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## Chapter 16: Equilibria in Solutions of Weak Acids and Bases

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**Weak Acids**• All weak acids behave the same way in aqueous solution: they partially ionize**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**HF + H2O ↔ H3O+ + F-**F- + H2O ↔ HF + OH- Consider the following: Solve Ka*Kb=**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?**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**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**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**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**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**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**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**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**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**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**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**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**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**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**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**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**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**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**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**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**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**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**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**What is the pH of a 0.5M HFsolution (at 250C)? Ka=7.1E-4 at**this temperature.**= 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**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**= 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****-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**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**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?**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?**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?**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?**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