1. Warm up 3/3/15  Only b.  Ka (for HCN) = 6.2x10 -10  pH=10.95

2. 15.1-15.5

3. 15.1 Solutions of Acids or Bases Containing a common Ion  We have the weak acid hydrofluoric acid (HF, K a = 7.2 x 10 -4 ) and its salt sodium fluoride (NaF).  The salt dissolves in water NaF(s)  Na + (aq) + F - (aq)  Since hydrofluoric acid is a weak acid and only slightly dissociated, the major species are HF, NA +, F - and H 2 O.  Common ion is F -

4. 15.1 Solutions of Acids or Bases Containing a common Ion Compare 1.0M HF and 1.0M HF + 1.0M NaF HF(aq) H + (aq) + F - (aq) in the second solution to be driven to the left by the presence of F - ions from the NaF. Dissociation of HF will be less in NaF HF(aq) H + (aq) + F - (aq) Added F- ions from NaF EQ shifts away from F-. Fewer H+ present

5. Common ion effect The shift in equilibrium position that occurs because of the addition of an ion already involved This effect makes a solution of NaF and HF less acidic

6. Common ion effect Solid NH 4 Cl added to a 1.0M NH 3 solution NH 4 Cl(s) ---> NH 4 + (aq) + Cl - (aq) The position of a ammonia-water to the left NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq)

7. Common ion effect Important in polyprotic acids The production of H+ by the 1 st step inhibits the succeeding steps.

8. 15.1 Acidic solution containing common ion 1) (Major species) HF, F -, Na +, and H 2 O The important species are HF and F - 2) (Equation) HF(aq) H + (aq) + F - (aq) 3) (K a ) K a = [H + ][F - ] = 7.2 x 10 -4 [HF]

9. [H+] and % dissociation 1.0M HF (Ka=7.2x10 -4 ), 1.0M NaF 4) (ICE) HFH + (aq)F - (aq) I (initial)1.00 C (change)-x+x E (equilibrium) 1.0-xx1.0+x

10. [H+] and % dissociation 1.0M HF (Ka=7.2x10 -4 ), 1.0M NaF K a = 7.2 x 10 -4 = [H + ][F - ] = (x)(1.0 + x) = (x)(1.0) [HF] 1.0 - x 1.0 Solving for x x = 1.0 (7.2 x 10 -4 ) = 7.2 x 10 -4 (The pH is 3.14) The percent dissociation of HF [H + ] x 100 = 7.2 x 10 -4 M x 100 = 0.072% [HF] 0 1.0M

11. [H+] and % dissociation 1.0M HF (Ka=7.2x10 -4 ), 1.0M NaF [H+] = 2.7 x 10 -2 M and the percent dissociation is 2.7% in 1.0M HF soln. The presence of the F - ions from the dissolved NaF greatly inhibits the dissociation of HF.

12. 15.2 Buffered Solutions ●A buffered solution resists a change in its pH when either hydroxide ions or protons are added. ● The most important practical example is our blood, which can absorb the acids and bases without changing its pH.

13. 15.2 Buffered Solutions ●A buffered solution may contain a weak acid and its salt (for example, HF and NaF) or a weak base and its salt (for example, NH 3 and NH 4 Cl). ●The best buffer contains large, equal amounts of the proton donor and the proton acceptor ●[HA]=[A - ] ●The best buffer, Ka=[H + ];pH=pK a

14. 15.2 The pH of a Buffered Solution I A buffered solution contains 0.50M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) and 0.50M sodium acetate (NaC 2 H 3 O 2 ). Calculate the pH of this solution. 1.Major species - HC 2 H 3 O 2, Na +, C 2 H 3 O 2 -, H 2 O the acetic acid will control the pH of the solution 2.Eq HC 2 H 3 O 2 (aq)  H+(aq) + C 2 H 3 O 2 - (aq)

15. 0.50M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) and 0.50M sodium acetate (NaC 2 H 3 O 2 ) Calculate the pH of this solution. 3. ICE 4. Solve for x 5. pH =4.74

16. 15.3 pH Changes in Buffered Solutions 1. The major species: HC 2 H 3 O 2, Na +, C 2 H 3 O 2, OH -, and H 2 O 2. EQ OH - +HC 2 H 3 O 2 → H 2 O + C 2 H 3 O 2 - The reaction proceed essentially to completion (until the OH - ions are consumed). 1.Assume that the reaction goes to completion, and carry out the stoichiometric calculations 2.Carry out the equilibrium calculations.

17. Calculate pH of 0.010 mol NaOH(s) + 1.0L of 0.50M buffered solution 1.Stoichiometry problem. 2.The equilibrium problem Major species: HC 2 H 3 O 2, Na +, C 2 H 3 O 2 -, and H 2 O The dominant equilibrium involves the dissociation of acetic acid.

