1. 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

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

3. 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 (K sp ) 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. Objectives - Lecture 2 Chemical Equilibrium

4. Electrical Conductivity Open Circuit Closed circuit

5. Electrolytes Non StrongWeak No ions in solution Many ions in solution Few ions in 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, CH 3 COOH) Non-electrolytes: their water solutions are nonconductors (e.g. sugar solution)

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

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 Mg 2+ and Ca 2+ –these ions form soap “scum” precipitates

10. http://www.jenolancaves.org.au CaCO 3 (s) + CO 2 (g) + H 2 O Ca 2+ (aq) + 2HCO 3 - (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 AgNO 3 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 CaCO 3 Intermediate solubility - ca. 1 g dissolves in 100 g of water (only a few of these) –e.g. 1.02 g of Ag(CH 3 COO) 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 conventionally as: AgNO 3 (aq) + NaCl (aq) precipitate AgCl (s) + NaNO 3 (aq) But AgNO 3, NaCl, and NaNO 3 are ionized in aqueous solution (AgCl is a solid - insoluble)

14. The conventional equation can be rewritten as: Net Ionic Equation: What is it? Ag + (aq) + NO 3 - (aq) + Na + (aq) + Cl - (aq) AgCl (s) + Na + (aq) + NO 3 - (aq) Here, the NO 3 - and Na + ions are unchanged. They are simply “spectator” ions - they do not participate in the reaction.

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

16. Spectator Ions Lead nitrate + Potassium chromate Lead chromate (s) + Potassium nitrate Pb 2+ + 2 NO 3 - 2 K + + CrO 4 2- PbCrO 4 (s) 2 K + + 2 NO 3 -

17. Spectator Ions So Net Ionic Equation is: Pb 2+ (aq) + CrO 4 2- (aq) PbCrO 4 (s)

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

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

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

21. The Solubility Product, K sp AgCl(s) Ag + (aq) + Cl - (aq) A b B a (s) bA a+ (aq) + aB b- (aq) [A a+ ] b [B b- ] a [A b B a ] K c = The concentration of [A b B a ] (a solid) is a constant, and is by convention set  1 in the definition of K c  K c = [A a+ ] b [B b- ] a = K sp

22. The Solubility Product, K sp K sp (AgCl) = [Ag + ] [Cl - ] K sp (Fe 2 S 3 ) = [Fe 3+ ] 2 [S 2- ] 3 K sp (BaSO 4 ) = [Ba 2+ ] [SO 4 2- ] Examples:

23. Calculation of Solubility K sp (AgCl) = [Ag + ] [Cl - ] = 1.8 x 10 -10 mol 2 dm -6 AgCl(s) Ag + (aq) + Cl - (aq) K sp = s x s = 1.8 x 10 -10 s = 1.34 x 10 -5 mol dm -3

24. Solubility Products at 25°C

25. Calculation of Solubility #2 K sp (PbF 2 ) = [Pb 2+ ] [F - ] 2 = 3.7 x 10 -8 M 3 PbF 2 (s) Pb 2+ (aq) + 2F - (aq) K sp = s x (2s) 2 = 3.7 x 10 -8 s= 2.1 x 10 -3 10 -3 M [Pb 2+ ] [Pb 2+ ] = 2.1 x 10 -3 10 -3 M, [F - ] [F - ] = 4.2 x 10 -3 10 -3 M

26. The Common Ion Effect [Pb 2+ ] = 2.1 x 10 -3 M, [F - ] = 4.2 x 10 -3 M [Pb 2+ ] = 2.1 x 10 -3 M, [F - ] = 4.2 x 10 -3 M The presence of an ion in solution which is common to the electrolyte will decrease the solubility: We’ve just seen that if PbF 2 is placed in water, the solubilities of the ions are: What happens if PbF 2 is placed in a solution of 0.02 M KF? (F - is the “common ion”)

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

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

29. Precipitation Reactions A solid will form (precipitate) if the concentrations of the ions exceed the solubility product. bA a+ (aq) + aB b- (aq) A b B a (s) First calculate the initial “reaction quotient”, Q o : Q o = [A a+ ] o b [B b- ] o a Then compare Q o with K sp

30. Precipitation Reactions #2 First calculate the initial “reaction quotient”, Q o : Q o = [A a+ ] o b [B b- ] o a Then compare Q o with K sp : If Q o > K sp Precipitation Occurs If Q o < K sp No Precipitation

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

32. Precipitation Reactions #4 Example 2. Electroplating with cadmium is a common industrial process. Before waste solutions can be discharged, [Cd 2+ ] 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: K sp (Cd(OH) 2 ) = 5.3 x 10 -15 M 3 What concentration of NaOH is required? Cd 2+ (aq) + 2OH - (aq) Cd(OH) 2 (s)

33. Precipitation Reactions #5 Example 2 cont... K sp (Cd(OH) 2 ) = 5.3 x 10 -15 M 3 Cd 2+ (aq) + 2OH - (aq) Cd(OH) 2 (s) K sp = 1 x 10 -6 x s 2 = 5.3 x 10 -15 s = 7.3 x 10 -5 Hence to reduce [Cd 2+ ] 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 CrO 4 2-. K sp (AgCl) = 2.8 x 10 -10, K sp (Ag 2 CrO 4 ) = 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. (Q o > K sp ) 2/ Precipitation of Ag 2 CrO 4 (s) will begin when [Ag + ] > 4.4 x 10 -5 M. (Q o > K sp )

35. Selective Precipitation 1/ Precipitation of AgCl(s) will begin when [Ag + ] > 2.8 x 10 -7 M. (Q o > K sp ) 2/ Precipitation of Ag 2 CrO 4 (s) will begin when [Ag + ] > 4.4 x 10 -5 M. (Q o > K sp ) 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. - End of Lecture 2 Chemical Equilibrium - End of 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 (K sp ) 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. Objectives Covered in Lecture 2