1. Acids and Bases Section 18.1: Calculations involving Acids and Bases Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 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 - ] 15.2 2 H 2 O (l ) H 3 O + (aq) + OH - (aq) K c = [H 3 O + ][OH - ] [H 2 O] 2 [H 2 O] = constant K c [H 2 O] 2 = K w = [H + ][OH - ]

3. The Ion Product of Water 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 when [H + ] = [OH - ] when [H + ] > [OH - ] when [H + ] < [OH - ] Solution Is neutral acidic basic 15.2

4. The Ion Product of Water 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 15.2 Note: The autoionization of water is an endothermic process. What will happen to the K w if the temperature increases? equil shifts to the right – more H + and OH - produced… K w increases… pH decreases water is still neutral… but pH is slightly less than 7 at higher temps ex: @ 50°C [H + ] = [OH - ] = 3.05 x 10 - 7 ;pH is 6.5 heat + H 2 O (l) H + (aq) + OH - (aq)

6. What is the concentration of OH - ions in a solution whose hydrogen ion concentration is 1.00 x 10 -4 M? 15.2

7. pH – A Measure of Acidity pH = - log [H + ] [H + ] = 10 -pH pH[H + ] 15.3 pOH = - log [OH - ] [OH - ] = 10 -pOH pH [OH - ] [H + ][OH - ]

8. 15.3 [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

9. What is the pH of a 2 x 10 -3 M HNO 3 solution? What is the pH of a 1.8 x 10 -2 M Ba(OH) 2 solution? 15.4

10. What is the pH of a 2 x 10 -3 M HNO 3 solution? HNO 3 is a strong acid – 100% dissociation. HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) pH = -log [H + ] = -log [H 3 O + ] = -log(0.002) = 2.7 Start End 0.002 M 0.0 M What is the pH of a 1.8 x 10 -2 M Ba(OH) 2 solution? Ba(OH) 2 is a strong base – 100% dissociation. Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) Start End 0.018 M 0.036 M0.0 M pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6 15.4

11. HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) 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 if K a strength of weak acid 15.5 if K a is small, acid is weak

14. What is the pH of a 0.5 M HF solution (at 25 0 C)? K a = Eqn: Initial (M) Change (M) Equilibrium (M) 0.500.00 -x-x+x+x +x+x 15.5

15. 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

16. 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. IB WILL NOT REQUIRE THIS!!! 15.5

17. NOTES: Solving weak acid ionization problems: 1.Identify the major species that can affect the pH. assume you can ignore the autoionization of water – this works in most cases (pH < 6), 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. 15.5

18. Solving weak acid ionization problems cont: 4.Solve for x by the approximation method. approximation assumes that the acid is quite weak… so not much dissociates… (<5%) [If approximation is not valid, solve for x exactly… NOT REQD FOR IB.] 5.Use the value for x to calculate concentrations of all species and/or pH of the solution. 15.5

19. What is the pH of a 0.122 M monoprotic acid whose K a is 2.5 x 10 -4 ? HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) K a = Ka  Ka  If K a << 1; do approximation 15.5

20. Your Turn: Practice Problem # 1 A 0.0100 M solution of a weak acid has a pH of 5.00. What is the dissociation constant of the acid? Answer: 1.00 x 10 -8 M

21. Ka is often expressed as pKa pKa = - log Ka Ka = 10 -pKa 15.3 acid strength pKa

23. Your Turn: Practice Problem # 2 Benzoic acid has a pKa of 4.2. What is the pH of a 0.100 M solution of this acid? Answer: 2.6

24. Your Turn: Practice Problem # 3a a. What concentration of hydrofluoric acid is required to give a solution of pH 2.00? Answer: 0.148 M

25. 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

26. Your Turn: Practice Problem # 3b b. What percentage of the hydrofluoric acid is dissociated at this pH, if the dissociation constant of the acid is 6.76 x 10 -4 M? Answer: 6.76 %

27. B (aq) + H 2 O (l) BH + (aq) + OH - (aq) Weak Bases and Base Ionization Constants K b = [BH + ][OH - ] [B] 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 + ].

28. NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) 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 + ].

30. Kb is often expressed as pKb pKb = - log Kb Kb = 10 -pKb 15.3 base strength pKb

31. Conjugate acidpKapKbConjugate base Stronger acid (but still weak) Weaker base H 3 PO 4 2.111.9H 2 PO 4 – HF3.310.7F–F– CH 3 COOH4.89.2CH 3 COO – H 2 CO 3 6.47.6HCO 3 – NH 4 + 9.34.7NH 3 Weaker acidStronger base (but still weak)

32. Your Turn: Practice Problem # 4 What is the pH of a 0.0500 M solution of ethylamine (pKb = 3.40)? Answer: 11.60

33. 15.7 Ionization Constants of Conjugate Acid-Base Pairs HA (aq) H + (aq) + A - (aq) A - (aq) + H 2 O (l) OH - (aq) + HA (aq) KaKa KbKb H 2 O (l) H + (aq) + OH - (aq) KwKw K a K b = K w Weak Acid and Its Conjugate Base Ka =Ka = KwKw KbKb Kb =Kb = KwKw KaKa pK a + pK b = 14.00

34. Homework Read Section 18.1 pp. 217-220 Do Ex 18.1 p 220-221 #1-5, 6a,6c,6e, 7, 8, 9, 10