1. 1

2. Buffers Titration Curves Solubility Equilibria Two important points: 1. Reactions with strong acids or strong bases go to completion. 2. Reactions with only weak acids and bases reach an equilibrium. 2

3. The Common Ion Effect 3

4. Weak acid: HA + H 2 O ⇌ H 3 O + + A - + Salt of conjugate Base: NaA → Na + (aq) + A - (aq) = two sources of A - Common Ion ! What affect does the addition of its conjugate base have on the weak acid equilibrium? On the pH? What affect does the addition of its conjugate base have on the weak acid equilibrium? On the pH? 4

5. Example: Using equilibrium constants from Appendix D, calculate the pH of the solution containing 0.085 M HC 3 H 5 O 2 (propanoic acid) HC 3 H 5 O 2 (aq) ↔ H +1 (aq) + C 3 H 5 O 2 -1 (aq) K a =1.3x10 -5 Initial: 0.085 0 0 change: -x +x +x Equilibrium: 0.085-x x x (X * X) / (0.085 – X) = 1.3 x 10 -5 X 2 / (0.085-X) = 1.3*10 -5 X 2 + 1.3 x 10 -5 X - 1.11 x 10 -6 = 0 X = 0.00104, pH = -log(1.04*10 -3 ) = 2.98

6. Example 2: Using equilibrium constants from Appendix D, calculate the pH of the solution containing 0.060 M KC 3 H 5 O 2 (Potassium Propionate) and 0.085 M HC 3 H 5 O 2 change:-x+x+x Equilibrium: 0.085-xx.060+ x x 2 +.060013x-(1.105*10 -6 )=0 x=1.84*10 -5 pH = -log(1.84*10 -5 ) = 4.74

7. The Common Ion Effect Consider the ionization of a weak acid, acetic acid: HC 2 H 3 O 2 (aq)  H + (aq) + C 2 H 3 O 2 – (aq) If we increase the [C 2 H 3 O 2 – ] ions by adding NaC 2 H 3 O 2, the equilibrium will shift to the left. ( Le Chatelier ) This reduces the [H + ] and raises the pH (less acidic) This phenomenon is called the common-ion effect. Common ion equilibrium problems are solved following the same pattern as other equilibrium problems (ICE charts) EXCEPT the initial concentration of the common ion must be considered (it is NOT zero).

8. Calculate the pH of a solution containing 1.0 M HF and 0.60 M KF. (Recall that the pH of 1.00 HF is 1.58.) Addition of F - shifts the equilibrium, reducing the [H + ]. 8

9. 9 Common Ion Effect

10. Example:Does the pH increase, decrease, or stay the same on addition of each of the following? (a) NaNO 2 to a solution of HNO 2 HNO 2  H + + NO 2 - pH, increases (b) (CH 3 NH 3 )Cl to a solution of CH 3 NH 2 CH 3 NH 2 + H 2 O  CH 3 NH 3 + + OH - pH, decreases (c) sodium formate to a solution of formic acid HCHO 2  H + + CHO 2 - pH, increases (d) potassium bromide to a solution of hydrobromic acid HBr  H + + Br - pH, no change (e) HCl to a solution of NaC 2 H 3 O 2 C 2 H 3 O 2 -1 + H 2 O  HC 2 H 3 O 2 + OH -1 pH, decreases

11. Buffer 11

12. What is a buffer? A buffer is a solution that resists changes in pH when small amounts of acid or base are added to it There are 2 types: ▫Acidic ▫Alkaline 12

13. Acidic buffers An acidic buffer has a pH less than 7 It is made from a equal molar concentrations of a weak acid and it’s conjugate base Example: acetic acid and acetic ion CH3COOH  CH3COO - + H + (weak acid) (conj. base) 13

14. Alkaline buffers An alkaline buffer has a pH greater than 7 It is made from a equal molar concentrations of a weak base and it’s conjugate acid Example: ammonia and ammonium ion NH3 + H2O   NH4 + + OH - (weak base) (conj. acid) 14

15. A BUFFER consists of a 1) mixture of a weak acid (HX) and its conjugate base (X – ) or 2) weak base (B) and its conjugate acid (HB + ) Thus a buffer contains both: an acidic species to neutralize added OH – When a small amount of OH – is added to the buffer solution, the OH – reacts with the acid in the buffer solution. a basic species to neutralize added H + When a small amount of H + is added to the buffer solution, the H + reacts with the base in the buffer solution.

