1. 1 Ch. 8: Acids and Bases Chem 20 El Camino College

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3. 3 Two Acid-Base Theories Arrhenius Theory An acid solution contains more H + ions than OH - ions A base solution contains more OH - ions than H + ions

4. 4 Two Acid-Base Theories Note--H + is a proton Bronsted-Lowry Theory An acid is a proton donor A base is a proton acceptor HBr(aq) + H 2 O( l )  H 3 O + (aq) + Br - (aq) NH 3 (aq) + H 2 O( l )  NH 4 + (aq) + OH - (aq) acidbase acidbase

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7. 7 Conjugate Acid-Base Pairs Conjugate Acid-Base Pairs differ by one H + The reactants side has the acid and the base The products side has the conjugate acid and the conjugate base HBr(aq) + H 2 O( l )  H 3 O + (aq) + Br - (aq) conjugate acid conjugate base acidbase

8. 8 Conjugate Acid-Base Pairs Acid: H + donor on left side Conjugate base: missing H + on right side Base: H + acceptor on left side Conjugate acid: received H + on right side

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10. 10 Conjugate Acid-Base Pairs Label and link conjugate acid-base pairs. NH 3 (aq) + CH 3 CO 2 H(aq)  NH 4 + (aq) + CH 3 CO 2 - (aq) conjugate acid conjugate base acidbase H 2 SO 4 (aq) + HSO 3 - (aq)  HSO 4 - (aq) + H 2 SO 3 (aq) conjugate acid conjugate base acid base

11. 11 The Water Equilibrium H 2 O( l ) H + (aq) + OH - (aq) Kw = (concentration H + in M)* (conc. OH - in M) Kw = [H + ][OH - ] = 1.0 x 10 -14 In aq. soln, if [H + ]=[OH - ], the soln is neutral. In aq. soln, if [H + ]>[OH - ], the soln is acidic. In aq. soln, if [H + ]<[OH - ], the soln is basic.

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13. 13 Examples Use this equation: [H + ][OH - ]=1.0 x 10 -14 Ex. If the conc. of H + is 3.5 x 10 -3 M, find [OH - ]. Is the solution acidic or basic? [OH - ] = 1 x 10 -14 [H + ] = 1 x 10 -14 3.5x10 -3 M = 2.9 x 10 -12 M acidic

14. 14 Examples Ex. If the conc. of H + is 9.9 x 10 -11 M, find [OH - ]. Is the solution acidic or basic? [OH - ] = 1 x 10 -14 [H + ] = 1 x 10 -14 9.9x10 -11 M = 1.0 x 10 -4 M Ex. If [OH - ] is 1.7 x 10 -10 M, find [H + ]. Is the solution acidic or basic? [H + ] = 1 x 10 -14 [OH - ] = 1 x 10 -14 1.7x10 -10 M = 5.9 x 10 -5 M basic acidic

15. 15 pH Values pH = 7 is a neutral solution pH < 7 is an acidic solution pH > 7 is a basic solution

16. 16 Table 18-2, p. 516

17. 17 pH = -log [H + ] Ex. If [H + ] = 1.5 x 10 -6 M, find [OH - ] and pH [OH - ] = 1 x 10 -14 [H + ] = 1 x 10 -14 1.5x10 -6 M = 6.7 x 10 -9 M pH = -log [H + ] = -log(1.5 x 10 -6 ) = 5.82

18. 18 pH = -log [H + ] Ex. If [OH - ] = 3.3 x 10 -4 M, find [H + ] and pH [H + ] = 1 x 10 -14 [OH - ] = 1 x 10 -14 3.3x10 -4 M = 3.0 x 10 -11 M pH = -log [H + ] = -log(3.0 x 10 -11 ) = 10.52

19. 19 pH = -log [H + ] Ex. If [H + ] = 8.5 x 10 -1 M, find [OH - ] and pH [OH - ] = 1 x 10 -14 [H + ] = 1 x 10 -14 8.5x10 -1 M = 1.2 x 10 -14 M pH = -log [H + ] = -log(8.5 x 10 -1 ) = 0.07

