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Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci Harwood Herring.

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Presentation on theme: "Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci Harwood Herring."— Presentation transcript:

1 Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition Chapter 19: Solubility and Complex-Ion Equilibria

2 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 2 of 34 Contents 19-1The Solubility Product Constant, K sp 19-2The Relationship Between Solubility and K sp 19-3The Common-Ion Effect in Solubility Equilibria 19-4Limitations of the K sp Concept 19-5Criteria for Precipitation and Its Completeness 19-6Fractional Precipitation 19-7Solubility and pH 19-8Equilibria Involving Complex Ions 19-9Qualitative Cation Analysis Focus On Shells, Teeth, and Fossils

3 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 3 of 34 19-1 The Solubility Product Constant, K sp CaSO 4 (s) Ca 2+ (aq) + SO 4 2- (aq) K sp = [Ca 2+ ][SO 4 2- ] = 9.1 · 10 -6 at 25°C The equilibrium constant for the equilibrium established between a solid solute and its ions in a saturated solution.

4 K sp vs. activities When the solubility of the salt is very low the activity coefficients, of the ions in solution are nearly equal to one. The activity of a pure solid is, by definition, equal to one. By setting activity coefficients of the ions in solution precisely to one the solubility product expression can be obtained: a(Ca 2+ ) = [Ca 2+ ] and thus a(Ca 2+ ) = [Ca 2+ ] General Chemistry: Chapter 19Slide 4 of 34

5 General expression for K sp For A a B b aA b+ + bB a- dissociation the K sp = [A] a [B] b, where a and b are the stoichiometric constants and the electrical charges omitted for simplicity of notation. General Chemistry: Chapter 19Slide 5 of 34

6 K sp values in Table 16 at 25 °C General Chemistry: Chapter 19Slide 6 of 34 Compounds Ksp AgBr Ag + + Br - 7.7 10 -13 CuBr Cu + + Br - 4.9 10 -9 Hg 2 Br 2 2 Hg + + 2Br - 4.6 10 -23 PbBr 2 Pb 2+ + 2 Br - 7.4 10 -5 Ag 2 CO 3 2 Ag + + CO 3 2- 6.5 10 -12 BaCO 3 Ba 2+ + CO 3 2- 8.0 10 -9 CaCO 3 Ca 2+ + CO 3 2- 4.8 10 -9 You will find many more values in the Table 16

7 Several examples General Chemistry: Chapter 19Slide 7 of 34 Lead(II) iodide Silver carbonate partially oxidized Silver bromide

8 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 8 of 34 The Relationship Between Solubility and K sp Molar solubility (S). –The molarity in a saturated aqueous solution. –Related to K sp g BaSO 4 /100 mL mol BaSO 4 /L [Ba 2+ ] and [SO 4 2- ] = K sp 1/2 = S K sp = 1.1 · 10 -10

9 The general formula for solubility For A a B b aA b+ + bB a- dissociation: K sp = [A] a [B] b Notice that [A] = a S, [B] = b S and General Chemistry: Chapter 19Slide 9 of 34

10 Examples at 25°C K sp = [Al 3+ ] · [OH - ] 3 = 3.0 · 10 -34 (a=1, b=3) – [Al 3+ ] = S, [OH - ] = 3S, K sp = S · (3S) 3 = 27 · S 4 – [Al 3+ ] = [Al(OH) 3 ] = S = 1.826 · 10 -9 M K sp = [Ca 2+ ] ·[F - ] 2 = 3.9 10 -11 (a=1, b=2) K sp = [Mg 2+ ] 3 · [PO 4 3- ] 2 = 1·10 -25 –K sp = (3S) 3 · (2S) 2 = 27 · 4 · S 5 –S =? (a=3, b=2) General Chemistry: Chapter 19Slide 10 of 34

11 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 11 of 34 19-3 The Common-Ion Effect in Solubility Equilibria

12 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 12 of 34 The Common-Ion Effect and Le Chatelliers Principle

13 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 13 of 34 19-4 Limitations of the K sp Concept K sp is usually limited to slightly soluble solutes. –For more soluble solutes we must use ion activities Activities (effective concentrations) become smaller than the measured concentrations. The Salt Effect (or diverse ion effect). –Ionic interactions are important even when an ion is not apparently participating in the equilibrium. Uncommon ions tend to increase solublity.

