1. Chapter 17 Additional Aspects of Aqueous Equilibria John D. Bookstaver St. Charles Community College Cottleville, MO Lecture Presentation © 2012 Pearson Education, Inc.

2. Buffers crash course

3. Aqueous Equilibria 17.1 The Common-Ion Effect Consider a solution of acetic acid: CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO  (aq) © 2012 Pearson Education, Inc. If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left. NaC 2 H 3 O 2 (s)  Na + (aq) + C 2 H 3 O 2 - (aq)

4. Aqueous Equilibria The Common-Ion Effect “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.” © 2012 Pearson Education, Inc.

5. Aqueous Equilibria Sample Exercise 17.1 Calculating the pH When a Common Ion is Involved Solution Analyze We are asked to determine the pH of a solution of a weak electrolyte (CH 3 COOH) and a strong electrolyte (NaCH 3 COO) that share a common ion, CH 3 COO - Plan In any problem in which we must determine the pH of a solution containing a mixture of solutes, it is helpful to proceed by a series of logical steps: 1. Consider which solutes are strong electrolytes and which are weak electrolytes, and identify the major species in solution. 2. Identify the important equilibrium that is the source of H + and therefore determines pH. 3. Tabulate the concentrations of ions involved in the equilibrium. 4. Use the equilibrium-constant expression to calculate [H + ] and then pH. What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution?

6. Aqueous Equilibria Notice where the “initial” concentrations come from:  The strong electrolyte 0.30 M NaCH 3 COO dissociates “immediately” (NaCH 3 COO (s)  Na + (aq) + CH 3 COO - (aq) ) so its concentrations are: *[Na + ] = 0.30 M [CH 3 COO - ] = 0.30 M  The weak electrolyte, 0.30 M CH 3 COOH starts entirely as CH 3 COOH molecules so the initial concentrations of CH 3 COOH, H + and CH 3 COO - for the acetic acid equilibrium, CH 3 COOH  CH 3 COO - + H + are, [CH 3 COOH] = 0.30 M [CH 3 COO - ] = 0 M [H + ] = 0 M * Na + is a spectator of course so it won’t be needed – of course!. This is the most important concept you can understand and apply in Ch17! What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution?

7. Aqueous Equilibria Solve First, because CH 3 COOH is a weak electrolyte and NaCH 3 COO is a strong electrolyte, the major species in the solution are CH 3 COOH (a weak acid), Na + (which is neither acidic nor basic and is therefore a spectator in the acid–base chemistry), and CH 3 COO – (which is the conjugate base of CH 3 COOH). Second, [H + ] and, therefore, the pH are controlled by the dissociation equilibrium of CH 3 COOH: [We have written the equilibrium using H + (aq) rather than H 3 O + (aq), but both representations of the hydrated hydrogen ion are equally valid.] Third, we tabulate the initial and equilibrium concentrations as we did in solving other equilibrium problems in Chapters 15 and 16. Let x = the H + concentration we’re looking for: Again notice Starting conc. of acetate comes from the NaC 2 H 3 O 2

8. Aqueous Equilibria The equilibrium concentration of CH 3 COO - (the common ion) is the initial concentration that is due to NaCH 3 COO (0.30 M) plus the change in concentration (x) that is due to the ionization of CH 3 COOH. Now we can use the equilibrium-constant expression: The dissociation constant for CH 3 COOH at 25  C is from Appendix D; addition of NaCH 3 COO does not change the value of this constant. Substituting the equilibrium constant concentrations from our table into the equilibrium expression gives: Because Ka is small, we assume that x is small compared to the original concentrations of CH 3 COOH and CH 3 COO – (0.30 M each). Thus, we can ignore the very small x relative to 0.30 M, giving:

9. Aqueous Equilibria The resulting value of x is indeed small relative to 0.30, justifying the approximation made in simplifying the problem. Finally, we calculate the pH from the equilibrium concentration of H + (aq): x = 1.8  10  5 M = [H + ] pH = –log(1.8  10  5 ) = 4.74 Comment In Section 16.6 we calculated that a 0.30 M solution of CH 3 COOH has a pH of 2.64, corresponding to [H + ] = 2.3  10  3. Thus, the addition of NaCH 3 COO has substantially decreased [H + ], as we expect from Le Châtelier’s principle. Practice Exercise Calculate the pH of a solution containing 0.085 M nitrous acid (HNO 2, K a = 4.5  10  4 ) and 0.10 M potassium nitrite (KNO 2 ). Answer: 3.42

10. Aqueous Equilibria Sample Exercise 17.2 Calculating Ion Concentrations When a Common Ion is Involved Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Solution Analyze We are asked to determine the concentration of F – and the pH in a solution containing the weak acid HF and the strong acid HCl. In this case the common ion is H +. Plan We can again use the four steps outlined in Sample Exercise 17.1.