20. 20 pH to [H + ] [H + ] = 10 -pH or antilog(-pH) The minus sign goes on the pH value first Ex. If the pH = 5.55, find [H + ] [H + ] = 10 -pH = 10 -5.55 = 2.8 x 10 -6 M

21. 21 pH to [H + ] Ex. If the pH = 8.88, find [H + ] [H + ] = 10 -pH = 10 -8.88 = 1.3 x 10 -9 M Ex. If the pH = 13.00, find [H + ] and [OH - ] [H + ] = 10 -pH = 10 -13.00 = 1.0 x 10 -13 M [OH - ] = 1 x 10 -14 [H + ] = 1 x 10 -14 1.0x10 -13 M = 1.0 x 10 -1 M

22. 22 Titration In the acid-base titration we’ll do in lab, a flask contains a mixture of acid and phenolphthalein (a ccs)titration Base is added by buret When the soln turns pale pink for 30 seconds, moles acid  moles base.

23. 23 Fig. 16-8, p. 457

24. 24 p. 459

25. 25 p. 459

26. 26 p. 459

27. 27 Titration: How many mL? Ex. The flask is filled with 25.00 mL of a 0.500 M HCl soln. How many mL of 0.300 M NaOH soln will neutralize the acid? HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O( l ) Start with volume of acid, convert to L Use molarity of acid as a conversion factor Use a mole ratio to convert mol acid to mol base Use molarity of base as a conversion factor, convert to mL.500 mol HCl 1 L = 41.7 mL 25.00mL HCl 1 mol HCl 1 mol NaOH.300 mol NaOH 1 L 1000 mL 1 L 1000 mL

28. 28 Titration: How many mL? Ex. The flask is filled with 75.00 mL of a 0.200 M HCl soln. How many mL of 0.150 M NaOH soln will neutralize the acid? Ex. The flask is filled with 55.00 mL of a 1.5 M HCl soln. How many mL of 0.90 M NaOH soln will neutralize the acid?.200 mol HCl 1 L = 100. mL 75.00mL HCl 1 mol HCl 1 mol NaOH.150 mol NaOH 1 L 1000 mL 1 L 1000 mL 1.5 mol HCl 1 L = 91.7 mL 55.00mL HCl 1 mol HCl 1 mol NaOH 0.90 mol NaOH 1 L 1000 mL 1 L 1000 mL

29. 29 Titration: Find Molarity Ex. The flask is filled with 30.00 mL of a 0.100 M HCl soln. What is the molarity of the NaOH soln if it takes 23.45 mL to neutralize the acid? Start with volume of acid, convert to L Use molarity of acid as a conversion factor Use a mole ratio to convert mol acid to mol base Solve for mol base ***Divide mol base by mL base to find molarity. Convert to L..100 mol HCl 1 L 30.00mL HCl 1 mol HCl 1 mol NaOH 1000 mL 1 L = 0.00300 mol NaOH = 0.128 M 1 L 1000 mL0.00300 mol NaOH 23.45 mL

30. 30 Titration: Find Molarity Ex. The flask is filled with 45.00 mL of a 0.996 M HCl soln. What is the molarity of the NaOH soln if it takes 52.33 mL to neutralize the acid?.996 mol HCl 1 L 45.00mL HCl 1 mol HCl 1 mol NaOH 1000 mL 1 L = 0.0448 mol NaOH =.856 M 1 L 1000 mL0.0448 mol NaOH 52.33 mL

31. 31 Titration: Find Molarity Ex. The flask is filled with 80.00 mL of a 2.30 M H 2 SO 4 soln. What is the molarity of the NaOH soln if it takes 70.00 mL to neutralize the acid? H 2 SO 4 (aq) + 2 NaOH(aq)  Na 2 SO 4 (aq) + 2 H 2 O( l ) 2.30 mol H 2 SO 4 1 L 80.00mL H 2 SO 4 1 mol H 2 SO 4 2 mol NaOH 1000 mL 1 L = 0.184 mol NaOH = 2.63 M 1 L 1000 mL0.184 mol NaOH 70.00 mL

32. 32 Strong and Weak Acids Strong acids completely break down into ions when dissolved in water Weak acids only break down into a few ions in water. Most of the weak acid molecules stay together in water.

