1. Chemistry 104 Chapter Nine Part 1 Acids, Bases, and Salts And other cooking secrets of the Culhane family

2. Why categorize & group substances? –M–Makes it easier to study: Similarities & differences Structures Reactive properties Make predictions Search for new chemicals w/ specific properties Categorizing substances: chemical families

3. Properties of acids and bases Acids have: –s–sour taste –c–corrosive to metals and other materials –g–good electrolytes

4. Bases have: – bitter taste – corrosive to many materials – slippery – emulsify fats – good electrolytes

5. Opener

6. Introduction to Acids and Bases 1884 definition by Arrhenius: An acid contains a H atom and dissolves in water to form a hydrogen ion, H + A base contains hydroxide and dissolves in water to form − OH. HCl(g) acid H + (aq) + Cl − (aq) NaOH(s) base Na + (aq) + − OH(aq)

7. Arrhenius Acid-Base Theory Arrhenius Acids: hydrogen-containing compound that, in water, produces H+ ions Arrhenius Base: hydroxide-containing compound that, in water, produces OH- ions Examples: –H–H–H–HNO3 (l) → H+ (aq) + NO3-(aq) –H–H–H–HCl (g) → H+ (aq) + Cl-(aq) –N–N–N–NaOH (s) → Na+ (aq) + OH-(aq) –K–K–K–KOH (s) → K+ (aq) + OH-(aq)

8. Arrhenius definition correctly predicts the behavior of many acids and bases. Definition is limited and sometimes inaccurate. e.g. H + does not exist in water. H + reacts with water forming a hydronium ion, H 3 O +. H 3 O + (aq)H + (aq) + H 2 O(l) hydrogen ion: does not really exist in solution hydronium ion: actually present in aqueous solution

9. Limitations of the Arrhenius theory Hydrochloric acid is neutralized by both sodium hydroxide solution or an ammonia solution. Forms a white salt: sodium chloride or ammonium chloride. Very similar reactions. The full equations are: HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) HCl(aq) + NH 3 (aq)  NH 4 Cl(aq) + H 2 O(l) Acid + Base  Salt

10. Limitations of the Arrhenius theory In NaOH case: H + ions from HCl acid react with OH - ions from the NaOH In ammonia’s (NH 3 ) case: where are the OH - ions?

11. Limitations of the Arrhenius theory Where are the OH - ions? You can say ammonia reacts with the water it’s dissolved in producing NH 4 + ions and OH - ions: HCl(aq) + NH 3 (aq)  NH 4 Cl(aq) + H 2 O(l)

12. But, when colorless gases HCl acid & NH 3 base mix: a white cloud of NH 4 Cl salt is produced There aren't any H + or OH - ions in solution – because there isn't any solution! Arrhenius’ theory says this is NOT an acid-base reaction, but the same product forms as when the 2 substances were in solution.

13. 1923 Bronsted and Lowry’s definition donation or acceptance of a hydrogen ion (H + = a proton) Includes more substances as acids & bases than just H + ion / OH - ion formation definition (especially non-hydroxide bases) An acid is: – A proton (H + ion) donor (gives them up) A base is: – A proton (H + ion) acceptor (grabber)

14. H+ ions in an aqueous solution react with water to form hydronium ions (H3O+) HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq) H 2 O is proton acceptor (base) HCl is the proton donor (acid) NH3(g) + HCl(g) → NH4+(aq) + Cl-(aq) proton acceptor: NH 3 proton donor: HCl Bronsted-Lowry Acid-Base Theory

15. Brønsted–Lowry definition is more widely used: A Brønsted–Lowry acid is a proton (H + ) donor. A Brønsted–Lowry base is a proton (H + ) acceptor. H 3 O + (aq) + Cl − (aq)HCl(g) + H 2 O(l) HCl is a Brønsted–Lowry acid because H 2 O is a Brønsted–Lowry base because

