1. © 2014 Pearson Education, Inc. Lecture Presentation Chapter 16 Aqueous Ionic Equilibrium chengkuan@mail.ntou.edu.tw

2. © 2014 Pearson Education, Inc. Aqueous Ionic Equilibrium 16.1 The Danger of Antifreeze 16.2 Buffers: Solution That Resist pH Change 16.3 Buffer Effectiveness: Buffer Range and Capacity 16.4 Titration and pH Curve 16.5 Solubility Equilibria and the Solubility Product Constant 16.6 Precipitation 16.7 Qualitative Chemical Analysis 16.8 Complex Ion Equilibria

3. © 2014 Pearson Education, Inc. The Danger of Antifreeze Each year, thousands of pets and wildlife species die from consuming antifreeze. Most brands of antifreeze contain ethylene glycol. –Sweet taste –Initial effect drunkenness Metabolized in the liver to glycolic acid –HOCH 2 COOH

4. © 2014 Pearson Education, Inc. Why Is Glycolic Acid Toxic? In high enough concentration in the bloodstream, glycolic acid overwhelms the buffering ability of the HCO 3 − in the blood, causing the blood pH to drop. Low blood pH compromises its ability to carry O 2. –Acidosis One treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol.

5. © 2014 Pearson Education, Inc. Buffer Solutions Resist changes in pH when an acid or base is added. Act by neutralizing acid or base that is added to the buffered solution. Contain either –Significant amounts of a weak acid and its conjugate base –Significant amounts of a weak base and its conjugate acid –Blood has a mixture of H 2 CO 3 and HCO 3 −

6. © 2014 Pearson Education, Inc. Making an Acidic Buffer Solution It must contain significant amounts of both a weak acid and its conjugate base.

7. © 2014 Pearson Education, Inc. Basic Buffers B: (aq) + H 2 O (l)  H:B + (aq) + OH − (aq) Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B + Cl − H 2 O (l) + NH 3 (aq)  NH 4 + (aq) + OH − (aq)

8. © 2014 Pearson Education, Inc. An Acidic Buffer Solution If a strong base is added, it is neutralized by the weak acid (HC 2 H 3 O 2 ) in the buffer. NaOH(aq) + HC 2 H 3 O 2 (aq)  H 2 O( l ) + NaC 2 H 3 O 2 (aq) If the amount of NaOH added is less than the amount of acetic acid present, the pH change is small. If a strong acid is added, it is neutralized by the conjugate base (NaC 2 H 3 O 2 ) in the buffer. HCl(aq) + NaC 2 H 3 O 2  HC 2 H 3 O 2 (aq) + NaCl(aq) If the amount of HCl is less than the amount of NaC 2 H 3 O 2 present, the pH change is small.

9. © 2014 Pearson Education, Inc. How Acid Buffers Work: Addition of Base HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) Buffers work by applying Le Châtelier’s principle to weak acid equilibrium. Buffer solutions contain significant amounts of the weak acid molecules, HA. These molecules react with added base to neutralize it. HA(aq) + OH − (aq) → A − (aq) + H 2 O(l)

10. © 2014 Pearson Education, Inc. Action of a Buffer Figure 16.3 pg 762

11. © 2014 Pearson Education, Inc. How Acid Buffers Work: Addition of Acid HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) The buffer solution also contains significant amounts of the conjugate base anion, A −. These ions combine with added acid to make more HA. H + (aq) + A − (aq) → HA(aq) After the equilibrium shifts, the concentration of H 3 O + is kept constant.

12. © 2014 Pearson Education, Inc. Common Ion Effect HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) Adding a salt containing the anion NaA, which is the conjugate base of the acid (the common ion), shifts the position of equilibrium to the left. This causes the pH to be higher than the pH of the acid solution, lowering the H 3 O + ion concentration.

13. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Calculate the pH of a buffer solution that is 0.100 M in HC 2 H 3 O 2 and 0.100 M in NaC 2 H 3 O 2 Example 16.1Calculating the pH of a Buffer Solution Solution 1.Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing the given concentrations of the acid and its conjugate base as the initial concentrations. Leave room in the table for the changes in concentrations and for the equilibrium concentrations. 2.Represent the change in the concentration of H 3 O + with the variable x. Express the changes in the concentrations of the other reactants and products in terms of x.

14. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Continued Example 16.1Calculating the pH of a Buffer Solution 3.Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. 4.Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization constant. In most cases, you can make the approximation that x is small. (See Sections 14.8 and 15.6 to review the x is small approximation.)

15. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Continued Example 16.1Calculating the pH of a Buffer Solution Substitute the value of the acid ionization constant (from Table 15.5) into the K a expression and solve for x. Confirm that x is small by calculating the ratio of x and the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Therefore the approximation is valid. 5.Determine the H 3 O + concentration from the calculated value of x and substitute into the pH equation to find pH.

16. © 2014 Pearson Education, Inc. Henderson–Hasselbalch Equation An equation derived from the K a expression that allows us to calculate the pH of a buffer solution. The equation calculates the pH of a buffer from the pK a and initial concentrations of the weak acid and salt of the conjugate base, as long as the “x is small” approximation is valid.

17. © 2014 Pearson Education, Inc. Deriving the Henderson–Hasselbalch Equation Consider the equilibrium expression for a generic acidic buffer: Solving for H 3 O + we have the following:

18. © 2014 Pearson Education, Inc. Deriving the Henderson–Hasselbalch Equation –log[H 3 O + ] = –logK a + log([A – ]/[HA) Since pH = –log[H 3 O + ] and pK a = –logK a pH = pK a + log([A – ]/[HA)

19. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Solution Equilibrium Approach Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table. Substitute the expressions for the equilibrium concentrations into the expression for the acid ionization constant. Make the x is small approximation and solve for x. Calculate the pH of a buffer solution that is 0.050 M in benzoic acid (HC 7 H 5 O 2 ) and 0.150 M in sodium benzoate (NaC 7 H 5 O 2 ). For benzoic acid, K a = 6.5 × 10 –5. Example 16.2Calculating the pH of a Buffer Solution as an Equilibrium Problem and with the Henderson– Hasselbalch Equation

20. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Since [H 3 O + ] = x, we calculate pH as follows: Confirm that the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). (See Sections 14.8 and 15.6 to review the x is small approximation.) The approximation is valid. Henderson–Hasselbalch Approach To find the pH of this solution, determine which component is the acid and which is the base and substitute their concentrations into the Henderson–Hasselbalch equation to calculate pH. Example 16.2Calculating the pH of a Buffer Solution as an Equilibrium Problem and with the Henderson– Hasselbalch Equation Continued

21. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro HC 7 H 5 O 2 is the acid and NaC 7 H 5 O 2 is the base. Therefore, we calculate the pH as follows: Confirm that the x is small approximation is valid by calculating the [H 3 O + ] from the pH. Since [H 3 O + ] is formed by ionization of the acid, the calculated [H 3 O + ] has to be less than 0.05 (or 5%) of the initial concentration of the acid in order for the x is small approximation to be valid. The approximation is valid. Example 16.2Calculating the pH of a Buffer Solution as an Equilibrium Problem and with the Henderson– Hasselbalch Equation Continued

22. © 2014 Pearson Education, Inc. The Henderson–Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product. The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product. B: + H 2 O  H:B + + OH − To apply the Henderson–Hasselbalch equation, the chemical equation of the basic buffer must be looked at like an acid reaction. H:B + + H 2 O  B: + H 3 O + Henderson–Hasselbalch Equation for Basic Buffers pK a + pK b = 14

23. © 2014 Pearson Education, Inc. Do I Use the Full Equilibrium Analysis or the Henderson–Hasselbalch Equation? The Henderson–Hasselbalch equation is generally good enough when the “x is small” approximation is applicable. Generally, the “x is small” approximation will work when both of the following are true: a)The initial concentrations of acid and salt are not very dilute. b)The K a is fairly small. For most problems, this means that the initial acid and salt concentrations should be over 100 to 1000 times larger than the value of K a.

