1. Acids and Bases Chapter 15 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Chapter 15 Semester 1/2009 15.1 Bronsted Acids and Bases 15.2 The Acid-Base Properties of Water 15.3 pH-A measure of Acidity 15.5 Weak Acids and Acid Ionization Constants 15.6 Weak Bases and Base Ionization Constants

3. 15.1 Bronsted Acids and Bases A Brønsted acid is a proton donor A Brønsted base is a proton acceptor acidbaseacidbase 15.1 acid conjugate base base conjugate acid

4. O H H+ O H H O H HH O H - + [] + 15.2 The Acid-Base Properties of Water H 2 O (l) H + (aq) + OH - (aq) H 2 O + H 2 O H 3 O + + OH - acid conjugate base base conjugate acid 15.2 autoionization of water

5. H 2 O (l) H + (aq) + OH - (aq) The Ion Product of Water K c = [H + ][OH - ] [H 2 O] [H 2 O] = constant K c [H 2 O] = K w = [H + ][OH - ] The ion-product constant (K w ) is the product of the molar concentrations of H + and OH - ions at a particular temperature. At 25 0 C K w = [H + ][OH - ] = 1.0 x 10 - 14 [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic 15.2

6. What is the concentration of OH - ions in a HCl solution whose hydrogen ion concentration is 1.3 M? K w = [H + ][OH - ] = 1.0 x 10 - 14 [H + ] = 1.3 M [OH - ] = KwKw [H + ] 1 x 10 -14 1.3 = = 7.7 x 10 -15 M 15.2

7. 15.3 pH – A Measure of Acidity pH = - log [H + ] [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic [H + ] = 1 x 10 -7 [H + ] > 1 x 10 -7 [H + ] < 1 x 10 -7 pH = 7 pH < 7 pH > 7 At 25 0 C pH[H + ] 15.3

8. pOH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

9. The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H + ion concentration of the rainwater? pH = - log [H + ] [H + ] = 10 -pH = 10 -4.82 = 1.5 x 10 -5 M The OH - ion concentration of a blood sample is 2.5 x 10 -7 M. What is the pH of the blood? pH + pOH = 14.00 pOH = -log [OH - ]= -log (2.5 x 10 -7 )= 6.60 pH = 14.00 – pOH = 14.00 – 6.60 = 7.40 15.3

10. HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) 15.5 Weak Acids (HA) and Acid Ionization Constants HA (aq) H + (aq) + A - (aq) K a = [H + ][A - ] [HA] K a is the acid ionization constant KaKa weak acid strength 15.5

12. What is the pH of a 0.5 M HF solution (at 25 0 C)? HF (aq) H + (aq) + F - (aq) K a = [H + ][F - ] [HF] = 7.1 x 10 -4 HF (aq) H + (aq) + F - (aq) Initial (M) Change (M) Equilibrium (M) 0.500.00 -x-x+x+x 0.50 - x 0.00 +x+x xx K a = x2x2 0.50 - x = 7.1 x 10 -4 Ka  Ka  x2x2 0.50 = 7.1 x 10 -4 0.50 – x  0.50 K a << 1 x 2 = 3.55 x 10 -4 x = 0.019 M [H + ] = [F - ] = 0.019 M pH = -log [H + ] = 1.72 [HF] = 0.50 – x = 0.48 M 15.5

13. When can I use the approximation? 0.50 – x  0.50 K a << 1 When x is less than 5% of the value from which it is subtracted. x = 0.019 0.019 M 0.50 M x 100% = 3.8% Less than 5% Approximation ok. What is the pH of a 0.05 M HF solution (at 25 0 C)? Ka  Ka  x2x2 0.05 = 7.1 x 10 -4 x = 0.006 M 0.006 M 0.05 M x 100% = 12% More than 5% Approximation not ok. Must solve for x exactly using quadratic equation or method of successive approximation. 15.5

14. Solving weak acid ionization problems: 1.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 + ]. 2.Use ICE to express the equilibrium concentrations in terms of single unknown x. 3.Write K a in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly. 4.Calculate concentrations of all species and/or pH of the solution. 15.5

15. What is the pH of a 0.122 M monoprotic acid whose K a is 5.7 x 10 -4 ? HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) 0.1220.00 -x-x+x+x 0.122 - x 0.00 +x+x xx K a = x2x2 0.122 - x = 5.7 x 10 -4 Ka  Ka  x2x2 0.122 = 5.7 x 10 -4 0.122 – x  0.122 K a << 1 x 2 = 6.95 x 10 -5 x = 0.0083 M 0.0083 M 0.122 M x 100% = 6.8% More than 5% Approximation not ok. 15.5

16. K a = x2x2 0.122 - x = 5.7 x 10 -4 x 2 + 0.00057x – 6.95 x 10 -5 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = x = 0.0081x = - 0.0081 HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) 0.1220.00 -x-x+x+x 0.122 - x 0.00 +x+x xx [H + ] = x = 0.0081 M pH = -log[H + ] = 2.09 15.5

17. percent ionization = Ionized acid concentration at equilibrium Initial concentration of acid x 100% For a monoprotic acid HA Percent ionization = [H + ] [HA] 0 x 100% [HA] 0 = initial concentration 15.5

18. NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) 15.6 Weak Bases and Base Ionization Constants K b = [NH 4 + ][OH - ] [NH 3 ] K b is the base ionization constant KbKb weak base strength 15.6 Solve weak base problems like weak acids except solve for [OH-] instead of [H + ].Solve for pOH instead of pH.