1. Unit 8 (Chp 15,17): Chemical & Solubility Equilibrium (K eq, K c, K p, K sp ) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

2. Dynamic (forward & reverse rxns occur simultaneously) double arrow Equilibrium:

3. Initially, forward & reverse rxns occur at different rates. (based on collisions, one slows down; other speeds up) Rate f Rate r At equilibrium: Rate forward = Rate reverse Dynamic Equilibrium:

4. A System at Equilibrium At equilibrium, concentrations (M, mol, P, etc.) of reactant and product remain constant.

5. HW p. 660 #2 …concentrations are ________ constant …forward and reverse rates are ________ equal At Equilibrium…

6. The Equilibrium Constant

7. Consider the reaction K eq = [C] c [D] d [A] a [B] b aA + bBcC + dD At equilibrium… Rate f = Rate r k f [A] a [B] b = k r [C] c [D] d k f [C] c [D] d k r [A] a [B] b = [products] [reactants]

8. The Equilibrium Constant The equilibrium constant expression (K eq ) is K c = [C] c [D] d [A] a [B] b [ ] is conc. in M K expressions do not include: pure solids(s) or pure liquids( l ) (b/c concentrations are constant) K = [products] [reactants] Consider the reaction aA + bBcC + dD

9. In three experiments at the same temperature, equilibrium was achieved & data were collected. The Equilibrium Constant Calculate K c for each experiment at this temperature and compare the values. Exp. Butanoic acid (moles) Ethanol (moles) Ethyl butanoate (moles) Water (moles) Products Reactants (Kc) 110.0 20.0 25.020.010.040.0 330.01.012.010.0 4 4 4 same ratio (constant)

10. What Does the Value of K Mean? If K > 1, the reaction is product-favored; more product at equilibrium. If K < 1, the reaction is is reactant-favored; more reactant at equilibrium. K = [products] [reactants] Reactants  Products

11. K p = (P C ) c (P D ) d (P A ) a (P B ) b The Equilibrium Constant Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written Example: 3 Fe (s) + 4 H 2 O (g) ↔ Fe 3 O 4 (s) + 4 H 2 (g)

12. Heterogeneous Equilibria The concentrations of solids and liquids do not appear in the equilibrium expression. K c = ? PbCl 2 (s) Pb 2+ (aq) + 2 Cl − (aq) K c = [Pb 2+ ] [Cl − ] 2

13. As long as some CaCO 3 or CaO remains, the amount of CO 2 above the solid be constant. CaCO 3 (s) CO 2 (g) + CaO (s) K c = [CO 2 ] K p = P CO 2 K c = ? K p = ?

14. K of reverse rxn = 1/K K c = = 4.0 [NO 2 ] 2 [N 2 O 4 ] N2O4N2O4 2 NO 2 K c = = 1. (4.0) [N 2 O 4 ] [NO 2 ] 2 N2O4N2O4 2 NO 2 ↔ ↔ K of multiplied reaction = K^ # (raised to power) K c = = (4.0) 2 [NO 2 ] 4 [N 2 O 4 ] 2 4 NO 2 2 N 2 O 4 ↔ Manipulating K

15. K of combined reaction = K 1 x K 2 … A + 3 D  3 B + 4 E K ovr = ? A  3 B + 2 CK 1 = 2.5 2 C + 3 D  4 EK 2 = 60 Manipulating K K ovr = (2.5)(60) HW p. 661 #14, 16, 20

16. Equilibrium Calculations

17. A closed system initially containing 0.100 M H 2 and 0.200 M I 2 at 448  C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M. Calculate K c at 448  C for the reaction taking place, which is H 2 (g) + I 2 (g) 2 HI (g) 1)find limiting reactant 2)mol reactant completely to mol product, but… NOW we still have some reactant left and some product formed at equilibrium.

