1. Chapter 16TMHsiung©2015Slide 1 of 64 Chapter 16 Aqueous Ionic Equilibrium 熊同銘 tmhsiung@gmail.com

2. Chapter 16TMHsiung©2015Slide 2 of 64 1.The Danger of Antifreeze 2.Buffers: Solutions That Resist pH Change 3.Buffer Effectiveness: Buffer Range and Buffer Capacity 4.Titrations and pH Curves 5.Solubility Equilibria and the Solubility Product Constant 6.Precipitation 7.Qualitative Chemical Analysis 8.Complex Ion Equilibria Contents

3. Chapter 16TMHsiung©2015Slide 3 of 64 1.The Danger of Antifreeze Each year, thousands of pets and wildlife species die from consuming antifreeze. Most brands of antifreeze contain ethylene glycol (sweet taste). Ethylene glycol (HOCH 2 CH 2 OH) metabolized in the liver to glycolic acid (HOCH 2 COOH). In high enough concentration in the bloodstream, glycolic acid overwhelms the buffering ability (pH 7.36-7.42) of the HCO 3 − in the blood, causing the blood pH to drop. The excess H + reduces the ability of hemoglobin (Hb) to carry oxygen: HbH + (aq) + O 2(g)  HbO 2(aq) + H + (aq) called acidosis

4. Chapter 16TMHsiung©2015Slide 4 of 64 2.Buffers: Solutions That Resist pH Change 1)Buffer solution: A solution that changes pH only slightly when small amounts of a strong acid or a strong base are added. The acid component of the buffer neutralize small added amounts of OH – : OH – + HA  H 2 O + A – The basic component neutralize small added amounts of H 3 O + : H 3 O + + A –  H 2 O + HA

5. Chapter 16TMHsiung©2015Slide 5 of 64 Buffering Action * For a 1L buffer prepared by 0,100 mol CH 3 COOH and 0.100 mol CH 3 COONa

6. Chapter 16TMHsiung©2015Slide 6 of 64 Buffer Preparation mixing a significant amounts of both weak acid with its salt (conjugate base), for example CH 3 COOH / CH 3 COONa mixing a significant amounts of both weak base with its salt (conjugate acid), for example NH 3 / NH 4 Cl *****

7. Chapter 16TMHsiung©2015Slide 7 of 64 2)Calculating the pH of a Buffer Solution Common ion effect: The tendency for a common ion to decrease the solubility of an ionic compound or to decrease the ionization of a weak acid or weak base. In this case, C 2 H 3 O 2 – is a common ion

8. Chapter 16TMHsiung©2015Slide 8 of 64 HA (aq) + H 2 O  H 3 O + (aq) + A – (aq) Therefore, (1) Henderson-Hasselbalch equation:(2) Either equation (1) or (2) available to estimate pH of the buffer In general: [HA]  C HA, [A – ]  C NaA Effective buffer range: pH = pK a  1 (i.e., [HA/A – ]: 0.1~ 10) The Henderson-Hasselbalch Equation

9. Chapter 16TMHsiung©2015Slide 9 of 64 Solution Equilibrium Approach (equation (1)) Calculate the pH of a buffer solution that is 0.050 M in benzoic acid (HC 7 H 5 O 2 ) and 0.150 M in sodium benzoate (NaC 7 H 5 O 2 ). For benzoic acid, K a = 6.5 × 10 –5. Example 16.2Calculating the pH of a Buffer Solution as an Equilibrium Problem and with the Henderson– Hasselbalch Equation OK

10. Chapter 16TMHsiung©2015Slide 10 of 64 Henderson–Hasselbalch Approach (equation (2)) OK

11. Chapter 16TMHsiung©2015Slide 11 of 64 Calculate the pH of a buffer solution that is 0.100 M in HC 2 H 3 O 2 and 0.100 M in NaC 2 H 3 O 2. For acetic acid, K a = 1.8 × 10 –5. Example 16.1Calculating the pH of a Buffer Solution Solution OK K a = 1.8 × 10 –5 *****