16. How do buffers work? 16

17. 17 Making an Acid Buffer

18. Composition of a Buffer - 4 ways to make a buffer solution: 1.)Weak acid + salt of the acid HCN and NaCN weak acid: HCN weak base: CN -1 2.)Weak base + salt of the base NH 3 and NH 4 Cl weak acid: NH 4 +1 weak base: NH 3 3.) EXCESS Weak acid + strong base 2 mol HCN + 1 mol NaOH  1 mol HCN + 1 mol NaCN + H 2 O weak acid: HCN weak base: CN -1 2 mol NH 4 Cl + 1 mol NaOH  1 mol NH 4 Cl + 1 mol NH 3 + NaCl weak acid: NH 4 +1 weak base: NH 3 4.)EXCESS Weak base + strong acid 2 mol NH 3 + 1 mol HCl  1 mol NH 3 and 1 mol NH 4 Cl weak acid: NH 4 +1 weak base: NH 3 2 mol NaF + 1 mol HCl  1 mol NaF + 1 mole HF + NaCl weak acid: HF weak base: F -1

19. Example: Explain why a mixture of HCl and KCl does not function as a buffer, whereas a mixture of HC 2 H 3 O 2 and NaC 2 H 3 O 2 does. HCl is a strong acid - Cl -1 is a negligible base and will NOT react with added H + - added H + will significantly change the pH of the solution HC 2 H 3 O 2 and C 2 H 3 O 2 -1 are a weak conjugate acid/base pair which act as a buffer HC 2 H 3 O 2 reacts with added base C 2 H 3 O 2 -1 reacts with added acid leaving the [H +1 ] and pH relatively unchanged

20. Henderson-Hasselbalch equation Just use the INITIAL concentrations

21. Example: Using equilibrium constants from Appendix D, calculate the pH of the solution containing 0.060 M KC 3 H 5 O 2 and 0.085 M HC 3 H 5 O 2

22. Example 5: Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11 M in sodium lactate

23. Example 6: A buffer is prepared by adding 20.0 g of acetic acid, HC 2 H 3 O 2 and 20.0 g of sodium acetate to enough water to form 2.00 L of solution. (a) Determine the pH of the buffer (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide are added to the buffer (b) C 2 H 3 O 2 -1 (aq) + H +1 (aq) + Cl -1 (aq)  HC 2 H 3 O 2 (aq) + Cl -1 (aq) (c) HC 2 H 3 O 2 (aq) + Na +1 (aq) + OH -1 (aq)  C 2 H 3 O 2 -1 (aq) + H 2 O (l) + Na +1 (aq)

24. pH range Buffers are only effective when the pH is within 1 unit from the pKa. pH = pKa ± 1 When this pH is exceeded, we have exceeded the buffer capacity. 24

25. Buffer Capacity Buffer capacity exceeded – when added acid or base totally consumes a buffer component Buffer capacity exceeded – when added acid or base totally consumes a buffer component 25

26. Understand Figures 26

27. Buffer Capacity Depends on pKa of CA/CB mix Depends on concentration of CA/CB 27

28. Blood Buffer 28

29. Titration Curves 29

30. Tro, Chemistry: A Molecular Approach 30 Acid-Base Titration

31. 31 Titration Curve: Unknown Strong Base Added to Strong Acid

32. Example – Titration Curves: Strong Acid-Strong Base Suppose that 50.00 mL of 0.200 M nitric acid is titrated against 0.100 M sodium hydroxide. What is the pH when: a.No NaOH is added? b.10.0 ml of 0.100 M NaOH is added? c.20.0 mL of 0.100 M NaOH is added? d.50.0 mL of 0.100 M NaOH is added? e.100.0 mL of 0.100 M NaOH is added? f.150.0 mL of 0.100 M NaOH is added? g.200.0 mL of 0.100 M NaOH is added?

