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# Ch14 z5e acid base

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• Z5e 14.1 658 The Nature of Acids and Bases Alternative link if can’t get through to YouTube: http://dailycackle.com/2007/01/20/science-experiment-gone-wrong/
• Z53 Fig. 14.1 659: reaction of HCl and H20
• Z5e 660 SE 14.1 a, c, e
• Z5e fig. 14.5 p. 661
• Z5e 661 14.2 Acid Strength; see Table 14.1
• For SE 14.2 -- you need to “remember” that HCl is a strong acid, so Cl - is a weak base (since id does not easily accept a proton (or a H + ). That is, acid strength is inversely proportional to its conjugate base strength.
• Z5e 663 SE 14.2
• Z5e 663 SE 14.2
• Z5e 663 SE 14.2
• Rf. Z5e 664 Note: in =m expression, H 2 O (l) is a pure liquid, so constant and drops out of the expression. K w called the ion-product constant (or the dissociation constant)
• Rf. Z5e 664 Note: in =m expression, H 2 O (l) is a pure liquid, so constant and drops out of the expression. K w called the ion-product constant (or the dissociation constant)
• Z5e 666 SE 14.4
• Z5e 664 Fig. 14.7
• Z5e 14.3 The pH Scale
• Z5e Fig. 14.8 667
• Z5e 668. SE 14.6
• Z5e 669 Section 14.4: Calculating the pH of Strong Acid Solutions
• Z5e 670 SE 14.7b
• Z5e 671 Section 14.5: Calculating the pH of Weak Acid Solutions
• [H + ] = .0060, so pH = 2.22 pOH = 11.78; [OH - ] = 1.7 x 10 -12 &amp; [H + ] = 6.0 x 10 -3
• Z5e 675 Major species: HF, HOC 6 H 5 , H 2 O Since HF is dominant producer of H + , use its Ka Set up “RICE” table, use approximations and 5% rule [H + ] = .030 &amp; pH = 1.53
• Z5e 675 Major species: HF, HOC 6 H 5 , H 2 O Since HF is dominant producer of H + , use its Ka Set up “RICE” table, use approximations and 5% rule [H + ] = .030 &amp; pH = 1.53
• Rf. Z5e 675 SE 14.9
• Z5e 677 [H] = 4.24 x 10 -3 vs. 1.34 x 10 -4 So % dissociation (aka % ionization) = 0.42% vs. 1.3%
• Rf. Z5e 680 SE 14.11 Since (X/0.100)(100%) = 8.1%, x = 8.1 x 10 -3 = [H + ] = [A - ], then plug into =m to solve for Ka. Ans. = 6.6 x 10 -4
• Z5e 680 fig. 14.10
• Z5e 681 Section 14.6 Bases
• See next slide for explanation
• [hydroxide] = 1.2 x 10 -10
• Z5e 688 Section 14.7 Polyprotic Acids
• [H + ] = 7.1 x 10 -2 [H 2 AsO 4 2- ] = 7.1 x 10 -2 also [HAsO 4 3- ] = 8.0 x 10 -8 = Ka 2 [AsO 4 3- ] = 6.8 x 10 -16 Remember: Must consider [H + ] from all sources
• [H + ] = 7.1 x 10 -2 [H 2 AsO 4 2- ] = 7.1 x 10 -2 also [HAsO 4 3- ] = 8.0 x 10 -8 = Ka 2 [AsO 4 3- ] = 6.8 x 10 -16 Remember: Must consider [H + ] from all sources
• [H + ] = 7.1 x 10 -2 [H 2 AsO 4 2- ] = 7.1 x 10 -2 also [HAsO 4 3- ] = 8.0 x 10 -8 = Ka 2 [AsO 4 3- ] = 6.8 x 10 -16 Remember: Must consider [H + ] from all sources
• [H + ] = 7.1 x 10 -2 [H 2 AsO 4 2- ] = 7.1 x 10 -2 also [HAsO 4 3- ] = 8.0 x 10 -8 = Ka 2 [AsO 4 3- ] = 6.8 x 10 -16 Remember: Must consider [H + ] from all sources
• [H + ] = 7.1 x 10 -2 [H 2 AsO 4 2- ] = 7.1 x 10 -2 also [HAsO 4 3- ] = 8.0 x 10 -8 = Ka 2 [AsO 4 3- ] = 6.8 x 10 -16 Remember: Must consider [H + ] from all sources
• Both [H + ] and [HSO4 - ] are 2 M , since completely dissociated.
• Both [H + ] and [HSO4 - ] are 2 M , since completely dissociated.