12. Chapter 16TMHsiung©2015Slide 12 of 64 Solution (a) A 1.0 L buffer solution contains 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2. (a) Calculate the new pH after adding 0.010 mol of solid NaOH to the buffer. (b) For comparison, calculate the pH after adding 0.010 mol of solid NaOH to 1.0 L of pure water. (Ignore any small changes in volume that might occur upon addition of the base.) Example 16.3Calculating the pH Change in a Buffer Solution after the Addition of a Small Amount of Strong Acid or Base OK *****

13. Chapter 16TMHsiung©2015Slide 13 of 64 Solution (b)

14. Chapter 16TMHsiung©2015Slide 14 of 64 Solution Use the Henderson–Hasselbalch equation to calculate the pH of a buffer solution that is 0.50 M in NH 3 and 0.20 M in NH 4 Cl. For ammonia, pK b = 4.75. Example 16.4Using the Henderson–Hasselbalch Equation to Calculate the pH of a Buffer Solution Composed of a Weak Base and Its Conjugate Acid NH 3(aq) + H 2 O  NH 4 + (aq) + OH – (aq) K b NH 4 + (aq) + H 2 O  NH 3(aq) + H 3 O + (aq) K a *****

15. Chapter 16TMHsiung©2015Slide 15 of 64 3.Buffer Effectiveness: Buffer Range and Buffer Capacity The effectiveness of a buffer depends on two factors: (1) the relative amounts of acid and base. (2) the absolute concentrations of acid and base. A buffer will be most effective when: (1) the [A – ]:[HA] = 1. (2) the [HA] and the [A – ] are large.

16. Chapter 16TMHsiung©2015Slide 16 of 64 1)Buffer Capacity: The amount of acid or base a buffer can neutralize. Quantitative buffer capacity: The number of moles of a strong acid or a strong base that causes 1.00 L of the buffer to undergo a 1.00 pH unit change. Buffer capacity character: The higher concentrations ( [HA] and [A – ]), the higher buffer capacity. Concentration ratio of [A – ]:[HA] = 1 has the higher buffer capacity. At pH  pK a, has the higher buffer capacity. *****

17. Chapter 16TMHsiung©2015Slide 17 of 64 2)Buffer range: the pH range the buffer can be effective. A buffer will be effective when 0.1 < [A – ]:[HA] < 10. Lowest pHHighest pH *The effective pH range of a buffer is pK a ± 1. *When choosing an acid to make a buffer, choose one whose pK a is closest to the pH of the buffer. *****

18. Chapter 16TMHsiung©2015Slide 18 of 64 Solution The best choice is formic acid because its pK a lies closest to the desired pH. You can calculate the ratio of conjugate base (CHO 2 – ) to acid (HCHO 2 ) required by using the Henderson–Hasselbalch equation as follows: Which acid would you choose to combine with its sodium salt to make a solution buffered at pH 4.25? For the best choice, calculate the ratio of the conjugate base to the acid required to attain the desired pH. chlorous acid (HClO 2 ) pK a = 1.95formic acid (HCHO 2 ) pK a = 3.74 nitrous acid (HNO 2 ) pK a = 3.34hypochlorous acid (HClO) pK a = 7.54 Example 16.5Preparing a Buffer *****

19. Chapter 16TMHsiung©2015Slide 19 of 64 4.Titrations and pH Curves 1)General description Equivalence point: The acid and base have been brought together in exact stoichiometric proportions. Endpoint: The point in titration at which indicator changes color. Titration curve: The graph of pH versus volume of titrant.

