1. CHE 116 CHAPTER 17 ADDITIONAL ASPECTS OF AQUEOUS EQUILIBRA

2. Common Ion Effect What happens if a salt is added to a solution of weak acid, and that salt contains a conjugate base to the weak acid? Weak acids only partially ionize in solution, with K a values of less than 1.0x10 -3. However, ionic compounds dissociate completely in aqueous solutions.

3. Common Ion Effect When salts dissociate – those ions can behave as conjugate acids or bases. Ions derived from a strong acid or base are weak conjugate acids/bases. They will not accept nor donate a proton. Ions that are derived from a weak acid or base will have a certain likelihood to accept or donate a proton – and therefore affect the pH.

4. Common Ion Effect Example: Acetic Acid HC 2 H 3 O 2 HC 2 H 3 O 2 (aq)  H + (aq) + C 2 H 3 O 2 - (aq) If sodium acetate is added: it will completely dissociate in solution NaC 2 H 3 O 2 (aq)  Na + (aq) + C 2 H 3 O 2 - (aq)

5. Common Ion Effect By adding the common ion, acetate, we have increased the concentration of the conjugate base of acetic acid but have not increased the concentration of H +. According to LeChatelier’s principle, the equilibrium will shift left to relieve the stress of adding the common ion. A shift to the left with eat up the H + and therefore decrease [H + ]; increasing pH.

6. Common Ion Effect Definition: The dissociation of a weak electrolyte is decreased by adding to the solution a strong electrolyte (salt) that has an ion in common with the weak electrolyte.

7. Common Ion Effect Example: NH 3 is a weak base that ionizes in H 2 O to produce NH 4 + and OH -. NH 3 (aq) + H 2 O(aq)  NH 4 + (aq) + OH - (aq) If a salt like (NH 4 ) 2 SO 4 were added: there would be an increase in NH 4 + which is the conjugate acid. As more NH 4 + were added, the [OH - ] would decrease and the pH would decrease. The SO 4 -2 ion would be considered a spectator and not affect the pH.

8. Common Ion Effect Example: Acetic acid (CH 3 COOH) is a weak acid with the following ionization reaction: CH 3 COOH + H 2 O  H 3 O + + CH 3 COO - K a = 1.8x10 -5 What is the pH of a solution that is 0.5M in acetic acid and 2.5M of sodium acetate: CH 3 COONa.

9. Common Ion Effect [CH 3 COO - ][H + ] K a = [CH 3 COOH] If X represents the amount of the acid that dissociates, it can be used to represent the concentrations of both the CH 3 COO - and H +. However, that is not the only source of CH 3 COO - ions. There is the additional 2.5M of a strong electrolyte that will produce 2.5M of CH 3 COO - ions as well.

10. Common Ion Effect Initial0.5M00 Change-x+x Equilibrium0.5 – xx2.5 + x CH 3 COOH  H 3 O + + CH 3 COO - Can we use the shortcut to the quadratic? Is the [CH 3 COOH] at least 100x bigger than K a ?

11. Common Ion Effect Because the change in concentration of acid is so small, we can assume 0.5 – x is approximate to 0.5M. In addition, if it is a negligible change in the concentration of the CH 3 COOH, than it will also be a negiligible addition to the concentration of the NaCH 3 COO. The terms in the equilibrium expression become 0.5M and 2.5M.

12. Common Ion Effect 1.8x10 -5 = (2.5)(x) 0.5 x = 3.60x10 -6 pH = -log(3.60x10 -6 ) = 5.44

13. Common Ion Effect What is the pH of an aqueous solution that contains 0.15M NH 3 and 0.05M (NH 4 ) 2 SO 4. The K b is 1.8x10 -5. Eq: NH 3 (aq) + H 2 O(aq)  NH 4 + (aq) + OH - (aq )

14. Common Ion Effect Initial0.15M00 Change-x+x Equilibrium0.15 – x0.1 + x+x By comparing [NH 3 ] to K b : we can skip the Quadratic. NH 3 (aq)  NH 4 + (aq) + OH - (aq )

15. Common Ion Effect 1.8x10 -5 = (0.1)(x) 0.15 x = 2.7x10 -5 pOH = -log(2.7x10 -5 ) = 4.57 pH = 9.43

16. Common Ion Effect Indicate whether the pH increases, decreases or remains the same when: 1. Ca(C 2 H 3 O 2 ) 2 is added to a solution of CH 3 COOH. 2. Ammonium Nitrate is added to a solution of NH 3. 3. NaNO 3 is added to a solution of NaOH.

