1. 16-1 CHEM 102, Fall 2014, LA TECH Instructor: Dr. Upali Siriwardane e-mail: upali@latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,W 8:00-9:00 & 11:00-12:00 am; Tu,Th,F 9:30 - 11:30 am. or by appointment. Test Dates : 9:30-10:45 am., CTH 328 Chemistry 102(01) Spring 2014 March 31, 2014 (Test 1): Chapter 13 April 23, 2014 (Test 2): Chapter 14 &15 May 19, 2014 (Test 3) Chapter 16 &17 May 21, 2014 (Make-up test) comprehensive: Chapters 13-17

2. 16-2 CHEM 102, Fall 2014, LA TECH Chapter 16. Aditional Aqueous Equilibria 17.1 Buffer Solutions 17.2 Acid-Base Titrations 17.3 Acid Rain 17.4 Solubility Equilibria and the Solubility Product Constant, K sp 17.5 Factors Affecting Solubility / Precipitation: Will It Occur?

3. 16-3 CHEM 102, Fall 2014, LA TECH Reaction of a basic anion or acidic cation with water is an ordinary Brønsted-Lowry acid-base reaction. CH 3 COO- (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) NH4 + (aq) + H 2 O (l) NH 3 (aq) + H 3 O + (aq) This type of reaction is given a special name.Hydrolysis The reaction of an anion with water to produce the conjugate acid and OH -. The reaction of a cation with water to produce the conjugate base and H 3 O +.Hydrolysis

4. 16-4 CHEM 102, Fall 2014, LA TECH Acid-Base Properties of Typical Ions

5. 16-5 CHEM 102, Fall 2014, LA TECH What salt solutions would be acidic, basic and neutral? 1)strong acid + strong base = neutral 2)weak acid + strong base = basic 3)strong acid + weak base = acidic 4) weak acid + weak base = neutral, basic or an acidic solution depending on the relative strengths of the acid and the base. basic or an acidic solution depending on the relative strengths of the acid and the base.

6. 16-6 CHEM 102, Fall 2014, LA TECH What pH? Neutral, basic or acidic? a)NaCla)NaCl neutral neutral b) NaC 2 H 3 O 2b) NaC 2 H 3 O 2 basic basic c) NaHSO 4c) NaHSO 4 acidic acidic d) NH 4 Cld) NH 4 Cl acidic acidic

7. 16-7 CHEM 102, Fall 2014, LA TECH 1) If the following substance is dissolved in pure water, will the solution be acidic, neutral, or basic? a) Solid sodium carbonate-(Na 2 CO 3 ): b) Sodium chloride- (NaCl): c) Sodium acetate- (NaC 2 H 3 O 2 ): d) Ammonium sulfate-((NH 4 ) 2 SO 4 ):

8. 16-8 CHEM 102, Fall 2014, LA TECH How do you calculate pH of a salt solution? Find out the pH, acidic or basic? If acidic it should be a salt of weak base If basic it should be a salt of weak acid if acidic calculate K a from K a = K w /K b if basic calculate K b from K b = K w /K a Do a calculation similar to pH of a weak acid or base

9. 16-9 CHEM 102, Fall 2014, LA TECH What is the pH of 0.5 M NH 4 Cl salt solution? (NH 3 ; K b = 1.8 x 10 -5 ) Find out the pH, acidic if acidic calculate K a from K a = K w /K b K a = K w /K b = 1 x 10 -14 /1.8 x 10 -5 ) K a = 5.56. X 10 -10 Do a calculation similar to pH of a weak acid

10. 16-10 CHEM 102, Fall 2014, LA TECH Continued NH 4 + + H 2 O H 3 + O + NH 3 NH 4 + + H 2 O H 3 + O + NH 3 [NH 4 + ] [H 3 + O ] [NH 3 ] [NH 4 + ] [H 3 + O ] [NH 3 ] Ini. Con. 0.5 M 0.0 M0.00 M Change -x x x Eq. Con. 0.5 - x x x [H 3 + O ] [NH 3 ] [H 3 + O ] [NH 3 ] K a (NH 4 + ) = --------------------= [NH 4 + ] [NH 4 + ] x 2 x 2 ---------------- ; appro.:0.5 - x. 0.5 (0.5 - x) (0.5 - x)