18. Reaction Initial Change Equilibrium H 2 I 2 2 HI + 0.100 M H 2 and 0.200 M I 2 at 448  C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M. Calculate K c at 448  C. RICE Tables

19. Reaction Initial Change Equilibrium H 2 I 2 2 HI + 0.100 M H 2 and 0.200 M I 2 at 448  C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M. [H 2 ] in = 0.100 M[I 2 ] in = 0.200 M What Do We Know? [HI] eq = 0.187 M [HI] in = 0 M 0 M 0.187 M 0.100 M 0.200 M

20. Initial0.100 M0 M Change+0.187 Equilibrium0.187 M Reaction Initial0.200 M Change Equilibrium [HI] Increases by 0.187 M H 2 I 2 2 HI + 0.100 M H 2 and 0.200 M I 2 at 448  C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M. Calculate K c at 448  C.

21. Reaction Initial0.100 M0.200 M0 M Change Equilibrium Initial Change–0.0935 +0.187 Equilibrium0.187 M Stoichiometry shows [H 2 ] and [I 2 ] decrease by half as much H 2 I 2 2 HI + 0.187 M HI x 1 mol H 2 = 0.0935 M H 2 2 mol HI

22. Reaction Initial Change Equilibrium Initial0.100 M0.200 M0 M Change–0.0935 +0.187 Equilibrium0.0065 M0.1065 M0.187 M We can now calculate the equilibrium concentrations of all three compounds… H 2 I 2 2 HI + Calculate K c at 448  C. Kc =Kc = [HI] 2 [H 2 ] [I 2 ] = (0.187) 2 (0.0065)(0.1065) = 51 HW p. 662 #27, 30, 40

23. ICE Practice given K, solve for x HW p. 663 #43 At 2000 o C the equilibrium constant for the reaction 2 NO (g) ↔ N 2 (g) + O 2 (g) is K c = 2.4 x 10 3. If the initial concentration of NO is 0.200 M, what are the equilibrium concentrations of NO, N 2, and O 2 ?

24. What Do We Know? Reaction Initial Change Equilibrium 2 NO N 2 O 2 + K c = 2.4 x 10 3 the initial concentration of NO is 0.200 M

25. What Do We Know? Reaction Initial0.200 M0 M Change Equilibrium 2 NO N 2 O 2 + K c = 2.4 x 10 3 the initial concentration of NO is 0.200 M What do we NOT know? ???

26. Initial0.2000 Change– 2x+ x Equilibrium 2 NO N 2 O 2 + Stoichiometry shows [NO] decreases by twice as much as [N 2 ] and [O 2 ] increases. Reaction Initial00 Change Equilibrium

27. Reaction Initial0 Change Equilibrium Initial0.20000 Change– 2x+ x Equilibrium0.200 – 2xxx We now have the equilibrium concentrations of all three compounds… (in terms of x) 2 NO N 2 O 2 + Now, what was the question again? what are the equilibrium concentrations of NO, N 2, and O 2 ?

28. Kc =Kc = [N 2 ] [O 2 ] [NO] 2 HW p. 663 # 44 49 = x (0.200 – 2x) x = 0.099 2.4 x 10 3 = (x) 2 (0.200 – 2x) 2 9.8 – 98x = x 9.8 = 99x [N 2 ] eq = 0.099 M [O 2 ] eq = 0.099 M [NO] eq = 0.0020 M Equilibrium0.200 – 2xxx (next slide)

29. ICE Practice given K, solve for x HW p. 663 #44 For the equilibrium Br 2 (g) + Cl 2 (g) ↔ 2 BrCl (g) at 400 K, K c = 7.0. If 0.30 mol of Br 2 and 0.30 mol of Cl 2 are introduced into a 1.0 L container at 400 K, what will be the equilibrium concentrations of Br 2, Cl 2, and BrCl?

