2. Section 14.1 Nature of Acids and Bases

3. Arrhenius Definition n Acids produce hydrogen ions in aqueous solution. –HCl (aq)  H + (aq) + Cl - (aq) n Bases produce hydroxide ions when dissolved in water. –NaOH (aq)  Na + (aq) + OH - (aq) n Limits what can be considered bases (as we’ll see with other definitions).

4. Lewis Definition n An acid is an electron pair acceptor. n A base is an electron pair donor. n Easy to see if Lewis structures are drawn: e - pair acceptor e - pair donor

5. Bronsted-Lowry Definition n This is the one we’ll focus on! n An acid is a proton (H + ) donor and a base is a proton acceptor. n HCl is an acid. n When it dissolves in water it gives its proton to water. n HCl(g) + H 2 O( l ) H 3 O + + Cl - n Water is a base since it accepts the H +. –In the Arrhenius definition water would not be considered a base!

6. Conjugate Acid-Base Pairs n General equation n HA(aq) + H 2 O( l ) H 3 O + (aq) + A - (aq) acid + base conjugate acid + conjugate acid + base conjugate acid + conjugate base base n This is an equilibrium. n Equilibrium favors the side with the weaker acid and base. n Refer to the handout for acid and base strength. n Note: conjugate bases of strong acids are weak, and conjugate bases of weak acids are strong. In other words, the stronger the acid/base, the weaker the conjugate base/acid and vice-versa.

7. Acid dissociation constant K a n Recall strong vs. weak acids/bases! –Strong = essentially completely dissociate in water (no equilibrium). –Weak = partially dissociate in water. Equilibrium is therefore present! n For a weak acid: HA(aq) + H 2 O( l ) H 3 O + (aq) + A - (aq) HA(aq) + H 2 O( l ) H 3 O + (aq) + A - (aq) n Water is often left out: HA (aq) H + (aq) + A - (aq) HA (aq) H + (aq) + A - (aq)

8. Acid dissociation constant K a n Since weak acids result in equilibrium, an equilibrium expression can be written: HA(aq) + H 2 O( l ) H 3 O + (aq) + A - (aq) HA(aq) + H 2 O( l ) H 3 O + (aq) + A - (aq) n K a = [H 3 O + ][A - ] [HA] n Acid dissociation constant = K a. n Note: technically all acids have a K a value. Size of K a indicates strength!

9. Acid dissociation constant K a n If water is left out: n HA(aq) H + (aq) + A - (aq) n K a = [H + ][A - ] [HA] Shown without H 3 O +

11. Section 14.2 Acid Strength

12. Types of Acids n Polyprotic Acids- more than 1 acidic hydrogen (diprotic, triprotic). –Ex: H 2 SO 4, H 3 PO 4 n Oxyacids - Proton is attached to the oxygen of an ion. –Ex: H 2 SO 4, HNO 3 n Organic acids contain the carboxyl group -COOH (acidic H + attached to O). –Ex: CH 3 COOH (acetic acid) n Generally very weak.

13. K a of Polyprotic Acids n The first K a value is much larger than the second, third, etc. n The second K a value is much larger than the third K a value, etc. n Since the first K a value is by far the largest, second, third, etc. K a values can be ‘ignored’ and the first K a value can usually be used for the acid dissociation constant of the acid.

14. Amphoteric n Can behave as either an acid or a base. n Water undergoes autonionization (it autoionizes). n 2H 2 O( l ) H 3 O + (aq) + OH - (aq) n K W = [H 3 O + ][OH - ]=[H + ][OH - ] n At 25ºC K W = 1.0 x10 -14 n Occurs in EVERY aqueous solution. n Neutral solution [H + ] = [OH - ]= 1.0 x10 -7 n Acidic solution [H + ] > [OH - ] n Basic solution [H + ] < [OH - ] Notice it’s very small!

15. Amphoteric Continued n Note: although water dissociation occurs in all solutions and contributes to [H + ], it is small in comparison and can be ignored when calculating [H + ]. n Adding other species to water can also allow water to act as an acid or a base.