17. Buffered Solutions Solutions of a weak conjugate acid base pair establish an equilibrium that is resistant to changes in pH. The acid is able to react with H+ ions and the base is able to react with OH- ions. However, the acid and base can’t react with one another in a neutralization reaction.

18. Buffered Solutions To prepare buffered solutions with usually mix a weak acid or base with a salt of that acid or base. By mixing CH 3 COOH with CH 3 COONa to ensure that concentrations of both the weak acid and the conjugate base: CH 3 COO - are present. In addition – there is no reaction that will occur between CH 3 COOH and CH 3 COO -.

19. Buffered Solutions If we look at a buffer system of a weak acid: HA and one of it’s salts: MA The equilibrium for this system can be written as: HA  H + (aq) + A - (aq) The acid dissociation constant can be expressed as:

20. Buffered Solutions We can rearrange that equation to solve for [H + ] because when considering buffer systems we are interested in how the [H + ] changes. The [H + ] and therefore the pH depends on two factors: - The K a - The ratio of concentration of acid to conjugate base

21. Buffered Solutions How will the pH change if H + or OH - ions are added to the system? If OH- ions are added to the system: they react with the buffer acid to produce H 2 O and A-: OH - (aq) + HA(aq)  H 2 O(l) + A - (aq) This causes the [HA] to decrease and [A-] to increase. This keeps the ratio of acid to conjugate base from changing that much. This keeps the [H + ] and pH from changing very much.

22. Buffered Solutions As long as the concentrations of [HA] and [A - ] are large in comparison to the amount of OH - added: the overall change in pH is very small and the solution is said to be buffered.

23. Buffered Solutions How do buffers resist pH changes when H + ions are added? The added H+ ions will react with the base component of the buffer: H + (aq) + A - (aq)  HA(aq) H 3 O + (aq) + A - (aq)  HA(aq) + H 2 O(l) This will result in [A-] to decreases and [HA} to increase: keeping the ratio relatively constant, therefore the pH is constant.

24. Buffered Solutions Henderson Hasselbalch Equation: used to calculate the pH of a buffered solution.

25. Buffered Solutions Example: What is the pH of a buffer that is 0.15M in CH 3 COOH and 0.05M in CH 3 COONa. The K a for acetic acid is 1.8x10 -5. pH = -log(1.8x10 -5 ) + log 0.05/0.15 pH = 4.74 – 0.477 pH = 4.26

26. Buffered Solutions Alternatively, you can use the ICEBOX method to solve. This will be more helpful when given the pH and asked to work backwards to determine the concentrations of acid and base.

27. Buffered Solutions Initial0.15M00.05 Change-x+x Equilibrium0.15 – xx0.05 + x CH 3 COOH  H 3 O + + CH 3 COO -

28. Buffered Solutions 1.8x10 -5 = (0.05 + x)(x) / (0.15 – x) The concentrations are more than 100x greater than K a so we can use the shortcut to solve for x. 1.8x10 -5 = 0.05x / 0.15 pH = 4.26

29. Buffered Solutions For the previous problem, what would be the pH be if 0.01M of NaOH is added to the buffer? NaOH will dissociate completely. While the Na + ion will not react with anything to affect the pH, the OH - will make the solution more basic.

30. Buffered Solutions Using the Henderson Hasselbach equation: pH = pKa + log ([base]/[acid]) The concentration of the base is 0.05M but the additional NaOH will increase the by 0.01. [Base] = 0.06M The concentration of the acid is 0.15 but the NaOH will react with the acid, and it will lose 0.01M. [Acid] = 0.14M.

31. Buffered Solutions pH = -log(1.8x10 -5 ) + log (0.06/0.14) pH = 4.37

32. Buffered Solutions Example: What is the pH of a buffer that is 0.12M of lactic acid HC 3 H 5 O 3 and 0.10M in sodium lactate: NaC 3 H 5 O 3. K a for lactic acid is 1.4x10 -4. ANS: 3.77

33. Buffered Solutions Preparing a Buffer Example How many moles of NH 4 Cl must be added to 2.0L of a 0.10M NH 3 to form a buffer with a pH of 9.0? The buffering action will depend on the conjugate acid base pair NH 3 and NH 4 +. Cl- is a spectator ion.

34. Buffered Solutions Equilibrium: NH 3 (aq) + H 2 O(l)  NH 4 + (aq) and OH - (aq) K b = 1.8x10 -5 [base] =[NH 3 ] = 0.10M [acid] = [NH 4 + ] = x

35. Buffered Solutions pH = pKa + log ([base]/[acid]) [acid] = 0.179M If the volume is 2.0L, then 0.36 moles of NH 4 Cl would be required.