33. Acid-Base Titrations – Titration Curves In an acid-base titration: A solution of base (or acid) of known concentration (called standard) is added to an acid (or base). Acid-base indicators or a pH meter are used to signal the equivalence point (when moles acid = moles base). The plot of pH versus volume during a titration is called a pH titration curve.

34. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Strong Acid-Strong Base Titration of HNO 3 with NaOH H + (aq) + OH - (aq)  H 2 O (l) V NaOH added (mL)pH 0.000.699 10.0 20.0 50.0 100.0 150.0 200.0

35. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Strong Acid-Strong Base Titration of HNO 3 with NaOH H + (aq) + OH - (aq)  H 2 O (l) V NaOH added (mL)pH 0.000.699 10.00.82 20.0 50.0 100.0 150.0 200.0

36. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Strong Acid-Strong Base Titration of HNO 3 with NaOH H + (aq) + OH - (aq)  H 2 O (l) V NaOH added (mL)pH 0.000.699 10.00.82 20.00.94 50.0 100.0 150.0 200.0

37. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Strong Acid-Strong Base Titration of HNO 3 with NaOH H + (aq) + OH - (aq)  H 2 O (l) V NaOH added (mL)pH 0.000.699 10.00.82 20.00.94 50.01.30 100.0 150.0 200.0

38. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Strong Acid-Strong Base Titration of HNO 3 with NaOH H + (aq) + OH - (aq)  H 2 O (l) V NaOH added (mL)pH 0.000.699 10.00.82 20.00.94 50.01.30 100.07.00 150.0 200.0

39. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Strong Acid-Strong Base Titration of HNO 3 with NaOH H + (aq) + OH - (aq)  H 2 O (l) V NaOH added (mL)pH 0.000.699 10.00.82 20.00.94 50.01.30 100.07.00 150.012.40 200.0

40. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Strong Acid-Strong Base Titration of HNO 3 with NaOH H + (aq) + OH - (aq)  H 2 O (l) V NaOH added (mL)pH 0.000.699 10.00.82 20.00.94 50.01.30 100.07.00 150.012.40 200.012.602

41. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Strong Acid-Strong Base Titration of HNO 3 with NaOH H + (aq) + OH - (aq)  H 2 O (l) V NaOH added (mL)pH 0.000.699 10.00.82 20.00.94 50.01.30 100.07.00 150.012.40 200.012.602 Major species at half-equivalence point: H +, NO 3 -, Na + pH = 1.30, V NaOH = 50.0 mL

42. Strong acid added to strong base equal moles of acid and base present Starts high Ends low Equivalence point = 7 start end

43. Strong base added to strong acid Starts low Ends high Equivalence point = 7 start end

44. Strong acid added to weak base Buffer area – in this area there is weak base and some salt of the weak base Actual pH depends on the salt formed but it will be < 7 Starts med-high Ends low Equivalence point < 7

45. Weak base added to strong acid Starts low Ends med-high Equivalence point < 7 start end

46. Weak acid added to strong base Actual pH depends on the salt formed but it will be > 7 Starts high Ends med-low Equivalence point > 7 start end

47. Weak acid added weak base Starts med-high Ends med-low Equivalence point = 7 start end

48. Strong base added to strong diprotic acid (H 2 SO 4 )

49. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Weak Acid-Strong Base Consider the titration of 50.0 mL of 0.10 M acetic acid (K a = 1.8 x 10 -5 ) with 0.10 M sodium hydroxide. What is the pH when: a.No NaOH is added? b.10.0 ml of 0.100 M NaOH is added? c.25.0 mL of 0.100 M NaOH is added? d.40.0 mL of 0.100 M NaOH is added? e.50.0 mL of 0.100 M NaOH is added? f.60.0 mL of 0.100 M NaOH is added? g.75.0 mL of 0.100 M NaOH is added?

50. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Weak Acid-Strong Base Titration of HC 2 H 3 O 2 with NaOH HC 2 H 3 O 2 (aq) + OH - (aq)  H 2 O (l) + C 2 H 3 O 2 - (aq) V NaOH added (mL)pH 0.002.87 10.0 25.0 40.0 50.0 60.0 75.0

51. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Weak Acid-Strong Base Titration of HC 2 H 3 O 2 with NaOH HC 2 H 3 O 2 (aq) + OH - (aq)  H 2 O (l) + C 2 H 3 O 2 - (aq) V NaOH added (mL)pH 0.002.87 10.04.14 25.0 40.0 50.0 60.0 75.0

52. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Weak Acid-Strong Base Titration of HC 2 H 3 O 2 with NaOH HC 2 H 3 O 2 (aq) + OH - (aq)  H 2 O (l) + C 2 H 3 O 2 - (aq) V NaOH added (mL)pH 0.002.87 10.04.14 25.04.74 40.0 50.0 60.0 75.0

53. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Weak Acid-Strong Base Titration of HC 2 H 3 O 2 with NaOH HC 2 H 3 O 2 (aq) + OH - (aq)  H 2 O (l) + C 2 H 3 O 2 - (aq) V NaOH added (mL)pH 0.002.87 10.04.14 25.04.74 40.05.35 50.0 60.0 75.0

54. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Weak Acid-Strong Base Titration of HC 2 H 3 O 2 with NaOH HC 2 H 3 O 2 (aq) + OH - (aq)  H 2 O (l) + C 2 H 3 O 2 - (aq) V NaOH added (mL)pH 0.002.87 10.04.14 25.04.74 40.05.35 50.08.73 60.0 75.0

55. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Weak Acid-Strong Base Titration of HC 2 H 3 O 2 with NaOH HC 2 H 3 O 2 (aq) + OH - (aq)  H 2 O (l) + C 2 H 3 O 2 - (aq) V NaOH added (mL)pH 0.002.87 10.04.14 25.04.74 40.05.35 50.08.73 60.011.96 75.0

56. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Weak Acid-Strong Base Titration of HC 2 H 3 O 2 with NaOH HC 2 H 3 O 2 (aq) + OH - (aq)  H 2 O (l) + C 2 H 3 O 2 - (aq) V NaOH added (mL)pH 0.002.87 10.04.14 25.04.74 40.05.35 50.08.73 60.011.96 75.012.30

57. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Weak Acid-Strong Base Titration of HC 2 H 3 O 2 with NaOH HC 2 H 3 O 2 (aq) + OH - (aq)  H 2 O (l) + C 2 H 3 O 2 - (aq) V NaOH added (mL)pH 0.002.87 10.04.14 25.04.74 40.05.35 50.08.73 60.011.96 75.012.30

58. Chapter 16 – Aqueous Ionic Equilibrium Example – Titration Curves: Weak Acid-Strong Base

59. 59 Titration Curve of a Weak Base with a Strong Acid

60. Tro, Chemistry: A Molecular Approach 60 Titration of a Polyprotic Acid Titration of 25.0 mL of 0.100 M H 2 SO 3 with 0.100 M NaOH H 2 SO 3 (aq) + OH - (aq)  HSO 3 - (aq) + H 2 O (l) HSO 3 - (aq) + OH - (aq)  SO 3 2- (aq) + H 2 O (l) Net Rxn: H 2 SO 3 (aq) + 2 OH - (aq)  SO 3 2- (aq) + 2 H 2 O (l)

61. Example: Predict whether the equivalence point of each of the following titrations is below, above or at pH 7: a) NaHCO 3 titrated with NaOH b) NH 3 titrated with HCl c) KOH titrated with HBr At the equivalence point, only products are present in solution, so determine the products of the reaction and then determine if the solution is acidic, basic or neutral a) NaHCO 3 + NaOH  Na 2 CO 3 + H 2 O weak acid strong base pH > 7 CO 3 -2 is basic, Na + is neutral, H 2 O is neutral b) NH 3 + HCl  NH 4 Cl weak base strong acidpH < 7 NH 4 +1 is acidic, Cl - is neutral c) KOH + HBr  KBr + H 2 O strong base strong acidpH = 7 K + and Br - are both neutral

