1. Acids and Bases Chapter 7 E-mail: benzene4president@gmail.combenzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/

2. Acids and Bases – ch. 7 1. Arrange the following solutions in order of most acidic to most basic. I. [OH – ] = 0.5 M II. [H + ] = 0.3 M III. pOH = 5.9 IV. pH = 1.2 V. [H + ] = 1.0x10 –4 M

3. Acids and Bases – ch. 7 [H 3 O + ] [OH - ] pH pOH K w = [H 3 O + ][OH - ] pK w = pH + pOH pH = -log[H 3 O + ] [H 3 O + ] = 10 -pH [OH - ] = 10 -pOH pOH = -log[OH - ]

4. Acids and Bases – ch. 7 2. At 25°C the water ionization constant is 1.0 x 10 –14 and at 98°C the water ionization constant is 6.8 x 10 –14. What is the pH of neutral water at both temperatures?

5. Acids and Bases – ch. 7 3. a. Which of the following is a stronger acid? HNO 2 (K a = 4.0 x 10 –4 ) or HCN (pK a = 9.21) b. Which is a stronger base? NO 2 – or CN –

6. Acids and Bases – ch. 7 As acid strength ↑ % ionization ↑ K a ↑ pK a ↓ As base strength ↑ % ionization ↑ K b ↑ pK b ↓ Conjugate pairs are inversely related As acid strength ↑ conjugate base strength ↓ K w = (K a )(K b ) As acid strength ↑ % ionization ↑ K a ↑ pK a ↓ As base strength ↑ % ionization ↑ K b ↑ pK b ↓ Conjugate pairs are inversely related As acid strength ↑ conjugate base strength ↓ K w = (K a )(K b )

7. Acids and Bases – ch. 7 4. Calculate the pH of the following solutions: a. 0.004 M HBr b. 0.25 moles of Ba(OH) 2 in 7.50 L of solution

8. Acids and Bases – ch. 7 5. Calculate the pH and the % ionization of the following solutions a. 0.05 M HClO (K a = 3.0 x 10 –8 ) b. 0.33 M CH 3 COOH (K a = 1.8 x 10 –5 ) mixed with 0.85 M HCN (K a = 6.2 x 10 –10 )

9. Acids and Bases – ch. 7 6. What concentration of HCOOH (K a = 1.77x10 -4 ) solution will have a pH of 2.2?

10. Acids and Bases – ch. 7 7. A solution with 0.5 M of weak mono-protic acid ionizes 11%. What is the K a for this acid?

11. Acids and Bases – ch. 7 8. What concentration of a weak acid that ionizes 0.1% will have a pH of 3.8?

12. Acids and Bases – ch. 7 9. What concentration of HNO 2 (K a = 4.0 x 10 –4 ) will have the same pH as 0.03M HNO 3 ?

13. Acids and Bases – ch. 7 10. If 0.1M solution of a weak acid has a pH of 3 what will be the pH of a 0.001M solution of the weak acid?

14. Acids and Bases – ch. 7 11. Calculate the pH and % ionization for the following: a. 0.01M NH 3 (K b = 1.8 x 10 -5 ) b. 0.5 M C 6 H 5 NH 2 (K b = 3.8 x 10 -10 )

15. Acids and Bases – ch. 7 12. If a 0.35 M solution of weak base has a pH of 11.2 what is the pH of a 0.08 M solution of the weak base?

16. Acids and Bases – ch. 7 13. Rank the following 0.1 M salt solutions in order of increasing pH. i. NaClO 4 ii. LiF iii. C 5 H 5 NHI iv. NH 4 Cl v. KCN

17. Acids and Bases – ch. 7 Salts ⇒ soluble ionic compounds that are the result of an acid base reaction ⇒ when a salt is dissolved in water each ion can potentially affect the pH of the solution ⇒ you need to analyze the ions individually Cations Group 1 and 2 metal ions ⇒ neutral All other cations ⇒ acids Cations Group 1 and 2 metal ions ⇒ neutral All other cations ⇒ acids Anions 1 st 7 ⇒ neutral All other anions ⇒ bases Anions 1 st 7 ⇒ neutral All other anions ⇒ bases Salts

18. Acids and Bases – ch. 7 14. Will a solution of NH 4 F be acidic, basic or neutral? NH 3 K b = 1.8 x 10 -5 HFK a = 7.2 x 10 -4

19. Acids and Bases – ch. 7 15. How many grams of KF must be dissolved in 500 mL of water in order to get a pH of 8.4? (K a of HF = 7.2 x 10 -4 )

20. Acids and Bases – ch. 7 16. When 0.100 mole of an unknown soluble salt is dissolved in 1.00 L of water, the pH of the solution is 8.07. Assume the volume of the solid salt is negligible. What is the identity of the salt? A) NaCN B) NaC 2 H 3 O 2 C) NaCl D) NaF E) NaOCl

21. Acids and Bases – ch. 7 You have completed ch. 7

22. Answer Key – ch. 7 1. Arrange the following solutions in order of most acidic to most basic. A) [OH – ] = 0.5 M ⇒ pOH = -log(0.5) = 0.3 ⇒ pH = 14 - 0.3 = 13.7 B) [H + ] = 0.3 M ⇒ pH = -log(0.3) = 0.52 C) pOH = 5.9 ⇒ pH = 14-5.9 = 8.1 D) pH = 1.2 E) [H + ] = 1.0x10 –4 M ⇒ pH = -log(1.0x10 –4 ) = 4 As a solution gets more acidic… [H + ]↑, pH↓, [OH – ]↓, pOH↑ B > D > E > C > A