11. 16-11 CHEM 102, Fall 2014, LA TECH x 2 x 2 K a (NH 4 + ) = ----------- = 5.56 x 10 -10 0. 5 0. 5 x 2 = 5.56 x 10 -10 x 0.5 = 2.78 x 10 -10 x 2 = 5.56 x 10 -10 x 0.5 = 2.78 x 10 -10 x= 2.78 x 10 -10 = 1.66 x 10 -5 [H + ] = x = 1.66 x 10 -5 M pH = -log [H + ] = - log 1.66 x 10 -5 pH = 4.77 pH of 0.5 M NH 4 Cl solution is 4.77 (acidic ) Continued

12. 16-12 CHEM 102, Fall 2014, LA TECH 2) What is the pH of a 0.05 M aqueous NH 4 Cl solution? (K b (NH 3 ) = 1.8 x 10 -5 ) a) equilibrium reaction for the hydrolysis of salt: b) K a for the conjugate acid NH 4 + : c) ICE set-up: I:___________________________________________C:__________________________________________E___________________________________________ d) Calculation of x, [H 3 O] + : e) pH of the solution:

13. 16-13 CHEM 102, Fall 2014, LA TECH 3) What is the pH of a 0.05 M aqueous NaC 2 H 3 O 2 solution? (K a (HC 2 H 3 O 2 ) = 1.8 x 10 -5 ) a) equilibrium reaction for the hydrolysis of salt: b) Kb for the conjugate base C 2 H 3 O 2 - : c) ICE set-up: I:___________________________________________C:__________________________________________E___________________________________________ d) Calculation of x, [OH] - : e) pOH and pH of the solution:

14. 16-14 CHEM 102, Fall 2014, LA TECH Acid-Base Chemistry of Some Antacids

15. 16-15 CHEM 102, Fall 2014, LA TECH 4) A 50.00-mL sample of 0.100 M KOH is being titrated with 0.100 M HNO3. Calculate the pH of the solution after 52.00 mL of HNO 3 is added. a) acid base reaction: b) moles of KOH: c) moles of HNO 3 : d) [H 3 O] + : e) pH of the solution:

16. 16-16 CHEM 102, Fall 2014, LA TECH Reaction of a basic anion with water is an ordinary Brønsted- Lowry acid-base reaction. CH 3 COO- (aq) + H 2 O (l) CH 3 COOH (aq) + OH- (aq) This type of reaction is given a special name.Hydrolysis The reaction of an anion with water to produce the conjugate acid and OH-. The reaction of a cation with water to produce the conjugate base and H 3 O+.Hydrolysis

17. 16-17 CHEM 102, Fall 2014, LA TECH Common Ion Effect Weak acid and salt solutions E.g. HC 2 H 3 O 2 and NaC 2 H 3 O 2 Weak base and salt solutions E.g. NH 3 and NH 4 Cl. H 2 O + C 2 H 3 O 2 - OH - + HC 2 H 3 O 2 (common ion) H 2 O + C 2 H 3 O 2 - OH - + HC 2 H 3 O 2 (common ion) H 2 O + NH 4 + H 3 + O + NH 3 H 2 O + NH 4 + H 3 + O + NH 3 (common ion) (common ion)

18. 16-18 CHEM 102, Fall 2014, LA TECH Solutions that resist pH change when small amounts of acid or base are added. Two types Mixture of weak acid and its salt Mixture of weak base and its salt HA (aq) + H 2 O (l) H 3 O+ (aq) + A- (aq) Add OH- Add H 3 O+ shift to right shift to left Based on the common ion effect.Buffers

19. 16-19 CHEM 102, Fall 2014, LA TECH The pH of a buffer does not depend on the absolute amount of the conjugate acid-base pair. It is based on the ratio of the two. Henderson-Hasselbalch equation. Easily derived from the Ka or Kb expression. Starting with an acid pH = pK a + log Starting with a base pH = 14 - ( pK b + log ) [HA] [A-] [HA]Buffers

20. 16-20 CHEM 102, Fall 2014, LA TECH Henderson-Hasselbalch Equation HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) [H 3 O + ] [A - ] [H 3 O + ] [A - ] K a = ---------------- K a = ---------------- [HA] [HA] [H 3 O + ] = K a  ([HA]/[A - ]) [H 3 O + ] = K a  ([HA]/[A - ]) pH = pK a + log([A - ]/[HA]) when the [A - ] = [HA] pH = pK a

