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General Chemistry II 2302102

General Chemistry II 2302102. Chemical Equilibrium for Gases and for Sparingly-Soluble Ionic Solids. Lecture 2. i.fraser@rmit.edu.au Ian.Fraser@sci.monash.edu.au. Chemical Equilibrium - 2 Lectures. Outline - 4 Subtopics. Equilibrium and Le Chatelier’s Principle (Completed)

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General Chemistry II 2302102

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  1. General Chemistry II2302102 Chemical Equilibrium for Gases and for Sparingly-Soluble Ionic Solids Lecture 2 i.fraser@rmit.edu.au Ian.Fraser@sci.monash.edu.au

  2. Chemical Equilibrium- 2 Lectures Outline - 4 Subtopics • Equilibrium and Le Chatelier’s Principle (Completed) • The Equilibrium Constant (Completed) • Temperature and Pressure Effects (Completed) • Sparingly-Soluble Ionic Compounds in Aqueous Solution

  3. Chemical Equilibrium Objectives - Lecture 2 • By the end of this lecture AND completion of the set problems, you should be able to: • Understand the concepts of: saturated and unsaturated solutions, freely soluble and sparingly soluble ionic compounds. • Understand and know several examples of precipitation reactions. • Understand the definition of the solubility product (Ksp) and its relationship to the solubility of sparingly soluble ionic compounds. • Understand the definition of the Common Ion Effect. • Calculate equilibrium concentrations in precipitation reactions.

  4. Electrical Conductivity Open Circuit Closed circuit

  5. Electrolytes Strong Weak Non Few ionsin solution Many ionsin solution No ionsin solution

  6. Electrolytes An electrolyte is a substance that conducts an electric current when dissolved in water (or in the the molten state)

  7. Electrolytes Examples:Most salts, Acids and bases Strong electrolytes: their water solutions are good conductors (e.g. NaCl solution) Weak electrolytes: their water solutions are poor conductors (e.g. vinegar - acetic acid, CH3COOH) Non-electrolytes: their water solutions are nonconductors (e.g. sugar solution)

  8. e.g. NaCl(s) Na+(aq) + Cl-(aq) AgCl(s) Ag+(aq) + Cl-(aq) Sparingly Soluble Salts • Some salts dissolve readily in water NaCl dissociates in solution to form ions • But other salts are only slightly soluble:

  9. Why are we interested in sparingly soluble salts? • Earth’s crust is dominated by these salts: • feldspars, gypsum, calcite, aluminosilicates, dolomite, and oxides & sulfides of metals • control major geochemical processes • Boiler “scale” - often iron & manganese oxides • “Hard” water - • Waters originating from limestone areas contain Mg2+ and Ca2+ • these ions form soap “scum” precipitates

  10. http://www.jenolancaves.org.au CaCO3(s) + CO2(g) + H2O Ca2+(aq) + 2HCO3-(aq) Caves are formed Stalactites and stalagmites are formed

  11. Terminology • Recall from an earlier lecture on phase equilibria, for solutions in equilibrium with solids: • The equilibrium concentration of a dissolved solid in solution is known as the solubility • Such a solution is called “saturated” • Undersaturated: More solute will dissolve. • Supersaturated: Solution has an excess of solute.

  12. Solubility Terminology Freely soluble - several g (or more) dissolve in 100 g of water • e.g. at 298K, 36 g of NaCl, 122 g of AgNO3 Sparingly soluble - << 1 g dissolves in 100 g of water • e.g. at 298K, 2.4 x 10-4 g of AgCl, 4.4 x 10-13 g of PbS, 9.3 x 10-4 g of CaCO3 Intermediate solubility - ca. 1 g dissolves in 100 g of water (only a few of these) • e.g. 1.02 g of Ag(CH3COO) at 293K, 0.19 g of Ca(OH)2 at 273K

  13. Net Ionic Equation: What is it? A reaction equation between two electrolytes, for example between silver nitrate and sodium chloride, can be written conventionallyas: AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) precipitate But AgNO3, NaCl, and NaNO3 are ionized in aqueous solution (AgCl is a solid - insoluble)

  14. Net Ionic Equation: What is it? The conventional equation can be rewritten as: Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq) AgCl (s) + Na+ (aq) + NO3- (aq) Here, the NO3- and Na+ ions are unchanged. They are simply “spectator” ions - they do not participate in the reaction.