20. Chapter 16TMHsiung©2015Slide 20 of 64 Acid-base indicators: A weak organic acid or a weak organic base whose undissociated form differs in color from its conjugate base or its conjugate acid. HIn + H 2 O  In – + H 3 O + Color in Color in acidic solution basic solution Generally, the indicator range is expressed as pH = pK a  1: Acid-base indicators are often used for applications in which a precise pH reading isn’t necessary. 2)Acid-Base Indicators

21. Chapter 16TMHsiung©2015Slide 21 of 64 Common Acid-Base Indicators

22. Chapter 16TMHsiung©2015Slide 22 of 64 (1)At beginning: [H 3 O + ] ≡ C HCl (2)Before the equivalence point: calculating net excess moles of HCl, then [H 3 O + ] (3)At the equivalence point: [H 3 O + ] = 1.0 x 10 −7, pH = 7.00 (4)After the equivalence point: calculating net excess moles of NaOH, then [OH – ], then, [H 3 O + ] 3)Titration of a Strong Acid with a Strong Base Example: Titration of HCl with NaOH *****

23. Chapter 16TMHsiung©2015Slide 23 of 64 Example: Titration of 25.0 mL of 0.100 M HCl with 0.100 M NaOH.

24. Chapter 16TMHsiung©2015Slide 24 of 64 Example Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 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 10.00 mL of 0.500 M NaOH (c) after the addition of 20.00 mL of 0.500 M NaOH (d) after the addition of 20.20 mL of 0.500 M NaOH Solution: (a)pH = –log[H 3 O + ] = –log(0.500) = 0.301

25. Chapter 16TMHsiung©2015Slide 25 of 64 (b)Initial amount of H 3 O + : 20.00 mL x 0.500 mmol H 3 O + /mL = 10.0 mmol H 3 O + Amount of OH – added: 10.00 mL x 0.500 mmol OH – /mL = 5.00 mmol OH – H 3 O + +OH – → 2 H 2 O Initial, mmol:10.05.00 Changes, mmol: –5.00 –5.00 Equilibrium, mmol:5.0– Total volume is 20.00 mL + 10.00 mL = 30.00 mL

26. Chapter 16TMHsiung©2015Slide 26 of 64 (c)Initial amount of H 3 O + : 20.00 mL x 0.500 mmol H 3 O + /mL = 10.0 mmol H 3 O + Amount of OH – added: 20.00 mL x 0.500 mmol OH – /mL = 10.0 mmol OH – [H 3 O + ][OH − ] = [H 3 O + ] 2 = 1.00 x 10 −14 [H 3 O + ] = 1.00 x 10 −7 pH = 7.00

27. Chapter 16TMHsiung©2015Slide 27 of 64 (d)Initial amount of H 3 O + : 20.00 mL x 0.500 mmol H 3 O + /mL = 10.0 mmol H 3 O + Amount of OH – added: 20.20 mL x 0.500 mmol OH – /mL = 10.1 mmol OH – H 3 O + +OH – → 2 H 2 O Initial, mmol:10.010.1 Changes, mmol:–10.0 –10.0 Equilibrium, mmol:–0.1 Total volume is 20.00 mL + 20.20 mL = 40.20 mL

28. Chapter 16TMHsiung©2015Slide 28 of 64 Titration curve for the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH Bromphenol blue, bromthymol blue, and phenolphthalein all change color at very nearly 20.0 mL

29. Chapter 16TMHsiung©2015Slide 29 of 64 4)Titration of a Strong Base with a Strong Acid Example: Titration of 25.0 mL of 0.100 M NaOH with 0.100 M HCl

30. Chapter 16TMHsiung©2015Slide 30 of 64 x = [H 3 O + ] 5)Titration of a Weak Acid with a Strong Base Example: Titration of HCHO 2 with NaOH (1)At the begining: HA + H 2 O  A – + H 3 O + C HA –x x x or (2)Before the equivalence point: Calculating net excess moles of HA and the produced A –, then, [HA], [A – ], then,

31. Chapter 16TMHsiung©2015Slide 31 of 64 x = [OH – ], then [H 3 O + ], then pH (3)At the equivalence point: Calculating the expected moles of A –, then, [A – ], then, A – + H 2 O  HA + OH – [A – ]–x x x (4)After the equivalence point: Calculating the net excess moles of OH –, then, [OH – ], then, [H 3 O + ], then pH

32. Chapter 16TMHsiung©2015Slide 32 of 64 Example Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 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 0.500 M NaOH (c) after the addition of 10.00 mL of 0.500 M NaOH (d) after the addition of 20.00 mL of 0.500 M NaOH (e) after the addition of 21.00 mL of 0.500 M NaOH