30. What Do We Know? Reaction Initial Change Equilibrium Br 2 Cl 2 2 BrCl + K c = 7.0 [Br 2 ] in = 0.30 M [Cl 2 ] in = 0.30 M the initial concentrations of Br 2 and Cl 2 are 0.30 M

31. What Do We Know? Reaction Initial0.30 M 0 M Change Equilibrium What do we NOT know? ??? Br 2 Cl 2 2 BrCl + K c = 7.0 [Br 2 ] in = 0.30 M [Cl 2 ] in = 0.30 M the initial concentrations of Br 2 and Cl 2 are 0.30 M

32. Reaction Initial0.30 M Change Equilibrium Initial0.30 M0 M Change– x + 2x Equilibrium Br 2 Cl 2 2 BrCl Stoichiometry shows [BrCl] increases by twice as much as [Br 2 ] and [Cl 2 ] decrease. +

33. Reaction Initial Change Equilibrium Initial0.30 M 0 M Change– x + 2x Equilibrium0.30 – x 2x We now have the equilibrium concentrations of all three compounds… (in terms of x) Now, what was the question again? what are the equilibrium concentrations of Br 2, Cl 2, and BrCl? Br 2 Cl 2 2 BrCl +

34. Kc =Kc = [BrCl] 2 [Br 2 ][Cl 2 ] 2.6 = 2x (0.30 – x) x = 0.17 7.0 = (2x) 2 (0.30 – x) 2 0.78 – 2.6x = 2x 0.78 = 4.6x [Br 2 ] eq = 0.13 M [Cl 2 ] eq = 0.13 M [BrCl] eq = 0.34 M Equilibrium0.30 – x 2x HW p. 664 #64 (next slide)

35. ICE Practice given K, solve for x HW p. 664 #64 For the equilibrium 2 IBr (g) ↔ I 2 (g) + Br 2 (g) K p = 8.5 x 10 –3 at 150 o C. If 0.025 atm of IBr is placed in a 2.0 L container, what is the partial pressure of this substance after equilibrium is reached?

36. What Do We Know? Reaction Initial Change Equilibrium 2 IBr I 2 Br 2 + K p = 8.5 x 10 –3 the initial pressure of IBr is 0.025 atm

37. What Do We Know? Reaction Initial0.025 atm0 atm Change Equilibrium 2 IBr I 2 Br 2 + What do we NOT know? ??? K p = 8.5 x 10 –3 the initial pressure of IBr is 0.025 atm

38. Reaction Initial0 atm Change Equilibrium Initial0.025 atm0 atm Change– 2x+ x Equilibrium 2 IBr I 2 Br 2 + Stoichiometry shows [NO] decreases by twice as much as [N 2 ] and [O 2 ] increases.

39. Reaction Initial0 atm Change Equilibrium Initial0.025 atm0 atm Change– 2x+ x Equilibrium0.025 – 2xxx We now have the equilibrium concentrations of all three compounds… (in terms of x) 2 IBr I 2 Br 2 + Now, what was the question again? What is the equilibrium partial pressure of IBr?

40. Kp =Kp = (P I 2 )(P Br 2 ) (P IBr ) 2 0.092 = x (0.025 – 2x) x = 0.0019 8.5 x 10 –3 = (x) 2 (0.025 – 2x) 2 0.0023 – 0.184x = x 0.0023 = 1.184x (P IBr ) eq = 0.021 atm Equilibrium0.025 – 2xxx HW p. 664 #66 (next slide)

41. ICE Practice HW p. 664 #66 Solid NH 4 HS is introduced into an evacuated flask at 24 o C. The following reaction takes place: NH 4 HS (s) ↔ NH 3 (g) + H 2 S (g) At equilibrium the total pressure in the container is 0.614 atm. What is K p for this equilibrium at 24 o C?

42. Reaction Initial Change Equilibrium Reaction Initial…0 atm Change Equilibrium What Do We Know? NH 4 HS (s) NH 3 H 2 S + At equilibrium, the total pressure (P T ) eq is 0.614 atm. K p = ? Solid NH 4 HS is introduced into an evacuated flask at 24 o C. 0.614 atm

43. What Do We Know? Reaction Initial…0 atm Change Equilibrium What do we NOT know? ??? NH 4 HS (s) NH 3 H 2 S + 0.614 atm At equilibrium, the total pressure (P T ) eq is 0.614 atm. K p = ? ??

