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Chapter 16: Equilibria in Solutions of Weak Acids and Bases PowerPoint Presentation
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Chapter 16: Equilibria in Solutions of Weak Acids and Bases

Chapter 16: Equilibria in Solutions of Weak Acids and Bases

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Chapter 16: Equilibria in Solutions of Weak Acids and Bases

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  1. Chapter 16: Equilibria in Solutions of Weak Acids and Bases

  2. Weak Acids • All weak acids behave the same way in aqueous solution: they partially ionize

  3. Ka is called the acid ionization constant • Table 18.1 and Appendix C list the Ka and pKa for a number of acids A “large” pKa,means a “small” value of Ka and only a “small” fraction of the acid molecules ionize

  4. 15.5

  5. Weak bases in water

  6. HF + H2O ↔ H3O+ + F- F- + H2O ↔ HF + OH- Consider the following: Solve Ka*Kb=

  7. Problem Solving • Calculate the pH of a 0.0200 M solution of a weak monoprotic acid which is 3% ionized at 25°C? What is Ka for the acid?

  8. Problem Solving • Calculate the pH of a 0.0300 M solution of a weak base that is 3% ionized at 25°C? What is Kb for the base

  9. If Ka for a weak acid is 2.9E-5, what is its pKa value?a) 4.54 b) 4.82 c) 5.29 d) 6.82 e) 7.89

  10. What is the value of Kb for the following weak conjugate bases? NaF NaCN 16.1. Ionization constants can be defined for weak acids and bases 18

  11. What is the value of Kb for the following weak conjugate bases? NaF Kw=Ka×KbKa for HF =6.8×10-4 1.47×10-11 NaCN Kw=Ka×KbKa for HCN =6.2×10-10 1.61×10-5 16.1. Ionization constants can be defined for weak acids and bases 19

  12. What is the value of Ka for the following weak conjugate acids? NH4Cl C6H5NH3NO3 16.1. Ionization constants can be defined for weak acids and bases 20

  13. What is the value of Ka for the following weak conjugate acids? NH4Cl Kw=Ka×KbKb for NH3 =1.8×10-5 5.56×10-10 C6H5NH3NO3 Kw=Ka×KbKa =4.1×10-10 2.44×10-5 16.1. Ionization constants can be defined for weak acids and bases 21

  14. Solving weak acid ionization problems: • Identify the major species that can affect the pH. • In most cases, you can ignore the autoionization of water. • Ignore [OH-] because it is determined by [H+]. • Use ICE to express the equilibrium concentrations in terms of single unknown x. • Write Kain terms of equilibrium concentrations. Solve for x w/ simplifying assumption. [A]>400*K If approximation is not valid, solve for x exactly • Calculate concentrations of all species and/or pH of the solution. 15.5

  15. HF (aq) H+(aq) + F-(aq) = 7.1 x 10-4 [H+][F-] Ka = [HF] What is the pH of a 0.5M HFsolution (at 250C)? Solved next page

  16. HF (aq) H+(aq) + F-(aq) = 7.1 x 10-4 [H+][F-] Ka = [HF] HF (aq) H+(aq) + F-(aq) What is the pH of a 0.5M HFsolution (at 250C)? Initial (M) 0.50 0.00 0.00 Change (M) -x +x +x Equilibrium (M) 0.50 - x x x Ka << 1 15.5

  17. HF (aq) H+(aq) + F-(aq) = 7.1 x 10-4 = 7.1 x 10-4 = 7.1 x 10-4 [H+][F-] x2 x2 Ka Ka = Ka = 0.50 - x 0.50 [HF] HF (aq) H+(aq) + F-(aq) What is the pH of a 0.5M HFsolution (at 250C)? Initial (M) 0.50 0.00 0.00 Change (M) -x +x +x Equilibrium (M) 0.50 - x x x Ka << 1 0.50 – x 0.50 x2 = 3.55 x 10-4 x = 0.019 M pH = -log [H+] = 1.72 [H+] = [F-] = 0.019 M [HF] = 0.50 – x = 0.48 M 15.5

  18. Determine the pH of 0.1M solutions of: HC2H3O2 Ka=1.8×10-5 HCN Ka=6.2×10-10 0.1M N/A 0 0 0.1M N/A 0 0 16.2. Calculations can involve finding or using Ka and Kb 26

  19. Determine the pH of 0.1M solutions of: HC2H3O2 Ka=1.8×10-5 HCN Ka=6.2×10-10 0.1M N/A 0 0 X=1.34(10-3)M -x -x +x +x pH=2.87 (0.1-x)≈0.1 N/A x x 0.1M N/A 0 0 -x -x +x +x X=7.87(10-6)M (0.1-x) ≈ 0.1 N/A x x pH=5.10 16.2. Calculations can involve finding or using Ka and Kb 27

  20. Determine the pH of 0.1M solutions of: N2H4 Kb=1.7×10-6 NH3 Kb=1.8×10-5 0.1M N/A 0 0 0.1M N/A 0 0 16.2. Calculations can involve finding or using Ka and Kb 28

  21. Determine the pH of 0.1M solutions of: N2H4 Kb=1.7×10-6 NH3 Kb=1.8×10-5 0.1M N/A 0 0 X=4.12(10-4)M -x -x +x +x pOH=3.38 pH=10.62 (0.1-x) ≈ 0.1 N/A x x 0.1M N/A 0 0 -x -x +x +x (0.1-x) ≈ 0.1 N/A x x X=1.34(10-3)M pOH=2.87 pH=11.13 16.2. Calculations can involve finding or using Ka and Kb 29