33. Chapter 16TMHsiung©2015Slide 33 of 64 Solution: (a) Assume x << 0.500

34. Chapter 16TMHsiung©2015Slide 34 of 64 (b)Addition of 8.00 mL of 0.500 M NaOH: CH 3 COOH+OH –  H 2 O+CH 3 COO – Initial, mmol:10.04.00 – Changes, mmol: –4.00 –4.00 +4.00 Equilib, mmol: 6.0 –4.00 [CH 3 COO – ] = 4.00 mmol/28.00 mL [CH 3 COOH] = 6.0 mmol/28.00 mL:

35. Chapter 16TMHsiung©2015Slide 35 of 64 (c)Addition of 10.00 mL of 0.500 M NaOH: CH 3 COOH+OH –  H 2 O+CH 3 COO – Initial, mmol:10.05.00 – Changes, mmol: –5.00 –5.00 +5.00 Equilib, mmol: 5.0 –5.00 10.005.00 mmol OH −

36. Chapter 16TMHsiung©2015Slide 36 of 64 (d)Addition of 20.00 mL of 0.500 M NaOH: CH 3 COOH+OH –  H 2 O+CH 3 COO – Initial, mmol:10.010.0 – Changes, mmol: –10.0 –10.0 +10.0 Equilib, mmol: ––10.0 20.0010.00 mmol OH − CH 3 COO – + H 2 O  CH 3 COOH+OH – Initial, M:0.25 – – Changes, M:–x +x +x Equilib, M: (0.25−x) xx [CH 3 COO − ] = 10 mmol/(20 mL + 20 mL)= 0.25 M

37. Chapter 16TMHsiung©2015Slide 37 of 64 Assume x << 0.25

38. Chapter 16TMHsiung©2015Slide 38 of 64 (e)Addition of 21.00 mL of 0.500 M NaOH: CH 3 COOH+OH –  H 2 O+CH 3 COO – Initial, mmol:10.0 10.5 – Changes, mmol: –10.0 –10.0 +10.0 Equilib, mmol: – 0.510.0 *****

39. Chapter 16TMHsiung©2015Slide 39 of 64 Titration curve for the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH The equivalence-point pH is NOT 7.00. It is > 7.00 for weak acid titrated with strong base Phenolphthalein is selected as the indicator for weak acid titrated with strong base Buffering occurred at pH ≈ pK a

40. Chapter 16TMHsiung©2015Slide 40 of 64 6)Titration of a Weak Base with a Strong Acid Example: Titration of 25 mL of 0.100 M NH 3 with 0.100 M HCl.

41. Chapter 16TMHsiung©2015Slide 41 of 64 Type of titrationExamplepH at Equivalence point Strong acid by Strong baseHCl/NaOH7 Weak acid by Strong baseCH 3 COOH/NaOH>7 Weak base by Strong acidNH 3 /HCl<7 7) pH values at Equivalence point

42. Chapter 16TMHsiung©2015Slide 42 of 64 8) The Titration of a Polyprotic Acid Example: Titration of 25 mL of 0.100 M H 2 SO 3 with 0.100 M NaOH.

43. Chapter 16TMHsiung©2015Slide 43 of 64 5. Solubility Equilibria and the Solubility Product Constant Solubility: The number of grams of solute in one liter of a saturated solution (g/L). Molar solubility (s): The number of moles of solute in one liter of a saturated solution (M). Solubility product constant (K sp ): The product of the molar concentrations of the constituent ions: A x B y(s)  xA m+ (aq) + yB n– (aq) K sp = [A m+ ] x [B n– ] y s x·s y·s Solubility equilibria calculations: Determining a value of K sp from experimental data Calculating equilibrium concentrations with known K sp

44. Chapter 16TMHsiung©2015Slide 44 of 64 Some K sp At 25 o C

45. Chapter 16TMHsiung©2015Slide 45 of 64 Calculate the molar solubility of PbCl 2 in pure water. For PbCl 2, K sp = 1.17 × 10 –5. Example 16.8Calculating Molar Solubility from K sp Solution