11. Aqueous Equilibria Again, notice where the “initial” concentrations come from:  The weak electrolyte, 0.20 M HF starts entirely as HF molecules so the initial concentrations of HF, H + and F - for the HF equilibrium, HF  F - + H + are, [HF] = 0.20 M [F - ] = 0 M [H + ] = 0 M  The strong electrolyte 0.10 M HCl dissociates “immediately” (HCl (aq)  H + (aq) + Cl - (aq) ) so its concentrations are: [H + ] = 0.10 M *[Cl - ] = 0.10 M * Cl - is a spectator of course so it won’t be needed – of course!.

12. Aqueous Equilibria Solve Because HF is a weak acid and HCl is a strong acid, the major species in solution are HF, H +, and Cl –. The Cl –, which is the conjugate base of a strong acid, is merely a spectator ion in any acid–base chemistry. The problem asks for [F – ], which is formed by ionization of HF. Thus, the important equilibrium is The common ion in this problem is the hydrogen (or hydronium) ion. Now we can tabulate the initial and equilibrium concentrations of each species involved in this equilibrium. Since we started with 0.10 M acid, let x be the F - and in this case the increase in H + due to the HF contribution: Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Again notice starting conc. of hydrogen comes from the immediate dissociation of the strong electrolyte HCl

13. Aqueous Equilibria The equilibrium constant for the ionization of HF, from Appendix D, is 6.8  10  4. Substituting the equilibrium-constant concentrations into the equilibrium expression gives If we assume that x is small relative to 0.10 or 0.20 M, this expression simplifies to This F – concentration is substantially smaller than it would be in a 0.20 M solution of HF with no added HCl. The common ion, H +, shifts the equilibrium to the left, and suppresses the ionization of HF. The concentration of H + (aq) is Thus, pH = 1.00 Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.

14. Aqueous Equilibria Comment Notice that for all practical purposes, the hydrogen ion concentration is due entirely to the HCl; the HF makes a negligible contribution by comparison.

15. Aqueous Equilibria Strategy! 1) Assume that strong electrolytes added dissociate immediately and completely 2) Assume that weak electrolytes dissociate next, and the extent will depend on the presence of the strong electrolyte already present 3) The equilibrium concentrations will depend on the weak electrolyte shift that occurs. In a problem involving mixing strong electrolytes with weak:

16. Aqueous Equilibria Strategy! 1) Assume that strong electrolytes added dissociate immediately and completely 2) Assume that weak electrolytes dissociate next, and the extent will depend on the presence of the strong electrolyte already present 3) The equilibrium concentrations will depend on the weak electrolyte shift that occurs. Repeat: The final concentration depends on the shift that occurs in the weak electrolyte

17. Aqueous Equilibria Strategy! Example: suppose you wish to find the final concentration of acetate ions (C 2 H 3 O 2 - ) in a solution of acetic acid (HC 2 H 3 O 2 ) when a strong acid like HCl is added.

18. Aqueous Equilibria Strategy! Example: adding HCl to a solution of HC 2 H 3 O 2 (strong acid) (weak acid) 1) Assume HCl completely dissociates giving off all of its H + ions.

19. Aqueous Equilibria Strategy! Example: adding HCl to a solution of HC 2 H 3 O 2 (strong acid) (weak acid) 2) Assume HC 2 H 3 O 2 will next begin to ionize. (The assumption is that acetic acid is waiting to ionize until after the strong acid is added. Silly I suppose but it works!)

20. Aqueous Equilibria Strategy! Example: adding HCl to a solution of HC 2 H 3 O 2 (strong acid) (weak acid) The final pH will depend on the weak electrolyte acetic acid (not on the HCl)

21. Aqueous Equilibria Strategy! Example: adding HCl to a solution of HC 2 H 3 O 2 (strong acid) (weak acid) You will need to set up and ICE table and K a expression to calculate the equilibrium concentration of H + that remains after the acetic acid shift!

22. Aqueous Equilibria Strategy! Try keeping this strategy in mind as you work problems involving strong and weak electrolyte mixtures.