21. 16-21 CHEM 102, Fall 2014, LA TECH Calcualtion of pH of Buffers Henderson Hesselbach Equation [ACID] [ACID] pH = pK a - log --------- [BASE] [BASE] pH = pK a + log --------- [ACID] [ACID]

22. 16-22 CHEM 102, Fall 2014, LA TECH Control of blood pH Oxygen is transported primarily by hemoglobin in the red blood cells. CO2 is transported both in plasma and the red blood cells. CO 2 (aq) + H 2 OH 2 CO 3 (aq) H + (aq) + HCO 3 - (aq) The bicarbonate buffer is essential for controlling blood pH The bicarbonate buffer is essential for controlling blood pH Buffers and blood

23. 16-23 CHEM 102, Fall 2014, LA TECH Buffer Capacity Refers to the ability of the buffer to retard changes in pH when small amounts of acid or base are added The ratio of [A - ]/[HA] determines the pH of the buffer whereas the magnitude of [A - ] and [HA] determine the buffer capacity

24. 16-24 CHEM 102, Fall 2014, LA TECH Adding an Acid or Base to a Buffer

25. 16-25 CHEM 102, Fall 2014, LA TECH Buffer Systems

26. 16-26 CHEM 102, Fall 2014, LA TECH 5) For the (buffer effect) of HC 2 H 3 O 2 /NaC 2 H 3 O 2 a) Acid dissociation reaction: b) Salt hydrolysis reaction: c) Common ions in both equilibria: d) Which way salt hydrolysis equilibrium move adding H 3 O + : e) Which way salt hydrolysis equilibrium move adding OH - :

27. 16-27 CHEM 102, Fall 2014, LA TECH 6) Describe the (buffer effect) of NH 3 /NH 4 Cl a) Buffer type: (weak acid or base)/soluble salt): b) Base dissociation reaction: c) Salt hydrolysis reaction: d) common ions in both equilibria: e) Which way salt hydrolysis equilibrium move adding H 3 O + : f) Which way salt hydrolysis equilibrium move adding OH - :

28. 16-28 CHEM 102, Fall 2014, LA TECH 7) What is the pH of a solution that is 0.2 M in acetic acid (K a = 1.8 x 10 -5 ) and 0.2 M in sodium acetate? a) Is it a acid, base, salt or buffer solution? b) Henderson-Hesselbalch equation: c) pK a : d) e) pH of the solution:

29. 16-29 CHEM 102, Fall 2014, LA TECH Titrations of Acids and Bases TitrationAnalyteTitrant analyte + titrant => products

30. 16-30 CHEM 102, Fall 2014, LA TECH Acid-base indicators are highly colored weak acids or bases. HIn In- + H+ color 1color 2 color 1 color 2 They may have more than one color transition. Example. Example. Thymol blue Red - Yellow - Blue One of the forms may be colorless - phenolphthalein (colorless to pink)Indicators

31. 16-31 CHEM 102, Fall 2014, LA TECH Acid-Base Indicator HIn + H 2 O H 3 O + + In - acid base acid base color color color color [H 3 O + ][In - ] Ka =Ka =Ka =Ka = [HIn] [HIn] They may have more than one color transition. Example. Thymol blue Red - Yellow - Blue Red - Yellow - Blue Weak acid that changes color with changes in pH

32. 16-32 CHEM 102, Fall 2014, LA TECH What is an Indicator? Indicator is an weak acid with different K a, colors to the acid and its conjugate base. E.g. phenolphthalein Hin H + + In - Hin H + + In - colorless pink colorless pink Acidic colorless Basic pink

33. 16-33 CHEM 102, Fall 2014, LA TECH Selection of an indicator for a titration Selection of an indicator for a titration a) strong acid/strong base a) strong acid/strong base b) weak acid/strong base b) weak acid/strong base c) strong acid/weak base c) strong acid/weak base d) weak acid/weak base d) weak acid/weak base Calculate the pH of the solution at he equivalence point or end point