62. Example 8: How many mL of 0.0850 M NaOH solution is required to titrate 40.0 mL of 0.0900 M HNO 3 ? ? mL 40.0 mL M NaOH V NaOH = M HNO3 V HNO3 0.0850 M 0.0900 M NaOH + HNO 3  H 2 O + NaNO 3 1 mole

63. Example: A 20.0 mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution. Calculate the pH of the solution after the following volumes of base solution have been added: (a) 15.0 mL (b) 19.9 mL (c) 20.0 mL (d) 20.1 mL (e) 35.0 mL mLmLmLmol H +1 mol OH -1 M ofpH HBrNaOHTotal(M) (V)(M) (V)excess ion (mol / tot vol) (a)20.015.035.00.004000.003000.0286 M H +1 1.544 (b)20.019.939.90.004000.003980.0005 M H +1 3.3 (c)20.020.040.00.004000.004001 x 10 -7 M H +1* 7.0 (d)20.020.140.10.004000.004020.0005 M OH -1 10.7 (e) 20.035.055.00.004000.007000.0545 M OH -1 12.736 When molarity of H + (or OH - ) is less than 10 -6 we must consider the autoionization of water! (H + = 1.0*10 -7 )

64. Example: Calculate the pH at the equivalence point for titrating 0.200 M solutions of each of the following bases with 0.200 M HBr: (a) NaOH (b) NH 2 OH (a) strong acid/strong base titration so pH = 7 (b) HBr + NH 2 OH  Br - + NH 2 OH 2 + strong acid weak base.200M all product at equivalence point – no excess & Br - is neutral and will have no affect on pH Volume doubles (equal molarity and 1:1 stoich ratio) so molarity halves [NH 2 OH 2 + ] = 0.200mol / 2 = 0.100 M NH 2 OH 2 +  H +1 + NH 2 OH K b = 1.1 x 10 -8 (appendix) I.100 0 0 C -x +x +x E0.100 – x x x K a = K w / K b = 1  10 -14 / 1.1  10 -8 = 9.1  10 -7 (x 2 ) / (0.100-x) = 9.1 x 10 -7 x = 3.0 x 10 -4 M = [H +1 ] pH = - log(3.0 x 10 -4 ) = 3.52

65. Effect of K a on Titration curve 65

66. 66 Monitoring pH During a Titration

67. Acid – Base Indicators marks the end point of a titration by changing color. Acid-base indicators are, weak acids (represented by HIn).

68. Acid – Base Indicators Bromthymol Blue Indicator In Acid In Base

69. Acid – Base Indicators Methyl Orange Indicator In Acid In Base

70. Acid – Base Indicators Phenolphthalein color at different pH values pH values 5 6 7 8 9

71. Acid – Base Indicators Consider a hypothetical indicator, HIn, a weak acid with K a =1.0x10 -8. It has a red color in acid and a blue color in base. HIn(aq)  H +1 (aq) + In -1 (aq) red blue

72. 72 Phenolphthalein

73. Tro, Chemistry: A Molecular Approach 73 Methyl Red

74. 74 Acid-Base Indicators

75. The pH curve for the titration of 100.0 mL of 0.10 M HCl with 0.10 M NaOH. Neither of the indicators shown would be useful for a titration. Bromthymol blue (pK a =7) would be useful.

76. The pH curve for the titration of 50 mL of 0.1 M HC 2 H 3 O 2 with 0.1 M NaOH. Here, phenolphthalein is the indicator of choice. It has a pK a value of about 9.