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34. 34 Strong and Weak Acids These are the only strong acids HCl hydrochloric acid HBr hydrobromic acid HI hydroiodic acid HNO 3 nitric acid H 2 SO 4 sulfuric acid HClO 3 chloric acid HClO 4 perchloric acid.

35. 35 Strong and Weak Acids All other acids are weak acids. Here are some examples HF hydrofluoric acid H 3 PO 4 phosphoric acid CH 3 CO 2 H acetic acid H 2 CO 3 carbonic acid.

36. 36 Buffers When a small amount of acid or base is added to pure water, pH changes dramatically A buffer resists change in pH when small amounts of acids or bases are added Blood is buffered in the body to a pH of 7.4 A buffer is a combination of a weak acid and its conjugate base (found in an ionic cmpd) A strong acid cannot make a buffer.

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39. 39 How Buffers Work This is a buffer made of CH 3 COOH and CH 3 COONa When acid is added, the extra H + reacts with CH 3 COO - to form more CH 3 COOH When base is added, the extra OH - reacts with CH 3 COOH to form more CH 3 COO -

40. 40 Buffers Which of the following represents a buffer system? HCl and NaCl HF and NaF HNO 3 and NaNO 3 CH 3 COOH and CH 3 COONa no yes no yes

41. 41 Practice: Conjugate Acid- Base Pairs Label and link conjugate acid-base pairs. HNO 3 (aq) + CH 3 CO 2 - (aq)  CH 3 CO 2 H (aq) + NO 3 - (aq) conjugate acid conjugate base acid HCl(aq) + SO 4 2 - (aq)  Cl - (aq) + HSO 4 - (aq) conjugate acid conjugate base acid base

42. 42 Practice: pH to [H + ] Ex. If the pH = 3.68, find [H + ] and [OH - ] in scientific notation. [H + ] = 10 -pH = 10 -3.68 = 2.1 x 10 -4 M [OH - ] = 1 x 10 -14 [H + ] = 1 x 10 -14 2.1x10 -4 M = 4.8 x 10 -11 M

43. 43 Practice: Titration Ex. The flask is filled with 20.50 mL of a 0.0996 M HCl soln. How many mL of 0.194 M NaOH soln will neutralize the acid?.0996 mol HCl 1 L = 10.5 mL 20.50mL HCl 1 mol HCl 1 mol NaOH.194 mol NaOH 1 L 1000 mL 1 L 1000 mL

44. 44 Practice: Titration Ex. The flask is filled with 25.00 mL of a 2.5 M HCl soln. How many mL of 0.50 M Ca(OH) 2 soln will neutralize the acid? 2 HCl(aq) + Ca(OH) 2 (aq)  CaCl 2 (aq) + 2 H 2 O( l ) = 62.5 mL 2.5 mol HCl 1 L 25.00mL HCl 2 mol HCl 1 mol Ca(OH) 2 0.50 mol Ca(OH) 2 1 L 1000 mL 1 L 1000 mL

45. 45 Practice: Titration Ex. The flask is filled with 20.60 mL of a 0.09662 M HCl soln. What is the molarity of the NaOH soln if it takes 20.92 mL to neutralize the acid?.09662 mol HCl 1 L 20.60mL HCl 1 mol HCl 1 mol NaOH 1000 mL 1 L = 0.001990 mol NaOH = 0.09512 M 1 L 1000 mL0.001990 mol NaOH 20.92 mL