16. Bronsted and Lowry acid must contain a hydrogen atom. Common B-L acids: HClH 2 SO 4 Hydrochloric acid Sulfuric acid HBrHNO 3 Hydrobromic acidNitric acid

17. B-L Acids

18. A monoprotic acid contains one acidic proton. HCl A diprotic acid contains two acidic protons. H 2 SO 4 A triprotic acid contains three acidic protons. H 3 PO 4 A Brønsted–Lowry acid may be neutral or it may carry a net positive or negative charge. HCl, H 3 O +, HSO 4 − Mono-, Di-, Triprotic Acids

19. Classify the following acids: HCl, HF, HNO 3 H 2 CO 3, H 2 SO 4 H 3 PO 4 CH 3 CH 2 COOH Some acids may appear to have numerous H + ions to donate, but be careful!

20. Mono-, Di-, Triprotic Acids Acetic acid: HC 2 H 3 O 2 has only one H to donate The other H atoms are strongly bonded to Carbon atoms (slightly polar bond) Acidic H atom is bonded to Oxygen (strongly polar) Water can act on that polar bond & release H + ion

21. Acetic acid, CH 3 COOH, contains several H atoms The H atom of the OH group is the acidic proton that is donated. H O | // H - C C | \ H O-H

23. Brønsted–Lowry Bases A Brønsted–Lowry base is a proton acceptor, so it must be able to form a bond to a proton. A base must contain a lone pair of electrons that can be used to form a new bond to the proton. N H H H + H 2 O(l) N H H H H + + OH - (aq) Brønsted–Lowry base

24. Brønsted–Lowry Bases( B: ) is a proton acceptor Must contain a lone pair of electrons that can be donated to form a new bond. Hydroxide (OH - ) which contains an oxygen atom with 3 lone pairs of electrons, is most common Brønsted–Lowry Base.

25. Brønsted–Lowry Bases Common Brønsted–Lowry Bases ( B) NaOH KOH Mg(OH) 2 Ca(OH) 2 H 2 O water NH 3 Lone pairs make these neutral compounds bases: The OH - is the base in each metal salt:

26. bases in consumer Products OH -

27. Proton Transfer The Reaction of a Brønsted–Lowry Acid with a Brønsted–Lowry Base HA+ B A − HB+B+ + gain of H + acidbase loss of H +

28. Conjugate (coupled) acid-base pairs: Chemical formulas differ by one H + ion Cation products are usually the conjugate acid Anion products are usually the conjugate base

29. Proton Transfer The Reaction of a Brønsted–Lowry Acid with a Brønsted–Lowry Base The product formed by loss of a proton from an acid is called its conjugate base. HA+ B A − HB+B+ + gain of H + acidbaseconjugate base conjugate acid loss of H + The product formed by gain of a proton by a base is called its conjugate acid.

30. Proton Transfer The Reaction of a Brønsted–Lowry Acid with a Brønsted–Lowry Base HBr+ + gain of H + acidbaseconjugate base conjugate acid loss of H + H2OH2O Br − H3O+H3O+ HBr and Br − are a conjugate acid–base pair. H 2 O and H 3 O + are a conjugate acid–base pair. The net charge must be the same on both sides of the equation.

31. Proton Transfer The Reaction of a Brønsted–Lowry Acid with a Brønsted–Lowry Base Amphoteric compound: HOH H 2 O as a base add H + HOH H + conjugate acid HOH H 2 O as an acid remove H + HO − conjugate base

32. Some chemicals can act as an acid or a base – depends on the chemicals they’re reacting with Water is a good example: Nitrous acid HNO 2 (aq) + H 2 O(l)  H 3 O + (aq) + NO 2 - (aq) acid base Ammonia NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) base acid Amphoteric substances