• Z5e 694 Section 14.8 Acid-Base properties of Salts
• Z5e 695 Write major species Set up =m between anion and water Since product includes hydroxide ion, must use K b so need to convert from K a Calculate hydroxide concentration, convert to pOH, subtract from 14 to get pH
• Z5e 695 Write major species Set up =m between anion and water Since product includes hydroxide ion, must use K b so need to convert from K a Calculate hydroxide concentration, convert to pOH, subtract from 14 to get pH
• Z5e 697
• Z5e 697
• Z5e 699 Table 14.5
• Z5e 701 Section 14.9 The Effect of Structure on Acid-Base Properties
• Z5e 702 Figure 14.11
• Z5e Section 14.10 Acid-Base Properties of Oxides
• Z5e 704 Section 14.11 The Lewis Acid-Base Model
• Al ion is Lewis acid and water is Lewis base
• Z5e 707 Section 14.12 Strategy for Solving Acid-Base Problems: A Summary
• Z5e 708; Tr 160.
• ### Ch14 z5e acid base

1. 2. Z5e 14.1 The Nature of Acids and Bases Arrhenius Definition <ul><li>Acids produce hydrogen ions in aqueous solution. </li></ul><ul><li>Bases produce hydroxide ions when dissolved in water. </li></ul><ul><li>Limited to aqueous solutions. </li></ul><ul><li>Only one kind of base (hydroxide). </li></ul><ul><li>NH 3 ammonia could not be an Arrhenius base. </li></ul><ul><li>http://www.youtube.com/watch?v=h2UczzWu-Qs&mode=related&search = </li></ul>
2. 3. Bronsted-Lowry Definitions <ul><li>An acid is an proton (H + ) donor and a base is a proton acceptor . </li></ul><ul><li>Acids and bases always come in pairs . </li></ul><ul><li>HCl is an acid. </li></ul><ul><li>When it dissolves in water it gives its proton to water (so H 2 O is a base). </li></ul><ul><li>HCl (g) + H 2 O (l)  H 3 O + + Cl - </li></ul><ul><li>Water is a base : makes hydronium ion </li></ul>
3. 4. Figure 14.1 The Reaction of HCI and H 2 O
4. 5. Bronsted-Lowry Acids <ul><li>Acids are hydrogen ion (H +) donors </li></ul><ul><li>Bases are hydrogen ion (H + ) acceptors </li></ul><ul><li>HCl + H 2 O H 3 O + + Cl - </li></ul><ul><li>donor acceptor + - </li></ul><ul><li>+ + </li></ul>
5. 6. NH 3 , A Bronsted-Lowry Base <ul><li>When NH 3 reacts with water, most of the reactants remain dissolved as molecules, but a few NH 3 reacts with water to form NH 4 + and hydroxide ion. </li></ul><ul><li>NH 3 + H 2 O NH 4 + (aq) + OH - (aq) </li></ul><ul><li>acceptor donor </li></ul><ul><li>+ + </li></ul>
6. 7. B. Definitions pp <ul><li>Brønsted-Lowry </li></ul>HCl + H 2 O  Cl – + H 3 O + <ul><ul><li>Acids are proton (H + ) donors. </li></ul></ul><ul><ul><li>Bases are proton (H + ) acceptors. </li></ul></ul>base acid conjugate acid conjugate base
7. 8. Conjugate Acid-Base Pairs
8. 9. B. Definitions H 2 O + HNO 3  H 3 O + + NO 3 – CB CA A B
9. 10. B. Definitions <ul><li>Amphoteric - can be an acid or a base. </li></ul>NH 3 + H 2 O NH 4 + + OH - CA CB B A
10. 11. B. Definitions <ul><li>F - </li></ul><ul><li>H 2 PO 4 - </li></ul><ul><li>H 2 O </li></ul><ul><li>HF </li></ul><ul><li>H 3 PO 4 </li></ul><ul><li>H 3 O + </li></ul><ul><li>Give the conjugate base for each of the following: </li></ul><ul><li>Polyprotic - an acid with more than one H + </li></ul>
11. 12. B. Definitions <ul><li>Br - </li></ul><ul><li>HSO 4 - </li></ul><ul><li>CO 3 2- </li></ul><ul><li>HBr </li></ul><ul><li>H 2 SO 4 </li></ul><ul><li>HCO 3 - </li></ul><ul><li>Give the conjugate acid for each of the following: </li></ul>
12. 13. Pairs pp <ul><li>General equation </li></ul><ul><li>HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) </li></ul><ul><li>Acid + Base Conjugate acid + Conjugate base </li></ul><ul><li>This is an equilibrium. </li></ul><ul><li>Competition for H + between H 2 O and A - </li></ul><ul><li>The stronger base controls direction. </li></ul><ul><li>If H 2 O is a stronger base it takes the H + </li></ul><ul><li>Equilibrium moves to right. </li></ul>