14 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 14 of 34 Effects on the Solubility of Ag 2 CrO 4

15 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 15 of 34 Ion Pairs

16 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 16 of 34 Incomplete Dissociation Assumption that all ions in solution are completely dissociated is not valid. Ion Pair formation occurs. –Some solute molecules are present in solution. –Increasingly likely as charges on ions increase. K sp (CaSO 4 ) = 2.3·10 -4 by considering solubility in g/100 mL Table 16: K sp = 4.9·10 -5 mol 2 L -2 Activities take into account ion pair formation and must be used.

17 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 17 of 34 Simultaneous Equilibria Other equilibria are usually present in a solution. –K w for example. –These must be taken into account if they affect the equilibrium in question.

18 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 18 of 34 19-5 Criteria for Precipitation and Its Completeness AgCl(s) Ag + (aq) + Cl - (aq) Mix AgNO 3 (aq) and KCl(aq) to obtain a solution that is 0.010 M in Ag + and 0.015 M in Cl -. Saturated, supersaturated or unsaturated? Q = [Ag + ][Cl - ] = (0.010)(0.015) = 1.5 · 10 -4 > K sp K sp = [Ag + ][Cl - ] = 1.77 · 10 -10

19 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 19 of 34 The Ion Product Q is generally called the ion product. Q > K sp Precipitation should occur. Q = K sp The solution is just saturated. Q < K sp Precipitation cannot occur.

20 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 20 of 34 Example 19-5 Applying the Criteria for Precipitation of a Slightly Soluble Solute. Three drops of 0.20 M KI are added to 100.0 mL of 0.010 M Pb(NO 3 ) 2. Will a precipitate of lead iodide form? (1 drop = 0.05 mL) PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) K sp = 8.5 · 10 -9 Determine the amount of I - in the solution: = 3·10 -5 mol I - n I - = 3 drops 1 drop 0.05 mL 1000 mL 1 L 0.20 mol KI 1 mol KI 1 mol I -

21 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 21 of 34 Example 19-5 [I - ] = 0.1000 L 3·10 -5 mol I - = 3·10 -4 M I - Determine the concentration of I - in the solution: Apply the Precipitation Criteria: Q = [Pb 2+ ][I - ] 2 = (0.010)(3·10 -4 ) 2 = 9·10 -10 < K sp = 8.5·10 -9

22 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 22 of 34 19-6 Fractional Precipitation A technique in which two or more ions in solution are separated by the proper use of one reagent that causes precipitation of both ions. Significant differences in solubilities are necessary.

23 General Chemistry: Chapter 19Slide 23 of 34 19-7 Solubility and pH Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH - (aq) K sp = 1.5·10 -11 OH - (aq) + H + (aq) H 2 O(aq) K = 1/K w = 1.0·10 14 2 OH - (aq) + 2 H + (aq) 2 H 2 O(aq) K' = (1/K w ) 2 = 1.0·10 28 Mg(OH) 2 (s) + 2 H + (aq) Mg 2+ (aq) + 2 H 2 O (aq) K = K sp (1/K w ) 2 = (1.5·10 -11 )(1.0 · 10 14 ) 2 = 1.5·10 17

24 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 24 of 34 19-8 Equilibria Involving Complex Ions AgCl(s) + 2 NH 3 (aq) [Ag(NH 3 ) 2 ] + (aq) + Cl - (aq)

25 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 25 of 34 Complex Ions Coordination compounds. –Substances which contain complex ions. Complex ions. –A polyatomic cation or anion composed of: A central metal ion. Ligands

26 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 26 of 34 Formation Constant of Complex Ions AgCl(s) + 2 NH 3 (aq) [Ag(NH 3 ) 2 ] + (aq) + Cl - (aq) AgCl(s) Ag + (aq) + Cl - (aq) Ag + (aq) + 2 NH 3 (aq) [Ag(NH 3 ) 2 ] + (aq) K sp = 1.8·10 -11 K f == 1.6·10 7 [Ag(NH 3 ) 2 ] + [Ag + ] [NH 3 ] 2

27 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 27 of 34 Table 19.2 Formation Constants for Some Complex Ions

28 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 28 of 34 Example 19-11 Determining Whether a Precipitate will Form in a Solution Containing Complex Ions. A 0.10 mol sample of AgNO 3 is dissolved in 1.00 L of 1.00 M NH 3. If 0.010 mol NaCl is added to this solution, will AgCl(s) precipitate? Ag + (aq) + 2 NH 3 (aq) [Ag(NH 3 ) 2 ] + (aq) Assume K f is large: Initial conc.0.10 M1.00 M0 M Change-0.10 M-0.20 M+0.10 M Eqlbrm conc.( ~ 0) M 0.80 M0.10 M