18. Calculate pH of 0.010 mol NaOH(s) + 1.0L of 0.50M buffered solution 3. ICE 4. Ka, solve K a = 1.8 x 10 -5 = [H + ][C 2 H 3 O 2 - ] = (x)(0.51 + x) ≈ (x)(0.51) [HC 2 H 3 O 2 ] 0.49 - x 0.49 x≈ 1.7 x 10 -5 [H + ] = x = 1.7 x 10 -5 and pH = 4.76

19. Calculate pH of 0.010 mol NaOH(s) + 1.0L of 0.50M buffered solution The change in pH 4.76 - 4.74 = +0.02 New solution Original solution The pH increased by 0.02 pH units Compare this with 0.01 M NaOH. [H - ] = K w = 1.0 x 10 -14 = 1.0 x 10 -12 [OH - ] 1.0 x 10 -2 The change in pH is12.00 - 7.00 = +5.00 New solution Pure water

20. Pay special attention to the following points: 1. Buffered solutions are simply solutions of weak acids or bases containing a common ion. The pH calculations on buffered solutions require common ion effect calculation 2. When a strong acid or base is added to a buffered solution, stoichiometry of the resulting reaction first. After the stoichiometric calculations, then consider the EQ calculations.

21. Copyright © Houghton Mifflin Company. All rights reserved.15–21 Strong Acid or Base Being Added to a Buffered Solution

22. Buffering: How does it work? Suppose a buffered solution contains relatively large quantities of a weak acid HA When hydroxide ions are added, the following reaction occurs: OH - + HA → A - + H 2 O The new result is that OH - ions are replaced by A - ions.

23. The stability of the pH: K a = [H + ][A - ] [HA] [H + ]= K a [HA] [A - ] If the amounts of HA and A - originally present are very large compared with the amount of OH- added, the change in the [HA]/[A - ] ratio will be small.

24. [HA]/[A - ] ratio and the [H + ] remain virtually constant To calculate [H + ] in a buffered solution containing 0.10 M HF (K a = 7.2x10 -4 ) and 0.30 M NaF: [H + ] = (7.2x10 -4 )(0.10/0.30) = 2.4x10 -4 M Another useful form: taking the negative log of both sides -log[H + ] = -log(K a ) - log([HA]/[A - ])

25. Henderson-Hasselbalch Equation pH = pK a - log([HA]/[A - ]) (useful when the ratio [HA]/[A - ] is known) For a particular buffering system (conjugate acid-base pair), all solutions that have the same ratio [A - ]/[HA] will have the same pH. (not a focus)

26. Warm up 3/4/15 Solve only d.

27. 15.4 The pH of a Buffered Solution II 1. major species: HC 3 H 5 O 3, Na +, C 3 H 5 O 3 -, and H 2 O 2. pH will be controlled by the lactic acid dissociation equilibrium: HC 3 H 5 O 3 (aq)← → H + (aq) + C 3 H 5 O 3 - (aq) 3. K a = [H + ][C 3 H 5 O 3 - ] = 1.4x10 -4 [HC 3 H 5 O 3 ]

28. [HC 3 H 5 O 3 ] (approx) = [HC 3 H 5 O 3 ] 0 = 0.75M [C 3 H 5 O 3 - ] (approx) = [C 3 H 5 O 3 - ] 0 = 0.25M [H + ] = K a ([HC 3 H 5 O 3 ]/ [C 3 H 5 O 3 - ] ) = (1.4x10 -4 )(0.75M/ 0.25M) = 4.2x10 -4 M pH = -log(4.2x10 -4 ) = 3.38 Alternative: Henderson-Hasselbalch pH = pK a + log([C 3 H 5 O 3 - ]/[HC 3 H 5 O 3 ]) = 3.85 + log(0.25M/0.75M) = 3.38

29. 15.5 The pH of a Buffered Solution III A buffered solution contains 0.25 M NH 3 (K b = 1.8 x 10 -5 ) and 0.40 M NH 4 Cl. Calculate the pH of this solution: 1. Major Species: NH 3, NH 4 -, Cl -, and H 2 O From dissolve NH 4 Cl 2. EQ NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) 3. K b = 1.8 x 10 -5 = [NH 4 ][OH - ] [NH 3 ]

30. 0.25 M NH 3 (K b = 1.8 x 10 -5 ) and 0.40 M NH 4 Cl. 3. ICE 4. Solve K b = 1.8 x 10 -5 = [NH 4 + ][OH - ] = (0.40 + x)(x) = (0.40)(x) [NH 3 ] 0.25 - x 0.25 [OH - ] = x = 1.1 x 10 -5 pOH = 4.95 pH = 14.00 - 4.95 = 9.05

31. Alternative solution Since the solution contains relatively large quantities of both NH 4 + and NH 3 NH 3 (aq) + H 2 O(l) NH 4 + (aq) - OH - (aq) Calculate [OH - ] then [H + ] from K w. Or we can use the dissociation equilibrium for NH 4 + NH 4 + (aq) NH 3 (aq) + H + (aq)