23. Answer Key – ch. 7 2. At 25°C the water ionization constant is 1.0 x 10 –14 and at 98°C the water ionization constant is 6.8 x 10 –14. What is the pH of neutral water at these temperatures? If a solution is neutral ⇒ [H 3 O + ] = [OH - ] For all aqueous solutions ⇒ [H 3 O + ]x[OH - ] = K w at 25°C ⇒ x 2 = 1.0 x 10 –14 ⇒ x = 1.0 x 10 -7 M = [H 3 O + ] = [OH - ] pH = -log(1.0 x 10 -7 M) pH = 7 at 98°C ⇒ x 2 = 6.8 x 10 –14 => x = 2.61x10 -7 M = [H 3 O + ] = [OH - ] pH = -log(2.61x10 -7 M) pH = 6.58

24. 3. a. Which of the following is a stronger acid? HNO 2 (K a = 4.0 x 10 –4 ) or HCN (pK a = 9.21) as K a ↑ % ionization ↑ or acid strength ↑ pK a ↓ K a =10 -pK a ⇒ K a for HCN =10 -9.21 = 6.2x 10 -10 Since K a for HNO 2 > K a for HCN ⇒ HNO 2 is a stronger acid b. Which is a stronger base? NO 2 – or CN – as acid strength ↑ conjugate base strength ↓ since HNO 2 is the stronger acid ⇒ NO 2 – is the weaker base ⇒ CN – is the stronger base Answer Key – ch. 7

26. 5. Calculate the pH and the % ionization of the following solutions a. 0.05 M HClO (K a = 3.0 x 10 –8 ) ⇒ weak acid HClOH2OH2O ⇌ H3O+H3O+ ClO 2 – I0.05N/A00 ∆- xN/A+ x Eq0.05- xN/Axx Insignificantly small

27. Answer Key – ch. 7 5. …continued b. 0.33 M CH 3 COOH (K a = 1.8 x 10 –5 ) mixed with 0.85 M HCN (K a = 6.2 x 10 –10 ) CH 3 COOH H2OH2O ⇌ H3O+H3O+ CH 3 COO - I0.33N/A00 ∆- xN/A+ x Eq0.33 - xN/Axx Insignificantly small

28. Answer Key – ch. 7 HCOOH H2OH2O ⇌ H3O+H3O+ I xN/A00 ∆ -0.0063N/A+0.0063 Eq x-0.0063N/A0.0063

29. Answer Key – ch. 7 HA H2OH2O ⇌ H3O+H3O+ A–A– I 0.5N/A00 ∆ -0.11(0.5)N/A+0.11(0.5) Eq 0.445N/A0.055 11%

30. Answer Key – ch. 7 8. What concentration of a weak acid that ionizes 0.1% will have a pH of 3.8? Since the pH is 3.8 ⇒ [H 3 O + ] = 10 –3.8 or 1.58x10 –4 M 1.58x10 –4 = 0.0001x x = 1.58M HA H2OH2O ⇌ H3O+H3O+ A–A– I xN/A00 ∆ -0.001xN/A+0.001x Eq x -0.001xN/A0.001x 0.1%

31. Answer Key – ch. 7 HNO 2 H2OH2O ⇌ H3O+H3O+ NO 2 – I xN/A00 ∆ -0.03N/A+0.03 Eq x – 0.03N/A0.03

32. Answer Key – ch. 7

33. NH 3 H2OH2O ⇌ OH - NH 4 + I 0.01N/A00 ∆ -xN/A+x Eq 0.01-xN/Axx

34. Answer Key – ch. 7 C 6 H 5 NH 2 H2OH2O ⇌ OH - C 6 H 5 NH 3 + I 0.5N/A00 ∆ – xN/A+x Eq 0.5 – xN/Axx

35. Answer Key – ch. 7

36. 13. Rank the following 0.1 M salt solutions in order of increasing pH. i. NaClO 4 ⇒ only contains spectator or neutral ions ii. LiF ⇒ F – is a weak base iii. C 5 H 5 NHI ⇒ C 5 H 5 NH + is a weak acid iv. NH 4 Cl ⇒ NH 4 + is a weak acid v. KCN ⇒ CN – is a weak base Since HF is a stronger acid than HCN ⇒ F – is a weaker base than CN – Since NH 3 is a stronger base than C 5 H 5 N ⇒ NH 4 + is a weaker acid than C 5 H 5 NH + C 5 H 5 NH + < NH 4 + < NaClO 4 < F – < CN –

37. Answer Key – ch. 7 14. Will a solution of NH 4 F be acidic, basic or neutral? NH 4 F is a salt that contains both a weak acid ( NH 4 + ) and a weak base ( F – ) Since HF is a stronger acid than NH 3 as a base ⇒ F – is a weaker base than NH 4 + as an acid pH of the salt will be acidic because NH 4 + is the stronger component

38. Answer Key – ch. 7 F–F– H2OH2O ⇌ OH – HF Ix N/A 00 ∆-2.51x10 –6 N/A +2.51x10 –6 Eq x-2.51x10 –6 N/A 2.51x10 –6

39. Answer Key – ch. 7 16. When 0.100 mole of an unknown soluble salt is dissolved in 1.00 L of water, the pH of the solution is 8.07. Assume the volume of the solid salt is negligible. What is the identity of the salt? A) NaCN B) NaC 2 H 3 O 2 C) NaCl D) NaF E) NaOCl You can eliminate NaCl because it only contains spectator or neutral ions ⇒ your table has a list of K a values ⇒ In order to get K a for the conjugate acid you need K b for the weak base ⇒ since the pH = 8.07 ⇒ pOH = 5.93 ⇒ [OH – ] = 1.17 x 10 –6 A–A– H2OH2O ⇌ OH – HA I0.1 N/A 00 ∆ +1.17 x 10 –6 Eq0.1N/A1.17 x 10 –6 continue to next slide…

40. Answer Key – ch. 7