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Chapter Fifteen 1 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Acids, Bases, and Acid–Base Equilibria.

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Presentation on theme: "Chapter Fifteen 1 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Acids, Bases, and Acid–Base Equilibria."— Presentation transcript:

1 Chapter Fifteen 1 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Acids, Bases, and Acid–Base Equilibria Chapter Fifteen

2 2 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The Brønsted–Lowry Theory Arrhenius theory: an acid forms H + in water; and a base forms OH – in water. But not all acid–base reactions involve water, and many bases (NH 3, carbonates) do not contain any OH –. Brønsted–Lowry theory defines acids and bases in terms of proton (H + ) transfer. A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry base is a proton acceptor. The conjugate base of an acid is the acid minus the proton it has donated. The conjugate acid of a base is the base plus the accepted proton.

3 Chapter Fifteen 3 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Ionization of HCl HCl acts as an acid by donating H + to H 2 O H 2 O is a base in this reaction because it accepts the H + Conjugate acid of H 2 O Conjugate base of HCl

4 Chapter Fifteen 4 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Ionization of Ammonia

5 Chapter Fifteen 5 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Water Is Amphiprotic H 2 O acts as an acid when it donates H +, forming the conjugate base ___ H 2 O acts as a base when it accepts H +, forming the conjugate acid ___ Amphiprotic: Can act as either an acid or as a base

6 Chapter Fifteen 6 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 15.1 Identify the Brønsted–Lowry acids and bases and their conjugates in: (a)H 2 S + NH 3 NH HS – (b)OH – + H 2 PO 4 – H 2 O + HPO 4 2–

7 Chapter Fifteen 7 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry K a and K b The equilibrium constant for a Brønsted acid is represented by K a, and that for a base is represented by K b. CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO – (aq) [H 3 O + ][CH 3 COO – ] K a = ––––––––––––––––– [CH 3 COOH] [NH 4 + ][OH – ] K b = ––––––––––––– [NH 3 ] NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH – (aq) Notice that H 2 O is not included in either equilibrium expression.

8 Chapter Fifteen 8 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Strength of Conjugate Acid–Base Pairs A stronger acid can donate H + more readily than a weaker acid. The stronger an acid, the weaker is its conjugate base. The stronger a base, the weaker is its conjugate acid. An acid–base reaction is favored in the direction from the stronger member to the weaker member of each conjugate acid–base pair.

9 Chapter Fifteen 9 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The stronger the acid … … the weaker the conjugate base. And the stronger the base … … the weaker the conjugate acid.

10 Chapter Fifteen 10 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Acid/Base Strength and Direction of Equilibrium In Table 15.1, HBr lies above CH 3 COOH in the acid column. Since HBr is a stronger acid than CH 3 COOH, the equilibrium for the reaction: lies to the left. Weaker acid  Stronger acid We reach the same conclusion by comparing the strengths of the bases (right column of Table 15.1). CH 3 COO – lies below Br – ; CH 3 COO – is the stronger base: Weaker base  Stronger base

11 Chapter Fifteen 11 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Strong Acids The “strong” acids—HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 4 —are considered “strong” because they ionize completely in water. The “strong” acids all appear above H 3 O + in Table The strong acids are leveled to the same strength—to that of H 3 O + —when they are placed in water.

12 Chapter Fifteen 12 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Periodic Trends in Acid Strength The greater the tendency for HX (general acid) to transfer a proton to H 2 O, the more the forward reaction is favored and the stronger the acid. A factor that makes it easier for the H + to leave will increase the strength of the acid. Acid strength is inversely proportional to H—X bond-dissociation energy. Weaker H—X bond => stronger acid. Acid strength is directly proportional to anion radius. Larger X radius => stronger acid.

13 Chapter Fifteen 13 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Periodic Trends in Acid Strength

14 Chapter Fifteen 14 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry

15 Chapter Fifteen 15 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry

16 Chapter Fifteen 16 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Strength of Oxoacids Acid strength increases with the electronegativity of the central atom, and with the number of terminal oxygen atoms.

17 Chapter Fifteen 17 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hypochlorous acid HOCl (OH)Cl pK a = 7.5 chlorous acid HClO 2 (OH)ClO pKa= 2.0 chloric acid HClO 3 (OH)ClO 2 pKa= -3 perchloric acid HClO 4 (OH)ClO 3 pKa= - 8

18 Chapter Fifteen 18 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry

19 Chapter Fifteen 19 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Strength of Carboxylic Acids Carboxylic acids all have the –COOH group in common. Differences in acid strength come from differences in the R group attached to the carboxyl group. In general, the more that electronegative atoms appear in the R group, the stronger is the acid.