32. Alternative solution To calculate [H + ]  obtain the K a value for NH 4 + from the given K b value for NH 3 since K a x K b = K w pH = pK a + log( [base] ) [acid] = 9.25 + log( 0.25 M ) = 9.25 - 0.20 = 9.05 0.40 M

33. 15.6 Adding strong Acid to a Buffered Solution Calculate the pH of the solution that results when 0.10 mol gaseous HCl is added to 1.0 L of the buffered solution from Sample exercise 15.5 1. Major Species : NH 3 NH 4 + Cl - H + and H 2 O NH 4 + is such a weak acid [K a = 5.6 x 10 -10 ] 2. EQ NH 3 will react with H + to form NH 4 + NH 3 (aq) + H + (aq) → NH 4 + (aq)

34. 0.10 mol gaseous HCl + 1.0 L of the buffered solution 3. Stoichiometry calculations Major Species : NH 3 NH 4 + Cl - and H 2 O

35. [NH 3 ] 0 = 0.15 mol = 0.15M 1.0L [NH 4 + ] 0 = 0.50 mol = 0.50 M 1.0 L Henderson-Hasselbalch equation pH = pK a + log( [NH 3 ] ) [NH 4 + ] = 9.25 + log ( 0.15M ) = 9.25 - 0.52 = 8.73 0.50M

36. 15.3 Buffering Capacity The buffering capacity of a buffered solution represents the amount of H+ or OH- the buffer can absorb without a significant change in pH. The pH of a buffered solution is determined by the ratio [A - ]/[HA].

37. Ex 15.7 Adding Strong Acid to a Buffered Solution pH = pK a + log( [C 2 H 3 O 2 - ] ) [HC 2 H 3 O 2 ] In each case, [C 2 H 3 O 2 - ] = [HC 2 H 3 O 2 ] The initial pH for both A and B pH = pK a + log(l) = pK a = -log(1.8 x 10 -5 ) = 4.74 Initial pH

38. Ex 15.7 Adding Strong Acid to a Buffered Solution After the addition of HCl the major species HC 2 H 3 O 2 Na + C 2 H 3 O 2 -, H +, Cl - and H 2 O H + (aq) + C 2 H 3 O 2 - (aq) → HC 2 H 3 O 2 (aq) Adding SA or SB, the SA/SB will react completely with the WA/WB.

39. A: 5.00M HC 2 H 3 O 2 and 5.00M NaC 2 H 3 O 2 add 0.010mol HCl gas Because HC 2 H 3 O 2 is a weak acid, this reaction runs to completion. For solution A: Stoichiometric table pH = pK a + log( [C 2 H 3 O 2 - ] ) [HC 2 H 3 O 2 ] = 4.74 + log ( 4.99 ) = 4.74 - 0.0017 = 4.74 5.01 No change

40. B: 0.050M HC 2 H 3 O 2 and 0.050M NaC 2 H 3 O 2 add 0.010mol HCl gas pH = 4.74 + log( 0.040 ) 0.060 = 4.74 - 0.18 = 4.56 Change in pH of B is higher than A. Solution A contains much larger quantities of buffering components Change in pH Less change in pH, better buffer

41. Better Buffered Solution Large changes in the ratio [A - ] / [HA] make large changes in pH. To avoid this situation for the most effective buffering, [HA] = [A - ] is the best buffer.

42. Adding Strong Acid to a Buffered Solution we want [A - ] / [HA] = 1 pH = pK a + log( [A - ] ) = pK a + log(l) = pK a [HA] The pK a of the weak acid should be as close as possible to the desired pH

43. Example pH = pK a + log([A - ]/[HA]) 4.00 = pK a + log(1) = pK a + 0 and pK a = 4.00 best choice of a weak acid pK a = 4.00 or K a = 1.0 x 10 -4

44. A pH of 4.30 corresponds to [H + ] = 10 -4.30 = antilog(-4.30) = 5.0 x 10 -5 M Use [H + ] = K a ([HA]/[A - ]) to find the ratio [HA]/[A - ] a. 3.7x10 -2 b. 3.8 c. 0.78d. 1.4x10 3 15.8 Preparing a Buffer

45. a. 3.7x10 -2 b. 3.8 c. 0.78d. 1.4x10 3 Since [HA]/[A - ] for benzoic acid is closest to 1, the system of benzoic acid and its sodium salt will be the best choice. 15.8 Preparing a Buffer

47. The stoichiometric(equivalence) point is often signaled by the color change of an indicator. M A V A =M B V B Titrant = the known concentration acid or base Analyte = the unknown concentration solution to be titrated 15.4 Titrations and pH Curves

48. The pH change looks like the graph on the right. x – volume of base added Y – pH The graph is called a pH curve or titration curve.