46. Chapter 16TMHsiung©2015Slide 46 of 64 The molar solubility of Ag 2 SO 4 in pure water is 1.2 × 10 –5 M. Calculate K sp. Example 16.9Calculating K sp from Molar Solubility Solution

47. Chapter 16TMHsiung©2015Slide 47 of 64 Followed Le Châtelier’s principle, i.e., the solubility of a slightly soluble ionic compound is lowered when a second solute that contains a common ion. Example: For equilibrated Ag 2 SO 4(s)  2Ag + (aq) + SO 4 2– (aq) (a)What is happened once adding Na 2 SO 4. (b)What is happened once adding AgNO 3. (c)What is happened once adding Ag 2 SO 4(s) Ans: (a)Shift left (b)Shift left (c)No effect 1)The Effect of a Common Ion on Solubility

48. Chapter 16TMHsiung©2015Slide 48 of 64 Solution What is the molar solubility of CaF 2 in a solution containing 0.100 M NaF? For CaF 2, K sp = 1.46 × 10 –10 Example 16.10Calculating Molar Solubility in the Presence of a Common Ion *****

49. Chapter 16TMHsiung©2015Slide 49 of 64 Example 1: AgCl (s)  Ag + (aq) + Cl – (aq) pH decrease ([H + ] increase), no effect on solubility Example 2: CaF 2(s)  Ca 2+ (aq) + 2F – (aq) Dissolution reaction 2H 3 O + (aq) + 2F – (aq)  2HF (aq) Acid-base reaction CaF 2(s) + 2H 3 O + (aq)  Ca 2+ (aq) + 2HF (aq) + 2H 2 O (l) pH decrease ([H + ] increase), solubility increase. Example 3: Mg(OH) 2(s)  Mg 2+ (aq) + 2OH – (aq) Dissolution reaction 2H 3 O + (aq) + 2OH – (aq)  4H 2 O (l) Acid-base reaction Mg(OH) 2(s) + 2H 3 O + (aq)  Mg 2+ (aq) + 4H 2 O (l) pH decrease ([H 3 O + ] increase), solubility increase. 2)The Effect Of pH On Solubility

50. Chapter 16TMHsiung©2015Slide 50 of 64 1)Whether Precipitation Occurs Calculating Q ip, the reaction quotient: Q > K sp precipitation occur Q = K sp saturated, i.e., at equilibrium Q < K sp precipitation cannot occur The effect of dilution when solutions are mixed must be considered. 6. Precipitation Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound. *****

51. Chapter 16TMHsiung©2015Slide 51 of 64 Solution Possible cross products: NaNO 3 soluble PbBr 2 K sp = 4.67 × 10 –6 Calculate Q and compare it to K sp. A precipitate will only form if Q > K sp. Q < Ksp; therefore no precipitate forms. A solution containing lead(II) nitrate is mixed with one containing sodium bromide to form a solution that is 0.0150 M in Pb(NO 3 ) 2 and 0.00350 M in NaBr. Does a precipitate form in the newly mixed solution? Example 16.12Predicting Precipitation Reactions by Comparing Q and K sp

52. Chapter 16TMHsiung©2015Slide 52 of 64 2)Selective Precipitation (a)When AgNO 3 (aq) is added to an aqueous solution containing Cl – and I –, the first precipitate to form yellow AgI (s). (b)Essentially most I – has precipitated before the precipitation of white AgCl (s) begins. Example AgCl (s)  Ag + (aq) +Cl – (aq) K sp = 1.8 x 10 –10 AgI (s)  Ag + (aq) +I – (aq) K sp = 8.5 x 10 –17 (a)(b)

53. Chapter 16TMHsiung©2015Slide 53 of 64 Solution The magnesium and calcium ions present in seawater ([Mg 2+ ] = 0.059 M and [Ca 2+ ] = 0.011 M) can be separated by selective precipitation with KOH. What minimum [OH – ] triggers the precipitation of the Mg 2+ ion? Mg(OH) 2 has a K sp of 2.06 x 10 -13 and that Ca(OH) 2 has a K sp of 4.68 x 10 -6. Example 16.13Finding the Minimum Required Reagent Concentration for Selective Precipitation *****