36. Buffered Solutions Buffering Capacity and pH The two most important characteristics of a buffer are: 1. The pH at which the buffer is most effective. This will be equal to the pKa for the buffer. 2. How much acid or base can be added before the pH begins to change.

37. Buffered Solutions Buffer Capacity is how much acid or base can be added and neutralize before the pH begins to change. The buffer capacity depends on the concentrations of the conjugate acid base pair that were used to prepare the buffer. The greater the concentrations of the conjugate acid – base pair: the more they are able to neutralize.

38. Buffered Solutions The best buffers resist changes in pH when the amount of the weak acid and conjugate base are nearly the same. From the H-H equation: when the concentration of base = concentration of acid, then pH = pKa. That is the optimal for any buffering system. General rule: buffers are useful for when pH = pKa +/- 1. Once the concentration of one component becomes 10x more concentrated than the other – the buffer is not as effective.

39. Buffered Solutions Addition of Strong Acids or Bases to Buffers When we add strong acids and bases, we assume that the additional H+ ions or OH- ions are completely consumed by reacting with components in the buffering system.

40. Buffered Solutions To determine how the the addition of a strong acid or base will affect pH: For a buffering system: HX and X- When a strong acid is added, it will react with X- to produce HX. We will get new amounts of HX (it will increase) and X- will decrease. When a strong base is added: the OH- added will react with HX to produce X-. HX will decrease and X- will increase.

41. Buffered Solutions

42. A buffer is made by adding 0.300 mol of CH 3 COOH and 0.300 mol of CH 3 COONa to enough water to make 1.00L of solution. The pH of the buffer is 4.74. a) Calculate the pH after 5.0 mL of 4.0M NaOH is added. b) Calculate the pH of a solution made by added 5.0 mL of 4.0M NaOH to pure water.

43. Buffered Solutions 1. First determine how the added base will affect the concentrations of the buffer. 2. Use resulting concentrations with the H-H equation to calculate the new pH.

44. Buffered Solutions CH 3 COOH + OH -  H 2 O + CH 3 COO - Initially, the amounts of acid and base are 0.300M. When 5.0 mL of 4.0M NaOH is added, 0.020 moles of OH ions are added to the system, and the total volume of the buffered system is now 1.005 L. To calculate the new molarity of the acid and base, we have to calculate the new moles of both acid and base, and use the new volume as well.

45. Buffered Solutions CH 3 COOHOH - CH 3 COO - Original Buffer 0.300 mol / 1 L 0 Addition0.020 mol New Buffer0.280 mol0 mol0.320 mol The [acid] = 0.280 mol/ 1.005 L The [base] = 0.320 mol/ 1.005 L To calculate the new pH, we need the pKa. In this case, the pKa = pH because the ratio of base to acid was 1:1

46. Buffered Solutions The new pH will be 4.80. B) pH of 0.020 mol of NaOH to 1.00L. pOH = = -log(0.020) pOH = 1.70 pH = 12.30

47. Acid Base Titrations A solution of a known concentration of base is added to an an acid until the acid is completely neutralized. Can also add acid to a base. The equivalence point is when a stoichiometrically equivalent amount of base has been added to the acid. Moles of acid = moles of base. An indicator is often used to signal when equivalence point is reached. Can also use a pH meter.

48. Acid Base Titrations If a pH meter is used to measure the pH changes as the acid and base are mixed together, the values can be plotted to create a titration curve. The titration curve data can also be used to determine the equivalence point, appropriate indicators and the K a or K b value.

49. Acid Base Titrations Depending upon the strength of the acid and the base, the titration curve will appear differently.

50. Acid Base Titrations Strong Acid – Strong Base

51. Acid Base Titrations The titration curve can be split into 4 regions, and we can calculate the pH at various stages. 1. Initial pH: pH before any base is added. Determined by initial concentration of the acid. 2. Between initial pH and equivalence point: pH increases slowly and then rises faster. pH of solution is determined by the amount of acid that has not yet been neutralized.

52. Acid Base Titrations 3. Equivalence point: equal number of moles of OH- and H+ have reacted. The cations and anions that remain are weak conjugates – so they will not affect pH. 4. After equivalence point: all of HCl has been neutralized, pH of determined by the NaOH that is in excess as you continue to add it to solution.