49. millimole(abbreviated mmol), thousandth of a mole. 1mmol = 1 mol/1000 = 10 -3 mol

50. millimoles per milliliter, as shown below: molarity = (mol solute / L solution) = [(mol solute) / 1000] / [(L solution / 1000)] = (mmol solute) / (mL solution) Number of mmol = volume (in mL) x molarity

51. Strong Acid-Strong Base Titrations H + (aq) + OH - (aq) → H 2 O(l) 1. Initial SA concentration (-log[H + ] based on the [Acid]) 2. Equivalence point (or endpoint) when moles of OH- = moles of H+. The pH is 7 3. Before and after the endpoint (calculate excess moles of H+ or OH-, divide by the total volume, and calculate pH)

52. Strong Acid-Strong Base Titrations

53. Weak Acid-Strong Base Titrations  The equivalence point is not at pH=7  A buffer region exists as you approach the endpoint  To calculate [H - ] after a certain amount of strong base has been added, we must deal with the weak acid dissociation equilibrium.  Even though the acid is weak, it reacts essentially to completion with hydroxide ion. 

54. Weak Acid-Strong Base Titrations ⇨ A stoichiometry problem. M A V A =M B V B  Initial weak acid concentration = use ICE table  Equivalence point = all weak acid has been neutralized or turned into the conjugate base. Calculate the [conjugate base], then ICE table w/ Kb=Kw/Ka  Calculate then[OH-], the pOH, pH  ⇨ 2. An equilibrium problem. The position of the weak acid equilibrium is determined and the pH is calculated.

55. Weak Acid-Strong Base Titrations  Half way to the equivalence point The pH = pKa At this point, there is a perfect buffer as the [HA]=[A-] ⇨ you can determine the Ka  Before and after the half way point The pH can be calculated using the H-H eqn or ICE. Use stoich to determine the [HA], [A-]  After the equivalence point The pH depends on the excess SB. Calculate excess moles of OH-, divide by the total volume. Calculate pOH, pH Effect by the A- is negligible

56. Weak Acid-Strong Base Titrations

57. Titrations of Weak Bases with Strong Acids ●The curve resembles an inverted WA-SB titration ●The pH at the equivalence point is less than 7 ●At the half way point, pH=pKa ●Acid-base indicator must be chosen with a Ka near to the [H+] of eq point.

58. Titrations of Weak Bases with Strong Acids

59. Titrations of Diprotic Acid – Strong Base ●They yield two H+ ●There are 2 eq points. ●The curve is not as distinct b/c of the various H+ donors and H+ acceptors ●At the first eq point; pH=pKa1 ●At the second eq point; pH=pKa2

60. Weak Diprotic Acid – Strong Base

61. Titration Calculation Steps to solve titration problems 1. Calculate volume at eq pt 2. Graph titration curve. Label eq pt and half way pt if WA/WB. 3. List major species 4. Determine the possible tables 1. Soich table for complete reaction (SA/SB) 2. ICE for eq (WA/WS only) 3. Both stoich/ICE (SA/SB, WB/WA) and leftover ions from the ICA

62. CASE STUDY: Strong Acid-Strong Base Titration The titration of 50.0 mL of 0.200 M HNO 3 with 0.100 M NaOH. 1. Equivalence point 1. (0.200M)(50.0mL)=(0.100M)(V) 2. V=100.mL 2. Titration curve

63. 50.0mL of 0.200MHNO 3 +0.100M NaOH A. No NaOH has been added major species: H +, NO 3 -, and H 2 O 0.200 M HNO 3 contains 0.200 M H + [H + ] = 0.200 M and pH = 0.699

64. 50.0mL of 0.200MHNO 3 +0.100M NaOH B. Add 10.0mL NaOH major species: H +, NO 3 -, Na+, OH-, and H 2 O. Stoich table [H + ] = (mmol H + left) / (volume of solution (mL)) = (9.00 mmol) / (50.0 + 10.0 mL) = 0.150 M pH = -log(0.150) = 0.824 H+H+ OH - H2OH2O Before (50.0mL)(0.200M) 10.0mmol (10.0mL)(0.100M) 1.00mmol After 10.0 mmol-1.00mmol=9.00 mmol1.00 mmol-1.00mmol=0 mmol

65. 50.0mL of 0.200MHNO 3 +0.100M NaOH C. 20.0 mL (total) of 0.100 M NaOH has been added major species: H +, NO 3 -, Na+, OH-, and H 2 O. stoich table [H + ] = (mmol H + left) / (volume of solution (mL)) = (8.00 mmol) / (50.0 + 20.0 mL) = 0.110 M pH = -log(0.11) = 0.942 H+H+ OH - H2OH2O Before (50.0mL)(0.200M) 10.0mmol (20.0mL)(0.100M) 2.00mmol After 10.0 mmol-2.00mmol=8.00 mmol2.00 mmol-2.00mmol=0 mmol