  15. Net Ionic Equation: What is it? It is known (observed) that the actual reaction occurs only between silver ions and chloride ions Ag+ (aq) + Cl- (aq) AgCl(s) The above equation is a net ionic equation.

  16. Spectator Ions Lead nitrate + Potassium chromate Pb2+ + 2 NO3- 2 K+ + CrO42- Lead chromate(s)+ Potassium nitrate PbCrO4(s) 2 K+ + 2 NO3-

  17. Spectator Ions So Net Ionic Equation is: Pb2+ (aq) + CrO42- (aq) PbCrO4(s)

  18. Equilibrium does NOT mean: • Reactions are not occurring Forward and reverse rates of reaction are equal - hence new products are formed at the same rate as they are broken down • The concentrations of all compounds are equal at equilibrium The concentrations of all compounds are determined by the position of the equilibrium - it may favour the products or the reactants

  19. [C]c [D]d K = [A]a [B]b Equilibrium Constant Generally for: aA + bB cC + dD

  20. AgCl(s) Ag+(aq) + Cl-(aq) [Ag+] [Cl-] K = A Saturated Solution is at Equilibrium The equilibrium constant expression is:

  21. AgCl(s) Ag+(aq) + Cl-(aq) AbBa(s) bAa+(aq) + aBb-(aq) [Aa+]b [Bb-]a Kc = [AbBa] The Solubility Product, Ksp The concentration of [AbBa] (a solid) is a constant, and is by convention set  1 in the definition of Kc [Aa+]b [Bb-]a = Ksp  Kc =

  22. The Solubility Product, Ksp Examples: Ksp(AgCl) = [Ag+] [Cl-] Ksp(Fe2S3) = [Fe3+]2 [S2-]3 Ksp(BaSO4) = [Ba2+] [SO42-]

  23. AgCl(s) Ag+(aq) + Cl-(aq) Calculation of Solubility Ksp(AgCl) = [Ag+] [Cl-] = 1.8 x 10-10 mol2 dm-6 Ksp = s x s = 1.8 x 10-10 s = 1.34 x 10-5 mol dm-3

  24. Solubility Products at 25°C

  25. PbF2(s) Pb2+(aq) + 2F-(aq) Calculation of Solubility #2 Ksp(PbF2) = [Pb2+] [F-]2= 3.7 x 10-8 M3 Ksp = s x (2s)2= 3.7 x 10-8 s = 2.1 x 10-3 M [Pb2+] = 2.1 x 10-3 M, [F-] = 4.2 x 10-3 M

  26. The Common Ion Effect The presence of an ion in solution which is common to the electrolyte will decrease the solubility: We’ve just seen that if PbF2 is placed in water, the solubilities of the ions are: [Pb2+] = 2.1 x 10-3 M, [F-] = 4.2 x 10-3 M What happens if PbF2 is placed in a solution of 0.02 M KF? (F- is the “common ion”)

  27. The Common Ion Effect What happens if PbF2 is placed in a solution of 0.02 M KF? (F- is the “common ion”) In this case, [F-] = 2s + 0.02 M Ksp = s x (2s + 0.02)2= 3.7 x 10-8 Ksp = 4s3+ 0.08s2+ 0.0004s = 3.7 x 10-8 Solve as a cubic, or if 0.02 >> s 3.7 x 10-8= s x (0.02)2 s = [Pb2+] = 9.3 x 10-5(c.f. 2.1 x 10-3 M)