53. Acid Base Titrations Calculate the pH when a) 49.0 mL and b) 51.0 mL of 0.100 M have been added to 50.0 mL of a 0.100 M HCl solution. To determine pH in both of these scenarios, we have to determine how many moles of H+ are present and how many moles of OH- have been added.

54. Acid Base Titrations a) 50.0 mL of 0.100 M HCl solution = 0.005 moles of H + When 49.0 mL of 0.100M NaOH is added: adding 0.0049 moles of OH -. At this point, there will be an addition 0.001 moles of H + remaining. The [H+] = 0.001 moles / 0.0990L = 1.0x10 -3 M. pH = 3.

55. Acid Base Titrations b) Because the volume of NaOH exceeds the HCl, we are past the equivalence point. Originally: 0.005 moles of H+ With the addition of 51 mL of 0.100 M NaOH, we add 0.0051 moles. All of the H+ will be consumed and there will be 0.0001 moles of OH- leftover. The [OH-] can be determined: 0.0001/0.1010L = 1.0x10 -3 M. pOH = 3, pH = 11.

56. Acid Base Titrations Something to note: The initial pH was 1, and when 49.0 mL of a strong base was added, the pH only jumped from 1 to 3, as calculated in part A. However, once the equivalence was surpassed: adding 51.0 mL – only an additional 2.0 mL – the pH jumps from 3 to 8. This quick spike in pH is a hallmark feature of titrations of strong acids and strong bases.

57. Acid Base Titrations Weak Acid – Strong Base Titrations

58. Acid Base Titrations 1. Initial pH: Use K a, cannot assume that the concentration of acid = [H + ]. 2. Between initial pH and equivalence point: Acid is being neutralized and the conjugate base is increasing in concentration. CH 3 COOH + OH  CH 3 COO - + H 2 O

59. Acid Base Titrations Calculating the pH here involves more complicated math: First thing to address is the neutralization reaction between CH 3 COOH and OH. We will be able to determine the [CH 3 COOH] still present as well as determine [CH 3 COO - ]. Essentially what is present during this step is a buffering system. We can then use the H-H equation to calculate the pH here.

60. Acid Base Titrations 3. Equivalence Point: Reached when 50.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH. If this were a strong acid/base system, the pH would be 7, because the cation/anion have no significant effect on pH. However, CH 3 COO - will affect the pH. The pH will be slightly above 7 because of the presence of the weak conjugate base.

61. Acid Base Titrations After equivalence point: Use the [OH-] from the excess to NaOH to calculate the pH. The pH will be calculated the same way as it was when it was a titration of a strong acid/base.

62. Acid Base Titrations Example: Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100M CH 3 COOH. K a = 1.8x10 -5. We have not yet reached the equivalence point because have not added equal volume of base to the acid.

63. Acid Base Titrations 1. Determine the moles of CH 3 COOH and CH 3 COO - present by determining how much CH3COOH is consumed by the addition of NaOH Initial Moles of CH 3 COOH: 0.005 Moles of NaOH added: 0.00450 Leftover moles of CH 3 COOH: 0.0005 mole

64. Acid Base Titrations When the NaOH is added to the CH 3 COOH, the neutralization reaction produced CH 3 COO – which is a conjugate base that will affect the pH. Must calculate how much CH 3 COO- is present. If 0.005 moles of CH 3 COOH react, and leave behind only 0.0005 moles. The other 0.00450 moles must be “converted” to CH 3 COO -.

65. Acid Base Titrations [CH 3 COOH] = 0.0005 / 0.0950 L [CH 3 COO-] = 0.00450 / 0.0950 L Can use either H-H equation or just K a for CH 3 COOH. pH = 5.70

66. Acid Base Titrations Example: Calculate the pH in the solution formed by 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (C 6 H 5 COOH, Ka = 6.3x10 -5 ). b) Calculate the pH in the solution by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH 3. A: 4.20 B: 9.26

67. Acid Base Titrations Calculating pH at equivalence point: Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. The moles of CH 3 COOH will be equal to the moles of NaOH. However, because this is a weak acid, we have to take into account how the CH 3 COO - will act as a base, to increase the pH slightly.