16. Strong Acids n HBr, HI, HCl, HNO 3, H 2 SO 4, HClO 4 n Completely dissociated n [H + ] = [HA] n Ex: What is [H + ] in a 0.10M HNO 3 solution? HNO 3  H + + NO 3 - HNO 3  H + + NO 3 - *Mole ratio is 1:1 for HNO 3 :H + *Mole ratio is 1:1 for HNO 3 :H + *Thus [H + ] = 0.10M *Thus [H + ] = 0.10M

17. Weak Acids n Ka will be small. n It will be an equilibrium problem to find [H + ]. n Determine whether most of the H + will come from the acid or the water. n Compare Ka and Kw. –Whichever is larger is the one that will donate more H + (this is usually the acid). n Rest is just like last chapter.

18. Weak Acids n Because you’re dealing with weak acids, it can be assumed that the acid won’t dissociate much. n Lets you make assumptions and simplify terms when solving for x. (If you forget and don’t simplify that’s OK- you will just need to use the quadratic formula). n Then calculate the [H + ].

19. Weak Acids n For weak acids, the concentrations of H + and A - may be found if both the initial concentration of the acid and K a are known. n Ex: What is the [H + ] in 0.300M acetic acid solution? K a = 1.8 x 10 -5 n Remember- this is an equilibrium problem! Set up your ICE table, then solve for x. –Answer: x = [H + ] = 2.3 x 10 -3 M

20. Practice Problem #1 n Pg. 675 #55: A 0.0560g sample of acetic acid is added to enough water to make 50.00mL of solution. Calculate the [H + ], A 0.0560g sample of acetic acid is added to enough water to make 50.00mL of solution. Calculate the [H + ], [CH 3 COO - ], and [CH 3 COOH]. The K a for acetic acid = 1.8 x 10 -5. [CH 3 COO - ], and [CH 3 COOH]. The K a for acetic acid = 1.8 x 10 -5.

21. Summary: Strong vs. Weak n Remember- strong acids: [HA] = [H + ] and weak acids = equilibrium problem to find [H + ]. n Otherwise, once the [H + ] is known, pH problems are the same.

22. Section 1 Homework n Pg. 672 # 19(a&c), 21, 24, 28, 29, 32

23. pH Scale n pH= -log[H + ] or pH = -log [H 3 O + ] n pH of pure water = -log(1.0 x 10 -7 ) = 7.00 –In other words, this is neutral. –pH 7.00 is basic –Notice that as pH decreases, [H + ] increases n Sig figs: only the digits after the decimal place of a pH are significant n [H + ] = 1.0 x 10 -8 pH= 8.00 2 sig figs

24. pH Scale Continued n The pH of a solution can be estimated if the [H + ] is known. –Look at the exponent to estimate the pH –Ex: [H + ] = 1 x 10 -5 ; pH = 5 n pOH can also be calculated:  pOH = -log[OH - ]

25. BasicAcidicNeutral 10 0 10 -1 10 -3 10 -5 10 -7 10 -9 10 -11 10 -13 10 -14 [H + ] 013579111314 pH 10 0 10 -1 10 -3 10 -5 10 -7 10 -9 10 -11 10 -13 10 -14 [OH - ] 013579111314 pOH

26. Important Relationships n K W = 1.0 x 10 -14 = [H + ][OH - ] n pK W = 14.00 = pH + pOH n pH = -log[H + ] n pOH = -log[OH - ] n [H + ],[OH - ],pH and pOH Given any one of these we can find the other three! n There are also often multiple ways to correctly solve these problems!

27. Example #1 n The [H + ] = 1.00 x 10 -5 M. What is the pOH for this solution? n Step 1: Find pH. n Step 2: Then use 14 = pH + pOH to solve for pOH. n Can you think of another approach? n Find [OH - ] using: 1.0x10 -14 = [H + ][OH - ]. Then use: pOH = -log[OH - ]. Answer: Either way you get pOH = 9.000 Note: if [H + ] is given, no need to worry if acid is strong or weak.

28. Example #2 n Calculate the pH of a 1.00M solution of HF, whose K a = 7.2x10 -4. (1) Small K a, so it’s a weak acid. (2) Major species: HF and H 2 O. (3) Which provides the most H + ions? K a = 7.2 x 10 -4 and K w = 1.0 x 10 -14 So HF provides more H + ions. (4) Equilibrium we’re investigating: HF (aq) H + (aq) + F - (aq) So, K a = 7.2 x 10 -4 = [H + ][F - ]/[HF] [H + ] is not given, so now we need to decide if the acid is strong or weak, and then solve for [H + ].