13. 14. Acid dissociation constant K a <ul><li>The equilibrium constant for the general equation. </li></ul><ul><li>HA(aq) + H 2 O(l) H 3 O 1+ (aq) + A 1- (aq) </li></ul><ul><li>K a = [H 3 O + ][A - ] [HA] </li></ul><ul><li>H 3 O + is often written H + ignoring the water in equation (it is implied). </li></ul>
14. 15. Acid Dissociation Reactions <ul><li>Write the dissociation reaction for the following (omitting water). </li></ul><ul><li>HCl . . . </li></ul><ul><li>HCl (aq) H 1+ (aq) + Cl 1- (aq) </li></ul><ul><li>NH 4 1+ . . . </li></ul><ul><li>NH 4 1+ (aq) H 1+ (aq) + NH 3(aq) </li></ul><ul><li>[Al(H 2 O) 6 ] 3+ . . . </li></ul><ul><li>[Al(H 2 O) 6 ] 3+ H 1+ + [Al(H 2 O) 5 ]OH 2+ </li></ul>
15. 16. Acid dissociation constant K a <ul><li>HA(aq) H + (aq) + A - (aq) </li></ul><ul><li>K a = [H + ][A - ] [HA] </li></ul><ul><li>We can write the expression for any acid. </li></ul><ul><li>Strong acids dissociate completely. </li></ul><ul><li>Equilibrium far to right. </li></ul><ul><li>Conjugate base must be weak. </li></ul>
16. 17. Figure 14.5 Acid Strength Versus Conjugate Base Strength z5e 661
17. 18. Z5e 14.2 Acid Strength <ul><li>Strong acids </li></ul><ul><li>K a is large </li></ul><ul><li>[H + ] is about equal to original [HA] </li></ul><ul><li>A - is a weaker base than water </li></ul><ul><li>Weak acids </li></ul><ul><li>K a is small </li></ul><ul><li>[H + ] <<< [HA] </li></ul><ul><li>A - is a stronger base than water </li></ul>
18. 19. Z5e Figure 14.4 p. 661 pp <ul><li>Graphic Representation of the Behavior of Acids of Different Strengths in Aqueous Solution </li></ul><ul><li>( a) = strong acid, complete dissociation. </li></ul><ul><li>( b) = weak acid, slight dissociation . </li></ul>
19. 20. <ul><li>Which of the following &quot;molecular&quot; pictures, “A” or “B”, best represents a concentrated solution of the weak acid HA with Ka = 10 -5 ? . . . </li></ul><ul><li>“ B”, because . . . </li></ul><ul><li>There are more molecules of undissociated form. </li></ul><ul><li>“ A” represents a strong acid. </li></ul>
20. 21. Types of Acids <ul><li>Polyprotic Acids - more than 1 acidic hydrogen (diprotic, triprotic). </li></ul><ul><li>Oxyacids - Proton is attached to the oxygen of an ion. </li></ul><ul><li>Organic acids contain the Carboxyl group -COOH with the H attached to O </li></ul><ul><li>Polyprotic acids are generally very weak (but, H 2 SO 4 is strong with 1 st H + and weak with 2 nd ). </li></ul>
21. 22. Relative Base Strength of Acids pp <ul><li>Using Table 14.2 p. 628, arrange the following in increasing base strength. H 2 O , F 1- , Cl 1- , NO 2 1- , CN 1- Steps . . . </li></ul><ul><li>Hint: Strongest acid will have weakest base. Which conjugate of the above bases would be the strongest acid? . . . </li></ul><ul><li>That’s right, HCl. So, Cl 1- must be the weakest base. </li></ul>
22. 23. Relative Base Strength of Acids pp <ul><li>Arrange H 2 O , F 1- , Cl 1- , NO 2 1- , CN 1- in increasing base strength. </li></ul><ul><li>Considering H 2 O , the others are conjugate bases of weak acids, so the above bases must be stronger than water acting as a base. So . . . </li></ul><ul><li>Cl - < water < conjugate bases of weak acids . . . </li></ul>
23. 24. Relative Base Strength of Acids pp <ul><li>Arrange H 2 O , F 1- , Cl 1- , NO 2 1- , CN 1- in increasing base strength. Cl - < water < conj. bases of weak acids. </li></ul><ul><li>Now, use Table 14.2 p. 628 to “order” the rest. Your answer is . . . </li></ul><ul><li>Cl - < H 2 O < F 1- < NO 2 1- < CN 1- </li></ul>