77. Aqueous Equilibria Titration Curves Information from the Curve: There are several things you can read from the titration curve itself. Consider this titration curve 1.This is a ___________ (strong/weak) acid titrated with a strong base. The acid is ________________ (monoprotic/diprotic). How would the other strength of acid look? 2.Place a dot (  ) on the curve at the equivalence point. The pH at the equivalence point is ____. Choose a good indicator for this titration. 3.What volume of base was used to titrate the acid solution? _______ mL 4.Place a box ( ) on the curve where the pH of the solution = the pK a of the acid. What is the pH at this point? _______ What is the pK a of the acid? _______ What is the K a of the acid? _____________ weak mono If it was a strong acid the curve would be more “sharp”. ~8.8 Phenolphthalein 25 ~4.8 1.58 x 10 -5 77

78. Aqueous Equilibria Titration Curves Information from the Curve: There are several things you can read from the titration curve itself. Consider this titration curve 1. Identify the solutions involved in this titration. 2.Place a dot (  ) on the curve at the equivalence point. The pH at the equivalence point is ____. Choose a good indicator for this titration. 3.What volume of base was used to titrate the acid solution? _______ mL 4.Place a box ( ) on the curve where the pH of the solution = the pK b of the base. What is the pH at this point? _______ What is the pK b of the acid? _______ What is the K b of the acid? _____________ A strong base in a weak acid ~5.8 Methyl red 20 ~8.8 1.58 x 10 -9 78

79. Solubility Product K sp 79

80. Solubility Equilibria Solubility – maximum amount of material that can dissolve in a given amount of solvent at a given temperature; units of g/100 g or M Solubility – maximum amount of material that can dissolve in a given amount of solvent at a given temperature; units of g/100 g or M Insoluble compound – compound with a solubility less than 0.01 M; also sparingly soluble. Insoluble compound – compound with a solubility less than 0.01 M; also sparingly soluble. 80

81. The Solubility-Product Constant, K sp equilibrium expression for the dissolving of BaSO 4. BaSO 4 (s)  Ba 2+ (aq) + SO 4 2– (aq) Since BaSO 4 (s) is a pure solid, the equilibrium expression depends only on the concentration of the ions. K sp = [Ba 2+ ][ SO 4 2– ] K sp is the equilibrium constant for the equilibrium between an ionic solid solute and its saturated aqueous solution. K sp is called the solubility-product constant

82. In general: the solubility product is equal to the product of the molar concentration of ions raised to powers corresponding to their stoichiometric coefficients. Al 2 (CO 3 ) 3 (S) → 2 Al +3 (aq) + 3 CO 3 -2 (aq) K sp = [Al +3 ] 2 [CO 3 -2 ] 3

83. Solubility Product Calculations In concentration tables, x = solubility In concentration tables, x = solubility Problem types Problem types 1. solubility → K sp 2. K sp → solubility 83

84. Know your solubility rules: SOLUBILITY GUIDELINES Soluble CompoundsExceptions NOT precipitatesPRECIPITATES NitratesNone AcetatesNone ChloratesNone ChloridesAg +1, Hg 2 +2, Pb +2 BromidesAg +1, Hg 2 +2, Pb +2 IodidesAg +1, Hg 2 +2, Pb +2 SulfatesCa +2, Sr +2, Ba +2, Hg 2 +2, Pb +2 Insoluble CompoundsExceptions PRECIPITATESNOT Precipitates SulfidesNH 4 +1, Li +1, Na +1, K +1, Ca +2, Sr +2, Ba +2 CarbonatesNH 4 +1, Li +1, Na +1, K +1 PhosphatesNH 4 +1, Li +1, Na +1, K +1 HydroxidesLi +1, Na +1, K +1, Ca +2, Sr +2, Ba +2 ChromatesNH 4 +1, Li +1, Na +1, K +1, Ca +2, Mg +2 We classify these based on the Solubility - maximum amount of solute that dissolves in water.

85. Solubility and K sp Solubility is the amount of substance that dissolves to form a saturated solution. This can be expressed as grams of solid that will dissolve per liter of solution. Molar solubility - the number of moles of solute that dissolve to form a liter of saturated solution. Solubility can be used to find K sp and K sp can be used to find solubility.

