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Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemistry FIFTH EDITION by Steven S. Zumdahl University of Illinois Chapter 15 Applications.

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Presentation on theme: "Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemistry FIFTH EDITION by Steven S. Zumdahl University of Illinois Chapter 15 Applications."— Presentation transcript:

1 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemistry FIFTH EDITION by Steven S. Zumdahl University of Illinois Chapter 15 Applications of Aqueous Equilibria Schedule Chapter 15 on Website

2 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 2 Section 15.6 Solubility Equilibria & the Solubility Product The typical ionic solid dissolves in water and Dissociates completely into separated hydrated Anions and cations. CaF 2 (s)  Ca 2+ (aq) + 2 F - (aq) BUT once the ions are in solution, they can collide and reform CaF 2 (s). THEREFORE, Equilibrium is reached!!!

3 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 3 CaF 2 (s)  Ca 2+ (aq) + 2 F - (aq) Equilibrium Expression: K = [Ca 2+ ] [F - ] 2 K = K sp (Solubility Product) [ ] = moles/L CaF 2 (s) not included in K.

4 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 4 Solubility Product For solids dissolving to form aqueous solutions. Bi 2 S 3 (s)  2Bi 3+ (aq) + 3S 2  (aq) K sp = solubility product constant and – K sp = [Bi 3+ ] 2 [S 2  ] 3

5 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 5 K sp Values at 25  C for Common Ionic Solids See Table 15.4 on page 759.

6 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 6 Solubility Product “Solubility” = s = concentration of Bi 2 S 3 that dissolves, which equals 1/2[Bi 3+ ] and 1/3[S 2  ]. or [Bi 3+ ] = 2s & [S 2- ] = 3s Note:K sp is constant (at a given temperature) s is variable (especially with a common – ion present)

7 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7 Important: 1.Experimentally determined solubility of an ionic solid can be used to calculate its K sp value. 2. Solubility of an ionic solid can be calculated if its K sp value is known. 3.Relative solubilities can be predicted by comparing K sp values only for salts that produce the same total # of ions. 4.Solubility of a solid is lowered if the solution already contains ions common to the solid.

8 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8 The pH of a solution can greatly affect a salt’s solubility. General Rule: If the anion, X -, is an effective base (i.e., HX is a weak acid) Then, the salt, MX, will show increased solubility in an acidic solution. These salts are much more soluble in an acidic solution than in pure water.

9 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 9 Added H + will react with the base X - forming the conjugate acid. As the base is removed, more of the salt will dissolve to replenish the basic anion, X -.

10 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 10 Let’s Do 81a & b, 85a, 87, 89 a & b, 91, 93, 95

11 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 11 Section 15.7 Precipitation & Qualitative Analysis So far solids dissolving in solutions Now--- Look at formation of solid from solution That is, the REVERSE Process.

12 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 12 Section 15.7 Precipitation & Qualitative Analysis Calculate Q, ion product Like K sp, but use initial conc. rather than equilibrium conc.

13 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 13 CaF 2 (s)  Ca 2+ (aq) + 2 F - (aq) Q = [Ca 2+ ] o [F - ] o 2 Q = Ion Product [ ] = moles/L CaF 2 (s) not included in Q.

14 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 14 If Q > K sp, then precipitation occurs. If Q < K sp, then no precipitation occurs. If precipitation occurs, you can also do calculations to determine equilibrium conc. in solution after precipitation. Let’s do: # 97 & 99

15 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 15 Selective Precipitation Method to separate a mixture of ions in which reagents are added to precipitate single ions or small groups of ions Example: Solution of Ba 2+ and Ag + ions Add NaCl only AgCl precipitates, Ba 2+ still in solution

16 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 16 Qualititative Analysis Scheme See Fig 15.11 & 15.12 on page 771 Read pages 770-772 Sulfide ion is often used to separate metal ions. See page 770. Let’s do: # 101

17 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 17 Section 15.8 Equilibria Involving Complex Ions Complex Ion: A charged species consisting of a metal ion surrounded by ligands. Ligand - a Lewis base - a molecule or ion having a lone electron pair that can be donated to an empty orbital on the metal ion to form a covalent bond. Common Ligands: H 2 O, NH 3, Cl -, CN -

18 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 18 Coordination Number: Number of ligands attached to a metal ion. (Most common are 6, 4 and 2.) Examples: Co(H 2 O) 6 2+, CoCl 4 2-, Ag(NH 3 ) 2 +

19 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 19 Formation (Stability) Constants: The equilibrium constants characterizing the stepwise addition of ligands to metal ions. Metal ions can add ligands one at a time in steps. Ag + + NH 3  Ag(NH 3 ) + K 1 = 2.1 x 10 3 Ag(NH 3 ) + + NH 3  Ag(NH 3 ) 2 + K 2 = 8.2 x 10 3 ________________________________________________________________ Ag + + 2NH 3  Ag(NH 3 ) 2 + K = K 1 x K 2

20 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 20 Let’s do Exercises # 103, 107, 109, 111, 115


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