24. 25. Amphoteric <ul><li>Behaves as acid & base -- autoionizes </li></ul><ul><li>2H 2 O(l) H 3 O + (aq) + OH - (aq) </li></ul><ul><li>K W = [H 3 O + ][OH - ]=[H + ][OH - ] </li></ul><ul><li>H 2 O (l) drops out of =m expression since it is a pure liquid </li></ul><ul><li>At 25ºC K W = 1.0 x10 -14 in EVERY aqueous solution. </li></ul><ul><li>K w = ion product constant (or dissociation constant) </li></ul><ul><li>So, we can use it to find [H + ] and [OH - ] </li></ul>
25. 26. Solutions <ul><li>In EVERY aqueous solution - 3 possibilities. </li></ul><ul><li>Neutral solution [H + ] = [OH - ]= 1.0 x10 -7 </li></ul><ul><li>Acidic solution [H + ] > [OH - ] </li></ul><ul><li>Basic solution [H + ] < [OH - ] </li></ul>
26. 27. Autoionization of Water problem <ul><li>K w = 1 x 10 -13 at 60ºC. Using Le Chatelier, predict if exo- or endothermic 2H 2 O (l) H 3 O 1+ (aq) + OH 1- (aq) Hint: compare K w at 25ºC. </li></ul><ul><li>Endothermic since K w increases with temperature so energy is a reactant. </li></ul>
27. 28. Autoionization of Water problem <ul><li>K w = 1 x 10 -13 at 60ºC. 2H 2 O (l) H 3 O 1+ (aq) + OH 1- (aq) </li></ul><ul><li>Calculate [H + ] & [OH - ] in neutral soln. </li></ul><ul><li>Hint: since neutral, they must be equal. </li></ul><ul><li>Answer . . . </li></ul><ul><li>3 x 10 -7 M since [H + ] [OH - ] = 1 x 10 -13 </li></ul>
28. 29. Figure 14.7 Two Water Molecules React to Form H 3 O + and OH -
29. 30. Z5e 14.3 The pH Scale <ul><li>pH= -log[H + ] </li></ul><ul><li>Used because [H + ] is usually very small </li></ul><ul><li>As pH decreases , [H + ] increases exponentially </li></ul><ul><li>Sig figs: Only the digits after the decimal place of a pH are significant! </li></ul><ul><li>[H + ] = 1.0 x 10 -8 pH= 8.00 2 sig figs </li></ul><ul><li>pOH= -log[OH - ] </li></ul><ul><li>pKa = -log K a </li></ul>
30. 31. <ul><li>A solution with pH = 5 is 100 times more acidic than a solution with a pH = ? A) 7 B) 3 C) 0.05 </li></ul><ul><li>7 because . . . </li></ul><ul><li>7 is 2 pH units more basic so 100 times less [H 3 O 1+ ] (or 100 x more [OH 1- ] </li></ul>
31. 32. Relationships pp <ul><li>K W = [H + ][OH - ] </li></ul><ul><li>-log K W = -log([H + ][OH - ]) </li></ul><ul><li>-log K W = -log[H + ]+ -log[OH - ] </li></ul><ul><li>pK W = pH + pOH </li></ul><ul><li>K W = 1.0 x10 -14 </li></ul><ul><li>14.00 = pH + pOH </li></ul><ul><li>[H + ] = 10 -pH and [OH - ] = 10 -pOH </li></ul><ul><li>[H + ],[OH - ],pH and pOH: Given any one of these we can find the other three. </li></ul>
32. 33. Figure 14.8 The pH Scale and pH Values of Some Common Substances
33. 34. Basic Acidic Neutral 10 0 10 -1 10 -3 10 -5 10 -7 10 -9 10 -11 10 -13 10 -14 [H + ] 0 1 3 5 7 9 11 13 14 pH Basic 10 0 10 -1 10 -3 10 -5 10 -7 10 -9 10 -11 10 -13 10 -14 [OH - ] 0 1 3 5 7 9 11 13 14 pOH
34. 35. Calculating pH of Solutions <ul><li>Always write down the major species in solution. If you forget you’ll probably get a wrong solution! </li></ul><ul><li>Remember, these are equilibria. </li></ul><ul><li>Remember the chemistry. </li></ul><ul><li>Don’t try to memorize, there is no one way to do this. </li></ul>
35. 36. Calculating pH of Solutions <ul><li>The pH of a blood sample was 7.41 at 25ºC. Calculate pOH, [H 1+ ] and [OH 1- ]. </li></ul><ul><li>Answers? . . . </li></ul><ul><li>pOH = 6.59 (14 - 7.41 = 6.59) [H 1+ ] = ? </li></ul><ul><li>[H 1+ ] = 3.9 x 10 -8 (10 -7.41 ) [OH 1- ] = ? </li></ul><ul><li>[OH 1- ] = 2.6 x 10 -7 Either 10 -6.59 or since K w = [H 1+ ][OH 1- ], [OH 1- ] = 10 -14 /3.9 x 10 -8 </li></ul>
36. 37. z5e 14.4 Calculating the pH of Strong Acids pp <ul><li>HBr, HI, HCl, HNO 3 , H 2 SO 4 , HClO 4 </li></ul><ul><li>ALWAYS WRITE THE MAJOR SPECIES </li></ul><ul><li>Above are completely dissociated (strong acids) </li></ul><ul><li>So, [H + ] = original [HA] </li></ul><ul><li>[OH - ] is going to be small because of equilibrium (essentially none since shifted so far to the right) </li></ul><ul><li>10 -14 = [H + ][OH - ] </li></ul><ul><li>If [HA] < 10 -7 then water contributes H + and must be taken into account!!! </li></ul>