33. Ionization involves taking covalently bonded molecule & splitting it in a solution to form ions (usually H 2 O is the solvent): HCl(s)  H + (aq) + Cl - (aq) Dissociation (think: opposite of “associate”) breaking ionic compounds down into ions when dissolved in a solution (usually H 2 O is the solvent): LiOH (s)  Li + (aq) + OH - (aq)

34. The difference between the aqueous solution processes of ionization and dissociation.

35. 9.3 Acid and Base Strength Relating Acid and Base Strength When a covalent acid dissolves in water, the H + transfer that forms H 3 O + is called dissociation. When a strong acid dissolves in water, 100% of the acid dissociates into ions. Common strong acids are HI, HBr, HCl, H 2 SO 4, and HNO 3. A single reaction arrow is used, because the product is greatly favored at equilibrium. H 3 O + (aq) + Cl − (aq)HCl(g) + H 2 O(l)

36. When a weak acid dissolves in water, only a small fraction of the acid dissociates into ions. Unequal reaction arrows are used, because the reactants are usually favored at equilibrium. H 3 O + (aq) + CH 3 COO − (aq)CH 3 COOH(l) + H 2 O(l) Common weak acids are H 3 PO 4, HF, H 2 CO 3, and HCN.

37. Figure 9.3

38. Table 9.1

39. Acid and Base Strength Relating Acid and Base Strength When a strong base dissolves in water, 100% of the base dissociates into ions. Na + (aq) + − OH(aq)NaOH(s) + H 2 O(l) Common strong bases are NaOH and KOH. When a weak base dissolves in water, only a small fraction of the base dissociates into ions. NH 4 + (aq) + − OH(aq)NH 3 (g) + H 2 O(l)

40. Acid and Base Strength Relating Acid and Base Strength A strong acid, HCl, is completely dissociated into H 3 O + (aq) and Cl − (aq). A weak acid contains mostly undissociated acid, CH 3 COOH.

41. Figure 9.5 OH -

42. Acid and Base Strength Relating Acid and Base Strength A strong acid readily donates a proton, forming a weak conjugate base. HCl strong acid Cl − weak conjugate base A strong base readily accepts a proton, forming a weak conjugate acid. OH - strong base H 2 O weak conjugate acid

43. Acid and Base Strength Using Acid Strength to Predict the Direction of Equilibrium The position of the equilibrium depends upon the strengths of the acids and bases. The stronger acid reacts with the stronger base to form the weaker acid and the weaker base. A Brønsted–Lowry acid–base reaction represents an equilibrium. HA + B A − HB+B+ + acidbaseconjugate base conjugate acid

44. Acid and Base Strength Predicting the Direction of Equilibrium When stronger acid + base are reactants (left side), the reaction readily occurs and the reaction proceeds to the right. HA+ B A − HB+B+ + stronger acid stronger base weaker base weaker acid A larger forward arrow means products are favored.

45. Acid and Base Strength Predicting the Direction of Equilibrium If an acid–base reaction forms stronger acid and base, equilibrium favors reactants & very little product forms. HA+ B A − HB+B+ + weaker acid weaker base stronger base stronger acid A larger reverse arrow means reactants are favored.

46. Acid and Base Strength HOW TO Predict the Direction of Equilibrium in an Acid–Base Reaction Example Are the reactants or products favored in the following acid–base reaction? Step [1] Identify the acid & base in reactants and the conjugate base & acid in the products. + + HCN(g) − OH(aq) − CN(aq) H 2 O(l)

47. Acid and Base Strength HOW TO Predict the Direction of Equilibrium in an Acid–Base Reaction Step [2] Determine relative strength of the acid and the conjugate acid. From Table 9.1, HCN is a stronger acid than H 2 O. Step [3] Equilibrium favors formation of the weaker acid. + + HCN(g)OH - (aq) CN - (aq) H 2 O(l) stronger acid weaker acid

48. Equilibrium & Acid Dissociation Constants For the reaction where an acid (HA) dissolves in water, H 3 O + (aq) + (aq)HA(g) + H 2 O(l)A − the following equilibrium constant can be written: K =

49. Equilibrium and Acid Dissociation Constants Multiplying both sides by [H 2 O] forms a new constant, called the acid dissociation constant, K a. K a = K[H 2 O] = [H 3 O + ][ ] A − [HA] acid dissociation constant

50. Equilibrium and Acid Dissociation Constants K a = [H 3 O + ][ ] A − [HA] The stronger the acid, the larger the value of K a. Equilibrium favors formation of the weaker acid: the acid with the smaller value of K a. What does a small K a mean in terms of reactants & products?