24. © 2014 Pearson Education, Inc. Action of a Buffer Figure 16.3 pg 762

25. © 2014 Pearson Education, Inc. How Much Does the pH of a Buffer Change When an Acid or Base Is Added? Calculating the new pH after adding acid or base requires breaking the problem into two parts: 1.A stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other. Added acid reacts with the A − to make more HA Added base reacts with the HA to make more A − 2.An equilibrium calculation of [H 3 O + ] using the new initial values of [HA] and [A − ]

26. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Solution Part I: Stoichiometry. The addition of the base converts a stoichiometric amount of acid to the conjugate base (adding base creates more base). Write an equation showing the neutralization reaction and then set up a table to track the changes. Part II: Equilibrium. Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table. Use the amounts of acid and conjugate base from part I as the initial amounts of acid and conjugate base in the ICE table. A 1.0 L buffer solution contains 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2. The value of K a for HC 2 H 3 O 2 is 1.8 × 10 –5. Because the initial amounts of acid and conjugate base are equal, the pH of the buffer is equal to pK a = –log(1.8 × 10 –5 ) = 4.74. Calculate the new pH after adding 0.010 mol of solid NaOH to the buffer. For comparison, calculate the pH after adding 0.010 mol of solid NaOH to 1.0 L of pure water. (Ignore any small changes in volume that might occur upon addition of the base.) Example 16.3Calculating the pH Change in a Buffer Solution after the Addition of a Small Amount of Strong Acid or Base

27. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Substitute the expressions for the equilibrium concentrations of acid and conjugate base into the expression for the acid ionization constant. Make the x is small approximation and solve for x. Calculate the pH from the value of x, which is equal to [H 3 O + ]. Continued Example 16.3Calculating the pH Change in a Buffer Solution after the Addition of a Small Amount of Strong Acid or Base

28. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Confirm that the x is small approximation is valid by calculating the ratio of x to the smallest number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). The approximation is valid. Part II: Equilibrium Alternative (using the Henderson–Hasselbalch equation). As long as the x is small approximation is valid, you can substitute the quantities of acid and conjugate base after the addition (from part I) into the Henderson–Hasselbalch equation and calculate the new pH. Continued Example 16.3Calculating the pH Change in a Buffer Solution after the Addition of a Small Amount of Strong Acid or Base

29. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro The pH of 1.0 L of water after adding 0.010 mol of NaOH is calculated from the [OH – ]. For a strong base, [OH – ] is simply the number of moles of OH – divided by the number of liters of solution. Check Notice that the buffer solution changed from pH = 4.74 to pH = 4.83 upon addition of the base (a small fraction of a single pH unit). In contrast, the pure water changed from pH = 7.00 to pH = 12.00, five whole pH units (a factor of 10 5 ). Notice also that even the buffer solution got slightly more basic upon addition of a base, as we would expect. To check your answer, always make sure the pH goes in the direction you expect: adding base should make the solution more basic (higher pH); adding acid should make the solution more acidic (lower pH). Continued Example 16.3Calculating the pH Change in a Buffer Solution after the Addition of a Small Amount of Strong Acid or Base

30. © 2014 Pearson Education, Inc. Buffering Effectiveness A good buffer should be able to neutralize moderate amounts of added acid or base. The effectiveness of a buffer depends on two factors (1) the absolute amounts of acid and base, and (2) the relative amounts of acid and base. The buffering capacity is the amount of acid or base a buffer can neutralize. The buffering range is the pH range the buffer can be effective.

31. © 2014 Pearson Education, Inc. Buffering Capacity A concentrated buffer can neutralize more added acid or base than a dilute buffer. HA(aq) + OH − (aq) → A − (aq) + H 2 O(l) H + (aq) + A − (aq) → HA(aq)

32. © 2014 Pearson Education, Inc. Buffering Capacity Buffering capacity is the amount of acid or base that can be added to a buffer without causing a large change in pH. The buffering capacity increases with increasing absolute concentration of the buffer components.

33. © 2014 Pearson Education, Inc. Effectiveness of Buffers A buffer will be most effective when the [base]:[acid] = 1. –Equal concentrations of acid and base A buffer will be effective when 0.1 < [base]:[acid] < 10. A buffer will be most effective when the [acid] and the [base] are large.

34. © 2014 Pearson Education, Inc. Buffering Range We have said that a buffer will be effective when 0.1 < [base]:[acid] < 10. Lowest pHHighest pH Therefore, the effective pH range of a buffer is pK a ± 1. When choosing an acid to make a buffer, choose one whose pK a is closest to the pH of the buffer.

35. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Solution The best choice is formic acid because its pK a lies closest to the desired pH. You can calculate the ratio of conjugate base (CHO 2 – ) to acid (HCHO 2 ) required by using the Henderson–Hasselbalch equation as follows: Which acid would you choose to combine with its sodium salt to make a solution buffered at pH 4.25? For the best choice, calculate the ratio of the conjugate base to the acid required to attain the desired pH. chlorous acid (HClO 2 ) pK a = 1.95formic acid (HCHO 2 ) pK a = 3.74 nitrous acid (HNO 2 ) pK a = 3.34hypochlorous acid (HClO) pK a = 7.54 Example 16.5Preparing a Buffer

36. © 2014 Pearson Education, Inc. Titration In an acid–base titration, a solution of known concentration (titrant) is slowly added to a solution of unknown concentration from a burette until the reaction is complete. –When the reaction is complete we have reached the endpoint of the titration. An indicator may be added to determine the endpoint. –An indicator is a chemical that changes color when the pH changes. When the moles of H 3 O + = moles of OH −, the titration has reached its equivalence point.

37. © 2014 Pearson Education, Inc. Titration

38. © 2014 Pearson Education, Inc. Titration Curve It is a plot of pH versus the amount of added titrant. The inflection point of the curve is the equivalence point of the titration. Prior to the equivalence point, the unknown solution in the flask is in excess, so the pH is closest to its pH. The pH of the equivalence point depends on the pH of the salt solution. –Equivalence point of neutral salt, pH = 7 –Equivalence point of acidic salt, pH < 7 (NH 4 + ) –Equivalence point of basic salt, pH > 7 (A - ) Beyond the equivalence point, the known solution in the burette is in excess, so the pH approaches its pH.

39. © 2014 Pearson Education, Inc. Titration Curve: Strong Base Added to Strong Acid

40. © 2014 Pearson Education, Inc. Titration Curve: Strong Base Added to Strong Acid

41. © 2014 Pearson Education, Inc. Titration of a Strong Base with a Strong Acid If the titration is run so that the acid is in the burette and the base is in the flask, the titration curve will be the reflection of the one just shown.

42. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Solution a.Begin by calculating the initial amount of NaOH (in moles) from the volume and molarity of the NaOH solution. Because NaOH is a strong base, it dissociates completely, so the amount of OH – is equal to the amount of NaOH. Calculate the amount of HNO 3 (in moles) added at 30.0 mL from the molarity of the HNO 3 solution. As HNO 3 is added to the solution, it neutralizes some of the OH –. Calculate the number of moles of OH – remaining by setting up a table based on the neutralization reaction that shows the amount of OH – before the addition, the amount of H 3 O + added, and the amounts left after the addition. A 50.0 mL sample of 0.200 M sodium hydroxide is titrated with 0.200 M nitric acid. Calculate pH: a.after adding 30.00 mL of HNO 3 b. at the equivalence point Example 16.6Strong Acid–Strong Base Titration pH Curve

43. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Calculate the OH – concentration by dividing the amount of OH – remaining by the total volume (initial volume plus added volume). Calculate the pOH from [OH – ]. Calculate the pH from the pOH using the equation pH + pOH = 14. b.At the equivalence point, the strong base has completely neutralized the strong acid. The [H 3 O + ] at 25 °C from the ionization of water is 1.00 × 10 –7 M and the pH is therefore 7.00. pH = 7.00 Continued Example 16.6Strong Acid–Strong Base Titration pH Curve

44. © 2014 Pearson Education, Inc. Titration of a Weak Acid with a Strong Base Titrating a weak acid with a strong base results in differences in the titration curve at the equivalence point and excess acid region. The initial pH is determined using the K a of the weak acid. The pH in the excess acid region is determined as you would determine the pH of a buffer. The pH at the equivalence point is determined using the K b of the conjugate base of the weak acid. The pH after equivalence is dominated by the excess strong base. –The basicity from the conjugate base anion is negligible.