29. Example #2 (5) Set up ICE table: HF (aq) H + (aq) + F - (aq) HF (aq) H + (aq) + F - (aq) I 1.00 0 0 C -x +x +x E 1.00-x x x 7.2 x 10 -4 = (x)(x) (1.00-x) (1.00-x) So: 7.2 x 10 -4 = x 2 Solve for x: x = 0.027 1.00 1.00 Simplification: since K a is small, can assume the initial [HF] won’t change noticeably, so at equilibrium [HF] ≈ 1.00.

30. Example #2 (5) Verify assumption was OK- x divided by the initial concentration of HF must be less than or equal to 5%: 0.027 x 100% = 2.7% 0.027 x 100% = 2.7% 1.00 1.00 Since 2.7% < 5%, the assumption that x was small enough to be neglected was valid. Since 2.7% < 5%, the assumption that x was small enough to be neglected was valid. (6) Calculate [H + ] and then the pH: [H + ] = x = 0.027M pH = -log(0.027) = 1.57

31. Practice Problem #1 n Calculate the pH of a 0.100M solution of hypochlorous acid (HOCl). The K a value = 3.5 x 10 -8. n [H + ] = 5.9 x 10 -5 M, so pH = 4.23

32. A mixture of Weak Acids n The process is the same. n Determine the major species. n The stronger will predominate (whichever has the largest K a value). n Doubt you’ll see this on the AP exam, but just in case!

33. Example n Calculate the pH of a solution that contains 1.00MHCN whose K a = 6.2x10 -10 and 5.00MHNO 2 whose K a = 4.0x10 -4. n Approach: same as before, now just need to consider THREE K values (the two above and water). n Since K a for HNO 2 is much larger than K w and the K a for HCN, this is the only one that needs to be used for finding [H + ]. n [H + ] = 4.5 x 10 -2, so pH = 1.35

34. AP Practice Question A 0.1M solution of acetic acid (CH 3 COOH) has a pH of about: a) 1 b) 3 c) 7 d) 10 You can answer this question without doing any math!

35. AP Practice Question What is the ionization constant, K a, for a weak monoprotic acid if a 0.30M solution has a pH of 4.0? a) 3.3 x 10 -8 c) 1.7 x 10 -6 b) 4.7 x 10 -2 d) 3.0 x 10 -4 Solve without using a calculator: -Estimate [H + ]: pH = 4.0, so [H + ] = 1 x 10 -4 -K a = (1 x 10 -4 ) 2 /0.30  ÷1 = 1 x 10 -8 so K a should be a little bigger

36. When K w Matters…

37. Percent Dissociation n = amount dissociated (M) x 100 initial concentration (M) n Example: Calculate the % dissociation of 1.00 M acetic acid, Ka = 1.8 x 10 -5. n Approach is the same as weak acid problems! Solve like an equilibrium problem for the necessary concentrations, then calculate % dissociation. n % dissociation = 0.42% [H + ] at equilibrium

38. Sections 2-3 Homework Problems Pg. 673 #17, 22, 40, 43, 47, 51, 57

39. Section 14.7 Polyprotic Acids

40. Polyprotic acids n Always dissociate stepwise. n The first H + comes off much easier than the second. n Ka for the first step is much bigger than Ka for the second, the second is bigger than the third, etc. –More difficult to lose the next H + because the negative charge increases. n Denoted Ka 1, Ka 2, Ka 3.

41. Polyprotic acid n H 2 CO 3 H + + HCO 3 - Ka 1 = 4.3 x 10 -7 n HCO 3 - H + + CO 3 -2 Ka 2 = 4.3 x 10 -10 n Conjugate base in first step is the acid in second. n In calculations we can normally ignore the second, third, etc. dissociation.

42. Sulfuric Acid is Special n In the first step it is a strong acid. –No K a value given- complete dissociation. n Second step is a weak acid. –Ka 2 = 1.2 x 10 -2 –Small, but not always small enough to ignore. –If the initial concentration of H 2 SO 4 is low enough, the second H + impacts pH!