37. 38. Strong Acids pp <ul><li>Calculate the pH of 1.0 x 10 -10 M HCl. Ans.? </li></ul><ul><li>pH = 7.00. Why? . . . </li></ul><ul><li>Always write the major species, which are . . . </li></ul><ul><li>H 2 O, H 1+ and Cl 1- </li></ul><ul><li>Both contribute H 1+ </li></ul><ul><li>But, the [HCl] is so small that it has no effect ! </li></ul><ul><li>Essentially all the H 1+ is coming from water. </li></ul><ul><li>So, pH = 7.00. </li></ul>
38. 39. Z5e 14.5 Calculating the pH of Weak Acids pp <ul><li>Ka will be small. </li></ul><ul><li>ALWAYS WRITE THE MAJOR SPECIES . </li></ul><ul><li>It will be an equilibrium problem from the start (and shifted to the left ). </li></ul><ul><li>Determine whether most of the H + will come from the acid or the water! </li></ul><ul><li>Compare to Ka or Kw per above </li></ul><ul><li>Rest is just like last chapter (see, summary next slide) </li></ul>
39. 40. Z5e p637 Solving Weak Acid Equilibrium Problems pp
40. 41. Example pp <ul><li>Calculate the pH of 2.0 M acetic acid HC 2 H 3 O 2 with a Ka 1.8 x10 -5 Steps. . . </li></ul><ul><li>Major species are . . . </li></ul><ul><li>HC 2 H 3 O 2 & H 2 O. pH answer is . . . (use RICE) </li></ul><ul><li>pH = 2.22 </li></ul><ul><li>Calculate pOH, [OH - ], [H + ] Ans. are . . . </li></ul><ul><li>pOH = 11.78 </li></ul><ul><li>[OH - ] = 1.7 x 10 -12 </li></ul><ul><li>[H + ] = 6.0 x 10 -3 </li></ul>
41. 42. A mixture of Weak Acids pp <ul><li>The process is the same. </li></ul><ul><li>Determine the major species. </li></ul><ul><li>The stronger will predominate. </li></ul><ul><li>So, use bigger Ka if concentrations are comparable ( i.e., can ignore smaller Ka) </li></ul>
42. 43. A mixture of Weak Acids pp <ul><li>Calculate the pH of a mixture of 1.20 M HF (Ka = 7.2 x 10 -4 ) and 3.4 M HOC 6 H 5 (Ka = 1.6 x 10 -10 ) </li></ul><ul><li>Major species: HF, HOC 6 H 5 , H 2 O </li></ul><ul><li>Since HF is the dominant producer of H + , use its Ka </li></ul><ul><li>Set up “RICE” table, use approximations and 5% rule to get . . . </li></ul><ul><li>[H + ] = .030 & pH = 1.53 </li></ul>
43. 44. A mixture of Weak Acids pp <ul><li>Calculate the [OC 6 H 5 1- ] of a mixture of 1.20 M HF (Ka = 7.2 x 10 -4 ) and 3.4 M HOC 6 H 5 (Ka = 1.6 x 10 -10 ) </li></ul><ul><li>Use Ka and “RICE” for HOC 6 H 5 , but remember that [H + ] concentration came from both acids </li></ul><ul><li>So, 1.6 x 10 -10 = ( 0.030 )(x)/(3.4) </li></ul><ul><li>[OC 6 H 5 1- ] = . . . </li></ul><ul><li>1.81 x 10 -8 </li></ul>
44. 45. Percent dissociation pp <ul><li>= amount dissociated x 100% initial concentration </li></ul><ul><li>For a weak acid percent dissociation increases as acid becomes more dilute . </li></ul><ul><li>Calculate the % dissociation of 1.00 M and 0.100 M Acetic acid (Ka = 1.8 x 10 -5 ). Ans. . . </li></ul><ul><li>% diss 1.00 M = 0.42%, 0.00100 M = 1.3% </li></ul><ul><li>As [HA] 0 decreases [H + ] decreases but % dissociation increases . </li></ul><ul><li>Le Chatelier’s principle operates here </li></ul>
45. 46. The other way pp <ul><li>What is the Ka of a weak acid that is 8.1 % dissociated as 0.100 M solution? Ans. . . </li></ul><ul><li>6.6 x 10 -4 Solving method . . . </li></ul><ul><li>Since (X/0.100)(100%) = 8.1%, </li></ul><ul><li>x = 8.1 x 10 -3 = [H + ] = [A - ] </li></ul><ul><li>Then plug into =m to solve for Ka. (8.1 x 10 -3 ) 2 ÷ 0.100 = 6.6 x 10 -4 (remember, divide by initial [ ]) </li></ul>
46. 47. Figure 14.10 p. 643. The Effect of Dilution on the Percent Dissociation and (H+) of a Weak Acid Solution