86. Example: a. If the molar solubility of CaF 2 at 35 o C is 1.24  10 -3 mol/L, what is K sp at this temperature? CaF 2  Ca +2 + 2 F -1.00124M actually dissolves K sp = [Ca +2 ] [F -1 ] 2 = (.00124 M)(.00248 M) 2 = 7.63 x 10 -9 b. It is found that 1.1  10 -2 g of SrF 2 dissolves per 100 mL of aqueous solution at 25 o C. Calculate the solubility product of SrF 2. [SrF 2 ] = (.011 g / 125.6 g/mole) /.100 L =.00088 M SrF 2  Sr +2 + 2 F -1.00088 M.00088 M 2(.00088) =.00176 M K sp = [Sr +2 ] [F -1 ] 2 = (.00088 M)(.00176) 2 = 2.7 x 10 -9.00124 M 2(.00124) =.00248 M

87. c. The K sp of Ba(IO 3 ) 2 at 25 o C is 6.0  10 -10. What is the molar solubility of Ba(IO 3 ) 2 ? Ba(IO 3 ) 2  Ba +2 + 2 IO 3 -1 x (x) (2x) 2 = 4 x 3 = 6.0 x 10 -10 x = 5.3 x 10 -4 M x 2 x

88. Factors That Affect Solubility Factors that have a significant impact on solubility are: - The presence of a common ion - The pH of the solution Common-Ion Effect “Solubility is decreased when a common ion is added.” This is an application of Le Châtelier’s principle: Consider the solubility of CaF 2 : CaF 2 (s)  Ca 2+ (aq) + 2F – (aq) If more F – is added (say by the addition of NaF), the equilibrium shifts left to offset the increase. Therefore, more CaF 2 (s) is formed (precipitation occurs).

89. Example: Using Appendix D, calculate the molar solubility of AgBr in (a) pure water (b) 3.0  10 -2 M AgNO 3 solution (c) 0.50 M NaBr solution (a) AgBr (s)  Ag +1 (aq) + Br -1 (aq) K sp = 5.0  10 -13 x x x 5.0  10 -13 = x 2 x = 7.1  10 -7 M (b) AgBr(s)  Ag +1 (aq) + Br -1 (aq) K sp = 5.0  10 -13 x.030 + x x 5.0  10 -13 = (.030+x) x x = 1.7  10 -11 M (c) AgBr(s)  Ag +1 (aq) + Br -1 (aq) K sp = 5.0  10 -13 x x.50 + x 5.0  10 -13 = x (.50+x) x = 1.0  10 -12 M notice the DECREASED solubility with the common ion in (b) and (c)

90. pH effects Consider: Mg(OH) 2 (s)  Mg 2+ (aq) + 2 OH – (aq) If OH – is removed, then the equilibrium shifts right and Mg(OH) 2 dissolves. OH – can be removed by adding a strong acid (lowering the pH): OH – (aq) + H + (aq)  H 2 O(aq) Another example: CaF 2 (s)  Ca 2+ (aq) + 2 F – (aq) If the F– is removed, then the equilibrium shifts right and CaF2 dissolves. F – can be removed by adding a strong acid (or lowering pH): F – (aq) + H + (aq)  HF(aq)

91. Example: Calculate the molar solubility of Mn(OH) 2 at (a) pH 7.0 (b) pH 9.5 (c) pH 11.8 the [OH -1 ] is set by the pH (or pOH) (a) pH = 7.0 so pOH = 7.0 so [OH -1 ] = 1.00  10 -7 Mn(OH) 2 (s)  Mn +2 (aq) + 2 OH -1 (aq)K sp = 1.6  10 -13 x x 1.00 x 10 -7 K sp = [Mn +2 ][OH -1 ] 2 1.6  10 -13 = (x) (1.00  10 -7 ) 2 x = 16 M (b) pH = 9.5 so pOH = 4.5 so [OH -1 ] = 3.16  10 -5 Mn(OH) 2 (s)  Mn +2 (aq) + 2 OH -1 (aq)K sp = 1.6  10 -13 x x 3.16 x 10 -5 K sp = [Mn +2 ][OH -1 ] 2 1.6  10 -13 = (x) (3.16  10 -5 ) 2 x = 1.7  10 -4 M (c) pH = 11.8 so pOH = 2.2 so [OH -1 ] = 6.31  10 -3 Mn(OH) 2 (s)  Mn +2 (aq) + 2 OH -1 (aq) K sp = 1.6  10 -13 x x 6.31 x 10 -3 K sp = [Mn +2 ][OH -1 ] 2 1.6  10 -13 = (x) (6.31  10 -3 ) 2 x = 4.0  10 -9 M “Common ion effect – increasing [OH - ] decreases solubility”