66. 50.0mL of 0.200MHNO 3 +0.100M NaOH D. 50.0 mL(total) of 0.100 M NaOH has been added major species: H +, NO 3 -, Na+, OH-, and H 2 O. Stoich table [H + ] = (mmol H + left) / (volume of solution (mL)) = (5.0 mmol) / (50.0 + 50.0 mL) = 0.0500 M pH = -log(0.11) = 1.301 H+H+ OH - H2OH2O Before (50.0mL)(0.200M) 10.0mmol (50.0mL)(0.100M) 5.00mmol After 10.0 mmol-5.00mmol=5.00 mmol5.00 mmol-5.00mmol=0 mmol

67. 50.0mL of 0.200MHNO 3 +0.100M NaOH E. 100.0 mL(total) of 0.100 M NaOH has been added. major species: H +, NO 3 -, Na+, OH-, and H 2 O. Stoich table [H + ] = [OH-]= 1.00x10 -7 M (neutralized) pH = 7.000 This is the stoichiometric point, or equivalence point. H+H+ OH - H2OH2O Before (50.0mL)(0.200M) 10.0mmol (100.0mL)(0.100M) 10.00mmol After 10.0 mmol-10.00mmol=0.00 mmol10.00 mmol-10.00mmol=0 mmol

68. 50.0mL of 0.200MHNO 3 +0.100M NaOH F. 150.0 mL(total) of 0.100 M NaOH has been added. major species: H +, NO 3 -, Na+, OH-, and H 2 O. Stoich table [OH - ] = (mmol OH - left) / (volume of solution (mL)) = (5.0 mmol) / (50.0 + 150.0 mL) = 0.0250 M OH- [H + ] = 1.0 x 10 -14 = 4.0 x 10 -13 M and pH 12.40 2.50 x 10 -2 H+H+ OH - H2OH2O Before (50.0mL)(0.200M) 10.0mmol (150.0mL)(0.100M) 15.00mmol After 10.0 mmol-10.00mmol=0.00 mmol15.00 mmol-10.00mmol=5.00 mmol

69. 50.0mL of 0.200MHNO 3 +0.100M NaOH G. 200.0 mL (total) of 0.100M NaOH has been added major species: H +, NO 3 -, Na+ OH-, and H 2 O. Stoich table [OH - ] = (mmol OH - left) / (volume of solution (mL)) = (10.0 mmol) / (50.0 + 200.0 mL) = 0.0400 M OH- [H + ] = 1.0 x 10 -14 = 2.5 x 10 -13 M and pH 12.60 4.00 x 10 -2 H+H+ OH - H2OH2O Before (50.0mL)(0.200M) 10.0mmol (200.0mL)(0.100M) 20.0mmol After 10.0 mmol-10.00mmol=0.00 mmol20.0 mmol-10.0mmol=10.0 mmol

70. CASE STUDY: Weak Acid-Strong Base Titration The titration of 50.0 mL of 0.10M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) with 0.10 M NaOH. 1. Equivalence point 1. (0.10M)(50.0mL)=(0.10M)(V) 2. V=50.0mL 2. Titration curve

71. 50.0 mL of 0.10M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) with 0.10 M NaOH. A. No NaOH has been added major species: H +, C 2 H 3 O 2 -, HC 2 H 3 O 2, and H 2 O Calculate pH =2.87

72. 50.0 mL of 0.10M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) with 0.10 M NaOH. B. 10.0 mL of 0.10M NaOH has been added. major species: OH -, Na +, HC 2 H 3 O 2, and H 2 O Strong base OH- react w/ strongest H+ donor. Equation OH- + HC 2 H 3 O 2  C 2 H 3 O 2 - + H 2 O OH - HC 2 H 3 O 2 C2H3O2-C2H3O2- Before (10.0mL)(0.10M) 1.0mmol (50.0mL)(0.10M) 5.0mmol 0 mmol After 1.0 mmol-1.00mmol=0 mmol5.0 mmol-1.0mmol=4.0 mmol 1.0 mmol

73. 50.0 mL of 0.10M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) with 0.10 M NaOH. B. 10.0 mL of 0.10M NaOH has been added. HC 2 H 3 O 2  H + + C 2 H 3 O 2 - OH - HC 2 H 3 O 2 C2H3O2-C2H3O2- Before (10.0mL)(0.10M) 1.0mmol (50.0mL)(0.10M) 5.0mmol 0 mmol After 1.0 mmol-1.00mmol=0 mmol5.0 mmol-1.0mmol=4.0 mmol 1.0 mmol

74. 50.0 mL of 0.10M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) with 0.10 M NaOH. B. 10.0 mL of 0.10M NaOH has been added. HC 2 H 3 O 2  H + + C 2 H 3 O 2 -

75. 50.0 mL of 0.10M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) with 0.10 M NaOH. C. 25.0 mL of 0.10M NaOH has been added. major species: OH -, Na +, HC 2 H 3 O 2, and H 2 O OH - HC 2 H 3 O 2 C2H3O2-C2H3O2- Before (25.0mL)(0.10M) 2.5mmol (50.0mL)(0.10M) 5.0mmol 0 mmol After 2.5 mmol-2.5mmol=0 mmol5.0 mmol-2.5mmol=2.5 mmol 2.5 mmol

76. C. 25.0 mL of 0.10M NaOH has been added. OH - HC 2 H 3 O 2 C2H3O2-C2H3O2- Before (25.0mL)(0.10M) 2.5mmol (50.0mL)(0.10M) 5.0mmol 0 mmol After 2.5 mmol-2.5mmol=0 mmol5.0 mmol-2.5mmol=2.5 mmol 2.5 mmol 50.0 mL of 0.10M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) with 0.10 M NaOH.