68. Acid Base Titrations Moles of CH 3 COOH: 0.005 Because the CH 3 COOH is completely neutralized, we can assume there is 0.005 moles of CH 3 COO- present in 1.00L. This quantity of CH 3 COO - will now act as a base: CH 3 COO - + H 2 O  CH 3 COOH + OH

69. Acid Base Titrations The Kb for CH 3 COO - can be calculated from the K a of 1.8x10 -5, and is found to be 5.6x10 -10. We want to use the equilibrium expression to calculate the concentration of OH that will be produced by the base activity of CH 3 COO -. K b = [CH 3 COOH][OH] [CH 3 COO]

70. Acid Base Titrations K b = [CH 3 COOH][OH] [CH 3 COO] 5.6x10 -10 = [x][x] 0.05 – x 5.6x10 -10 = [x][x] 0.05 x = [OH] = 5.3x10-6, pOH = 5.28, pH = 8.72.

71. Acid Base Titrations Calculate the pH at the equivalence point when a) 40.0 mL of 0.025 M benzoic acid is titrated with 0.050 M NaOH b) 40.0 mL of 0.100 M NH3 is titrated with 0.100 M HCl. A: 8.21 B: 5.28

72. Acid Base Titrations The titration for a weak acid – strong base system is visually different from a strong acid – strong base titration: 1. The solution of weak acid that you begin the titration with will have a higher pH than a strong acid with the same concentration. 2. The change in pH during the rapid rise portion of the curve is more gradual, less of a change than a strong acid titration. 3. The pH at the equivalence point will be higher than 7 for a weak acid titration.

73. Acid Base Titrations

74. Titrating with Polyprotics: Some weak acids are polyprotics, which means that the neutralization with OH- ions occurs in multiple steps. This will result in multiple equivalence points in the titration curve when the neutralization steps are relatively far apart. Example: H 3 PO 3 + OH -  H 2 PO 3 - + H 2 O H 2 PO 3 - + OH -  HPO 3 -2 + H 2 O

75. Acid Base Titrations At the first equivalence point: all of the H + ions from the first dissociation have been reacted with OH - ions from the base. At the second equivalence point: the H + ions from both dissociations have reacted, which is twice as many H + ions, which requires a larger volume of OH - ions.

76. Acid Base Titrations

77. Rather than pH meters, indicators are used to signal the equivalence point. For strong acid – strong base titrations, adding a very small volume of the acid or base can change the pH dramatically. The rapid rise portion of the curve represents the equivalence point. An indicator the changes color within that region will give a decent estimate of the volume of base needed to neutralize the H + ions. When the indicator changes color is known as the endpoint of the titration.

78. Acid Base Titrations For the titration of NaOH and HCl:

79. Acid Base Titrations The indicators chosen for a strong acid strong base system may not be appropriate for a weak acid – strong base base titration. The pH change for these systems is much smaller, so you need an indicator that will be effective in a much smaller “window” than strong acid – strong base titration.

80. Acid Base Titrations

81. trations

82. Solubility Equilibria The remainder of the chapter will be looking at equilibrium systems that are heterogeneous – both the dissolution and precipitation of compounds. The K sp value can give us an idea of how soluble a solid is in solution rather than the general solubility rules that were discussed in Chapter 4.

83. Solubility Equilibria When solids are dissolved in water to form saturated solutions, the ions are said to be in equilibrium with the solid. The rate of crystallization is equal to the rate of dissolution. Equilibrium constants can be written and calculated for these processes, which will give us an indication of the extent of which the solid dissolves. For solubility equilibrium we can determine the K sp ; which is the solubility product constant.

84. Solubility Equilibria When writing the expression for the dissolution of BaSO 4 : BaSO 4 (s)  Ba +2 (aq) + SO 4 -2 (aq) K sp = [Ba +2 ][SO 4 -2 ] = 1.1x10 -10 : very small, not very soluble. Recall that solids are not included in the equilibrium expression. In general: the solubility product constant is the product of the concentration of the ions raised to the power of its coefficient in the equation. The coefficient is equal to the subscript from the original chemical formula.

85. Solubility Equilibria Solubility vs. K sp : Solubility: the amount of a substance that dissolves to produce a saturated solution and is expressed as molarity of g/L. Ksp: a measure of how much of the solid dissolves to produce a measurable concentration of dissolved ions. Unitless.

86. Solubility Equilibria The solubility of a substance often depends on the conditions of the system such as pH, presence of other common ions or other species. The K sp value is constant for a specific solute at a specific temperature.