43. Sulfuric Acid Example n Calculate the pH of a 0.0100M H 2 SO 4 solution. Ka 2 = 0.012. n The first acidic proton (H + ) dissociates completely (H 2 SO 4 is strong at first): H 2 SO 4  H + + HSO 4 - H 2 SO 4  H + + HSO 4 - Thus: [H 2 SO 4 ] = [H + ] = [HSO 4 - ] = 0.0100M Thus: [H 2 SO 4 ] = [H + ] = [HSO 4 - ] = 0.0100M n Now, we need to consider the second acidic proton.

44. Sulfuric Acid Example n The second acidic proton comes from HSO 4 -, which is a weak acid. So this is treated like an equilibrium problem: HSO 4 - H + + SO 4 -2 HSO 4 - H + + SO 4 -2 I 0.0100 0.0100 0 C -x +x +x E 0.0100-x 0.0100+x x n Plug into Ka 2 expression: 0.012 = (x)(0.0100+x)/(0.0100-x)

45. Sulfuric Acid Example n We can try to use the simplification that x is negligible with respect to HSO 4 - and H + : 0.012 = (x)(0.0100)/(0.0100), however the value for x = 0.012, which does not make sense. Thus the simplification is not valid! however the value for x = 0.012, which does not make sense. Thus the simplification is not valid! n Use quadratic formula to solve! n x = 0.0045 n [H + ] = 0.0100+0.0045 = 0.0145M n pH = 1.84

46. Notice that most Ka 1 values are significantly larger than the Ka 2 values, and thus typically do not impact the [H + ] or the pH. Typically, exceptions are sulfuric acid and oxalic acid.

47. Section 7 Homework Pg. 676 #93, 96, 97, 98

48. Lesson Essential Question: How do calculations with acids differ from calculations with bases?

49. Section 14.6 Bases

50. Bases n The OH - is a strong base. n Hydroxides of the alkali metals are strong bases because they dissociate completely when dissolved. n The hydroxides of alkaline earth metals (Ca(OH) 2, etc.) are also strong, but they don’t dissolve well in water. n Used as antacids because [OH - ] can’t build up.

51. Bases without OH - n Bases are proton acceptors: n NH 3 + H 2 O NH 4 + + OH - n It is the lone pair on N that accepts the proton. Many weak bases contain N n Use K b instead: n B (aq) + H 2 O( l ) HB + (aq) + OH - (aq) n K b = [ HB + ][ OH - ] [ B ]

52. Strength of Bases n Hydroxides are strong. n Others are weak. n Smaller K b = weaker base.

54. Strong Base Example  Calculate the pH of a 0.050M solution NaOH.  Treat it like a strong acid problem, just use pOH.  pOH = 1.3; pH = 14 – 1.3 = 12.7

55. Weak Base Example  Calculate the pH for a 1.0M solution of CH 3 NH 2, whose K b =1.8x10 -5.  Treat it like a weak acid problem, just with OH - instead of H +.  [OH - ] = 0.016M; pOH = 1.8; pH = 12.2

56. Another Important Relationship  For conjugate acid-base pairs, K a x K b = K w.  This can be particularly useful if you’re dealing with a weak base and only the K a value of the conjugate acid is known.  The K b for the base can be calculated and the equilibrium concentrations can then be found.

57. Example  0.400M of the acetate ion is dissolved in solution. What is the pH of the solution? K a = 1.8x10 -5 for acetic acid.  Write the equilibrium equation. Then find K b for acetate and solve.  [OH - ] = 1.5x10 -5 M; pH = 9.18

58. Section 14.8 Acid-Base Properties of Salts

59. Salts Can Affect pH n Salts are ionic compounds. n Most salts are soluble in water, and completely dissociate. n Depending upon the ions making up the salt, they COULD affect the pH once dissociated.

60. Neutral Salts n Salts whose ions could react with water to produce a strong acid or base DO NOT effect the pH. –Ex: Add NaCl to water. Two possible reactions: Na + + H 2 O NaOH + H + Cl - + H 2 O HCl + OH - n Both possible reactions favor the reactants, meaning no H + or OH - is produced, so pH remains neutral. Strong acids and bases dissociate completely

61. Acidic Salts n If the cation of a salt reacts with water to produce a weak base and H +, the solution is acidic. n Ex: NH 4 Cl has two possible reactions: n We already know Cl - has no effect. n NH 4 + NH 3 + H + n NH 3 is a weak base, so equilibrium does NOT favor the reactants; thus the formation of H + is significant, and pH is effected.