77. C. 25.0 mL of 0.10M NaOH has been added. Halfway point: Half of the equivalence pt. [H+]=Ka and pH=pKa 50.0 mL of 0.10M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) with 0.10 M NaOH.

78. E. 50.0 mL (total) of 10.0 M NaOH has been added Equivalence point major species: Na +, C 2 H 3 O 2 -, and H 2 O OH - HC 2 H 3 O 2 C2H3O2-C2H3O2- Before (50.0mL)(0.10M) 5.0mmol (50.0mL)(0.10M) 5.0mmol 0 mmol After 5.0 mmol-5.0mmol=0 mmol 5.0 mmol

79. E. 50.0 mL (total) of 10.0 M NaOH has been added major species: Na +, C 2 H 3 O 2 -, and H 2 O C 2 H 3 O 2 - (aq)+H 2 O(l)  HC 2 H 3 O 2 (aq)+OH - (aq) A weak base reaction characterized by K b. OH - HC 2 H 3 O 2 C2H3O2-C2H3O2- After 5.0 mmol-5.0mmol=0 mmol 5.0 mmol 50.0 mL of 0.10M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) with 0.10 M NaOH.

80. E. 50.0 mL (total) of 10.0 M NaOH has been added 50.0 mL of 0.10M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) with 0.10 M NaOH. C2H3O2-C2H3O2- OH - HC 2 H 3 O 2 I 5.0mmol/100.0mL=0.050M0mmol0 M C -x+x E 0.050M – xx0.050+x

81. Weak Acid-Strong Base Titration The pH of the equivalence point of a titration of a weak acid with a strong base is always greater than 7 because the anion of the acid, which remains in solution at the equivalence point, is a base.

82. 50.0 mL of 0.10M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) with 0.10 M NaOH. F. 60.0 mL (total) of 10.0 M NaOH has been added Equivalence point major species: Na +, C 2 H 3 O 2 -, OH -, and H 2 O The pH is determined by the excess OH - OH - HC 2 H 3 O 2 C2H3O2-C2H3O2- Before (60.0mL)(0.10M) 6.0mmol (50.0mL)(0.10M) 5.0mmol 0 mmol After 6.0 mmol-5.0mmol=1.0 mmol5.0 mmol-5.0mmol=0 mmol5.0 mmol

83. 50.0 mL of 0.10M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) with 0.10 M NaOH. F. 60.0 mL (total) of 10.0 M NaOH has been added

84. Weak Acid-Strong Base Titration The shapes of the strong and weak acid curves are the same after equivalence points because excess OH - controls the pH

85. Weak Acid-Strong Base Titration  A weak acid, the pH of the equivalence point is greater than 7.  The equivalence point in an acid-base titration is defined by the stoichiometry, not by the pH.  The equivalence point occurs when enough titrant has been added to react exactly with all the acid or base being titrated

86. Case Study: Weak Base-Strong Acid Titration The titration of 100.0mL of 0.050M NH 3 with 0.10M HCl Before the addition of any HCl: 1. Calculate equivalent pt (0.050M)(100.0mL)=(0.10M)V V=50.0mL 2.Titration curve

87. The titration of 100.0mL of 0.050M NH 3 with 0.10M HCl A. No HCl has been added 1.Major species: NH 3 and H 2 O 2.No reactions occur that go to completion 3.NH 3 (aq) + H 2 O(l) ←→ NH 4 + (aq) + OH - (aq) ICE table first (EQ) Use K b =1.8x10 -5 to calculate [OH - ] NH 3 OH - NH 4 + I 0.050M0 M C -x+x E 0.050M – xxx

88. The titration of 100.0mL of 0.050M NH 3 with 0.10M HCl A. No HCl has been added Use K b to calculate [OH - ] [OH-]=9.5x10 -4, pOH=3.02, pH=10.98

89. The titration of 100.0mL of 0.050M NH 3 with 0.10M HCl B. 10.0 mL HCl added 1.Major species: NH 3, H +, Cl -, and H 2 O (H + and Cl - are from added HCl) 2.NH 3 (aq) + H + (aq)←→NH 4 + (aq) NH 3 H+H+ NH 4 + Before (0.050M)(100.0mL) =5.0mmol (0.10M)(10.0mL) =1.0mmol 0mmol After 5.0mmol-1.0mmol=4.0mmol1.0mmol-1.0mmol=0X mmol

