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

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

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

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

16 Table 18-2, p. 516

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

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

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

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 = = 2.8 x M

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

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 Fig. 16-8, p. 457

24 p. 459

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26 p. 459

27 Titration: How many mL? Ex. The flask is filled with mL of a M HCl soln. How many mL of 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 Titration: How many mL? Ex. The flask is filled with mL of a M HCl soln. How many mL of M NaOH soln will neutralize the acid? Ex. The flask is filled with 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 Titration: Find Molarity Ex. The flask is filled with mL of a M HCl soln. What is the molarity of the NaOH soln if it takes 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 = mol NaOH = M 1 L 1000 mL mol NaOH mL

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

31 Titration: Find Molarity Ex. The flask is filled with mL of a 2.30 M H 2 SO 4 soln. What is the molarity of the NaOH soln if it takes 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 = mol NaOH = 2.63 M 1 L 1000 mL0.184 mol NaOH mL

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 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 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 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 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 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 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 (aq)  Cl - (aq) + HSO 4 - (aq) conjugate acid conjugate base acid base

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

43 Practice: Titration Ex. The flask is filled with mL of a M HCl soln. How many mL of 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 Practice: Titration Ex. The flask is filled with 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) mol Ca(OH) 2 1 L 1000 mL 1 L 1000 mL

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