62. Basic Salts n If the anion of a salt reacts with water to produce a weak acid and OH -, the solution is basic. n Ex: NaF has two possible reactions: n We already know Na + has no effect. n F - + H 2 O HF + OH - n HF is a weak acid, so equilibrium does NOT favor the reactants; thus the formation of OH - is significant, and pH is effected.

63. Acidic & Basic Salts n It is possible that a salt could contain a cation that forms a weak base and an anion that forms a weak acid. n If this is the case, K a and K b must be compared. n The larger value dictates the pH. If K a is larger, the solution is acidic. If K b is larger, the solution is basic. If both are equal, the pH is neutral. –Ex: NH 4 CN in water: K a & K b = 5.6 x 10 -10

66. Practice Given the following salts, predict what the pH would be if they were added to water (acidic, basic, or neutral). 1) KNO 3 2) Na 2 CO 3 3) NH 4 Br (note: HBr = strong) 4) NH 4 F (note: HF = weak) neutral basic acidic Comparison of K a and K b needed.

67. Sample Problem: n Calculate the pH of a 0.10M NH 4 Cl solution. K b = 1.8 x 10 -5 for NH 3. n Only NH 4 + contributes to pH change: NH 4 + NH 3 + H + NH 4 + NH 3 + H + n Need K a because NH 4 + is an acid! n K w = K a x K b for a conjugate acid/base pair. n So: K a = (1.0x10 -14 )/(1.8x10 -5 ) K a = 5.6x10 -10 K a = 5.6x10 -10

68. Sample Problem: n Then, treat calculations like an equilibrium problem: NH 4 + NH 3 + H + NH 4 + NH 3 + H + I 0.10 0 0 C -x +x +x E 0.10-x x x n 5.6 x 10 -10 = (x)(x) 0.10-x 0.10-x 5.6 x 10 -10 = x 2 /0.10 => x = 7.5x10 -6

69. Sample Problem: n Simplification: 0.10 – x ≈ 0.10 5.6 x 10 -10 = x 2 /0.10 => x = 7.5x10 -6 5.6 x 10 -10 = x 2 /0.10 => x = 7.5x10 -6 n Verify: 7.5x10 -6 x 100 = 0.0075% 7.5x10 -6 x 100 = 0.0075% 0.10 0.10 n Verification is acceptable. n [H + ] = 7.5x10 -6 M n pH = 5.12

70. Section 8 Homework Pg. 676 #99, 101, 104, 111

71. Section 15.2-15.3 Buffers

72. Buffers n Solutions that maintain a constant pH when an acid or base is added. –In other words, change in pH is resisted. n One species present can consume/react with H + and another species present can consume OH -. n Very useful and important in many ways, especially for life. –Blood contains buffers to maintain a constant pH!

73. Buffers n Typically achieved by mixing a weak acid with its conjugate base, or a weak base with its conjugate acid. –The conjugate acid or base is usually added as a salt. n Examples: –Acetic acid mixed with sodium acetate. –Ammonia mixed with ammonium chloride. Notice these species will not effect the pH.

74. Acetic Acid & Sodium Acetate n HC 2 H 3 O 2  H + + C 2 H 3 O 2 - n Sodium acetate completely dissolves. n Addition of acetate ion shifts equilibrium to the left. n At this point there is a lot of acetic acid and the acetate ion. n Acetic acid (and H + ) can neutralize any base added, and the acetate ion can neutralize any acid added.

75. Ammonia & Ammonium Chloride n NH 3 + H 2 O  NH 4 + + OH - n Ammonium chloride completely dissolves. n Addition of ammonium ion shifts equilibrium to the left. n At this point there is a lot of ammonia and the ammonium ion. n Ammonia (and OH - ) can neutralize any acid added, and the ammonium ion can neutralize any base added.

76. Why Weak Acids & Bases? n You probably noticed that only weak acids and bases are used for buffer solutions. Why is this so? n Consider the strong acid HCl and adding NaCl to it. n HCl completely dissociates; no equilibrium: HCl  H + + Cl - HCl  H + + Cl - n Addition of NaCl doesn’t cause any shift in equilibrium, and only acid (H + ) remains. This could only neutralize the addition of a base, not the addition of an acid.