34. 16-34 CHEM 102, Fall 2014, LA TECH pH and Color of Indicators

35. 16-35 CHEM 102, Fall 2014, LA TECH Red Cabbage as Indicator

36. 16-36 CHEM 102, Fall 2014, LA TECH Acid-base indicators are weak acids that undergo a color change at a known pH. phenolphthalein pH Indicator examples

37. 16-37 CHEM 102, Fall 2014, LA TECH 8) If 50 ml of a 0.01 M HCl solution is titrated with a 0.01 M NaOH solution, what will be the concentration of salt (NaCl) the pH at the endpoint? a) NaCl solution acidic, basic or neutral? b) Concentration of [NaCl]: c) pH of the solution? d) Suitable indicator for the titration:

38. 16-38 CHEM 102, Fall 2014, LA TECH Titration Apparatus Burette delivering base to a flask containing an acid. The pink color in the flask is due to the phenolphthalein indicator.

39. 16-39 CHEM 102, Fall 2014, LA TECH Endpoint vs. Equivalence Point Endpoint point where there is a physical change, such as color change, with the indicator Equivalence Point # moles titrant = # moles analyte #moles titrant =(V  M) titrant #moles analyte =(V  M) analyte

40. 16-40 CHEM 102, Fall 2014, LA TECH 9) If 50 mL of a 0.01 M HCl solution is titrated with a 0.01 M NH 3 (K b = 1.8 x 10 -5 ) solution, what will be a) The initial pH (0.01 M NH 3 ): b) Concentration of NH 4 Cl at the endpoint: c) pH at the endpoint: d) Suitable indicator for the titration:

41. 16-41 CHEM 102, Fall 2014, LA TECH 10) If 50 ml of a 0.01 M HC 2 H 3 O 2 solution is titrated with a 0.01 M NaOH solution, what will be the a) Molarity of NaC 2 H 3 O 2 at the endpoint: b) The pH at the endpoint: c) What indicator would be most suitable for this titration:

42. 16-42 CHEM 102, Fall 2014, LA TECH Polyprotic Acids

43. 16-43 CHEM 102, Fall 2014, LA TECH Organic or Carboxylic Acids

44. 16-44 CHEM 102, Fall 2014, LA TECH FCH 2 CO 2 H (strongest acid) > ClCH 2 CO 2 H > BrCH 2 CO 2 H (weakest acid). FCH 2 CO 2 H (strongest acid) > ClCH 2 CO 2 H > BrCH 2 CO 2 H (weakest acid). Acid K a pK a HCOOH (formic acid) 1.78 X 10 -43 0.75 CH 3 COOH (acetic acid) 1.74 X 10 -54 0.76 CH 3 CH 2 COOH (propanoic acid)1.38 x 10 -5 4.86 Organic or Carboxylic Acids

45. 16-45 CHEM 102, Fall 2014, LA TECH Acid-Base in the Kitchen vinegar - acetic acid lemon juice (citrus juice) - citric acid baking soda - NaHCO 3 milk - lactic acid baking powder - H 2 PO 4 - & HCO 3 -

46. 16-46 CHEM 102, Fall 2014, LA TECH Household Cleaners

47. 16-47 CHEM 102, Fall 2014, LA TECH Dishwashing Detergent

48. 16-48 CHEM 102, Fall 2014, LA TECH Acid-Base Indicator Behavior acid color shows when [In - ] 1  [HIn] 10 [H 3 O + ][In - ] 1 =  [H 3 O + ] = K a [HIn] 10 base color shows when [In-] 1  [HIn] 10 [H 3 O+][In-] = 10  [H 3 O+] = K a [HIn]

49. 16-49 CHEM 102, Fall 2014, LA TECH Indicator pH Range acid color shows when pH + 1 = pK a and base color shows when pH - 1 = pK a Color change range is pK a = pH  1 or pH = pK a  1

50. 16-50 CHEM 102, Fall 2014, LA TECH Titration curves Acid-base titration curve A plot of the pH against the amount of acid or base added during a titration. Plots of this type are useful for visualizing a titration. It also can be used to show where an indicator undergoes its color change.