87. Solubility Equilibria A solution of copper(I)chloride is made such that a solid amount remains after equilibrium. The concentration of Cu + (aq) ion is determined to be 1.1x10 -3 M. A) What is the value of K sp ? B) What is the solubility of CuCl in g/L? A) Dissociation: CuCl  Cu + (aq) + Cl - (aq) Ksp = [Cu + ][Cl - ] = [1.1x10 -3 ][1.1x10 -3 ] = 1.21x10 -6

88. Solubility Equilibria B) If the molar concentration of Cu + and Cl - is 1.10x10 -3 mol/L, it must be converted to g/L. Cu+: 1.10x10 -3 mol/L x 63.5 g/mol = 0.0699g/L Cl-: 1.10x10 -3 mol/L x 35.4 g/mol = 0.0389 g/L The total solubility of CuCl = 0.0699 + 0.0389 or 0.109 g/L of CuCl will dissolve.

89. Solubility Equilibria A saturated solution of Mg(OH) 2 in contact with undissolved Mg(OH) 2 is prepared at 25 o C. The pH of the solution is 10.17. Assuming that Mg(OH) 2 dissociates completely in water and there is no other simultaneous equilibria involving the Mg +2 and OH - ions, calculate K sp. ANS: 1.6x10 -12

90. Solubility Equilibria The Ksp for CaF 2 is 3.9x10 -11 at 25 o C. Assuming the CaF 2 dissociates completely, calculate the solubility of CaF 2 in grams per liter. CaF 2  Ca +2 (aq) + 2F - (aq) ICEBOX!!!! 3.9x10 -11 = x(2x) 2 = x(4x 2 ) = 4x 3 X = 2.1x10 -4 mol/L

91. Solubility Equilibria X represents the amount of ions dissolved, which came from the dissolution of CaF 2. Therefore we can convert 2.1x10 -4 mol/L to grams per liter using the GFM of CaF 2. The solubility of CaF 2 will be 1.6x10 -2 g/L.

92. Solubility Equilibria Example: the K sp for LaF 3 is 2.0x10 -19. What is the solubility in mol/L? ANS: 9x10 -6 mol/L.

93. Factors Affecting Solubility The K sp value varies with temperature so solubility is also influenced by temperature. However there are other factors that will affect solubility without changing the value of K sp. These factors are: common ion effect, pH and the presence of complexing agents.

94. Factors Affecting Solubility Common Ion Effect: For the equilibrium CaF 2  Ca +2 (aq) + 2F - (aq) If there is another source of one of these ions, the equilibrium will shift to the left, reducing the solubility. This will favor the formation of the solid and reduce the concentration of dissolved ions. K sp is not changed. Rule: The solubility of a slightly soluble salt is decreased with the addition of a second solute that produces a common ion.

95. Factors Affecting Solubility Ex: Calculate the molar solubility of CaF 2 at 25 o C in a solution that is a) 0.010M in Ca(NO 3 ) 2 and b)0.010M in NaF. CaF 2  Ca +2 (aq) + 2F - (aq) CaF 2 Ca +2 2F - Initial0.010M0 Change+x+2x Equilibrium0.010 + x2x

96. Factors Affecting Solubility K sp = 3.9x10 -11 = [Ca +2 ][F - ] 2 = (0.010 + x)(2x) 2 Because Ksp is so small compared to the concentration of the Ca(NO 3 ) 2 added, we can assume the change will be neglible. 3.9x10 -11 = 0.010(2x) 2 x = 3.1x10 -5 M

97. Factors Affecting Solubility When F - is the common ion, Ca +2 will be set equal to x and F - will be 0.010 + 2x. 3.9x10 -11 = x(0.010 + 2x) 2 = x(0.010) 2 x = 3.9x10 -7 M This means 3.9x10 -7 moles of CaF 2 will dissolve in 1 liter of 0.010 M NaF solution.

98. Factors Affecting Solubility Solubility and pH pH will affect the solubility of any salt that contains a basic anion such as Mg(OH) 2. Mg(OH) 2 (s)  Mg 2+ (aq) + 2OH - (aq) K sp = [Mg 2+ ] [OH - ] 2 = 1.8 x 10 -11

99. Factors Affecting Solubility We need to calculate the [OH] in order to see how the pH of a solution will affect the solubility. For the equilibrium: Mg(OH) 2 (s)  Mg 2+ (aq) + 2OH - (aq) 1.8x10 -11 = [X][2x] 2 1.8x10 -11 = 4x 3 1.04x10 -4 M = x The [OH] = 2x = 2.08x10 -4 M. The pOH is then equal to 3.68 and the pH is 10.3. If we put this solution in a buffer with a pH of 9.0 – how will the solubility change?