51. Using Table 9.2 to find the K a for each acid, then determine which is a stronger acid: HCN or HSO 4 - ? HCNKa = 4.9 x 10 -10 HSO 4 - Ka = 1.2 x 10 -2 HSO 4 - is the stronger acid

52. Using Table 9.2 to find the K a for each acid, then determine which is a stronger acid: CH 3 COOH or NH 4 + ? CH 3 COOHKa = 1.8 x 10 -5 NH 4 + Ka = 5.6 x 10 -10 CH 3 COOH is the stronger acid

53. If vitamin C (ascorbic acid) has a K a of 7.9 x 10 -5, are reactants or products favored in this neutralization reaction: C 6 H 8 O 6(aq) + NH 3(aq) C 6 H 7 O 6(aq) + NH 4 + (aq) Vitamin C Ammonia Conjugate Conjugate acid base base acid K a of NH 4 + = 5.6 x 10 -10 (smaller K a = weaker acid) Products are favored

54. Chemistry 104 Chapter Nine Part 2 Acids, Bases, and Salts And other party favors for small children

55. Dissociation of Water H OH base HOH H + conjugate acid HOH conjugate base HO − ++ loss of H + gain of H + Water is Amphoteric: can act as a Brønsted–Lowry acid or a base. Two water molecules can react together in an acid–base reaction:

56. Dissociation of Water [H 3 O + ][ − OH] [H 2 O] 2 From rxtn of 2 water molecules: 2 H 2 O  OH - + H 3 O + The following equilibrium constant expression can be written: K =

57. Dissociation of Water [H 3 O + ][ − OH] [H 2 O] 2 Multiplying both sides by [H 2 O] 2 yields K w, the ion-product constant for water. K w = [H 3 O + ][ − OH] K =

58. Dissociation of Water Experimentally it can be shown that [H 3 O + ] = [ − OH] = 1.0 x 10 −7 M at 25 o C K w = [H 3 O + ] [ − OH] K w = (1.0 x 10 −7 ) x (1.0 x 10 −7 ) K w = 1.0 x 10 −14 K w is constant: for all aqueous solutions at 25 o C.

59. Dissociation of Water To calculate [OH - ] when [H 3 O + ] is known: To calculate [H 3 O + ] when [OH - ] is known: K w = [H 3 O + ][OH - ] [ − OH] = 1 x 10 −14 [H 3 O + ] [ − OH] = KwKw [H 3 O + ] [OH - ] = 1 x 10 −14 [H 3 O + ] [OH - ] = KwKw [H 3 O + ]

60. Dissociation of Water If the [H 3 O + ] in a cup of coffee is 1.0 x 10 −5 M, then the [ − OH] can be calculated as follows : [ − OH] = KwKw [H 3 O + ] == In this cup of coffee, [H 3 O + ] > [ – OH], and the solution is acidic.

61. Dissociation of Water

62. Calculate the value of [H 3 O + ] and [OH - ] in a 0.01 M NaOH solution. NaOH is a strong base (completely dissociates into Na + and OH - ions), So conc of NaOH will = conc of OH - ions Use the [OH - ] to calculate [H 3 O + ] from K w. [H 3 O + ] = K w = 1.0 x 10 -14 [OH - ] 1.0 x 10 -2 [H 3 O + ] = 1.0 x 10 -12 M

63. 9.6A The pH Scale: Calculating pH pH = −log [H 3 O + ] The lower the pH, the higher the concentration of H 3 O + (acid). Acidic solution: pH 1 x 10 −7 Basic solution: pH > 7  [H 3 O + ] < 1 x 10 −7 Neutral solution: pH = 7  [H 3 O + ] = 1 x 10 −7 Why?