20 Chapter Fifteen 20 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Aspirin: acetylsalicylic acid Vitamin C: Ascorbic acid

21 Chapter Fifteen 21 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Respiratory System - Ventilation can affect carbon dioxide, and therefore carbonic acid 1. Action of Carbonic Anhydrase CO2 + H2 O H2 CO3 Carbon Dioxide Water Carbonic anhydrase Carbonic acid 2. Effect of Respiration on pH resp up---->"'blows off" CO2 ---->CO2 down ----> H2CO3 down----> pH up resp down----> CO2 accumulates ----> CO2 up----> H2CO3 up----> pH down

22 Chapter Fifteen 22 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 15.2 Select the stronger acid in each pair: (a) nitrous acid, HNO 2, and nitric acid, HNO 3 (b) Cl 3 CCOOH and BrCH 2 COOH

23 Chapter Fifteen 23 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Strengths of Amines as Bases Aromatic amines are much weaker bases than aliphatic amines. This is due in part to the fact that the  electrons in the benzene ring of an aromatic molecule are delocalized and can involve the nitrogen atom’s lone-pair electrons in the resonance hybrid. As a result, the lone-pair electrons are much less likely to accept a proton. Electron-withdrawing groups on the ring further diminish the basicity of aromatic amines relative to aniline.

24 Chapter Fifteen 24 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 15.3 Select the weaker base in each pair:

25 Chapter Fifteen 25 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry

26 Chapter Fifteen 26 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Self-Ionization of Water Even pure water conducts some electricity. This is due to the fact that water self-ionizes: The equilibrium constant for this process is called the ion product of water (K w ). At 25 °C, K w = 1.0 x 10 –14 = [H 3 O + ][OH – ] This equilibrium constant is very important because it applies to all aqueous solutions—acids, bases, salts, and nonelectrolytes—not just to pure water.

27 Chapter Fifteen 27 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The pH Scale Concentration of H 3 O + can vary over a wide range in aqueous solution, from about 10 M to about 10 –14 M. A more convenient expression for H 3 O + is pH. pH = –log [H 3 O + ] and so [H 3 O + ] = 10 –pH The “negative logarithm” function of pH is so useful that it has been applied to other species and constants. pOH = –log [OH – ] and so [OH – ] = 10 –pOH pK w = –log K w At 25 °C, pK w = pK w = pH + pOH = 14.00

28 Chapter Fifteen 28 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The pH Scale Since pH is a logarithmic scale, cola drinks (pH about 2.5) are about ____ times as acidic as tomatoes (pH about 4.5)

29 Chapter Fifteen 29 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 15.4 By the method suggested in Figure 15.5, a student determines the pH of milk of magnesia, a suspension of solid magnesium hydroxide in its saturated aqueous solution, and obtains a value of What is the molarity of Mg(OH) 2 in its saturated aqueous solution? The suspended, undissolved Mg(OH) 2 (s) does not affect the measurement. Example 15.5 A Conceptual Example Is the solution 1.0 x 10 –8 M HCl acidic, basic, or neutral?

30 Chapter Fifteen 30 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Equilibrium in Solutions of Weak Acids and Weak Bases These calculations are similar to the equilibrium calculations performed in Chapter An equation is written for the reversible reaction. 2.Data are organized, often in an ICE format. 3.Changes that occur in establishing equilibrium are assessed. 4.Simplifying assumptions are examined (the “5% rule”). 5.Equilibrium concentrations, equilibrium constant, etc. are calculated.

31 Chapter Fifteen 31 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry

32 Chapter Fifteen 32 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Types of Weak Bases C 2 H 3 O H 2 O = HC 2 H 3 O 2 + OH -

33 Chapter Fifteen 33 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry K w = K a K b K w / K a = K b 1.0 X / 1.8 X = 5.6 X What is the pH of a 1.0 X M NaC 2 H 3 O 2 solution? C 2 H 3 O H 2 O = HC 2 H 3 O 2 + OH - Initial1.0 X * Change-x-x+x+x+x+x Equilibriu m 1.0 X – xXX Since [ C 2 H 3 O 2 - ] initial > 400 X K b, we can neglect x compared to 1.0 X X = x 2 / 1.0 X x = [OH - ] = 2.4 X M=> [H + ] = K w / [OH - ] = 4.2 X pH = - Log 4.2 X = 8.38 (you see, I told you it was a base).