51. 16-51 CHEM 102, Fall 2014, LA TECH Indicator and Titration Curve 0.1000 M HCl vs 0.1000 M NaOH

52. 16-52 CHEM 102, Fall 2014, LA TECH EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. at 0.00 mL of NaOH added, initial point [H 3 O + ] = [HCl] = 0.1000 M pH = 1.0000

53. 16-53 CHEM 102, Fall 2014, LA TECH EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. at 15.00 mL of NaOH added V a  M a > V b  M b thus (V a  M a ) - (V b  M b ) [H3O+] = (V a + V b ) ((35.00mL)  (0.1000M)) - ((15.00mL)  (0.1000M)) = (35.00 + 15.00)mL = 4.000  10 -2 M pH = 1.3979

54. 16-54 CHEM 102, Fall 2014, LA TECH EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. at 35.00 mL of NaOH added V a  M a = V b  M b, equivalence point at equivalence point of a strong acid - strong base titration pH  7.0000

55. 16-55 CHEM 102, Fall 2014, LA TECH EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. at 50.00 mL of NaOH added Vb  Mb > Va  Ma, post equvalence point (Vb  Mb) - (Va  Ma) [OH-] = (Va + Vb) ((50.00mL)  (0.1000M)) - ((35.00mL)  (0.1000M)) = (35.00 + 50.00)mL = 1.765  10 -2 M pOH = 1.7533 pH = 14.00 - 1.7533 = 12.25

56. 16-56 CHEM 102, Fall 2014, LA TECH equivalence pointx Titration of Strong Acid with Strong Base

57. 16-57 CHEM 102, Fall 2014, LA TECH Titration of Weak Acid with Strong Base

58. 16-58 CHEM 102, Fall 2014, LA TECH Effect of Acid Strength on Titration Curve

59. 16-59 CHEM 102, Fall 2014, LA TECH Titration of Weak Base with Strong Acid

60. 16-60 CHEM 102, Fall 2014, LA TECH Titration of Diprotic Weak Acid with Strong Base

61. 16-61 CHEM 102, Fall 2014, LA TECH pH range of Indicators pH range of Indicators litmus (5.0-8.0) bromothymole blue (6.0-7.6) methyl red (4.8-6.0) thymol blue (8.0-9.6) phenolphthalein (8.2-10.0) thymolphthalein (9.4-10.6) thymolphthalein (9.4-10.6)

62. 16-62 CHEM 102, Fall 2014, LA TECH Acid Rain acid rain is defined as rain with a pH < 5.6 pH < 5.6 pH = 5.6 for rain in equilibrium with atmospheric carbon dioxide

63. 16-63 CHEM 102, Fall 2014, LA TECH Sulfuric Acid from Sulfur burning SO 2 S + O 2 => SO 2 SO 3 2 SO 2 + O 2 => 2 SO 3 Sulfuric Acid SO 3 + H 2 O => H 2 SO 4

64. 16-64 CHEM 102, Fall 2014, LA TECH Nitric Acid 2 NO 2(g) + H 2 O (l) => HNO 3(aq) + HNO 2(aq)

65. 16-65 CHEM 102, Fall 2014, LA TECH How Acid Precipitation Forms

66. 16-66 CHEM 102, Fall 2014, LA TECH Acid Precipitation in U.S.

67. 16-67 CHEM 102, Fall 2014, LA TECH Solubility Product solubility-product - the product of the solubilities solubility-product constant => K sp constant that is equal to the solubilities of the ions produced when a substance dissolves

68. 16-68 CHEM 102, Fall 2014, LA TECH Solubility Product Constant In General: A x B y  xA +y + yB -x [A +y ]x [B -x ]y K = [A x B y ] [A x B y ] K = K sp = [A +y ] x [B -x ] y K sp = [A +y ] x [B -x ] y For silver sulfate Ag 2 SO 4  2 Ag + + SO 4 -2 K sp = [Ag + ] 2 [SO 4 -2 ]

69. 16-69 CHEM 102, Fall 2014, LA TECH Solubility Product Constant Values

70. 16-70 CHEM 102, Fall 2014, LA TECH Dissolving Slightly Soluble Salts Using Acids Insoluble salts containing anions of Bronsted-Lowry bases can be dissolved in solutions of low pH

71. 16-71 CHEM 102, Fall 2014, LA TECH Calcium Carbonate Dissolved in Acid Limestone Dissolving in Ground Water CaCO 3(S) + H 2 O + CO 2 => Ca +2 (aq) + 2 HCO 3 - (aq) Stalactite and Stalagmite Formation Ca +2 (aq) + 2 HCO 3 - (aq) => CaCO 3(S) + H 2 O + CO 2