77. Why Weak Acids & Bases? n So… EQUILIBRIUM THAT EXISTS FOR WEAK ACIDS AND BASES IS KEY FOR DEVELOPING A SUCCESSFUL BUFFER! n It’s the only way for both acid & base to exist together!

78. pH & pOH of Buffers n Henderson-Hasselbalch equation can be used to find pH or pOH of buffers. –Must know K a or K b and initial concentrations of the acid/conjugate base or base/conjugate acid. n pH = pK a + log([A - ]/[HA]) n pOH = pK b + log([HB + ]/[B]) –Note: pK a or pK b means you take –log of the K value.

79. pH & pOH of Buffers n pH = pK a + log([A - ]/[HA]) n pOH = pK b + log([HB + ]/[B]) –Also notice above that the desired pH or pOH can be ‘fine-tuned’ by adjusting the ratio of weak acid/base to its conjugate base/acid. –Ex: the more A - you add, and the less HA you add, the higher the buffer pH will be. n Can also find pH as we’ve been doing. –Set up as an equilibrium problem and solve.

80. AP Practice Question The following questions refer to aqueous solutions containing 1:1 mole ratios of the following pairs of substances. Assume all concentrations are 1M. a) NH 3 and H 3 CCOOH (acetic acid) b) KOH and NH 3 c) HCl and KCl d) H 3 PO 4 and KH 2 PO 4 e) NH 3 and NH 4 Cl

81. AP Practice Question Cont. The solution with the highest pH. a) NH 3 and H 3 CCOOH (acetic acid) b) KOH and NH 3 c) HCl and KCl d) H 3 PO 4 and KH 2 PO 4 e) NH 3 and NH 4 Cl

82. AP Practice Question Cont. The solution with the lowest pH. a) NH 3 and H 3 CCOOH (acetic acid) b) KOH and NH 3 c) HCl and KCl d) H 3 PO 4 and KH 2 PO 4 e) NH 3 and NH 4 Cl

83. AP Practice Question Cont. The solution with the pH closest to neutral. a) NH 3 and H 3 CCOOH (acetic acid) b) KOH and NH 3 c) HCl and KCl d) H 3 PO 4 and KH 2 PO 4 e) NH 3 and NH 4 Cl

84. AP Practice Question Cont. A buffer at an alkaline pH. a) NH 3 and H 3 CCOOH (acetic acid) b) KOH and NH 3 c) HCl and KCl d) H 3 PO 4 and KH 2 PO 4 e) NH 3 and NH 4 Cl

85. AP Practice Question Cont. A buffer at an acidic pH. a) NH 3 and H 3 CCOOH (acetic acid) b) KOH and NH 3 c) HCl and KCl d) H 3 PO 4 and KH 2 PO 4 e) NH 3 and NH 4 Cl

86. Buffer Calculation Example What is the pH of a solution that has 2.00mol NH 3 and 3.00mol NH 4 Cl in 1.00L of solution? The K b for NH 3 = 1.81 x 10 -5. Henderson-Hasselbalch: pOH = -log(1.81 x 10 -5 ) + log(3.00/2.00) pOH = 4.918, so pH = 14 – 4.918 = 9.082 * Can verify answer by solving as an equilibrium problem. Write equilibrium for NH 3 producing NH 4 + since you’re given K b.

87. Buffer Calculation Practice What is the pH of a solution of 0.75M lactic acid (K a = 1.4 x 10 -4 ) and 0.25M sodium lactate? Lactic acid (HC 3 H 5 O 3 ) is often a component of of biologic systems (such as milk and muscle tissue during exercise). Use Henderson-Hasselbalch; answer: pH = 3.38

88. Homework n Pg. 740 # 21 & 23

89. Buffer Capacity n More concentrated acid-base pairs = more acid or base can be neutralized = less change in pH. n This is the buffer capacity: how well a buffer solution can maintain a constant pH.

90. Making Buffers n To make a buffer with an acidic pH: weak acid + salt containing conjugate base. n To make a buffer with a basic pH: weak base + salt containing conjugate acid. n When preparing a buffer solution you must choose an acid whose pK a value is close to the desired pH. –Look at exponent for easy estimation!