90. B. 10.0 mL HCl added [OH-]=8.0x10 -4 pOH=3.09, pH=10.91 The titration of 100.0mL of 0.050M NH 3 with 0.10M HCl ( K b =1.8x10 -5 ) NH 3 OH - NH 4 + I (4.00mmol)/110.0mL) =0.036M 0 M C -x+x E 0.036M – xxx

91. Note after the reaction of NH 3 with H + is run to completion, the solution contain the following major species: NH 3, NH 4 +, Cl -, and H 2 O (NH 4 + formed from titration reaction) NH 4 + (aq)←→NH 3 (aq) + H + (aq) or NH 3 (aq) + H 2 O(l)←→NH 4 + (aq) + OH - (aq)

92. The titration of 100.0mL of 0.050M NH 3 with 0.10M HCl C. 25.0 mL HCl added 1.Major species: NH 3, H +, Cl -, and H 2 O (H + and Cl - are from added HCl) 2.NH 3 (aq) + H + (aq)←→NH 4 + (aq) NH 3 H+H+ NH 4 + Before (0.050M)(100.0mL) =5.0mmol (0.10M)(25.0mL) =2.5mmol 0mmol After 5.0mmol-2.5mmol=2.5mmol2.5mmol-2.5mmol=0X mmol

93. C. 25.0 mL HCl added [OH-]=6.0x10 -4 pOH=3.22, pH=10.78 The titration of 100.0mL of 0.050M NH 3 with 0.10M HCl ( K b =1.8x10 -5 ) NH 3 OH - NH 4 + I (2.5mmol)/125mL) =0.02M 0 M C -x+x E 0.02M – xxx

94. At the equivalence point: 1. the equivalence point occurs when all the original NH 3 is converted to NH 4 + Major species: NH 4 +, Cl -, and H 2 O 1. No reactions go to completion 2. The dominant equilibrium is the dissociation of the weak acid NH 4 +, for which K a = K w K b (for NH 3 )

95. The titration of 100.0mL of 0.050M NH 3 with 0.10M HCl D. 50.0 mL HCl added 1.Major species: NH 3, H +, Cl -, and H 2 O (H + and Cl - are from added HCl) 2.NH 3 (aq) + H + (aq)←→NH 4 + (aq) NH 3 H+H+ NH 4 + Before (0.050M)(100.0mL) =5.0mmol (0.10M)(50.0mL) =5.0mmol 0mmol After 5.0mmol-5.0mmol=0mmol5.0mmol-5.0mmol=0X mmol

96. D. 50.0 mL HCl added K a = K w = 1.0x10 -14 =5.6x10 -10 K b (for NH 3 ) 1.8x10 -5 (5.0mmol)/(150.0mL)=0.033M NH 4 + The titration of 100.0mL of 0.050M NH 3 with 0.10M HCl ( K b =1.8x10 -5 ) NH 3 OH - NH 4 + I (2.5mmol)/125mL) =0.02M 0 M C -x+x E 0.02M – xxx

97. D. 50.0 mL HCl added (0.033M NH 4 + ) K a =5.6x10 -10 NH 4 +  NH 3 + H + X=4.3x10 -6 M pH=5.36 The titration of 100.0mL of 0.050M NH 3 with 0.10M HCl ( K b =1.8x10 -5 )

98. Beyond the equivalence point: 1. Excess HCl has been added Major Species: H +, NH 4 +, Cl -, and H 2 O 1. No reaction goes to completion 2. [H + ] will be determined by the excess H + : [H + ] = mmol H + in excess mL solution

99. The titration of 100.0mL of 0.050M NH 3 with 0.10M HCl E. 60.0 mL HCl added 1.Major species: H +, NH 4 +, Cl -, and H 2 O 2.NH 3 (aq) + H + (aq)←→NH 4 + (aq) 3.[H+]=1.0mmol/160mL=6.3x10 -3 M 4.pH=2.20 NH 3 H+H+ NH 4 + Before (0.050M)(100.0mL) =5.0mmol (0.10M)(60.0mL) =6.0mmol 0mmol After 5.0mmol-5.0mmol=0mmol6.0mmol-5.0mmol=1.0mmolX mmol

100. Copyright © Houghton Mifflin Company. All rights reserved.15–100 Treat the Stoichiometry and Equilibrium Problems Separately

101. Copyright © Houghton Mifflin Company. All rights reserved.15–101 Figure 15.4 The pH Curves for the Titrations of 50.0-mL Samples pf 0.10 M Acids with Various Ka Values with 0.10 M NaOH

102. 15.5 Acid-Base Indicators Determining the equivalence point: 1. Use a pH meter 2. Use an acid-base indicator. Although the equivalence point of a titration is defined by stoichiometry, it is not necessarily the same as the end point (where the indicator changes color)

