- (aq) Ksp • Zn2+ (aq) + 4NH3 (aq) ⇌ Zn(NH3)42+ (aq) Kf • Net: Zn(OH)2 (s) + 4NH3 (aq)  Zn(NH3)42+ (aq) + 2OH- (aq) • Knet = KspKf = 4 X 10-17 X 3.6 X 108 = 1 X 10-8

  47. Table 16.2

  48. Visualizing Dissolving by Complex Formation

  49. Example 16.10

  50. Example 16.10, (Cont'd)

More Related