29 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 29 of 34 Example 19-11 [Ag + ] is small but not 0, use K f to calculate [Ag+]: Ag + (aq) + 2 NH 3 (aq) [Ag(NH 3 ) 2 ] + (aq) Initial concs.0 M0.80 M0.10 M Changes+x M+2x M-x M Eqlbrm conc.x M 0.80 + 2x M0.10 - x M 0.10 (1.6 ·10 7 )(0.80) 2 x = [Ag + ] = = 9.8 · 10 -9 M = 1.6·10 7 [Ag(NH 3 ) 2 ] + [Ag + ] [NH 3 ] 2 0.10-x x(0.80 + 2x) 2 0.10 x(0.80) 2 = ~ K f =

30 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 30 of 34 Example 19-11 Compare Q sp to K sp and determine if precipitation will occur: = (9.8·10 -9 )(1.0·10 -2 ) = 9.8·10 -11 [Ag + ][Cl - ]Q sp = K sp = 1.8·10 -10 Q sp < K sp AgCl does not precipitate.

31 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 31 of 34 19-9 Qualitative Cation Analysis An analysis that aims at identifying the cations present in a mixture but not their quantities. Think of cations in solubility groups according to the conditions that causes precipitation chloride group hydrogen sulfide group ammonium sulfide group carbonate group. –Selectively precipitate the first group of cations then move on to the next.

32 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 32 of 34 Qualitative Cation Analysis

33 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 33 of 34 Chloride Group Precipitates (a)Group precipitate Wash ppt with hot water PbCl 2 is slightly soluble. Test aqueous solution with CrO 4 2-. (c) Pb 2+ (aq) + CrO 4 2- PbCrO 4 (s) Test remaining precipitate with ammonia. (b) AgCl(s) + 2 NH 3 (aq) Ag(NH 3 ) 2 (aq) + Cl - (aq) (b) Hg 2 Cl 2 (a) + 2 NH 3 Hg(l) + HgNH 2 Cl(s) + NH 4 + (aq) + Cl - (aq)

34 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 34 of 34 Hydrogen Sulfide Equilibria H 2 S(aq) + H 2 O(l) HS - (aq) + H 3 O + (aq)K a1 = 1.0·10 -7 HS - (aq) + H 2 O(l) S 2- (aq) + H 3 O + (aq) K a2 = 1.0·10 -19 S 2- is an extremely strong base and is unlikely to be the precipitating agent for the sulfide groups.

35 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 35 of 34 Lead Sulfide Equilibria PbS(s) + H 2 O(l) Pb 2+ (aq) + HS - (aq) + OH - (aq) K sp = 3·10 -28 H 3 O + (aq) + HS - (aq) H 2 S(aq) + H 2 O(l) 1/K a1 = 1.0/1.0·10 -7 H 3 O + (aq) + OH - (aq) H 2 O(l) + H 2 O(l) 1/K w = 1.0/1.0·10 -14 PbS(s) + 2 H 3 O + (aq) Pb 2+ (aq) + H 2 S(aq) + 2 H 2 O(l) K spa = = 3·10 -7 K sp K a1 ·K w 3·10 -28 1.0·10 -7 ·1.0·10 -14 =

36 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 36 of 34 Dissolving Metal Sulfides Several methods exist to re-dissolve precipitated metal sulfides. –React with an acid. FeS readily soluble in strong acid but PbS and HgS are not because their K sp values are too low. –React with an oxidizing acid. 3 CuS(aq) + 8 H + (aq) + 2 NO 3 - (aq) 3 Cu 2+ (aq) + 3 S(s) + 2 NO(g) + 4 H 2 O(l)

37 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 37 of 34 A Sensitive Test for Copper(II) [Cu(H 2 O) 4 ] 2+ (aq) + 4 NH 3 (aq) [Cu(NH 3 ) 4 ] 2+ (aq) + 4 H 2 O(l)

38 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 38 of 34 Focus On Shells, Teeth and Fossils Ca 2+ (aq) + 2 HCO 3 - (aq) CaCO 3 (s) + H 2 O(l) + CO 2 (g) Calcite Ca 5 (PO 4 ) 3 OH(s) + 4 H 3 O + (aq) 5 Ca 2+ (s) + 5 H 2 O(l) + 3 HPO 4 2- (aq) Fluoroapatite Ca 5 (PO 4 ) 3 F(s) Hydroxyapatite Ca 5 (PO 4 ) 3 OH(s)

39 Prentice-Hall © 2002General Chemistry: Chapter 19Slide 39 of 34 Chapter 19 Questions Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding. Choose a variety of problems from the text as examples. Practice good techniques and get coaching from people who have been here before.


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