72. 16-72 CHEM 102, Fall 2014, LA TECH The Common Ion Effect common ion second source which is completely dissociated In the presence of a second source of the ion, there will be less dissolved than in its absence common ion effect a salt will be less soluble if one of its constitutent ions is already present in the solution

73. 16-73 CHEM 102, Fall 2014, LA TECH EXAMPLE: What is the molar solubility of AgCl in pure water and in 1.0 M NaCl? AgCl  Ag + + Cl - K sp = [Ag + ][Cl - ] = 1.82  10 -10 M 2 let x = molar solubility = [Ag + ] = [Cl - ] (x)(x) = K sp = [Ag + ][Cl - ] = 1.82  10 -10 M 2 x = 1.35  10 -5 M

74. 16-74 CHEM 102, Fall 2014, LA TECH EXAMPLE: What is the molar solubility of AgCl in pure water and in 1.0 M NaCl? AgCl  Ag + + Cl - K sp = [Ag + ][Cl - ] = 1.82  10 -10 M 2 let x = molar solubility = [Ag + ] [Cl - ] = 1.0 M K sp = [Ag + ][Cl - ] = (x)(1.0M) = 1.82  10 -10 M 2 x = 1.82  10 -10 M

75. 16-75 CHEM 102, Fall 2014, LA TECH Formation of Complexes ligand - Lewis base complexes - product of Lewis acid-base reaction Ag + (aq) + 2 NH 3(aq)  [Ag(NH 3 ) 2(aq) ] + Ag + (aq) + Cl - (aq)  AgCl (s) AgCl (s) + 2 NH 3(aq)  [Ag(NH 3 ) 2(aq) ] + + Cl - (aq)

76. 16-76 CHEM 102, Fall 2014, LA TECH Sodium Thiosulfate Dissolves Silver Bromide

77. 16-77 CHEM 102, Fall 2014, LA TECH 11) For a saturated solution of a) AgCl in water: i) Solubility equilibrium reaction: ii) K sp expression:

78. 16-78 CHEM 102, Fall 2014, LA TECH 11) For a saturated solution of b) CaF 2 in water: i) Solubility equilibrium reaction: ii) K sp expression: c) Fe 2 S 3 in water i) Solubility equilibrium reaction: ii) Ksp expression:

79. 16-79 CHEM 102, Fall 2014, LA TECH 12) Which of following has the highest molar solubility (mole/L)? Salt K sp Salt K sp a) CaCO 3 5 × 10 -9 b) PbCO3 1.4 × 10 -13 c) Li 2 CO 3 2 × 10 -3 d) NiCO 3 1.2 × 10 -7

80. 16-80 CHEM 102, Fall 2014, LA TECH Formation Constants

81. 16-81 CHEM 102, Fall 2014, LA TECH Amphoterism

82. 16-82 CHEM 102, Fall 2014, LA TECH Reactant Quotient, Q ion product of the initial concentration same form as solubility product constant Q < K sp - no precipitate forms an unsaturate solution Q > K sp - precipitate may form to restore condition of saturated solution Q = K sp - no precipitate forms, saturated solution

83. 16-83 CHEM 102, Fall 2014, LA TECH Will Precipitation Occur?

84. 16-84 CHEM 102, Fall 2014, LA TECH 13) For Li 2 CO 3, K sp is 2 × 10 -3 M 3. What is the concentration of Li + in a saturated solution of Li 2 CO 3 ?

85. 16-85 CHEM 102, Fall 2014, LA TECH 14) Chemical analysis gave [Pb 2+ ] = 0.012 M, and [Br - ] = 0.024 M in a solution. From a table, you find K sp for PbBr 2 has a value of 4 x 10 -5 M 3. a) Solubility equilibrium reaction: b) Q sp : c) K sp : d) Q sp K sp ? e) Is the solution saturated, oversaturated or unsaturated?