23. Aqueous Equilibria Practice Exercise Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid (HCOOH, K a = 1.8  10  4 ) and 0.10 M in HNO 3. Answer: [HCOO – ] = 9.0  10  5, pH = 1.00

24. Aqueous Equilibria 17.2 Buffers Buffers are solutions of a weak conjugate acid–base pair. They are particularly resistant to pH changes, even when strong acid or base is added. © 2012 Pearson Education, Inc.

25. Aqueous Equilibria Buffers crash course

26. Aqueous Equilibria Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, the HF reacts with the OH  to make F  and water. HF + OH -  F - + H 2 O © 2012 Pearson Education, Inc.

27. Aqueous Equilibria Buffers Similarly, if acid is added, the F  reacts with it to form HF and water. H 3 O + + F -  H 2 O + HF © 2012 Pearson Education, Inc.

28. Aqueous Equilibria Sample Exercise 17.3 Calculating the pH of a Buffer What is the pH of a buffer that is 0.12 M in lactic acid [HC 3 H 5 O 3 ] and 0.10 M in sodium lactate [NaC 3 H 5 O 3 ]? For lactic acid, K a = 1.4  10  4. Solution Analyze We are asked to calculate the pH of a buffer containing lactic acid (HC 3 H 5 O 3 ) and its conjugate base, the lactate ion (C 3 H 5 O 3 – ). Plan We will first determine the pH using the method described in Section 17.1. Because HC 3 H 5 O 3 is a weak electrolyte and NaC 3 H 5 O 3 is a strong electrolyte, the major species in solution are HC 3 H 5 O 3, Na +, and C 3 H 5 O 3 –. The Na + ion is a spectator ion. The HC 3 H 5 O 3 –C 3 H 5 O 3 – conjugate acid–base pair determines [H + ] and, thus, pH; [H + ] can be determined using the acid-dissociation equilibrium of lactic acid.

29. Aqueous Equilibria Notice where the “initial” concentrations come from:  The weak electrolyte, 0.12 M lactic acid starts entirely as HC 3 H 5 O 3 molecules so the initial concentrations of HC 3 H 5 O 3, H + and C 3 H 5 O 3 - for the HC 3 H 5 O 3 equilibrium, HC 3 H 5 O 3  C 3 H 5 O 3 - + H + are, [ HC 3 H 5 O 3 ] = 0.12 M [ C 3 H 5 O 3 - ] = 0 M [H + ] = 0 M  The strong electrolyte 0.10 M sodium lactate dissociates “immediately” ( NaC 3 H 5 O 3 (aq)  Na + (aq) + C 3 H 5 O 3 - (aq) ) so its concentrations are: *[Na + ] = 0.10 M [ C 3 H 5 O 3 - ] = 0.10 M Notice: This will give you all of the “ I” values for your ICE table * Na + is a spectator of course so it won’t be needed – of course!.

30. Aqueous Equilibria Solve The initial and equilibrium concentrations of the species involved in this equilibrium are The equilibrium concentrations are governed by the equilibrium expression: Because K a is small and a common ion is present, we expect x to be small relative to either 0.12 or 0.10 M. Thus, our equation can be simplified to give

31. Aqueous Equilibria Solving for x gives a value that justifies our approximation: Practice Exercise Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. Appendix D: benzoic acid HC 7 H 5 O 2 k a =6.3 x 10 -5 Answer: 4.42

32. Aqueous Equilibria pH Range The pH range is the range of pH values over which a buffer system works effectively. It is best to choose an acid with a pK a close to the desired pH. © 2012 Pearson Education, Inc.

33. Aqueous Equilibria Sample Exercise 17.4 Preparing a Buffer Solution Analyze We are asked to determine the amount of NH 4 + ion required to prepare a buffer of a specific pH. How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH is 9.00? (Assume that the addition of NH 4 Cl does not change the volume of the solution.) Plan The major species in the solution will be NH 4 +, Cl –, and NH 3. Of these, the Cl – ion is a spectator (it is the conjugate base of a strong acid). Thus, the NH 4 + –NH 3 conjugate acid–base pair will determine the pH of the buffer. The equilibrium relationship between NH 4 + and NH 3 is given by the base- dissociation reaction for NH 3 : The key to this exercise is to use this K b expression to calculate [NH 4 + ].