34 Chapter Fifteen 34 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry

35 Chapter Fifteen 35 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 15.6 Ordinary vinegar is approximately 1 M CH 3 COOH and as shown in Figure 15.6, it has a pH of about 2.4. Calculate the expected pH of 1.00 M CH 3 COOH(aq), and show that the calculated and measured pH values are in good agreement.

36 Chapter Fifteen 36 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 15.7 What is the pH of M ClCH 2 COOH(aq)?

37 Chapter Fifteen 37 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example 15.8 What is the pH of M NH 3 (aq)? Example 15.9 The pH of a M aqueous solution of dimethylamine is What are the values of K b and pK b ? The ionization equation is (CH 3 ) 2 NH + H 2 O (CH 3 ) 2 NH OH – K b = ? Dimethylamine Dimethylammonium ion Example A Conceptual Example Without doing detailed calculations, indicate which solution has the greater [H 3 O + ], M HCl or M CH 3 COOH.

38 Chapter Fifteen 38 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Polyprotic Acids A monoprotic acid has one ionizable H atom per molecule. A polyprotic acid has more than one ionizable H atom per molecule. –Sulfuric acid, H 2 SO 4 Diprotic –Carbonic acid, H 2 CO 3 Diprotic –Phosphoric acid, H 3 PO 4 Triprotic The protons of a polyprotic acid dissociate in steps, each step having a value of K a. Values of K a decrease successively for a given polyprotic acid. K a1 > K a2 > K a3, etc. Simplifying assumptions may be made in determining the concentration of various species from polyprotic acids.

39 Chapter Fifteen 39 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example Calculate the following concentrations in an aqueous solution that is 5.0 M H 3 PO 4 : (a) [H 3 O + ] (b) [H 2 PO 4 – ] (c) [HPO 4 2– ] (d) [PO 4 3– ] Example What is the approximate pH of 0.71 M H 2 SO 4 ?

40 Chapter Fifteen 40 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Ions as Acids and Bases HCl is a strong acid, therefore Cl – is so weakly basic in water that a solution of chloride ions (such as NaCl) is virtually neutral. Acetic acid, CH 3 COOH, is a weak acid, so acetate ion, CH 3 COO –, is significantly basic in water. A solution of sodium acetate (which dissociates completely into sodium and acetate ions in water) is therefore slightly basic: CH 3 COO – + H 2 O  CH 3 COOH + OH –

41 Chapter Fifteen 41 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Carbonate Ion as a Base A carbonate ion accepts a proton from water, leaving behind an OH – and making the solution basic.

42 Chapter Fifteen 42 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Ions as Acids and Bases (cont’d) Salts of strong acids and strong bases form neutral solutions: NaCl, KNO 3 Salts of weak acids and strong bases form basic solutions: KNO 2, NaClO Salts of strong acids and weak bases form acidic solutions: NH 4 NO 3 Salts of weak acids and weak bases form solutions that may be acidic, neutral, or basic; it depends on the relative strengths of the cations and the anions: NH 4 NO 2, CH 3 COONH 4.

43 Chapter Fifteen 43 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Aqueous Cations [Fe(H 2 O) 6 ]+ 3 + H 2 O  [Fe(H 2 O) 5 OH]+ 2 + H 3 O +

44 Chapter Fifteen 44 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example A Conceptual Example (a) Is NH 4 I(aq) acidic, basic,or neutral? (b) What conclusion can you draw from Figure 15.8d about the equilibrium constants for the hydrolysis reactions in CH 3 COONH 4 (aq)?

45 Chapter Fifteen 45 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Ions as Acids and Bases (cont’d) In order to make quantitative predictions of pH of a salt solution, we need an equilibrium constant for the hydrolysis reaction. The relationship between K a and K b of a conjugate acid–base pair is: K a x K b = K w If instead we have values of pK a or pK b : pK a + pK b = pK w = (at 25 °C)

46 Chapter Fifteen 46 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example Calculate the pH of a solution that is 0.25 M CH 3 COONa(aq). Example What is the molarity of an NH 4 NO 3 (aq) solution that has a pH = 4.80?