47. 48. Z5e 14.6 Bases <ul><li>The OH - is a strong base. </li></ul><ul><li>Hydroxides of the alkali metals are strong bases because they dissociate completely when dissolved. </li></ul><ul><li>The hydroxides of alkaline earths Ca(OH) 2 etc. are strong dibasic bases, but they don’t dissolve well in water. </li></ul><ul><li>So, used as antacids because [OH - ] can’t build up with them, which would hurt the stomach lining.. </li></ul>
48. 49. Bases without OH - <ul><li>B-L bases are proton acceptors . </li></ul><ul><li>NH 3 + H 2 O NH 4 + + OH - </li></ul><ul><li>It is the lone pair on nitrogen (N:) that accepts the proton. </li></ul><ul><li>Many weak bases contain N </li></ul><ul><li>B (aq) + H 2 O(l) BH + (aq) + OH - (aq) </li></ul><ul><li>K b = [ BH + ][ OH - ] [ B ] </li></ul>
49. 50. Strength of Bases <ul><li>Hydroxides are strong. </li></ul><ul><li>Others are weak. </li></ul><ul><li>Smaller K b = weaker base. </li></ul><ul><li>Calculate the pH of a solution of 4.0 M pyridine (K b = 1.7 x 10 -9 ). Answer . . . </li></ul>N :
50. 51. Strength of Bases <ul><li>Calculate the pH of a solution of 4.0 M pyridine (K b = 1.7 x 10 -9 ) Did you get pH = 4.08? </li></ul><ul><li>Oops!!! </li></ul><ul><li>You got p OH = 4.08 ( from [OH 1- ] = 1.2 x 10 -10 -- you were given K b ) </li></ul><ul><li>pH = 14 - pOH = 9.92 </li></ul><ul><li>This makes sense since pyridine is a weak base ! (so, high pH) </li></ul>
51. 52. Z5e 14.7 Polyprotic acids pp <ul><li>These always dissociate stepwise . </li></ul><ul><li>The first H + comes off much easier than the second. </li></ul><ul><li>So, the K a for the first step is much bigger than Ka for the second. </li></ul><ul><li>Denoted Ka 1 , Ka 2 , Ka 3 </li></ul><ul><li>Overall K = Ka 1 •Ka 2 •Ka 3 . . . (AP quest) </li></ul><ul><li>For a typical polyprotic acid in H 2 0, only the 1st dissociation step is important in determining the pH </li></ul>
52. 53. Polyprotic acid <ul><li>H 2 CO 3 H + + HCO 3 - Ka 1 = 4.3 x 10 -7 </li></ul><ul><li>HCO 3 - H + + CO 3 -2 Ka 2 = 4.3 x 10 -10 </li></ul><ul><li>The base in first step is acid in second. </li></ul><ul><li>In pH calculations we can normally ignore the second dissociation. </li></ul><ul><li>But, you may have to calculate Ka 1 and Ka 2 in Lab 14, depending on your acid. </li></ul>
53. 54. Calculate the Concentration pp <ul><li>Of all the ions in a solution of 1.00 M Arsenic acid H 3 AsO 4 Ka 1 = 5.0 x 10 -3 Ka 2 = 8.0 x 10 -8 Ka 3 = 6.0 x 10 -10 Steps. . . </li></ul><ul><li>What’s the first step? . . . </li></ul><ul><li>Write the major species, which are . . . </li></ul><ul><li>H 2 O and H 3 AsO 4 </li></ul>
54. 55. Calculate the Concentration pp <ul><li>Of all the ions in a solution of 1.00 M Arsenic acid H 3 AsO 4 </li></ul><ul><li>Write the =m expression for H 3 AsO 4 . . . </li></ul><ul><li>H 3 AsO 4 H 1+ + H 2 AsO 4 1- </li></ul><ul><li>x 2 / [H 3 AsO 4 ] = K a1 = 5.0 x 10 -3 </li></ul><ul><li>x = 7.1 x 10 -2 = [H 1+ ] and [H 2 AsO 4 1- ] </li></ul>
55. 56. Calculate the Concentration pp <ul><li>Of all the ions in a solution of 1.00 M Arsenic acid H 3 AsO 4 </li></ul><ul><li>Now, write =m expression for H 2 AsO 4 1- </li></ul><ul><li>H 2 AsO 4 1- H 1+ + HAsO 4 2- </li></ul><ul><li>We cannot use x 2 / [H 2 AsO 4 1- ] = K a2 </li></ul><ul><li>Why? </li></ul><ul><li>[H 1+ ] ≠ [HAsO 4 2- ] because we have [H 1+ ] of 7.1 x 10 -2 from the first =m. </li></ul>
56. 57. Calculate the Concentration pp <ul><li>Of all the ions in a solution of 1.00 M Arsenic acid H 3 AsO 4 </li></ul><ul><li>H 2 AsO 4 1- H 1+ + HAsO 4 2- </li></ul><ul><li>[H 1+ ][HAsO 4 2- ]/[H 2 AsO 4 1- ] = . . . </li></ul><ul><li>(7.1 x 10 -2 )(x) = k a2 = 8.0 x 10 -8 7.1 x 10 -2 </li></ul><ul><li>So, x = [HAsO 4 2- ] = k a2 </li></ul><ul><li>This is important for titrations . </li></ul>