45. © 2014 Pearson Education, Inc. Titrating Weak Acid with a Strong Base

46. © 2014 Pearson Education, Inc. Titrating Weak Acid with a Strong Base The initial pH is that of the weak acid solution. –Calculate like a weak acid equilibrium problem Before the equivalence point, the solution becomes a buffer. –Calculate mol HA init and mol A − init using reaction stoichiometry –Calculate pH with Henderson–Hasselbalch using mol HA init and mol A − init Half-neutralization pH = pK a

47. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Solution a.The equivalence point occurs when the amount (in moles) of added base equals the amount (in moles) of acid initially in the solution. Begin by calculating the amount (in moles) of acid initially in the solution. The amount (in moles) of KOH that must be added is equal to the amount of the weak acid. Calculate the volume of KOH required from the number of moles of KOH and the molarity. A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate: a.the volume required to reach the equivalence point b.the pH after adding 5.00 mL of KOH c.the pH at one-half the equivalence point Example 16.7Weak Acid–Strong Base Titration pH Curve

48. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro b.Use the concentration of the KOH solution to calculate the amount (in moles) of OH – in 5.00 mL of the solution. Prepare a table showing the amounts of HNO 2 and NO 2 – before and after the addition of 5.00 mL KOH. The addition of the KOH stoichiometrically reduces the concentration of HNO 2 and increases the concentration of NO 2 –. Since the solution now contains significant amounts of a weak acid and its conjugate base, use the Henderson– Hasselbalch equation and pK a for HNO 2 (which is 3.34) to calculate the pH of the solution. Continued Example 16.7Weak Acid–Strong Base Titration pH Curve

49. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro c.At one-half the equivalence point, the amount of added base is exactly one-half the initial amount of acid. The base converts exactly half of the HNO 2 into NO 2 –, resulting in equal amounts of the weak acid and its conjugate base. The pH is therefore equal to pK a. Continued Example 16.7Weak Acid–Strong Base Titration pH Curve

50. © 2014 Pearson Education, Inc. Titration Curve of a Weak Base with a Strong Acid

51. © 2014 Pearson Education, Inc. Titration of a Polyprotic Acid If K a1 >> K a2, there will be two equivalence points in the titration. –The closer the K a ’s are to each other, the less distinguishable the equivalence points are. Titration of 25.0 mL of 0.100 M H 2 SO 3 with 0.100 M NaOH

52. © 2014 Pearson Education, Inc. Monitoring pH during a Titration The general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H 3 O + ]. –Using a probe that specifically measures just [H 3 O + ]. The endpoint of the titration is reached at the equivalence point in the titration—at the inflection point of the titration curve. If you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator.

53. © 2014 Pearson Education, Inc. Monitoring pH During a Titration Figure 16.10 pg 780

54. © 2014 Pearson Education, Inc. Indicators Many dyes change color depending on the pH of the solution. These dyes are weak acids, establishing an equilibrium with the H 2 O and H 3 O + in the solution. HIn (aq) + H 2 O (l)  In  (aq) + H 3 O + (aq) The color of the solution depends on the relative concentrations of In  :Hin.

55. © 2014 Pearson Education, Inc. Monitoring a Titration with an Indicator For most titrations, the titration curve shows a very large change in pH for very small additions of titrant near the equivalence point. An indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH. –pK a of HIn ≈ pH at equivalence point

56. © 2014 Pearson Education, Inc. Phenolphthalein

57. © 2014 Pearson Education, Inc. Methyl Red

58. © 2014 Pearson Education, Inc. Acid–Base Indicators Table 16.1 pg 783

59. © 2014 Pearson Education, Inc. Solubility Equilibria All ionic compounds dissolve in water to some degree. –However, many compounds have such low solubility in water that we classify them as insoluble. We can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water.

60. © 2014 Pearson Education, Inc. Solubility Product The equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, K sp. For an ionic solid M n X m, the dissociation reaction is M n X m (s)  nM m+ (aq) + mX n− (aq). The solubility product would be K sp = [M m+ ] n [X n− ] m. For example, the dissociation reaction for PbCl 2 is PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq). And its equilibrium constant is K sp = [Pb 2+ ][Cl − ] 2.

61. © 2014 Pearson Education, Inc. Solubility Product

62. © 2014 Pearson Education, Inc. Molar Solubility Solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature. The molar solubility is the number of moles of solute that will dissolve in a liter of solution. –The molarity of the dissolved solute in a saturated solution For the general reaction M n X m (s)  nM m+ (aq) + mX n− (aq)

63. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Calculate the molar solubility of PbCl 2 in pure water. Example 16.8Calculating Molar Solubility from K sp Solution Begin by writing the reaction by which solid PbCl 2 dissolves into its constituent aqueous ions and write the corresponding expression for K sp. Refer to the stoichiometry of the reaction and prepare an ICE table, showing the equilibrium concentrations of Pb 2+ and Cl – relative to S, the amount of PbCl 2 that dissolves. Substitute the equilibrium expressions for [Pb 2+ ] and [Cl – ] from the previous step into the expression for K sp.

64. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Continued Example 16.8Calculating Molar Solubility from K sp Solve for S and substitute the numerical value of K sp (from Table 16.2) to calculate S.

65. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro The molar solubility of Ag 2 SO 4 in pure water is 1.2 × 10 –5 M. Calculate K sp. Example 16.9Calculating K sp from Molar Solubility Solution Begin by writing the reaction by which solid Ag 2 SO 4 dissolves into its constituent aqueous ions, then write the corresponding expression for K sp. Use an ICE table to define [Ag + ] and [SO 4 2– ] in terms of S, the amount of Ag 2 SO 4 that dissolves. Substitute the expressions for [Ag + ] and [SO 4 2– ] from the previous step into the expression for K sp. Substitute the given value of the molar solubility for S and calculate K sp.

66. © 2014 Pearson Education, Inc. The Effect of Common Ion on Solubility Addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt. For example, addition of NaCl to the solubility equilibrium of solid PbCl 2 decreases the solubility of PbCl 2. PbCl 2 (s)  Pb 2+ (aq) + 2Cl − (aq) Addition of Cl − shifts the equilibrium to the left.

67. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Solution Begin by writing the reaction by which solid CaF 2 dissolves into its constituent aqueous ions. Write the corresponding expression for K sp. Use the stoichiometry of the reaction to prepare an ICE table showing the initial concentration of the common ion. Fill in the equilibrium concentrations of Ca 2+ and F – relative to S, the amount of CaF 2 that dissolves. Substitute the equilibrium expressions for [Ca 2+ ] and [F – ] from the previous step into the expression for K sp. Since K sp is small, you can make the approximation that 2S is much less than 0.100 and will therefore be insignificant when added to 0.100 (this is similar to the x is small approximation in equilibrium problems). What is the molar solubility of CaF 2 in a solution containing 0.100 M NaF? Example 16.10Calculating Molar Solubility in the Presence of a Common Ion

68. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Solve for S and substitute the numerical value of Ksp (from Table 16.2) to calculate S. Note that the calculated value of S is indeed small compared to 0.100; our approximation is valid. Continued Example 16.10Calculating Molar Solubility in the Presence of a Common Ion For comparison, the molar solubility of CaF 2 in pure water is 3.32 × 10 –4 M, which means CaF 2 is over 20,000 times more soluble in water than in the NaF solution. (Confirm this for yourself by calculating its solubility in pure water from the value of K sp ).

69. © 2014 Pearson Education, Inc. The Effect of pH on Solubility For insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide. For insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility. M 2 (CO 3 ) n (s)  2 M n+ (aq) + nCO 3 2− (aq) H 3 O + (aq) + CO 3 2− (aq)  HCO 3 − (aq) + H 2 O(l)

70. © 2014 Pearson Education, Inc. Precipitation Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound. If we compare the reaction quotient, Q, for the current solution concentrations to the value of K sp, we can determine if precipitation will occur. –Q = K sp, the solution is saturated, no precipitation. –Q < K sp, the solution is unsaturated, no precipitation. –Q > K sp, the solution would be above saturation, the salt above saturation will precipitate. Some solutions with Q > K sp will not precipitate unless disturbed; these are called supersaturated solutions.

71. © 2014 Pearson Education, Inc. Precipitation occurs if Q > K sp. A supersaturated solution will precipitate if a seed crystal is added. Image top right of page 789 Precipitation

72. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Solution First, determine the possible cross products and their K sp values (Table 16.2). Any cross products that are soluble will not precipitate (see Table 4.1). A solution containing lead(II) nitrate is mixed with one containing sodium bromide to form a solution that is 0.0150 M in Pb(NO 3 ) 2 and 0.00350 M in NaBr. Does a precipitate form in the newly mixed solution? Example 16.12Predicting Precipitation Reactions by Comparing Q and K sp

73. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Possible cross products: NaNO 3 soluble PbBr 2 K sp = 4.67 × 10 –6 Calculate Q and compare it to K sp. A precipitate will only form if Q > K sp. Q < Ksp; therefore no precipitate forms. Continued Example 16.12Predicting Precipitation Reactions by Comparing Q and K sp

74. © 2014 Pearson Education, Inc. Selective Precipitation A solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others. A successful reagent can precipitate with more than one of the cations, as long as their K sp values are significantly different.

75. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Solution The precipitation commences when the value of Q for the precipitating compound just equals the value of K sp. Set the expression for Q for magnesium hydroxide equal to the value of K sp, and solve for [OH – ]. This is the concentration above which Mg(OH) 2 precipitates. When Q = K sp, The magnesium and calcium ions present in seawater ([Mg 2+ ] = 0.059 M and [Ca 2+ ] = 0.011 M) can be separated by selective precipitation with KOH. What minimum [OH – ] triggers the precipitation of the Mg 2+ ion? Example 16.13Finding the Minimum Required Reagent Concentration for Selective Precipitation

76. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Solution First, find the OH – concentration at which Ca 2+ begins to precipitate by writing the expression for Q for calcium hydroxide and substituting the concentration of Ca 2+ from Example 16.13. Set the expression for Q equal to the value of K sp for calcium hydroxide and solve for [OH – ]. This is the concentration above which Ca(OH) 2 precipitates. When Q = K sp, Example 16.14Finding the Concentrations of Ions Left in Solution after Selective Precipitation You add potassium hydroxide to the solution in Example 16.13. When the [OH – ] reaches 1.9 × 10 –6 M (as you just calculated), magnesium hydroxide begins to precipitate out of solution. As you continue to add KOH, the magnesium hydroxide continues to precipitate. However, at some point, the [OH – ] becomes high enough to begin to precipitate the calcium ions as well. What is the concentration of Mg 2+ when Ca 2+ begins to precipitate?

77. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Find the concentration of Mg 2+ when OH – reaches the concentration you just calculated by writing the expression for Q for magnesium hydroxide and substituting the concentration of OH – that you just calculated. Then set the expression for Q equal to the value of K sp for magnesium hydroxide and solve for [Mg 2+ ]. This is the concentration of Mg 2+ that remains when Ca(OH) 2 begins to precipitate. When Q = K sp, As you can see from the results, the selective precipitation worked very well. The concentration of Mg 2+ dropped from 0.059 M to 4.9 × 10 –10 M before any calcium began to precipitate, which means that we separated 99.99% of the magnesium out of the solution. Continued Example 16.14Finding the Concentrations of Ions Left in Solution after Selective Precipitation

78. © 2014 Pearson Education, Inc. Qualitative Analysis An analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme. –Wet chemistry A sample containing several ions is subjected to the addition of several precipitating agents. Addition of each reagent causes one of the ions present to precipitate out.

79. © 2014 Pearson Education, Inc. Figure 16.16 pg 118 A General Qualitative Analysis Scheme

80. © 2014 Pearson Education, Inc. Group 1 Group one cations are Ag +, Pb 2+, and Hg 2 2+. All these cations form compounds with Cl − that are insoluble in water. –As long as the concentration is large enough –PbCl 2 may be borderline Molar solubility of PbCl 2 = 1.43 × 10 −2 M Precipitated by the addition of HCl.

81. © 2014 Pearson Education, Inc. Group 2 Group two cations are Cd 2+, Cu 2+, Bi 3+, Sn 4+, As 3+, Pb 2+, Sb 3+, and Hg 2+. All these cations form compounds with HS − and S 2− that are insoluble in water at low pH. Precipitated by the addition of H 2 S in HCl.

82. © 2014 Pearson Education, Inc. Group 3 Group three cations are Fe 2+, Co 2+, Zn 2+, Mn 2+, Ni 2+ precipitated as sulfides, as well as Cr 3+, Fe 3+, and Al 3+ precipitated as hydroxides. All these cations form compounds with S 2− that are insoluble in water at high pH. Precipitated by the addition of H 2 S in NaOH. Group 4 Group four cations are Mg 2+, Ca 2+, Ba 2+. All these cations form compounds with PO 4 3− that are insoluble in water at high pH. Precipitated by the addition of (NH 4 ) 2 HPO 4.