86. 16-86 CHEM 102, Fall 2014, LA TECH Kidney Stones Kidney stones are normally composed of: calcium oxalate calcium phosphate magnesium ammonium phsphate

87. 16-87 CHEM 102, Fall 2014, LA TECH Calcium Oxalate Precipitate formed from calcium ions from food rich in calcium, dairy products, and oxalate ions from fruits and vegetables Ca +2 + C 2 O 4 -2  CaC 2 O 4

88. 16-88 CHEM 102, Fall 2014, LA TECH PRECIPITATION REACTIONS

89. 16-89 CHEM 102, Fall 2014, LA TECH Analysis of Silver Group These salts formed are insoluble, they do dissolve to some SLIGHT extent. AgCl(s) Ag + (aq) + Cl - (aq) When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED.

90. 16-90 CHEM 102, Fall 2014, LA TECH Analysis of Silver Group AgCl(s) Ag + (aq) + Cl - (aq) When solution is SATURATED, expt. shows that [Ag + ] = 1.67 x 10 -5 M. (SOLUBILITY) of AgCl. What is [Cl - ]?

91. 16-91 CHEM 102, Fall 2014, LA TECH Analysis of Silver Group AgCl(s) Ag + (aq) + Cl - (aq) Saturated solution has [Ag + ] = [Cl - ] = 1.67 x 10 -5 M Use this to calculate K sp K sp = [Ag + ] [Cl - ] K sp = [Ag + ] [Cl - ] = (1.67 x 10 -5 )(1.67 x 10 -5 ) = (1.67 x 10 -5 )(1.67 x 10 -5 ) = 2.79 x 10 -10 = 2.79 x 10 -10

92. 16-92 CHEM 102, Fall 2014, LA TECH Because this is the product of “solubilities”, we call it K sp = solubility product constant See Table in the Text K sp = solubility product

93. 16-93 CHEM 102, Fall 2014, LA TECH Lead(II) Chloride PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq) K sp = 1.9 x 10 -5

94. 16-94 CHEM 102, Fall 2014, LA TECH Consider PbI 2 dissolving in water PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) Calculate K sp if solubility = 0.00130 M Solution 1.Solubility = [Pb 2+ ] = 1.30 x 10 -3 M [I - ] = 2 x [Pb 2+ ] = 2.60 x 10 -3 M [I - ] = 2 x [Pb 2+ ] = 2.60 x 10 -3 M 2.K sp = [Pb 2+ ] [I - ] 2 = [Pb 2+ ] {2 [Pb 2+ ]} 2 = [Pb 2+ ] {2 [Pb 2+ ]} 2 = 4 [Pb 2+ ] 3 = 4 (solubility) 3 = 4 [Pb 2+ ] 3 = 4 (solubility) 3 K sp = 4 (1.30 x 10 -3 ) 3 = 8.8 x 10 -9 K sp = 4 (1.30 x 10 -3 ) 3 = 8.8 x 10 -9 Solubility of Lead(II) Iodide

95. 16-95 CHEM 102, Fall 2014, LA TECH Precipitating an Insoluble Salt Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 If [Hg 2 2+ ] = 0.010 M, what [Cl - ] is req’d to just begin the precipitation of Hg 2 Cl 2 ? (maximum [Cl - ] in 0.010 M Hg 2 2+ without forming Hg 2 Cl 2 ?)

96. 16-96 CHEM 102, Fall 2014, LA TECH Precipitating an Insoluble Salt Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 Recognize that K sp = product of maximum ion concs. Precip. begins when product of ion Concs. EXCEEDS the K sp.

97. 16-97 CHEM 102, Fall 2014, LA TECH Precipitating an Insoluble Salt Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 Solution [Cl - ] that can exist when [Hg 2 2+ ] = 0.010 M, If this conc. of Cl - is just exceeded, Hg 2 Cl 2 begins to precipitate.

98. 16-98 CHEM 102, Fall 2014, LA TECH Precipitating an Insoluble Salt Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 Now raise [Cl - ] to 1.0 M. What is the value of [Hg 2 2+ ] at this point? Solution [Hg 2 2+ ] = K sp / [Cl - ] 2 = K sp / (1.0) 2 = 1.1 x 10 -18 M = K sp / (1.0) 2 = 1.1 x 10 -18 M The concentration of Hg 2 2+ has been reduced by 10 16 !

