1. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 1 of 45 Chapter 17: Additional Aspects of Acid-Base Equilibria CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci Harwood Herring Madura

2. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 2 of 45 Contents 17-1The Common-Ion Effect in Acid-Base Equilibria 17-2Buffer Solutions 17-3Acid-Base Indicators 17-4Neutralization Reactions and Titration Curves 17-5Solutions of Salts of Polyprotic Acids 17-6Acid-Base Equilibrium Calculations: A Summary

3. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 3 of 45 17-1 The Common-Ion Effect in Acid- Base Equilibria  The Common-Ion Effect describes the effect on an equilibrium by a second substance that furnishes ions that can participate in that equilibrium.  The added ions are said to be common to the equilibrium.

4. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 4 of 45 Solutions of Weak Acids and Strong Acids  Consider a solution that contains both 0.100 M CH 3 CO 2 H and 0.100 M HCl. CH 3 CO 2 H + H 2 O CH 3 CO 2 - + H 3 O + HCl + H 2 O Cl - + H 3 O + (0.100-x) M x M x M 0.100 M [H 3 O + ] = (0.100 + x) M essentially all due to HCl

5. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 5 of 45 Acetic Acid and Hydrochloric Acid 0.1 M CH 3 CO 2 H 0.1 M CH 3 CO 2 H + 0.1 M CH 3 CO 2 Na 0.1 M HCl + 0.1 M CH 3 CO 2 H

6. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 6 of 45 Demonstrating the Common-Ion Effect: Solution of a weak Acid and a Strong Acid. (a) Determine [H 3 O + ] and [CH 3 CO 2 - ] in 0.100 M CH 3 CO 2 H. (b) Then determine these same quantities in a solution that is 0.100 M in both CH 3 CO 2 H and HCl. CH 3 CO 2 H + H 2 O → H 3 O + + CH 3 CO 2 - Recall Example 17-6 (p 680): [H 3 O + ] = [CH 3 CO 2 - ] = 1.3  10 -3 M EXAMPLE 17-1

7. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 7 of 45 CH 3 CO 2 H + H 2 O → H 3 O + + CH 3 CO 2 - Initial concs. weak acid0.100 M0 M0 M strong acid0 M0.100 M0 M Changes-x M+x M+x M Equilibrium(0.100 - x) M (0.100 + x) M x M Concentration Assume x << 0.100 M, 0.100 – x  0.100 + x  0.100 M EXAMPLE 17-1

8. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 8 of 45 Eqlbrm conc.(0.100 - x) M (0.100 + x) M x M Assume x << 0.100 M, 0.100 – x  0.100 + x  0.100 M CH 3 CO 2 H + H 2 O → H 3 O + + CH 3 CO 2 - [H 3 O + ] [CH 3 CO 2 - ] [C 3 CO 2 H] Ka=Ka= x · (0.100 + x) (0.100 - x) = x · (0.100) (0.100) = = 1.8  10 -5 [CH 3 CO 2 - ] = 1.8  10 -5 M compared to 1.3  10 -3 M. Le Châtelier’s Principle EXAMPLE 17-1

9. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 9 of 45 Suppression of Ionization of a Weak Acid

10. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 10 of 45 Suppression of Ionization of a Weak Base

11. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 11 of 45 Solutions of Weak Acids and Their Salts

12. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 12 of 45 Solutions of Weak Bases and Their Salts

13. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 13 of 45 17-2 Buffer Solutions  Two component systems that change pH only slightly on addition of acid or base.  The two components must not neutralize each other but must neutralize strong acids and bases.  A weak acid and it’s conjugate base.  A weak base and it’s conjugate acid

14. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 14 of 45 Pure Water Has No Buffering Ability

15. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 15 of 45 Buffer Solutions  Consider [CH 3 CO 2 H] = [CH 3 CO 2 - ] in a solution. [H 3 O + ] [CH 3 CO 2 - ] [C 3 CO 2 H] Ka=Ka= = 1.8  10 -5 = [CH 3 CO 2 - ] [C 3 CO 2 H] KaKa [H 3 O + ] = pH = -log[H 3 O + ] = -logK a = -log(1.8  10 -5 ) = 4.74

16. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 16 of 45 How A Buffer Works

17. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 17 of 45 Preparing a Buffer Solution

18. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 18 of 45 The Henderson-Hasselbalch Equation  A variation of the ionization constant expression.  Consider a hypothetical weak acid, HA, and its salt NaA: HA + H 2 O A - + H 3 O + [H 3 O + ] [A - ] [HA] Ka=Ka=[H 3 O + ] [HA] Ka=Ka= [A - ] -log[H 3 O + ]-log [HA] -logK a = [A - ]

19. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 19 of 45 Henderson-Hasselbalch Equation -log[H 3 O + ] - log [HA] -logK a = [A - ] pH - log [HA] pK a = [A - ] pK a + log [HA] pH = [A - ] pK a + log [acid] pH = [conjugate base]

20. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 20 of 45 Henderson-Hasselbalch Equation  Only useful when you can use initial concentrations of acid and salt.  This limits the validity of the equation.  Limits can be met by: 0.1 < [HA] < 10 [A - ] [A - ] > 10  K a and [HA] > 10  K a pK a + log [acid] pH= [conjugate base]

21. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 21 of 45 Preparing a Buffer Solution of a Desired pH. What mass of NaC 2 H 3 O 2 must be dissolved in 0.300 L of 0.25 M HC 2 H 3 O 2 to produce a solution with pH = 5.09? (Assume that the solution volume is constant at 0.300 L) HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 - + H 3 O + Equilibrium expression: [H 3 O + ] [HC 2 H 3 O 2 ] Ka=Ka= [C 2 H 3 O 2 - ] = 1.8  10 -5 EXAMPLE 17-5

22. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 22 of 45 [H 3 O + ] [HC 2 H 3 O 2 ] Ka=Ka= [C 2 H 3 O 2 - ] = 1.8  10 -5 [H 3 O + ] = 10 -5.09 = 8.1  10 -6 [HC 2 H 3 O 2 ] = 0.25 M Solve for [C 2 H 3 O 2 - ] [H 3 O + ] [HC 2 H 3 O 2 ] = K a [C 2 H 3 O 2 - ] = 0.56 M 8.1  10 -6 0.25 = 1.8  10 -5 EXAMPLE 17-5

23. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 23 of 45 1 mol NaC 2 H 3 O 2 82.0 g NaC 2 H 3 O 2 mass C 2 H 3 O 2 - = 0.300 L [C 2 H 3 O 2 - ] = 0.56 M 1 L 0.56 mol 1 mol C 2 H 3 O 2 - 1 mol NaC 2 H 3 O 2   = 14 g NaC 2 H 3 O 2 EXAMPLE 17-5

24. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 24 of 45 Six Methods of Preparing Buffer Solutions

25. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 25 of 45 Calculating Changes in Buffer Solutions

26. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 26 of 45 Buffer Capacity and Range  Buffer capacity is the amount of acid or base that a buffer can neutralize before its pH changes appreciably.  Maximum buffer capacity exists when [HA] and [A - ] are large and approximately equal to each other.  Buffer range is the pH range over which a buffer effectively neutralizes added acids and bases.  Practically, range is 2 pH units around pK a.

27. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 27 of 45 17-3 Acid-Base Indicators  Color of some substances depends on the pH. HIn + H 2 O In - + H 3 O + In the acid form the color appears to be the acid color. In the base form the color appears to be the base color. Intermediate color is seen in between these two states. The complete color change occurs over about 2 pH units.

28. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 28 of 45 Indicator Colors and Ranges Slide 27 of 45General Chemistry: Chapter 17Prentice-Hall © 2007

29. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 29 of 45 Testing the pH of a Swimming Pool

30. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 30 of 45 17-4 Neutralization Reactions and Titration Curves  Equivalence point:  The point in the reaction at which both acid and base have been consumed.  Neither acid nor base is present in excess.  End point:  The point at which the indicator changes color.  Titrant:  The known solution added to the solution of unknown concentration.  Titration Curve:  The plot of pH vs. volume.

31. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 31 of 45 The millimole  Typically:  Volume of titrant added is less than 50 mL.  Concentration of titrant is less than 1 mol/L.  Titration uses less than 1/1000 mole of acid and base. L/1000 mol/1000 = M = L mol mL mmol =

32. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 32 of 45 Titration of a Strong Acid with a Strong Base

33. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 33 of 45 Titration of a Strong Acid with a Strong Base  The pH has a low value at the beginning.  The pH changes slowly:  until just before the equivalence point.  The pH rises sharply:  perhaps 6 units per 0.1 mL addition of titrant.  The pH rises slowly again.  Any Acid-Base Indicator will do.  As long as color change occurs between pH 4 and 10.

34. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 34 of 45 Titration of a Strong Base with a Strong Acid

35. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 35 of 45 Titration of a Weak Acid with a Strong Base

36. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 36 of 45 Titration of a Weak Acid with a Strong Base

37. Prentice-Hall © 2007 General Chemistry: Chapter 17 Slide 37 of 45 17-6 Acid-Base Equilibrium Calculations: A Summary  Determine which species are potentially present in solution, and how large their concentrations are likely to be.  Identify possible reactions between components and determine their stoichiometry.  Identify which equilibrium equations apply to the particular situation and which are most significant.