57. 58. Calculate the Concentration pp <ul><li>Of all the ions in a solution of 1.00 M Arsenic acid H 3 AsO 4 . </li></ul><ul><li>You do it for the last K a3 = 6.0 x 10 -10 </li></ul><ul><li>[AsO 4 3- ] = ??? </li></ul><ul><li>6.8 x 10 -16 from (x)(7.1 x 10 -2 ) = 6.0 x 10 -10 8.0 x 10 -8 </li></ul>
58. 59. Sulfuric acid is special <ul><li>In first step it is a strong acid. </li></ul><ul><li>In the second step, Ka 2 = 1.2 x 10 -2 </li></ul><ul><li>Calculate the [ ]s of the major species in a 3.0 M solution of H 2 SO 4 . Answer . . . </li></ul><ul><li>Both [H + ] and [HSO4 - ] are 3.0 M , since completely dissociated. </li></ul><ul><li>We can neglect the effect of the 2nd step because it is so small (<5%) compared to the 1st step complete disassociation. </li></ul>
59. 60. Sulfuric acid is special <ul><li>In first step it is a strong acid. In the second step, Ka 2 = 1.2 x 10 -2 </li></ul><ul><li>Calculate the concentrations in a 2.0 x 10 -3 M solution of H 2 SO 4 (tricky: why?) </li></ul><ul><li>Since dilute , the 2nd step does contribute H + , so quadratic expression must be used because Ka 2 is not small enough to simplify the expression. </li></ul><ul><li>A pre-test question requires you to use the quadratic for an H 2 SO 4 problem. </li></ul>
60. 61. Z5e 14.8 Salts as acids and bases <ul><li>Salts are ionic compounds. </li></ul><ul><li>Salts of the cation of strong bases and the anion of strong acids are neutral. </li></ul><ul><li>for example, NaCl, KNO 3 </li></ul><ul><li>There is no equilibrium for strong acids and bases. </li></ul><ul><li>So, we can ignore the reverse reaction. </li></ul>
61. 62. Basic Salts <ul><li>If the anion of a salt is the conjugate base of a weak acid  basic solution. </li></ul><ul><li>In an aqueous solution of NaF the major species are Na + , F - , and H 2 O </li></ul><ul><li>F - + H 2 O HF + OH - </li></ul><ul><li>K b =[HF][ OH - ] [F - ] </li></ul><ul><li>but K a = [ H + ][F - ] [HF] </li></ul>
62. 63. Basic Salts <ul><li>K b x K a = [HF][OH - ] x [H + ][F - ] [F - ] [HF] </li></ul>
63. 64. Basic Salts <ul><li>K a x K b = [HF][OH - ] x [H + ][F - ] [F - ] [HF] </li></ul>
64. 65. Basic Salts <ul><li>K a x K b = [HF][OH - ] x [H + ][F - ] [F - ] [HF] </li></ul>
65. 66. Basic Salts <ul><li>K a x K b = [HF][OH - ] x [H + ][F - ] [F - ] [HF] </li></ul><ul><li>K a x K b =[OH - ] [H + ] </li></ul>
66. 67. Basic Salts <ul><li>K a x K b = [HF][OH - ] x [H + ][F - ] [F - ] [HF] </li></ul><ul><li>K a x K b = [OH - ] [H + ] </li></ul><ul><li>Since [OH - ] [H + ] = K W </li></ul><ul><li>K a x K b = K W </li></ul>
67. 68. K a tells us K b pp <ul><li>Calculate the pH of 1.00 M NaCN. K a of HCN is 6.2 x 10 -10 Answer . . . </li></ul><ul><li>Answer: pH = 11.60. Here’s why . . . </li></ul><ul><li>The anion of a weak acid is a weak base. Why? Consider the above: </li></ul><ul><li>2 reactions: (1) dissociation of HCN (2) CN - reacting with H 2 O. </li></ul><ul><li>The CN - ion competes with OH - for H + </li></ul><ul><li>Base strength = OH - > CN - > H 2 O. </li></ul>
68. 69. K a tells us K b pp <ul><li>Calculate the pH of a solution of 1.00 M NaCN. K a of HCN is 6.2 x 10 -10 Ans. pH = 11.60 </li></ul><ul><li>Write major species: Na 1+ , CN 1- and H 2 O </li></ul><ul><li>Set up =m between anion and water CN 1- + HOH HCN + OH 1- </li></ul><ul><li>Since product includes hydroxide ion, must use K b so need to get from K a (using K w ) </li></ul><ul><li>Calculate hydroxide concentration, convert to pOH, subtract from 14 to get pH </li></ul>
69. 70. Acidic salts <ul><li>A salt with the cation of a weak base and the anion of a strong acid will be acidic . </li></ul><ul><li>The same development as bases leads to </li></ul><ul><li>K a x K b = K W </li></ul><ul><li>Other acidic salts are those of highly charged metal ions (e.g., aluminum ion -- 3+). </li></ul><ul><li>More on this later. </li></ul>