100. Factors Affecting Solubility At a pH of 9.0, the pOH is 5.0. Therefore the [OH] is 1.0x10 -5 M. 1.8x10 -11 = [Mg +2 ][OH - ] 2 1.8x10 -11 = [Mg +2 ][1.0x10 -5 ] 2 [Mg +2 ] = 0.18 M. If you compare this to the concentration of Mg +2 in the original – 1.04x10 -4 – the solubility of the Mg +2 increased: more salt dissolved at the lower pH.

101. Factors Affecting Solubility By lowering the pH, the amount of OH present was decreased from 2.08x10 -4 M to 1.0x10 -5 M. This shifted the equilibrium to the right, which result in more of the salt to dissolve. To make this salt dissolve completely, the solution could be made more acidic. In general: the solubility of a compound contain a basic anion such as OH will increase as the solution becomes more acidic. When other basic anions such as F- are in solution, the solubility of their salts will increase as well. The more basic the anion, the more the solubility of the salt will be affected by the pH. Anions from strong acids with neglible basicity will be unaffected by the pH.

102. Factors Affecting Solubility Example: CaF 2  Ca +2 + 2F - The F - ion is a weak base and can combine with a H + ion to produce a weak acid HF. 2F - + 2H +  2HF These two reactions are occurring simultaneously in an aqueous environment to produce the net ionic equation: CaF 2 (s) + 2H + (aq)  Ca +2 (aq) + 2HF(aq) As H + increases, the reaction is driven to the right and more salt dissolves.

103. Factors Affecting Solubility Ex: Which the following substances are more soluble in an acidic solution than in a basic solution: a) Ni(OH 2 ) b) CaCO 3 c)BaF 2 d) AgCl? ANS: Choices A, B and C will be more soluble in an acidic environment. Choice D will be unaffected.

104. Factors Affecting Solubility Formation of Complex Ions Recall from chapter 16 that metal ions often act as Lewis acids when interacting with water. The metal ions will accept a non-bonding pair of electrons from the water molecule with acts as the Lewis base. Other compounds that can act with Lewis bases will similarly interact with metal ions and affect their solubility. This affects transition metal ions greatly. AgCl will have a very low solubility in H 2 O, but will dissolve readily in the presence of NH 3 because ammonia acts as a Lewis base.

105. Factors Affecting Solubility 1. AgCl(s)  Ag + (aq) + Cl - (aq) 2. Ag + (aq) + 2NH 3 (aq)  Ag(NH 3 ) 2 + (aq) Net Ionic: AgCl(s) + 2NH 3 (aq)  Ag(NH 3 ) 2 + (aq) + Cl - (aq) The addition of NH 3 drives the reaction to the right, increasing the solubility of Cl - (aq).

106. Factors Affecting Solubility In order for the NH 3 molecules or any other Lewis base to increase the solubility of metal ion compound – the Lewis base must have a greater interaction with the metal ion than the water molecules will. When the metal ion is assembled onto the Lewis base it is called a complex ion.

107. Factors Affecting Solubility In this case: Ag + (aq) + 2 NH 3 (aq)  Ag(NH 3 ) 2 + (aq) How stable the complex ion is can be evaluated by looking at the K f value.

108. Factors Affecting Solubility General Rule: The solubility of metal salts are increased with the presence of good Lewis bases such as NH 3, CN - or OH -. This depends on how well the metal ion will form a complex ion with the Lewis base, which can be evaluated by looking at the K f value for the complex ion.

109. Factors Affecting Solubility What is the concentration of Ag + (aq) in a 0.010 M solution of AgNO 3 (s) at equilibrium is NH 3 (aq) is added to give a concentration of NH 3 of 0.20 M? When the NH 3 is added, it reacts with Ag+ to form the complex ion Ag(NH 3 ) 2 +. We want to know how much of the Ag + remains after the complex ion is formed.

110. Factors Affecting Solubility Ag + (aq) + 2NH 3 (aq)  Ag(NH 3 ) 2 + (aq) K f = 1.7 x 10 7 Things we know: [NH 3 ] = 0.20M [AgNO 3 ] = 0.010M If all of the AgNO 3 dissociates, then we can assume that [Ag +1 ] = 0.010M. Because the Kf value quite large, we can assume that nearly all of the Ag + ions get converted to the complex ion.

111. Factors Affecting Solubility We can use ICEBOX to calculate the equilibrium concentration of Ag + ions. Ag + NH 3 Ag(NH 3 ) 2 + Initial0.010 M ChangeX-X EquilibriumX0.200.010 – X

112. Factors Affecting Solubility x = 1.47 x 10 -8 M This means that the presence of the complex ion drastically reduces the equilibrium concentration of Ag+ ions, which drives the dissolution of AgNO 3.