44. Initial…0 atm Change– x+ x Equilibrium Reaction Initial0 atm Change Equilibrium Stoichiometry shows [NH 3 ] increases by the same amount as [H 2 S] increases. NH 4 HS (s) NH 3 H 2 S + 0.614 atm

45. Initial…0 atm Change– x+ x Equilibrium…xx Reaction Initial0 atm Change Equilibrium We now have the equilibrium concentrations of all three compounds… (in terms of x) Now, what was the question again? What is K p ? NH 4 HS (s) NH 3 H 2 S + 0.614 atm

46. K p = (P NH 3 )(P H 2 S ) K p = (x) 2 K p = (0.307) 2 K p = 0.0942 Equilibrium…xx WS Equil Calc’s III P T = P NH 3 + P H 2 S 0.614 = x + x 0.614 = 2x P T = 0.614 atm x = 0.307

47. The Reaction Quotient (Q) A Q expression gives the same ratio as the equilibrium (K) expression, but … …for a system that is NOT at equilibrium. Kc = Kc = [C] c [D] d [A] a [B] b aA + bBcC + dD Calculate Q by substituting INITIAL (current) concentrations into the Q expression. Q = [C] c [D] d [A] a [B] b NOT (given on exam)

48. K Q rate f = rate r R P KQ Q = [P] [R] K Q If Q = K, system is at equilibrium (K). Q = KQ = K = K

49. If Q < K, too much reactant, system will shift right faster to reach equilibrium (K). rate f > rate r R P rate f = rate r R P Q =Q = [P] [R] K Q KQK Q Q < KQ < K Q =Q = [P] [R]

50. If Q > K, too much product, system will shift left faster to reach equilibrium (K). rate f < rate r R P rate f = rate r R P Q = [P] [R] K Q KQK Q Q > KQ > K HW p. 662 #36, 38 Q =Q = [P] [R]

51. Le Châtelier’s Principle

52. rate f > rate r R P Le Châtelier’s Principle: “Systems at equilibrium disturbed by a change amount (M, PP) of reactants or products volume (V) of container temperature (T) (changes K value) …will shift (  or  ) to counteract the change.” Because… …the rxn goes faster in one direction (Q  K) until rate f = rate r (Q = K again, R P ). R P (that affects collision frequency) like: rate f < rate r or

53. The conversion of nitrogen (N 2 ) & hydrogen (H 2 ) into ammonia (NH 3 ) is tremendously significant in agriculture, where ammonia-based fertilizers are of utmost importance. The Haber Process N 2 (g) + 3 H 2 (g)  2 NH 3 (g)

54. K =K = [NH 3 ] 2 [N 2 ][H 2 ] 3 Q =Q = [NH 3 ] 2 [N 2 ] [H 2 ] 3 K =K = [NH 3 ] 2 [N 2 ] [H 2 ] 3 Q < KQ < K Q = K (same K) Q = K Add reactant: (changes Q)

55. This apparatus pushes the equilibrium to the right by removing the product ammonia (NH 3 ) from the system as a liquid. Remove product: N 2 (g) + 3 H 2 (g)  2 NH 3 (g) (changes Q)

56. Change Volume (moles of gas) ↓ V ( ↑ P T ) shifts to ↑ V ( ↓ P T ) shifts to V has no shift if fewer mol of gas (↓n gas ) more mol of gas (↑n gas ) N 2 (g) + 3 H 2 (g)  2 NH 3 (g) equal mol of gas R & P 1. 2 NO (g)  N 2 (g) + O 2 (g), ↑V will shift ____. 2. N 2 O 4 (g)  2 NO 2 (g), ↓V will shift ____. 3. CaCO 3 (s)  CO 2 (g) + CaO (s), shift  by _V. NOT  ↓ ↑ P T by adding noble gas? no shift b/c same R & P pressures (changes Q)