99. 16-99 CHEM 102, Fall 2014, LA TECH Separating Metal Ions Cu 2+, Ag +, Pb 2+ K sp Values AgCl 1.8 x 10 -10 PbCl 2 1.7 x 10 -5 PbCrO 4 1.8 x 10 -14 K sp Values AgCl 1.8 x 10 -10 PbCl 2 1.7 x 10 -5 PbCrO 4 1.8 x 10 -14 Cu 2+ Ag + Pb 2+ Cl - Insoluble PbCl 2 AgCl Soluble CuCl 2 Heat Insoluble AgCl Soluble PbCl 2 CrO 4 -2 Insoluble PbCrO 4

100. 16-100 CHEM 102, Fall 2014, LA TECH Separating Salts by Differences in K sp A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate Ag 2 CrO 4 (red) and PbCrO 4 (yellow). Which precipitates first? K sp for Ag 2 CrO 4 = 9.0 x 10 -12 K sp for PbCrO 4 = 1.8 x 10 -14 Solution The substance whose K sp is first exceeded precipitates first. The ion requiring the smaller amount of CrO 4 2- ppts. first.

101. 16-101 CHEM 102, Fall 2014, LA TECH Separating Salts by Differences in K sp [CrO 4 2- ] to ppt. PbCrO 4 = K sp / [Pb 2+ ] = 1.8 x 10 -14 / 0.020 = 9.0 x 10 -13 M = 1.8 x 10 -14 / 0.020 = 9.0 x 10 -13 M [CrO 4 2- ] to ppt. Ag 2 CrO 4 = K sp / [Ag + ] 2 = 9.0 x 10 -12 / (0.020) 2 = 2.3 x 10 -8 M = 9.0 x 10 -12 / (0.020) 2 = 2.3 x 10 -8 M PbCrO 4 precipitates first. PbCrO 4 precipitates first. Solution Calculate [CrO 4 2- ] required by each ion.

102. 16-102 CHEM 102, Fall 2014, LA TECH How much Pb 2+ remains in solution when Ag + begins to precipitate? Solution We know that [CrO 4 2- ] = 2.3 x 10 -8 M to begin to ppt. Ag 2 CrO 4. What is the Pb 2+ conc. at this point? [Pb 2+ ] = K sp / [CrO 4 2- ] = 1.8 x 10 -14 / 2.3 x 10 -8 M = 7.8 x 10 -7 M = 7.8 x 10 -7 M Lead ion has dropped from 0.020 M to < 10 -6 M Separating Salts by Differences in Ksp

103. 16-103 CHEM 102, Fall 2014, LA TECH Common Ion Effect Adding an Ion “Common” to an Equilibrium PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) NaClNa + (aq) + Cl - (aq)

104. 16-104 CHEM 102, Fall 2014, LA TECH Calculate the solubility of BaSO 4 in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution a) Solubility in pure water = [Ba 2+ ] = [SO 4 2- ] = x K sp = [Ba 2+ ] [SO 4 2- ] = x 2 x = (K sp ) 1/2 = 1.1 x 10 -5 M The Common Ion Effect

105. 16-105 CHEM 102, Fall 2014, LA TECH BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) K sp = 1.1 x 10 -10 Solution b) Now dissolve BaSO 4 in water already containing 0.010 M Ba 2+. Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSO 4 be less than or greater than in pure water?___ The Common Ion Effect

106. 16-106 CHEM 102, Fall 2014, LA TECH BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution [Ba 2+ ][SO 4 2- ] [Ba 2+ ][SO 4 2- ]initialchangeequilib. The Common Ion Effect

107. 16-107 CHEM 102, Fall 2014, LA TECH Calculate the solubility of BaSO 4 in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution [Ba 2+ ][SO 4 2- ] [Ba 2+ ][SO 4 2- ] initial0.010 0 change+ y + y equilib.0.010 + y y The Common Ion Effect

108. 16-108 CHEM 102, Fall 2014, LA TECH K sp = [Ba 2+ ] [SO 4 2- ] = (0.010 + y) (y) Because y << x (1.1 x 10 -5 M) 0.010 + y  0.010. Therefore, K sp = 1.1 x 10 -10 = (0.010)(y) y = 1.1 x 10 -8 M = solubility in presence of added Ba 2+ ion. Le Chatelier’s Principle is followed! The Common Ion Effect