113. Factors Affecting Solubility Amphoterism: Certain metal oxides or metal hydroxides can act as both an acid or a base. They are called amphoteric oxides/hydroxides. They will dissolve in acidic solutions because their anions will react with the H+ ions to form water. They also readily dissolve in basic solutions as their metal ion is able to react with the OH- ions to form highly complex ions.

114. Factors Affecting Solubility The metal hydroxides that are commonly amphoteric involve Al +3, Cr +3, Zn +2 and Sn +2. The extent to which they will react with the acid or base solution to dissolve primarily depends on the metal ion in the salt. Ca +2, Fe +2 and Fe +3 are not amphoteric in nature.

115. Precipitation of Ions When mixing solutions in a metathesis reaction, the two new compounds formed may or may not be soluble. If a compound is insoluble, it is helpful to determine whether or not a precipitate will form. For salts that are even insoluble, depending on their K sp values, we may not see a precipitate. We will be able to answer that question by calculating Q values for the specific mixture and comparing it to the K sp value, we can determine whether or not a precipitate will actually.

116. Precipitation of Ions If Q > K sp : precipitation occurs. It will continue until the concentration of ions decreases until Q = K sp If Q = K sp : equilibrium exists If Q < K sp : Solid dissolves and ion concentration will increase until Q = K sp

117. Precipitation of Ions Example: Does a precipitate form when 0.10 L of 8.0x10 -3 M Pb(NO 3 ) 2 is added to 0.40 L of 5.0x10 -3 M Na 2 SO 4 ? 1. Determine what species will form an insoluble compound. 2. Determine the concentration of each ion in the new solution. 3. Calculate Q and then compare to K sp.

118. Precipitation of Ions In this mixture the Pb +2 ion and the SO 4 -2 ion will form an insoluble compound. The K sp for this compound is 6.3x10 -7. [Pb +2 ] = 8.0x10 -4 mol / 0.50 L [SO4 -2 ] = 2.0x10 -3 mol / 0.50 L Q = 6.4x10 -6 … Q > K sp : Precipitate forms

119. Precipitation of Ions We can also choose to mix solutions together to precipitate certain ions out of solution. If a solution containing Ag + and Cu +2 is then mixed with HCl, the Ag + ion will precipitate but the Cu +2 ion will not.

120. Precipitation of Ions A solution contains 1.0x10 -2 M of Ag + and 2.0x10 -2 M Pb +2. When Cl- is added, both AgCl (K sp = 1.8x10 -10 ) and PbCl 2 (K sp = 1.7x10 -5 ) can precipitate. What concentration of Cl - is necessary to begin the precipitation of each salt? Which salt precipitates first? Ans: AgCl will precipitate first, it requires any amount more than 1.8x10 -8 M of Cl - ions to precipitate. The Pb +2 will need more than 2.9x10 -2 M to precipitate.

121. Precipitation of Ions A solution containing 0.050 M Mg +2 and Cu +2. Which ion precipitates first as OH - is added? What concentration of OH - is necessary to begin the precipitation of each cation? Mg(OH) 2 Ksp = 1.8x10 -11, Ksp for Cu(OH) 2 is 4.8x10 -20. ANS: The Cu +2 will precipitate first when the [OH - ] exceeds 9.8x10 -20 M.

122. Qualitative Analysis Using a systematic approach to a solution that would allow us to detect the presence of metal ions. By adding reagents in a specific way and based on the solubility rules of the metal cations, we are able to precipitate certain ions out to identify what was in the original mixture.

123. Qualitative Analysis 1. Insoluble chlorides: When HCl is added to a mixture of cations, if Ag +, Pb +2 and Hg 2 +2 are present, they will form a precipitate. The other cations will be left in solution which can be decanted and treated with a different reagent which will target them.

124. Qualitative Analysis 2. The solution is slightly acid and then treated with H 2 S, which will result in the precipitation of the highly insoluble sulfide salts. Many have very small Ksp values. 3. (NH 4 ) 2 S is added to target some of the remaining cation-sulfide salts that are more soluble. At this point, so much sulfide ion has been added, that the Ksp value is exceeded and they will begin to drop out of solution.

125. Qualitative Analysis 4. After the chlorides and sulfides salts have been precipitated out, the only metal cations that could remain will be the Group 1 and Group 2. (NH 4 ) 2 HPO 4 will precipitate the Group 2 ions. 5. All that will remain are the Group 1 metals and the NH 4 + ion. These can be tested individually.