57. Change Temperature Changing temp. is the ONLY way to change the value of ___. K Why? K1 =K1 = [P] [R] K2 =K2 = [P] [R] K2 =K2 = [P] [R] (larger) (smaller) (original)  T (add heat) shifts…  T (add heat) shifts… …in the _____thermic direction to ______ heat

58. Change Temperature CoCl 4 2– + 6 H 2 O (l)  4 Cl – + Co(H 2 O) 6 2+ ∆H = __ heat + + heat Remove heat Add heat ice Remove product:Add product: –

59. increase rate f and rate r. R P Equilibrium occurs faster, but… at no shift (composition [P]/[R] is same). [P][R][P][R] K =K = (same) Catalysts

60. N 2 (g) + 3 H 2 (g)  2 NH 3 (g) ∆H = –92 Le Châtelier’s Principle Change in external factor Shift to restore equilibrium Reason Increase pressure (decrease volume) Increase temp. Increase [N 2 ] Increase [NH 3 ] Add a catalyst (practice) Right   Left Right   Left No Shift ↑P (↓V) shifts to side of fewer moles of gas ∆H = –, adding heat shifts in the endo- dir. Adding reactant shifts right to consume it Adding product shifts left to consume it Catalysts change how fast, but not how far. + heat ∆H = –

61. Le Châtelier’s Principle Change in external factor Shift to restore equilibrium Reason ________ pressure (_______ volume) _________ temp. _________ P O 2 _________ P H 2 O Decrease H 2 O 2  Left Right   Left No Shift ↓P (↑V) shifts to side of more moles of gas ∆H = +, remove heat shifts in the exo- dir. Adding reactant shifts right to consume it Removing reactant shifts left to produce it (s) & (l) do not affect Q & K (usually) 2 H 2 O (g) + O 2 (g)  2 H 2 O 2 ( l ) ∆H = +108 (practice) Decrease Increase Decrease heat + Increase Decrease ∆H = +

62. shift away faster (consume) shift toward faster (replace ) fewer mol of gas (↓n gas ) Le Châtelier’s Principle more mol of gas (↑n gas ) R  P (summary) no shift (H + R  P) (R  P + H) WS Eq Pract. 2 #4 (changes K) (P total ) (M, PP R, PP P ) in endo dir. to use up heat in exo dir. to make more heat

63. Unit 8 (Chp 15, 17): Chemical & Solubility Equilibria (K eq, K c, K p, K sp ) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

64. K sp = [X + ] a [Y – ] b Solubility Product Constant (K sp ) Write the K expression for a saturated solution of PbI 2 in water: PbI 2 (s)  Pb 2+ (aq) + 2 I − (aq) X a Y b (s)  a X + (aq) + b Y − (aq) K = For “insoluble” solids, the equilibrium constant, K sp, is the solubility product constant, or the… …product of M’s of dissolved ions at equilibrium. [Pb 2+ ][I − ] 2 sp

65. grams of solid (s) dissolved in 1 L (g/L) mol of solid (s) dissolved in 1 L (M) product of conc.’s (M) of ions (aq) at equilibrium solubility: molar solubility: K sp : X a Y b  a X + + b Y – [X a Y b ] [X + ] [Y – ] molar solubility molar conc.’s of ions K sp = [X + ] a [Y – ] b K sp is NOT the Solubility

66. 4.2 g/L 0.027 M CaCrO 4 0.027 M Ca 2+ 0.027 M CrO 4 2– CaCrO 4 (s)  Ca 2+ + CrO 4 2– [CaCrO 4 ] [Ca 2+ ] [CrO 4 2– ] molar mass: 156.08 g/mol 0.00073 HW p. 763 #46 K sp = [Ca 2+ ][CrO 4 2– ] grams of solid (s) dissolved in 1 L (g/L) mol of solid (s) dissolved in 1 L (M) product of conc.’s (M) of ions (aq) at equilibrium solubility: molar solubility: K sp :