70. 71. Acidic salts pp <ul><li>Calculate the pH of a solution of 0.40 M NH 4 Cl (the K b of NH 3 1.8 x 10 -5 ). Steps </li></ul><ul><li>How do we know the correct =m rxn to use? . . . </li></ul><ul><li>Write the major species! NH 4 1+ , Cl 1- , H 2 O </li></ul><ul><li>Cl - is not conj. base of WA, but of SA (no equilibrium), so look at NH 4 + which is the CA of WB. </li></ul><ul><li>Since NH 4 + is a major species and NH 3 isn’t, the =m must be: NH 4 + NH 3 + H + and use K a </li></ul><ul><li>So, don’t use NH 3 + OH 1- NH 4 1+ or K b !! </li></ul><ul><li>Use K a x K b = K w to calculate K a </li></ul><ul><li>Answer for pH is . . . </li></ul><ul><li>[H + ] = 1.5 x 10 -5 and pH = 4.83 </li></ul>
71. 72. Anion of weak acid, cation of weak base <ul><li>K a > K b acidic </li></ul><ul><li>K a < K b basic </li></ul><ul><li>K a = K b Neutral </li></ul><ul><li>See Table 14.6, p. 660 for summary </li></ul>
72. 73. Acid-base Properties of Aqueous Ions
73. 74. Z5e 14.9 Structure & Acid base Properties <ul><li>A molecule with an H in it is a potential acid. </li></ul><ul><li>The stronger the X-H bond the less acidic ; i.e. , more basic (compare bond dissociation energies). </li></ul><ul><li>The more polar the X-H bond the stronger the acid (use electronegativities). </li></ul><ul><li>The more polar H-O-X bond - stronger acid. </li></ul>
74. 75. Strength of oxyacids <ul><li>The more oxygen hooked to the central atom, the more acidic the hydrogen. </li></ul><ul><li>HClO 4 > HClO 3 > HClO 2 > HClO </li></ul><ul><li>Remember that the H is attached to an oxygen atom ( not the central atom). </li></ul><ul><li>The oxygens are electronegative. </li></ul><ul><li>So, they pull electrons away from hydrogen </li></ul>
75. 76. Strength of oxyacids Electron Density Cl O H
76. 77. Strength of oxyacids Electron Density O Cl O H
77. 78. Strength of oxyacids O O Electron Density Cl O H
78. 79. Strength of oxyacids O O O Electron Density Cl O H
79. 80. Hydrated metals <ul><li>Highly charged metal ions pull the electrons of surrounding water molecules toward them. </li></ul><ul><li>Makes it easier for H + to come off. </li></ul>Al +3 O H H
80. 81. Metal Ion Surrounded by Water Molecules
81. 82. Z5e 14.10 Acid-Base Properties of Oxides <ul><li>Non-metal oxides dissolved in water can make acids. </li></ul><ul><li>SO 3 (g) + H 2 O(l) H 2 SO 4 (aq) </li></ul><ul><li>Ionic (e.g., metal) oxides dissolve in water to produce bases. </li></ul><ul><li>CaO(s) + H 2 O(l) Ca(OH) 2 (aq) </li></ul>
82. 83. Z5e 14.11 Lewis Acids and Bases <ul><li>Most general definition. </li></ul><ul><li>Acids are electron pair acceptors. </li></ul><ul><li>Bases are electron pair donors. </li></ul>B F F F : N H H H
83. 84. Lewis Acids and Bases <ul><li>Boron trifluoride wants more electrons. </li></ul>B F F F : N H H H
84. 85. Lewis Acids and Bases <ul><li>Boron trifluoride wants more electrons. </li></ul><ul><li>BF 3 is Lewis base NH 3 is a Lewis Acid. </li></ul>B F F F N H H H
85. 86. B. Definitions <ul><li>Lewis </li></ul><ul><ul><li>Acids are electron pair acceptors. </li></ul></ul><ul><ul><li>Bases are electron pair donors. </li></ul></ul>Lewis base Lewis acid
86. 87. Lewis Acids and Bases Al +3 ( H H O Al ( H H O + 6 +3 ) ) 6
87. 88. Z5e 14.12 Strategy for solving Acid-Base Problems pp <ul><li>What major species are present? </li></ul><ul><li>Does a reaction occur that can be assumed to go to completion? </li></ul><ul><li>What equilibrium dominates the solution? </li></ul><ul><li>See guide on next slide and page 667. </li></ul>
88. 89. Z5e Solving Acid-Base Problems p. 667 pp