64. 9.6A The pH Scale: Calculating pH pH = −log [H 3 O + ] A logarithm is a power of ten: log 1 = 0 because 10 0 = 1 log 10 = 1 because 10 1 = 10 log 100 = 2 because 10 2 = 100, etc A negative logarithm is a negative power of ten: -log 0.1 = 1 because 10 -1 = 0.1 -log 0.01 = 2 because 10 -2 = 0.01 -log 0.001 = 3 because 10 -3 = 0.001, etc The lower the pH, the higher the concentration of H 3 O +. Why?

65. 9.6A The pH Scale: Calculating pH pH = −log [H 3 O + ] The lower the pH, the higher the concentration of H 3 O +. Acidic solution: e.g. pH 5  [H 3 O + ] = 1 x 10 −5 M pH 5 = 0.00001 Moles H 3 O + / Liter Basic solution: e.g. pH 9  [H 3 O + ] = 1 x 10 −9 M pH 9 = 0.000000001 Moles H 3 O + / Liter Neutral solution: e.g. pH 7  [H 3 O + ] = 1 x 10 −7 M pH 7 = 0.0000001 Moles H 3 O + / Liter

66. Everyday substances and their pH

67. Indicators are chemicals that change color in the presence of an acid or a base. Purple cabbage is an indicator. A pH probe may also be used. It measures electrical conductivity differences. Remember: Acids and bases are good electrolytes.

68. The pH Scale Calculating pH from [H 3 O + ] If [H 3 O + ] = 1.2 x 10 –5 M for a solution, what is its pH? pH = –log [H 3 O + ] = –log (1.2 x 10 –5 ) = –(–4.92) = 4.92 Check the answer: It should be between The solution is acidic because the pH < 7.

69. The pH Scale Calculating [H 3 O + ] from pH If the pH of a solution is 8.50, what is the [H 3 O + ]? pH = −log [H 3 O + ] 8.50 = −log [H 3 O + ] −8.50 = log [H 3 O + ] antilog (−8.50 ) = [H 3 O + ] [H 3 O + ] = 3.2 x 10 −9 M

70. The pH Scale Calculating pH A logarithm has same # of digits to the right of the decimal point as are contained in the coefficient of the original #. [H 3 O + ] = 3.2 x 10 −9 M two significant figures pH = 8.50 two digits after decimal point pH = 8.50

71. Overview

72. Focus on the Human Body The pH of Body Fluids

73. Common Acid–Base Reactions Reaction of Acids with Hydroxide Bases Neutralization reaction: An acid-base reaction that produces a salt and water as products. HA(aq) + MOH(aq) acid base OH(l) + MA(aq) H watersalt The acid HA donates a proton (H + ) to the − OH base to form The anion A − from the acid combines with the cation M + from the base to form

74. Common Acid–Base Reactions HOW TO Draw a Balanced Equation for a Neutralization Reaction Between HA and MOH Example Write a balanced equation for the reaction of Mg(OH) 2 with HCl. Step [1] Identify the acid and base in the reactants and draw H 2 O as one product. HCl(aq) + Mg(OH) 2 (aq) acid base H 2 O(l) + water salt

75. Common Acid–Base Reactions HOW TO Draw a Balanced Equation for a Neutralization Reaction between HA and MOH Step [2] Determine the structure of the salt. The salt is formed from the parts of the acid and base that are not used to form H 2 O. HCl H + reacts to form H 2 O Cl − used to form salt Mg(OH) 2 Mg 2+ used to form salt 2 − OH react to form water Mg 2+ and Cl − combine to form MgCl 2.