47 Chapter Fifteen 47 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The Common Ion Effect Consider a solution of acetic acid. If we add acetate ion as a second solute (i.e., sodium acetate), the pH of the solution increases: LeChâtelier’s principle: What happens to [H 3 O + ] when the equilibrium shifts to the left?

48 Chapter Fifteen 48 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry The Common Ion Effect (cont’d) The common ion effect is the suppression of the ionization of a weak acid or a weak base by the presence of a common ion from a strong electrolyte. Acetic acid solution at equilibrium: a few H 3 O + ions and a few CH 3 COO – ions When acetate ion is added, and equilibrium reestablished: more acetate ions, fewer H 3 O + ions

49 Chapter Fifteen 49 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry HC 2 H 3 O 2 = H + + C 2 H 3 O 2 - We calculated before that a 1.0 X M solution would have [H + ] = 4.2 X M and pH = 3.37 How does Le Chatelier suggest the above equilibrium would shift if NaC 2 H 3 O 2 was added? How would the pH change? Say we added enough NaC 2 H 3 O 2 to make the solution 1.0 X M in C 2 H 3 O 2 -. What would the pH be Initial1.0 X *1.0 X Change-x-x+x+x+x+x Equilibriu m 1.0 X – xX1.0 X x K a = x (1.0 X x)Assume x is negligible compared to 1.0 X M (1.0 X – x ) x = 1.8 X = [H + ] => pH = 4.74

50 Chapter Fifteen 50 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example Calculate the pH of an aqueous solution that is both 1.00 M CH 3 COOH and 1.00 M CH 3 COONa.

51 Chapter Fifteen 51 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Buffer Solutions Many industrial and biochemical reactions—especially enzyme-catalyzed reactions—are sensitive to pH. To work with such reactions we often need a solution that maintains a nearly-constant pH. A buffer solution is a solution that changes pH only slightly when small amounts of a strong acid or a strong base are added. A buffer contains significant concentrations of both –a weak acid and its conjugate base, or –a weak base and its conjugate acid.

52 Chapter Fifteen 52 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Buffer Solutions (cont’d) The acid component of the buffer neutralizes small added amounts of OH –, forming the weaker conjugate base which does not affect pH much: HA + OH –  H 2 O + A – The base component neutralizes small added amounts of H 3 O +, forming the weaker conjugate acid which does not affect pH much. A – + H 3 O +  H 2 O + HA Pure water does not buffer at all …

53 Chapter Fifteen 53 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Pure water increases in pH by about 5 pH units when the OH – is added, and decreases by about 5 pH units when the H 3 O + is added. In contrast, the same amounts of OH – and H 3 O + added to a buffer solution barely change the pH.

54 Chapter Fifteen 54 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry An Equation for Buffer Solutions In certain applications, there is a need to repeat the calculations of the pH of buffer solutions many times. This can be done with a single, simple equation, but there are some limitations. The Henderson–Hasselbalch equation: To use this equation, the ratio [conjugate base]/[weak acid] must have a value between 0.10–10 and both concentrations must exceed K a by a factor of 100 or more. [conjugate base] pH = pK a + log –––––––––––––– [weak acid]

55 Chapter Fifteen 55 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Comparing three weak acids HF K a = 6.8 X pKa = 3.17 HC 2 H 3 O x HCN 4.9 X As you can see, the stronger the acid, the smaller pKa. You will have a similar trend with pK b.

56 Chapter Fifteen 56 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Buffer Capacity and Buffer Range There is a limit to the ability of a buffer solution to neutralize added acid or base. This buffer capacity is reached before either buffer component has been consumed. In general, the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize. As a rule, a buffer is most effective if the concentrations of the buffer acid and its conjugate base are equal or nearly so. Therefore, a buffer is most effective when the desired pH of the buffer is very near pK a of the weak acid of the buffer.

57 Chapter Fifteen 57 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Human blood is a buffered solution Source: Visuals Unlimited

58 Chapter Fifteen 58 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Molecular model: HC 2 H 3 O 2, C 2 H 3 O 2 -

59 Chapter Fifteen 59 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example A buffer solution is 0.24 M NH 3 and 0.20 M NH 4 Cl. (a) What is the pH of this buffer? (b) If mol NaOH is added to L of this solution, what will be the pH? Example What concentration of acetate ion in M CH 3 COOH(aq) produces a buffer solution with pH = 5.00?