34. Aqueous Equilibria Solve We obtain [OH  ] from the given pH: and so Because K b is small and the common ion [NH 4 + ] is present, the equilibrium concentration of NH 3 essentially equals its initial concentration: pOH = 14.00 – pH = 14.00 – 9.00 = 5.00 [OH – ] = 1.0  10  5 M [NH 3 ] = 0.10 M ReactionNH 3 H2OH2ONH 4 + OH - Initial0.10 M-x0 Change- 10 -5 -+ 10 -5 Equilibrium0.10 M-x+ 10 -5 10 -5 M

35. Aqueous Equilibria We now use the expression for K b to calculate [NH 4 + ]: Thus, for the solution to have pH = 9.00, [NH 4 + ] must equal 0.18 M. The number of moles of NH 4 Cl needed to produce this concentration is given by the product of the volume of the solution and its molarity: (2.0 L)(0.18 mol NH 4 Cl/L) = 0.36 mol NH 4 Cl

36. Aqueous Equilibria Practice Exercise Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (C 6 H 5 COOH) to produce a pH of 4.00. Answer: Notice: sodium benzoate (strong electrolyte with the conjugate base) with benzoic acid (weak conjugate acid) 0.13 M

37. Aqueous Equilibria When Strong Acids or Bases Are Added to a Buffer When strong acids or bases are added to a buffer, it is safe to assume that all of the strong acid or base is consumed in the reaction. © 2012 Pearson Education, Inc.

38. Aqueous Equilibria Addition of Strong Acid or Base to a Buffer 1.Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. 2.Use traditional K a methods to determine the final pH

39. Aqueous Equilibria Sample Exercise 17.5 Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol CH 3 COOH and 0.300 mol CH 3 COONa to enough water to make 1.000 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1). (a) Calculate the pH of this solution after 5.0 mL of 4.0 M NaOH(aq) solution is added. (b) For comparison, calculate the pH of a solution made by adding 5.0 mL of 4.0 M NaOH(aq) solution to 1.000 L of pure water.

40. Aqueous Equilibria Solution Analyze We are asked to determine the pH of a buffer after addition of a small amount of strong base and to compare the pH change with the pH that would result if we were to add the same amount of strong base to pure water. Plan Solving this problem involves the two steps outlined in Figure 17.3. First we do a stoichiometry calculation to determine how the added OH – affects the buffer composition. Then we use the resultant buffer composition and the equilibrium-constant expression for the buffer to determine the pH. A buffer is made by adding 0.300 mol CH 3 COOH and 0.300 mol CH 3 COONa to enough water to make 1.000 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1). (a) Calculate the pH of this solution after 5.0 mL of 4.0 M NaOH(aq) solution is added.

41. Aqueous Equilibria Solve (a) Stoichiometry: Prior to this neutralization reaction, there are 0.300 mol each of CH 3 COOH and CH 3 OO –. The amount of base added is 0.0050 L  4.0 mol/L = 0.020 mol. The OH – provided by NaOH reacts with CH 3 COOH, the weak acid component of the buffer. Neutralizing the 0.020 mol OH – requires 0.020 mol of CH 3 COOH. Consequently, the amount of CH 3 COOH decreases by 0.020 mol, and the amount of the product of the neutralization, CH 3 COO –, increases by 0.020 mol. 0.020 moles OH - 0.020 moles CH 3 COOH + forms 0.020 moles CH 3 COO - This table summarizes the changes occurring during the neutralization: ICFICF -0.020+0.020

42. Aqueous Equilibria Equilibrium Calculation: We now turn our attention to the equilibrium for the ionization of acetic acid, the relationship that determines the buffer pH: Notice that all of the added OH - has been neutralized. Also, our quantities in this table are in moles, not Molarity. The final volume will be the combined volumes of the two solutions, which will determine final molarities K a = [H + ][CH 3 COO - ] [CH 3 COOH] Converting moles to Molarity: [CH 3 COO - ] = 0.320 moles/1.005 L = 0.318 M [CH 3 COOH] = 0.280 moles/1.005 L = 0.279 M

43. Aqueous Equilibria And substituting into our eq expression: [H + ][0.318] [0.279] [H + ] = 1.58 x 10 -5 M pH = - log [H+ ] = -log (1.58 x 10 -5 ) = 4.80 ReactionCH 3 COOH H+ H+ CH 3 COO - Initial0.279 M0 0.318 Change- x- x+ x Equilibrium0.279 M -xx 0.318 -x “Since acetic acid is a weak acid (Ka is a low 10 -5 ) we assume that x is insignificant compared to our equilibrium concentrations of CH 3 COOH and CH 3 COO - so 0.279 M –x = 0.279 M and 0.318 –x = 0.318 M”

44. Aqueous Equilibria Method 2: Using the quantities of CH 3 COOH – and CH 3 COO remaining in the buffer, we determine the pH using the Henderson–Hasselbalch equation. pH = - log(1.8 x 10 -5 ) + log [CH 3 COO - ] [CH 3 COOH] pH = 4.74 + log [0.318] [0.279] pH = 4.74 + 0.057 = 4.80 Does this seem easier? We must do most of the same steps to get to this point?