76. Common Acid–Base Reactions HOW TO Draw a Balanced Equation for a Neutralization Reaction between HA and MOH Step [3] Balance the equation. HCl(aq) + Mg(OH) 2 (aq) acid base H 2 O(l) + water salt MgCl 2 2 Place a 2 to balance Cl. Place a 2 to balance O and H. 2

77. Common Acid–Base Reactions Reaction of Acids with Hydroxide Bases A net ionic equation contains only the species involved in a reaction. HCl(aq) + NaOH(aq) H—OH(l) + NaCl(aq) Written as individual ions: H + (aq) + Cl − (aq) + Na + (aq) + − OH(aq) H—OH(l) + Na + (aq) + Cl − (aq) Omit the spectator ions, Na + and Cl –. H + (aq) + − OH(aq)H—OH(l) What remains is the net ionic equation:

78. Common Acid–Base Reactions Reaction of Acids with Bicarbonate Bases A bicarbonate base, HCO 3 −, reacts with one H + to form carbonic acid, H 2 CO 3. H + (aq) + HCO 3 − (aq) Carbonic acid then decomposes into H 2 O and CO 2. H 2 O(l) + CO 2 (g) H 2 CO 3 (aq) HCl(aq) + NaHCO 3 (aq) H 2 O(l) + CO 2 (g) NaCl(aq) + H 2 CO 3 (aq) For example:

79. Common Acid–Base Reactions Reaction of Acids with Bicarbonate Bases A carbonate base, CO 3 2–, reacts with two H + to form carbonic acid, H 2 CO 3. 2 H + (aq) + CO 3 2– (aq) H 2 O(l) + CO 2 (g) H 2 CO 3 (aq) 2 HCl(aq) + Na 2 CO 3 (aq) H 2 O(l) + CO 2 (g) 2 NaCl(aq) + H 2 CO 3 (aq) For example:

80. The Acidity and Basicity of Salt Solutions A salt can form an acidic, basic, or neutral solution depending on whether its cation and anion are derived from a strong or weak acid and base. For the salt M + A − : The cation M + comes from the base. The anion A − comes from the acid HA. NaCl Na + from NaOH strong base Cl − from HCl strong acid A salt derived from a strong acid and strong base forms a neutral solution (pH = 7).

81. The Acidity and Basicity of Salt Solutions NaHCO 3 Na + from NaOH strong base HCO 3 − from H 2 CO 3 weak acid A salt derived from a strong base and a weak acid forms a basic solution (pH > 7). NH 4 Cl NH 4 + from NH 3 weak base Cl − from HCl strong acid A salt derived from a weak base and a strong acid forms an acidic solution (pH < 7).

82. The Acidity and Basicity of Salt Solutions The ion derived from the stronger acid or base determines whether the solution is acidic or basic.

83. Sea Water pH 8.5

84. To determine concentration of an acid or base in a solution, we carry out a titration. Want to know: concentration of an acid solution, a base of known concentration is added slowly until the acid is neutralized. When the acid is neutralized: # of moles of acid = # of moles of base This is called the end point of the titration. Titration

85. Use an indicator that turns color when the end point is reached

86. Titration Determining an unknown molarity from titration data requires three operations: Moles of base Moles of base Molarity of acid solution Molarity of acid solution mole–mole conversion factor mole–mole conversion factor M (mol/L) conversion factor M (mol/L) conversion factor Moles of acid Moles of acid Volume of base solution Volume of base solution M (mol/L) conversion factor M (mol/L) conversion factor [1] [2] [3]

87. Titration HOW TO Determine the Molarity of an Acid Solution from Titration Example What is the molarity of an HCl solution if 22.5 mL of a 0.100 M NaOH solution are needed to titrate a 25.0 mL sample of the acid? volume of base (NaOH) 22.5 mL conc. of base (NaOH) 0.100 M volume of acid (HCl) 25.0 mL conc. of acid (HCl) ?