60 Chapter Fifteen 60 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Acid–Base Indicators An acid–base indicator is a weak acid or base. The acid form (HA) of the indicator has one color, the conjugate base (A – ) has a different color. One of the “colors” may be colorless. In an acidic solution, [H 3 O + ] is high. Because H 3 O + is a common ion, it suppresses the ionization of the indicator acid, and we see the color of HA. In a basic solution, [OH – ] is high, and it reacts with HA, forming the color of A –. Acid–base indicators are often used for applications in which a precise pH reading isn’t necessary.

61 Chapter Fifteen 61 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Different indicators have different values of K a, so they exhibit color changes at different values of pH …

62 Chapter Fifteen 62 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Figure 8.10: The pH curve for the titration of 50.0

63 Chapter Fifteen 63 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Table 8.4: A Summary of Various Points in the Titration of a Triprotic Acid

64 Chapter Fifteen 64 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Figure 8.11: A summary of the important equilibria at various points in the titration of a triprotic acid

65 Chapter Fifteen 65 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example A Conceptual Example Explain the series of color changes of thymol blue indicator produced by the actions pictured in Figure 15.14: (a) A few drops of thymol blue are added to HCl(aq). (b) Some sodium acetate is added to solution (a). (c) A small quantity of sodium hydroxide is added to solution (b). (d) An additional, larger quantity of sodium hydroxide is added to solution (c).

66 Chapter Fifteen 66 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Neutralization Reactions At the equivalence point in an acid–base titration, the acid and base have been brought together in precise stoichiometric proportions. The endpoint is the point in the titration at which the indicator changes color. Ideally, the indicator is selected so that the endpoint and the equivalence point are very close together. The endpoint and the equivalence point for a neutralization titration can be best matched by plotting a titration curve, a graph of pH versus volume of titrant.

67 Chapter Fifteen 67 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Titration Curve, Strong Acid with Strong Base Bromphenol blue, bromthymol blue, and phenolphthalein all change color at very nearly 20.0 mL At about what volume would we see a color change if we used methyl violet as the indicator?

68 Chapter Fifteen 68 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example Calculate the pH at the following points in the titration of mL of M HCl with M NaOH: H 3 O + + Cl – + Na + + OH –  Na + + Cl – + 2 H 2 O (a) before the addition of any NaOH (b) after the addition of mL of M NaOH (c) after the addition of mL of M NaOH (d) after the addition of mL of M NaOH

69 Chapter Fifteen 69 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Titration Curve, Weak Acid with Strong Base The equivalence-point pH is NOT 7.00 here. Why not?? Bromphenol blue was ok for the strong acid/strong base titration, but it changes color far too early to be useful here.

70 Chapter Fifteen 70 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Figure 8.4: The pH curves for the titrations of 50.0

71 Chapter Fifteen 71 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Figure 8.6: The indicator phenolphthalein is pink in basic solution and colorless in acidic solution.

72 Chapter Fifteen 72 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example Calculate the pH at the following points in the titration of mL of M CH 3 COOH with M NaOH: CH 3 COOH + Na + + OH –  Na + + CH 3 COO – + H 2 O (a) before the addition of any NaOH (b) after the addition of 8.00 mL of M NaOH (c) after the addition of mL of M NaOH (d) after the addition of mL of M NaOH (e) after the addition of mL of M NaOH

73 Chapter Fifteen 73 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Example A Conceptual Example This titration curve shown in Figure involves 1.0 M solutions of an acid and a base. Identify the type of titration it represents.

74 Chapter Fifteen 74 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Lewis Acids and Bases In organic chemistry, Lewis acids are often called electrophiles (“electron-loving”) and Lewis bases are often called nucleophiles (“nucleus-loving”). There are reactions in nonaqueous solvents, in the gaseous state, and even in the solid state that can be considered acid–base reactions which Brønsted–Lowry theory is not adequate to explain. A Lewis acid is a species that is an electron-pair acceptor and a Lewis base is a species that is an electron-pair donor. Sulfur accepts an electron pair from the oxygen of CaO CaO(s) + SO 2 (g)  CaSO 3 (s)

75 Chapter Fifteen 75 Hall © 2005 Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Cumulative Example A M aqueous solution of cyanoacetic acid, CNCH 2 COOH, has a freezing point of –0.11 °C. Calculate the freezing point of a M aqueous solution of cyanoacetic acid.


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