67. K sp Calculations PbBr 2 (s)  Pb 2+ + 2 Br – 0.010 M 0 M 0 M –0.010 +0.010 +0.020 0 M 0.010 M 0.020 M If solubility (or molar solubility) is known, solve for K sp. [PbBr 2 ] is 0.010 M at 25 o C. HW p. 763 #48a K sp = (0.010)(0.020) 2 K sp = 4.0 x 10 –6 1 PbBr 2 dissociates into… 1 Pb 2+ ion & 2 Br – ions R I C E K sp = [Pb 2+ ][Br – ] 2 (maximum that can dissolve) (all dissolved = saturated) (any excess solid is irrelevant)

68. R I C E K sp Calculations PbI 2 (s)  Pb 2+ + 2 I – 0.0012 M 0 M 0 M –0.0012 +0.0012 +0.0024 0 M 0.0012 M 0.0024 M K sp = [Pb 2+ ][I – ] 2 Solubility (or molar solubility) is known, solve for K sp. [PbI 2 ] = 0.0012 M HW p. 763 #50 K sp = (0.0012)(0.0024) 2 K sp = 6.9 x 10 –9 0.54 g x 1 mol 461 g =0.0012 mol solubility: 0.54 g/L Molar solubility: 0.0012 M

69. If only K sp is known, solve for x (M). AgCl (s)  Ag + + Cl – x 0 M 0 M –x +x +x 0 M x x K sp = [Ag + ][Cl – ] K sp = x 2 1.8 x 10 –10 = x 2 √1.8 x 10 –10 = x 1.3 x 10 –5 = x K sp for AgCl is 1.8 x 10 –10. [AgCl] = 1.3 x 10 –5 M [Ag + ] = 1.3 x 10 –5 M [Cl – ] = 1.3 x 10 –5 M (molar solubility) K sp Calculations R I C E

70. If only K sp is known, solve for x (M). CaSO 4 (s)  Ca 2+ + SO 4 2– x 0 M 0 M –x +x +x 0 M x x K sp = [Ca 2+ ][SO 4 2– ] K sp = x 2 2.4 x 10 –5 = x 2 √2.4 x 10 –5 = x 0.0049 = x K sp for CaSO 4 is 2.4 x 10 –5. [CaSO 4 ] = 0.0049 M [Ca 2+ ] = 0.0049 M [SO 4 2– ] = 0.0049 M HW p. 663 #47 (molar solubility) K sp Calculations R I C E

71. PbCl 2 (s)  Pb 2+ + 2 Cl – x 0 M 0 M –x +x +2x 0 M x 2x K sp = [Pb 2+ ][Cl – ] 2 K sp = (x)(2x) 2 K sp = 4x 3 [PbCl 2 ] = 0.016 M [Pb 2+ ] = 0.016 M [Cl – ] = 0.032 M 1.6 x 10 –5 = 4x 3 3 √4.0 x 10 –6 = x 0.016 = x If only K sp is known, solve for x (M). K sp for PbCl 2 is 1.6 x 10 –5. (molar solubility) K sp Calculations R I C E

72. Cr(OH) 3 (s)  Cr 3+ + 3 OH – x 0 M 0 M –x +x +3x 0 M x 3x K sp = [Cr 3+ ][OH – ] 3 K sp = (x)(3x) 3 K sp = 27x 4 [Cr(OH) 3 ] = 1.6 x 10 –8 M [Cr 3+ ] = 1.6 x 10 –8 M [OH – ] = 4.8 x 10 –8 M 1.6 x 10 –30 = 27x 4 4 √5.9 x 10 –32 = x 1.6 x 10 –8 = x If only K sp is known, solve for x (M). K sp for Cr(OH) 3 is 1.6 x 10 –30. (molar solubility) K sp Calculations R I C E