103. Most common acid-base indicators: ● they exhibit one color when the proton is attached (ex: HIn, WA) ● they exhibit a different color when the proton is absent (ex: H + + In - ) HIn(aq)←→H + (aq) + In - (aq) K a = [H + ][In - ] [HIn] [H + ] = K a x [HIn] [In - ] When [HIn]/[In-] = 10/1, color will change

104. Common Indicators IndicatorpH RangeColor ChangeKaKa Thymol blue1.2 – 2.8red → yellow Methyl red4.4 – 6.2red → yellow5.0 x 10 -6 Litmus5.0 – 8.0red → blue Bromothymol blue6.2 – 7.6yellow → blue1.0 x 10 -7 Phenolphthalein8.0 – 10.0colorless → pink1.0 x 10 -8

105. Copyright © Houghton Mifflin Company. All rights reserved.15–105 Figure 15.6 The Acid and Base Forms of the Indicator Phenolphthalein

106. Copyright © Houghton Mifflin Company. All rights reserved.15–106 Figure 15.8 The Useful pH Ranges for Several Common Indicators

107. Copyright © Houghton Mifflin Company. All rights reserved.15–107 Figure 15.9 The pH Curve for the Titration of 100.0 mL of 0.10 M of HCl with 0.10 M NaOH

108. Copyright © Houghton Mifflin Company. All rights reserved.15–108 Figure 15.10 The pH Curve for the Titration of 50 mL of 0.1 M HC 2 H 3 O 2 with 0.1 M NaOH; Phenolphthalein Will Give an End Point Very Close to the Equivalence Point of the Titration

110. More skipped items

111. Pg.705 Ex. 15.9 Titration of a Weak Acid Hydrogen cyanide gas (HCN) a powerful respiratory inhibitor is highly toxic. It is a very weak acid (K a = 6.2 x 10 -10 ) when dissolved in water. If a 50.0mL sample of 0.100M HCN is titrated with 0.100 M NaOH calculate the pH of solution a. after 8.00 mL of 0.100 M NaOH has been added b. at the halfway point of the titration c. at the equivalence point of the titration

112. Titration of a weak acid a.The stoichiometry problem The equilibrium problem, the major species : HCN, CN -, Na +, and H 2 O The position of the acid dissociation equilibrium HCN(aq) H + (aq) + CN - (aq) will determine the pH

113. Titration of a Weak Acid The corresponding ICE table is

114. Titration of a Weak Acid 6.2 x 10 -10 = K a = [H - ][CN - ] = x (0.80 / 58.0 + x) [HCN] 4.2 / 58.0 - x = x (0.80 / 58.0) = x( 0.80 / 4.2) 4.2 / 58 x = 3.3 x 10 -9 M = [H + ] and pH = 8.49

115. Titration of a Weak Acid b. At the halfway point of the titration, HCN 50.0 mL x 0.100 M = 5.00 mmol The halfway point occurs when 2.50 mmol OH - has been added Volume of NaOH x 0.100M = 2.50 mmol OH - Volume of NaOH = 25.0 mL At the halfway point [HCN] is equal to [CN - ] and pH is equal to pK a. After 25.0 mL of 0.100 M NaOH has been added. pH = pK a = -log(6.2 x 10 -10 ) = 9.21

116. c. At the equivalence point. When a total of 5.00 mmol OH - has been added. NaOH solution is 0.100 M, the equivalence point 50.0 mL NaOH has been added. From 5.00 mmol CN -. Major species CN -, Na +, and H 2 O

117. The reaction will control the pH involves the basic cyanide ion extracting a proton from water. CN - (aq) + H 2 O (l) ⇋ HCN (aq) + OH - (aq) K b = K w = 1.0 x 10 -14 = 1.6 x 10 -5 = [HCN][OH - ] K a 6.2 x 10 -10 [CN - ]

118. [OH - ] = x = 8.9 x 10 -4 [H + ] = 1.1 x 10 -11 and pH = 10.96 Amount of acid determines the equivalence point acetic acid the pH 8.72; hydrocyanic acid 10.96. Because the CN - ion is a much stronger base than the C 2 H 3 O 2 - ion.

119. The pH at the halfway point occurs in each case when the same volume of 0.10M NaOH has been added the shapes of the curves are dramatically different. The weaker the acid, the greater the pH value at the equivalence point.

120. 15.10 Calculating K a

121. The stoichiometry problem The equilibrium problem The major species HA, A -, Na +, and H 2 O HA(aq) ←→ H + (aq) + A - (aq) K a = [H + ][A - ] [HA]

122. ICE Table ●x is known here because the pH at this point is 6.00 x = [H + ] = antilog(-pH) = 1.0x10 -6 M ●substitute

123. Easier Way ●The original solution contained 2.00 mmol of HA ●20.0 mL of added 0.0500M NaOH contains 1.0 mol OH - ●this is the halfway point (where [HA] is qual to [A - ] [H] = K a = 1.0x10 -6