83. © 2014 Pearson Education, Inc. Group 5 Group five cations are Na +, K +, NH 4 +. All these cations form compounds that are soluble in water—they do not precipitate. They are identified by the color of their flame.

84. © 2014 Pearson Education, Inc. Complex Ion Formation Transition metals tend to be good Lewis acids. They often bond to one or more H 2 O molecules to form a hydrated ion. –H 2 O is the Lewis base, donating electron pairs to form coordinate covalent bonds. Ag + (aq) + 2 H 2 O(l)  Ag(H 2 O) 2 + (aq) Ions that form by combining a cation with several anions or neutral molecules are called complex ions. The attached ions or molecules are called ligands.

85. © 2014 Pearson Education, Inc. Complex Ion Equilibria If a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand. Ag(H 2 O) 2 + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) + 2 H 2 O (l) –Generally, H 2 O is not included, because its complex ion is always present in aqueous solution. Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq)

86. © 2014 Pearson Education, Inc. Formation Constant The reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction. Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) The equilibrium constant for the formation reaction is called the formation constant, K f.

87. © 2014 Pearson Education, Inc. Formation Constant

88. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro You mix a 200.0 mL sample of a solution that is 1.5 × 10 –3 M in Cu(NO 3 ) 2 with a 250.0 mL sample of a solution that is 0.20 M in NH 3. After the solution reaches equilibrium, what concentration of Cu 2+ (aq) remains? Example 16.15Complex Ion Equilibria Solution Write the balanced equation for the complex ion equilibrium that occurs and look up the value of K f in Table 16.3. Since this is an equilibrium problem, you have to create an ICE table, which requires the initial concentrations of Cu 2+ and NH 3. Calculate those concentrations from the given values.

89. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Continued Construct an ICE table for the reaction and write down the initial concentrations of each species. Example 16.15Complex Ion Equilibria

90. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Since the equilibrium constant is large and the concentration of ammonia is much larger than the concentration of Cu 2+, you can assume that the reaction will be driven to the right so that most of the Cu 2+ is consumed. Unlike previous ICE tables, where you let x represent the change in concentration in going to equilibrium, here you let x represent the small amount of Cu 2+ that remains when equilibrium is reached. Substitute the expressions for the equilibrium concentrations into the expression for K f and solve for x. Continued Example 16.15Complex Ion Equilibria

91. © 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Continued Confirm that x is indeed small compared to the initial concentration of the metal cation. Since x = 2.7 × 10 –13 << 6.7 × 10 –4, the approximation is valid. The remaining [Cu 2+ ] = 2.7 × 10 –13 M. The remaining Cu 2+ is very small because the formation constant is very large. Example 16.15Complex Ion Equilibria

92. © 2014 Pearson Education, Inc. The Effect of Complex Ion Formation on Solubility The solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands. AgCl (s)  Ag + (aq) + Cl − (aq) K sp = 1.77 × 10 −10 Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) K f = 1.7 × 10 7 Adding NH 3 to a solution in equilibrium with AgCl (s) increases the solubility of Ag +.

93. © 2014 Pearson Education, Inc. Figure 16.18

94. © 2014 Pearson Education, Inc. Solubility of Amphoteric Metal Hydroxides Many metal hydroxides are insoluble. All metal hydroxides become more soluble in acidic solution. –Shifting the equilibrium to the right by removing OH − Some metal hydroxides also become more soluble in basic solution. –Acting as a Lewis base forming a complex ion Substances that behave as both an acid and base are said to be amphoteric. Some cations that form amphoteric hydroxides include Al 3+, Cr 3+, Zn 2+, Pb 2+, and Sb 2+.

95. © 2014 Pearson Education, Inc. Al 3+ Al 3+ is hydrated in water to form an acidic solution. Al(H 2 O) 6 3+ (aq) + H 2 O (l)  Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) Addition of OH − drives the equilibrium to the right and continues to remove H from the molecules. Al(H 2 O) 5 (OH) 2+ (aq) + OH − (aq)  Al(H 2 O) 4 (OH) 2 + (aq) + H 2 O (l) Al(H 2 O) 4 (OH) 2 + (aq) + OH − (aq)  Al(H 2 O) 3 (OH) 3(s) + H 2 O (l)

96. © 2014 Pearson Education, Inc. Solubility of an Amphoteric Hydroxide