91. AP Practice Question What is a solution with an initial H 3 PO 2 concentration of 1M and an initial KH 2 PO 2 concentration of 1M? a) solution with pH > 7, which is a buffer b) solution with pH < 7, which isn’t a buffer c) solution with pH < 7, which is a buffer d) solution with pH > 7, which isn’t a buffer

92. AP Practice Question What is a solution with an initial KCOOH concentration of 1M and an initial KH 2 PO 2 concentration of 1M? a) solution with pH > 7, which is a buffer b) solution with pH < 7, which isn’t a buffer c) solution with pH < 7, which is a buffer d) solution with pH > 7, which isn’t a buffer

93. AP Practice Question Using the following information, choose the best answer for preparing a pH = 8 buffer. H 3 PO 4 : K a = 7.2 x 10 -3 H 2 PO 4 - : K a = 6.3 x 10 -8 HPO 4 -2 : K a = 4.2 x 10 -13 a) K 2 HPO 4 + KH 2 PO 4 b) H 3 PO 4 c) K 2 HPO 4 + K 3 PO 4 d) K 3 PO 4

94. Adding Acid/Base to a Buffer Calculate the pH of a solution when 0.10mol of gaseous HCl is added to 1.0L of a solution containing 0.25M NH 3 and 0.40M NH 4 Cl. n Acid is being added, so the base NH 3 will react with it. n When an acid reacts with a base, the reaction goes to completion: all dissociated H + reacts with NH 3. n Use stoichiometry to calculate moles of species left after the reaction.

95. Adding Acid/Base to a Buffer Calculate the pH of a solution when 0.10mol of gaseous HCl is added to 1.0L of a solution containing 0.25M NH 3 and 0.40M NH 4 Cl. n Set up an ‘IF’ table (for a reaction that goes to completion- initial and final values). –Don’t have to do this; IF table just helps set up your given information to solve the problem. NH 3 + H +  NH 4 + NH 3 + H +  NH 4 + I 0.25mol 0.10mol 0.40mol F Must be in moles for stoichiometry !

96. Adding Acid/Base to a Buffer Calculate the pH of a solution when 0.10mol of gaseous HCl is added to 1.0L of a solution containing 0.25M NH 3 and 0.40M NH 4 Cl. n Once you have the initial moles, determine which reactant is limiting (usually the added strong acid or base). Then find moles of all species remaining using stoichiometry. NH 3 + H +  NH 4 + NH 3 + H +  NH 4 + I 0.25mol 0.10mol 0.40mol F 0.15mol 0mol 0.50mol LR

97. Adding Acid/Base to a Buffer Calculate the pH of a solution when 0.10mol of gaseous HCl is added to 1.0L of a solution containing 0.25M NH 3 and 0.40M NH 4 Cl. n Use final mole values to find final concentrations: [NH 3 ] = 0.15M; [NH 4 + ] = 0.50M n Then use final concentrations of acid and base left in the Henderson-Hasselbalch equation. n Note: K b for NH 3 = 1.8 x 10 -5 n pOH = -log(1.8 x 10 -5 ) + log(0.50/0.15) = 5.27 n pH = 14 – 5.27 = 8.73 n pH of original buffer solution = 9.05. Not much change!

98. Adding Acid/Base to a Buffer Calculate the pH of a solution when 0.10mol of gaseous HCl is added to 1.0L of a solution containing 0.25M NH 3 and 0.40M NH 4 Cl. n Alternative calculation: instead of using K b, you could also have used K a. n K a = (1 x 10 -14 )/(1.8 x 10 -5 ) = 5.6 x 10 -10 n pH = -log(5.6 x 10 -10 ) + log(0.15/0.50) = 8.73 n Note the change in the log taken of the ratio! n Note also that the pH is the same!

99. Practice What is the pH of a solution containing 5.00M HC 2 H 3 O 2 and 5.00M NaC 2 H 3 O 2 when 0.010mol of HCl gas is added to 1.0L of this solution? K a = 1.8 x 10 -5 for acetic acid. *Reaction: C 2 H 3 O 2 - + H +  HC 2 H 3 O 2 *Set up IF table! Find final concentrations. n LR = H + added n Final concentrations: [C 2 H 3 O 2 - ] = 4.99M and [HC 2 H 3 O 2 ] = 5.01M n Final pH = 4.74

100. Homework n Pg. 741 #27 & 29 (d) only