88. Titration Step [1] Determine the number of moles of base used to neutralize the acid. Volume of base solution Volume of base solution M (mol/L) conversion factor M (mol/L) conversion factor mL–L conversion factor mL–L conversion factor 22.5 mL NaOH x 1 L 1000 mL x 0.100 mol NaOH 1 L = 0.00225 mol NaOH HOW TO Determine the Molarity of an Acid Solution from Titration

89. Titration HOW TO Determine the Molarity of an Acid Solution from Titration Step [2] Determine the number of moles of acid that react from the balanced chemical equation. HCl(aq) + NaOH(aq) H 2 O(l) + NaCl(aq) 0.00225 mol NaOH x mole–mole conversion factor mole–mole conversion factor 1 mol HCl 1 mol NaOH = 0.00225 mol HCl

90. Titration HOW TO Determine the Molarity of an Acid Solution from Titration Step [3] Determine the molarity of the acid from the number of moles and the known volume. M= mol L = 0.00225 mol HCl 25.0 mL solution x 1000 mL 1 L =0.0900 M HCl Answer mL–L conversion factor mL–L conversion factor

91. Sample Problem 9.18

92. Page 287_01

93. Buffers A buffer is a solution whose pH changes very little when acid or base is added. Most buffers are solutions composed of roughly equal amounts of a weak acid the salt of its conjugate base The buffer resists change in pH because added base, − OH, reacts with the weak acid added acid, H 3 O +, reacts with the conjugate base

95. Buffers General Characteristics of a Buffer CH 3 COOH(aq) + H 2 O(l)H 3 O + (aq) + CH 3 COO − (aq) weak acid conjugate base If an acid is added to the following buffer equilibrium, Adding more product… …drives the reaction to the left. then the excess acid reacts with the conjugate base, so the overall pH does not change much.

96. Buffers General Characteristics of a Buffer CH 3 COOH(aq) + − OH(aq) H 2 O(l) + CH 3 COO − (aq) weak acid conjugate base If a base is added to the following buffer equilibrium, Adding more reactant… …drives the reaction to the right. then the excess base reacts with the weak acid, so the overall pH does not change much.

97. Buffers Common Buffers

98. Buffers Calculating the pH of a Buffer The effective pH range of a buffer depends on its K a. [H 3 O + ][ ] A − [HA] K a = Rearranging this expression to solve for [H 3 O + ]: H 3 O + (aq) + (aq)HA(aq) + H 2 O(l)A −

99. Figure 9.11

100. Focus on the Human Body Buffers in the Blood Normal blood pH is between 7.35 and 7.45. The principle buffer in the blood is carbonic acid/ bicarbonate (H 2 CO 3 /HCO 3 − ). CO 2 (g) + H 2 O(l) H 2 CO 3 (aq) H2OH2O H 3 O + (aq) + HCO 3 − (aq) CO 2 is constantly produced by metabolic processes in the body. The amount of CO 2 is related to the pH of the blood.

101. Focus on the Human Body Buffers in the Blood Respiratory acidosis results when the body fails to eliminate enough CO 2, due to lung disease or failure. CO 2 (g) + 2 H 2 O(g) H 3 O + (aq) + HCO 3 − (aq) A lower respiratory rate increases [CO 2 ]. This drives the reaction to the right, increasing [H 3 O + ]. Blood then has a higher [H 3 O + ] and a lower pH.

103. Focus on the Human Body Buffers in the Blood Respiratory alkalosis is caused by hyperventilating; very little CO 2 is produced by the body. CO 2 (g) + 2 H 2 O(g) H 3 O + (aq) + HCO 3 − (aq) A higher respiratory rate decreases [CO 2 ]. This drives the reaction to the left, decreasing [H 3 O + ]. Blood then has a lower [H 3 O + ] and a higher pH.