  28. PbF2(s) Pb2+(aq) + 2F-(aq) Application of Le Chatelier’s Principle • Recall that: if possible, systems react to an imposed change by reducing the impact of that change • Increase [F-] (by adding KF - soluble) • Force equilibrium to left, and therefore reduce [Pb2+]

  29. bAa+(aq) + aBb-(aq) AbBa(s) Qo = [Aa+]ob [Bb-]oa Precipitation Reactions • A solid will form (precipitate) if the concentrations of the ions exceed the solubility product. First calculate the initial “reaction quotient”, Qo : Then compare Qo with Ksp

  30. Qo = [Aa+]ob [Bb-]oa Precipitation Reactions #2 First calculate the initial “reaction quotient”, Qo : Then compare Qo with Ksp : If Qo> KspPrecipitation Occurs If Qo< KspNo Precipitation

  31. Qo = [Ag+]o [Cl-]o = 2 x 10-4x1 x 10-5 = 2 x 10-9 M2 Precipitation Reactions #3 Example 1. 4 x 10-4 M AgNO3 is mixed with an equal volume of 2 x 10-5 M NaCl. Will a precipitate form? First calculate the initial “reaction quotient”, Qo : Then compare Qo with Ksp (1.8 x 10-10 M2): Is Qo> Ksp Yes! Precipitation Occurs

  32. Cd2+(aq) + 2OH-(aq) Cd(OH)2(s) Precipitation Reactions #4 Example 2.Electroplating with cadmium is a common industrial process. Before waste solutions can be discharged, [Cd2+] must be reduced to < 1 x 10-6 M. This may be achieved by adding sodium hydroxide to precipitate the cadmium as the hydroxide salt: Ksp(Cd(OH)2) = 5.3 x 10-15 M3 What concentration of NaOH is required?

  33. Precipitation Reactions #5 Example 2 cont... Ksp(Cd(OH)2) = 5.3 x 10-15 M3 Cd2+(aq) + 2OH-(aq) Cd(OH)2(s) Ksp = 1 x 10-6x s2= 5.3 x 10-15 s = 7.3 x 10-5 Hence to reduce [Cd2+] < 1 x 10-6 M, [OH-] must be > 7.3 x 10-5 M

  34. Selective Precipitation Problem: Ag+(aq) as silver nitrate is added to a mixture containing 0.001 M Cl- and 0.001 M CrO42-. Ksp (AgCl) = 2.8 x 10-10, Ksp (Ag2CrO4) = 1.9 x 10-12. What will happen as [Ag+] is increased? 1/ Precipitation of AgCl(s) will begin when [Ag+] > 2.8 x 10-7 M. (Qo > Ksp) 2/ Precipitation of Ag2CrO4(s) will begin when [Ag+] > 4.4 x 10-5 M. (Qo > Ksp)

  35. Selective Precipitation 1/ Precipitation of AgCl(s) will begin when [Ag+] > 2.8 x 10-7 M. (Qo > Ksp) 2/ Precipitation of Ag2CrO4(s) will begin when [Ag+] > 4.4 x 10-5 M. (Qo > Ksp) But in order for the value of [Ag+] to exceed 4.4 x 10-5 it is necessary that [Cl-] < 6.4 x 10-6 Thus on addition of silver nitrate, AgCl will precipitate until the concentration of the chloride ion remaining is less than 6.4 x 10-6 M, at which stage precipitation of silver chromate will occur.

  36. Chemical Equilibrium - End of Lecture 2 Objectives Covered in Lecture 2 • After studying this lecture should be able to: • Understand the concepts of: saturated and unsaturated solutions, freely soluble and sparingly soluble ionic compounds. • Understand and know several examples of precipitation reactions. • Understand the definition of the solubility product (Ksp) and its relationship to the solubility of sparingly soluble ionic compounds. • Understand the definition of the Common Ion Effect. • Calculate equilibrium concentrations in precipitation reactions.

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