  22. What is the pH of a 0.30 M solution of phenol (C6H5OH), an ingredient in some older mouthwashes? Ka=1.3×10-10? 9.2 0.52 9.4 none of these 16.2. Calculations can involve finding or using Ka and Kb 5.20 30

  23. Determine the % ionization of 0.2M solution of HC2H3O2 Ka=1.8×10-5 0.2M N/A 0 0 16.2. Calculations can involve finding or using Ka and Kb 31

  24. Determine the % ionization of 0.2M solution of HC2H3O2 Ka=1.8×10-5 0.2M N/A 0 0 -x -x +x +x (0.2-x) ≈ 0.2 N/A x x x=1.90×10-3M 0.95 % ionized 16.2. Calculations can involve finding or using Ka and Kb 32

  25. Determine the % ionization of 0.1M solution of HC2H3O2 Ka=1.8×10-5 0.1M N/A 0 0 -x -x +x +x (0.1-x) ≈ 0.1 N/A x x x=1.34 x 10-3M 1.3 % ionized 16.2. Calculations can involve finding or using Ka and Kb 33

  26. Determine the % ionization of 0.1M solution of HC2H3O2 Ka=1.8×10-5 0.1M N/A 0 0 16.2. Calculations can involve finding or using Ka and Kb 34

  27. What is the % ionization of 0.10M HOCl? (Ka=3.5×10-8) 0.6 % 0.06% 3.5×10-6 % none of these 16.2. Calculations can involve finding or using Ka and Kb 35

  28. What is the % ionization of 0.50M HOCl? (Ka=3.5×10-8) 0.13 % 7.0×10-6 % 0.06 % none of these 16.2. Calculations can involve finding or using Ka and Kb 0.3% 36

  29. The pH of a 0.50 M solution of an acidic drug, HD, is 3.5. What is the value of the Ka for the drug? 16.2. Calculations can involve finding or using Ka and Kb 37

  30. The pH of a 0.50 M solution of an acidic drug, HD, is 3.5. What is the value of the Ka for the drug? Given the pH, we find the value of x=10-pH x=10-3.5 =3.16×10-4 Ka=2.0×10-7 16.2. Calculations can involve finding or using Ka and Kb 38

  31. What is the pH of a 0.5M HFsolution (at 250C)? Ka=7.1E-4 at this temperature.

  32. = 7.1 x 10-4 0.006 M 0.019 M x2 x 100% = 12% x 100% = 3.8% 0.05 M 0.50 M Ka 0.05 When can I use the approximation? Ka << 1 0.50 – x 0.50 When x is less than 5% of the value from which it is subtracted. Less than 5% Approximation ok. x = 0.019 What is the pH of a 0.05M HFsolution (at 250C)? x = 0.006 M More than 5% Approximation not ok. Must solve for x exactly using quadratic equation or method of successive approximation. 15.5

  33. The method of successive approximations

  34. HA (aq) H+(aq) + A-(aq) What is the pH of a 0.122M monoprotic acid whose Ka is 5.7 x 10-4? Solved next page

  35. = 5.7 x 10-4 = 5.7 x 10-4 0.0083 M x2 x2 x 100% = 6.8% 0.122 M Ka Ka = 0.122 - x 0.122 HA (aq) H+(aq) + A-(aq) What is the pH of a 0.122M monoprotic acid whose Ka is 5.7 x 10-4? Initial (M) 0.122 0.00 0.00 Change (M) -x +x +x Equilibrium (M) 0.122 - x x x Ka << 1 0.122 – x 0.122 x2 = 6.95 x 10-5 x = 0.0083 M More than 5% Approximation not ok. 15.5

  36. -b ± b2 – 4ac = 5.7 x 10-4 x = 2a x2 Ka = 0.122 - x HA (aq) H+(aq) + A-(aq) Initial (M) 0.122 0.00 0.00 Change (M) -x +x +x Equilibrium (M) 0.122 - x x x x2 + 0.00057x – 6.95 x 10-5 = 0 ax2 + bx + c =0 x = 0.0081 x = - 0.0081 pH = -log[H+] = 2.09 [H+] = x = 0.0081 M 15.5

  37. Few substances are more effective in relieving intense pain than morphine. Morphine is an alkaloid – an alkali-like compound obtained from plants – and alkaloids are all weak bases. In 0.010 M morphine, the pH is 10.10. Calculate the K b and pK b

  38. Nicotinic acid, HC2H4NO2, is a B vitamin. It is also a weak acid with Ka = 1.4e-5. What is the [H+], pH and percent ionization of a 0.050 M solution of nicotinic acid?

  39. Pyridine, C5H5N, is a bad smelling liquid for which Kb = 1.5e-9. What is the pH and percent ionization of a 0.010 M aqueous solution of pyridine?

  40. Phenol is an organic compound that in water has Ka = 1.3e-10. What is the pH and percent ionization of a 0.15M solution of phenol in water?

  41. Codeine, a cough suppressant extracted from crude opium, is a weak base with a pKb of 5.79. What will be the pH of a 0.020 M solution of codeine?

  42. What is [H3O+] in a 1.1 E-1 M solution of HCN?Ka for HCN is 4.0E-10.a) 4.0E-10 Mb) 3.6E-9 Mc) 6.0E-5 Md) 6.6E-6 Me) 1.1E-1 M