73. LaF 3 (s)  La 3+ + 3 F – x 0 M 0 M –x +x +3x 0 M x 3x K sp = [La 3+ ][F – ] 3 K sp = (x)(3x) 3 2 x 10 –19 = 27x 4 x = 9 x 10 –6 M LaF 3 HW p. 763 #52a K sp Calculations (in pure H 2 O) #52b (in 0.010 M KF) LaF 3 (s)  La 3+ + 3 F – x 0 M –x +x +3x 0 M x 0.010 + 3x K sp = [La 3+ ][F – ] 3 K sp = (x)(0.010 + 3x) 3 2 x 10 –19 = (x)(0.010) 3 x = 2 x 10 –13 M LaF 3 0.010 M ≈ 0.010 b/c K <<<1 Solubility is lower in _________________ sol’n w/ common ion

74. Factors Affecting Solubility Common-Ion Effect (more Le Châtelier) –I–If a common ion is added to an equilibrium solution, the equilibrium will shift left and the solubility of the salt will decrease. BaSO 4 (s)  Ba 2+ (aq) + SO 4 2− (aq) BaSO 4 would be least soluble in which of these 1.0 M aqueous solutions? Na 2 SO 4 BaCl 2 Al 2 (SO 4 ) 3 NaNO 3 most soluble? adding common ion shifts left (less soluble) OR WS K sp #1-2

75. Factors Affecting Solubility Mg(OH) 2 (s)  Mg 2+ (aq) + 2 OH − (aq) Greater solubility because… Added acid (H + ) reacts with OH – thereby removing product causing a shift to the right to dissolve more solid Mg(OH) 2. Addition of HCl (acid) to this solution would cause…

76. Basic anions, more soluble in acidic solution. Mg(OH) 2 (s)  Mg 2+ (aq) + 2 OH − (aq) H+H+ H + NO Effect on: Cl –, Br –, I –, NO 3 –, SO 4 2–, ClO 4 – Adding H + would cause… shift , more soluble. HW p. 763 #55

77. Factors Affecting Solubility Complex Ions –Metal ions can and form complex ions with lone e – pairs in the solvent.

78. forming complex ions… …increases solubility AgCl (s)  Ag + (aq) + Cl − (aq) NH 3 Ag(NH 3 ) 2 + p. 765 #59

79. Will a Precipitate Form? In a solution, –If Q = K sp, at equilibrium (saturated). –If Q < K sp, more solid will dissolve (unsaturated) until Q = K sp. (products too small, proceed right→) –If Q > K sp, solid will precipitate out (saturated) until Q = K sp. (products too big, proceed left←) HW p. 764 #62b, 66 Q = [X + ] a [Y − ] b WS K sp #4 K sp = [X + ] a [Y – ] b X a Y b (s)  a X + (aq) + b Y − (aq)

80. AgIO 3 (s)  Ag + + IO 3 – K sp = [Ag + ][IO 3 – ] K sp = 3.1 x 10 –8 HW p. 764 #62b (OR…is Q > K ?) Q = [Ag + ][IO 3 – ] Q = 100 mL of 0.010 M AgNO 3 10 mL of 0.015 M NaIO 3 [Ag + ] = ________ (0.010 M)(100 mL) = M 2 (110 mL) (mixing changes M and V) M 1 V 1 = M 2 V 2 0.0091 M [IO 3 – ] = ________ (0.015 M)(10 mL) = M 2 (110 mL) 0.0014 M Will a Precipitate Form? Q = (0.0091)(0.0014) Q = 1.3 x 10 –5 Q > K, so… rxn shifts left prec. will form

81. BaSO 4 (s)  Ba 2+ + SO 4 2– HW p. 764 #66a (BaSO 4 )(SrSO 4 ) SrSO 4 (s)  Sr 2+ + SO 4 2– K sp = [Sr 2+ ][SO 4 2– ] When Will a Precipitate Form? K sp = [Ba 2+ ][SO 4 2– ] 1.1 x 10 –10 = (0.010)(x) 3.2 x 10 –7 = (0.010)(x) x = 1.1 x 10 –8 [SO 4 2– ] = 1.1 x 10 –8 M x = 3.2 x 10 –5 [SO 4 2– ] = 3.2 x 10 –5 M #66b Ba 2+ will precipitate first b/c… less SO 4 2– is needed to reach equilibrium (K sp ).