54. Chapter 16TMHsiung©2015Slide 54 of 64 Solution Use the value of K sp for calcium hydroxide and solve for [OH – ]. This is the concentration above which Ca(OH) 2 precipitates. Example 16.14Finding the Concentrations of Ions Left in Solution after Selective Precipitation You add potassium hydroxide to the solution in Example 16.13. When the [OH – ] reaches 1.9 × 10 –6 M (as you just calculated), magnesium hydroxide begins to precipitate out of solution. As you continue to add KOH, the magnesium hydroxide continues to precipitate. However, at some point, the [OH – ] becomes high enough to begin to precipitate the calcium ions as well. What is the concentration of Mg 2+ when Ca 2+ begins to precipitate?

55. Chapter 16TMHsiung©2015Slide 55 of 64 7.Qualitative Chemical Analysis 1)Classical qualitative inorganic analysis Identify inorganic ions by acid-base chemistry, precipitation reactions, oxidation-reduction, and complex-ion formation etc. “Qualitative” interest in what is present, not how much is present. Classical qualitative analysis relevant to all the basic concepts of equilibria in aqueous solutions. Quantitative analysis is concerned with quantity, or the amounts of substances in a solution or mixture.

56. Chapter 16TMHsiung©2015Slide 56 of 64 2)Procedures of Qualitative Analysis

57. Chapter 16TMHsiung©2015Slide 57 of 64 3)A General Qualitative Analysis Scheme

58. Chapter 16TMHsiung©2015Slide 58 of 64 1) Complex-Ion Consists of a central metal atom or ion bonded with ligands. Metal center, Lewis acid, accepts electron pairs Ligands, Lewis bases, donate electron pairs. At least one lone pair of electron in the Lewis structures of ligands (Lewis base). The ligands and metal center are jointed by coordinate covalent (dative) bonds. The constant of formation reaction of a complex ion is called a formation constant (or stability constant). 8.Complex Ion Formation

59. Chapter 16TMHsiung©2015Slide 59 of 64 Common ligands examples Complex ion formation example

60. Chapter 16TMHsiung©2015Slide 60 of 64 Some Formation Constants

61. Chapter 16TMHsiung©2015Slide 61 of 64 Example (from other textbook) Calculate the concentration of Ag +, in an aqueous solution prepared as 0.10 M AgNO 3 and 3.0 M NH 3. Ag + (aq) + 2NH 3(aq)  [Ag(NH 3 ) 2 ] + (aq) K f = 1.6 x 10 7 Assume all shift right firstly Solution: *****

62. Chapter 16TMHsiung©2015Slide 62 of 64 1)The Effect of Complex Ion Equilibria on Solubility AgCl (s) in NH 3(aq) solution for example: AgCl (s)  Ag + (aq) + Cl – (aq) K sp = 1.77 x 10 –10 Ag + (aq) + 2NH 3(aq)  [Ag(NH 3 ) 2 ] + (aq) K f = 1.7 x 10 7 AgCl (s) + 2NH 3(aq)  [Ag(NH 3 ) 2 ] + (aq) + Cl – (aq) K = K sp x K f = 3.0 x 10 –3 In the presence of a high concentration of NH 3, a significant amount of AgCl dissolves.

63. Chapter 16TMHsiung©2015Slide 63 of 64 2)The Solubility of Amphoteric Metal Hydroxides Amphoteric hidroxides: Al(OH) 3, Zn(OH) 2, and Cr(OH) 3 etc. Example: Al(OH) 3(s) + 3H 3 O + (aq) → [Al(H 2 O) 6 ] 3+ (aq) Al(OH) 3(s) + OH – (aq) → [Al(OH) 4 ] – (aq) Dissolution by forming complex

64. Chapter 16TMHsiung©2015Slide 64 of 64 End of Chapter 16