92. Example: Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) ZnCO 3 (b) ZnS (c) BiI 3 (d) AgCN (e) Ba 3 (PO 4 ) 2 If the anion of the salt is the conjugate base of a weak acid, it will combine with H +1, reducing the concentration of the anion and making the salt more soluble ZnCO 3  Zn +2 + CO 3 -2 the CO 3 -2 ion will react with the added H + CO 3 -2 + H +  HCO 3 -1 Le Chatelier effect of removing CO 3 -2 more soluble in acid: ZnCO 3, ZnS, AgCN, Ba 3 (PO 4 ) 2

93. Precipitation and Separation of Ions Consider the following: BaSO 4 (s)  Ba 2+ (aq) + SO 4 2– (aq) At any instant in time, Q = [Ba 2+ ][ SO 4 2– ] If Q > K sp, (too many ions) precipitation occurs until Q = K sp. If Q = K sp equilibrium exists (saturated solution) If Q < K sp, (not enough ions) solid dissolves until Q = K sp.

94. 94 precipitation occurs if Q > K sp

95. Selective Precipitation of Ions Removal of one metal ion from a solution of two or more metal ions is called selective precipitation. Ions can be separated from each other based on the solubilities of their salt compounds. Example: If HCl is added to a solution containing Ag + and Cu 2+, the silver precipitates (as AgCl) while the Cu 2+ remains in solution Generally, the less soluble ion is removed first!

96. 96 Qualitative Analysis

97. 97

98. Example: Will Ca(OH) 2 precipitate if the pH of a 0.050 M solution of CaCl 2 is adjusted to 8.0? if Q > than K sp then precipitation will occur pH = 8.0 so pOH = 6.0 so [OH -1 ] = 1.0  10 -6 M Ca(OH) 2 (s)  Ca +2 (aq) + 2 OH -1 (aq) K sp = 6.5  10 -6 --.050 1.00 x 10 -6 Q = [Ca +2 ] [OH -1 ] 2 Q = (.050)(1.0  10 -6 ) 2 = 5.0  10 -14 Q < K so no precipitation occurs

99. Example: A solution contains 0.00020 M Ag +1 and 0.0015 M Pb +2. If NaI is added, will AgI or PbI 2 precipitate first? Specify the [I -1 ] needed to begin precipitation for each cation. the cation needing the lower [I -1 ] will precipitate first AgI (s)  Ag +1 (aq) + I -1 (aq) K sp = 8.3 x 10 -17 --.000200 x K sp = [Ag +1 ][x] 8.3  10 -17 = (.00020)[x] 4.2  10 -13 = x = [I -1 ] PbI 2 (s)  Pb +2 (aq) + 2 I -1 (aq)K sp = 1.4 x 10 -8 -- 0.0015 x K sp = [Pb +2 ][x] 2 1.4  10 -8 = (.0015)[x] 2 3.1  10 -3 = x = [I -1 ] AgI will precipitate first at an [I -1 ] = 4.2  10 -13

100. Comparing Salt Solubilities Generally: solubility ↑ K sp ↑ Can only compare K sp values if the salts produce the same number of ions Can only compare K sp values if the salts produce the same number of ions If different numbers of ions are produced, solubility must be compared. If different numbers of ions are produced, solubility must be compared. 100

101. Factors that Affect Solubility 1. Common-Ion Effect LeChatelier’s Principle revisited LeChatelier’s Principle revisited Addition of a product ion causes the solubility of the solid to decrease, but the K sp remains constant. 2. pH LeChatelier’s Principle again! LeChatelier’s Principle again! Basic salts are more soluble in acidic solution. Acidic salts are more soluble in basic solution. Environmental example: CaCO 3 – limestone Stalactites and stalagmites form due to changing pH in the water and, thus, change the solubility of the limestone. 101

102. Common-ion Effect 102

103. Effect of pH on Solubility 103