45. Aqueous Equilibria (b) To determine the pH of a solution made by adding 0.020 mol of NaOH to 1.000 L of pure water, we first determine the concentration of OH – ions in solution, [OH – ] = 0.020 mol/(1.005 L) = 0.020 M We use this value in Equation 16.18 to calculate pOH and then use our calculated pOH value in Equation 16.20 to obtain pH: pOH = –log[OH – ] = –log(0.020) = +1.70 pH = 14 – (+1.70) = 12.30 Comment Note that the small amount of added NaOH changes the pH of water significantly. In contrast, the pH of the buffer changes very little when the NaOH is added, as summarized in Figure 17.4.

46. Aqueous Equilibria Practice Exercise Determine (a) the pH of the original buffer described in Sample Exercise 17.5 after the addition of 0.020 mol HCl and (b) the pH of the solution that would result from the addition of 0.020 mol HCl to 1.000 L of pure water. Answers : Sample Exercise 17.5 Calculating pH Changes in Buffers Continued (a) 4.68, (b) 1.70

47. Aqueous Equilibria 17.3 Titration In this technique, a known concentration of base (or acid) is slowly added to a solution of acid (or base). © 2012 Pearson Education, Inc.

48. Aqueous Equilibria

49. Aqueous Equilibria

50. Aqueous Equilibria Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base. © 2012 Pearson Education, Inc.

51. Aqueous Equilibria Titration of a Strong Acid with a Strong Base

52. Aqueous Equilibria Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly.

53. Aqueous Equilibria Just before (and after) the equivalence point, the pH increases rapidly. © 2012 Pearson Education, Inc. Titration of a Strong Acid with a Strong Base

54. Aqueous Equilibria At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid. Titration of a Strong Acid with a Strong Base

55. Aqueous Equilibria As more base is added, the increase in pH again levels off. Titration of a Strong Acid with a Strong Base

56. Aqueous Equilibria Sample Exercise 17.6 Calculating pH for a Strong Acid–Strong Base Titration Solution Analyze We are asked to calculate the pH at two points in the titration of a strong acid with a strong base. The first point is just before the equivalence point, so we expect the pH to be determined by the small amount of strong acid that has not yet been neutralized. The second point is just after the equivalence point, so we expect this pH to be determined by the small amount of excess strong base. Calculate the pH when (a) 49.0 mL and (b) 51.0 mL of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. Plan (a) As the NaOH solution is added to the HCl solution, H + (aq) reacts with OH – (aq) to form H 2 O. Both Na + and Cl – are spectator ions, having negligible effect on the pH. To determine the pH of the solution, we must first determine how many moles of H + were originally present and how many moles of were added. We can then calculate how many moles of each ion remain after the neutralization reaction. To calculate [H + ], and hence pH, we must also remember that the volume of the solution increases as we add titrant, thus diluting the concentration of all solutes present.

57. Aqueous Equilibria Solve The number of moles of H + in the original HCl solution is given by the product of the volume of the solution and its molarity: Likewise, the number of moles of OH – in 49.0 mL of 0.100 M NaOH is Continued Sample Exercise 17.6 Calculating pH for a Strong Acid–Strong Base Titration

58. Aqueous Equilibria Because we have not reached the equivalence point, there are more moles of H + present than OH –. Each mole of reacts OH – with 1 mol of H +. Using the convention introduced in Sample Exercise 17.5, we have the neutralization reaction: The volume of the reaction mixture increases as the NaOH solution is added to the HCl solution. Thus, at this point in the titration, the volume in the titration flask is 50.0 mL + 49.0 mL = 99.0 mL = 0.0990 L Thus, the concentration of H + (aq) in the flask is The corresponding pH is –log(1.0  10  3 ) = 3.00

59. Aqueous Equilibria Continued Sample Exercise 17.6 Calculating pH for a Strong Acid–Strong Base Titration Plan (b) We proceed in the same way as we did in part (a) except we are now past the equivalence point and have more OH – in the solution than H +. As before, the initial number of moles of each reactant is determined from their volumes and concentrations. The reactant present in smaller stoichiometric amount (the limiting reactant) is consumed completely, leaving an excess of hydroxide ion.

60. Aqueous Equilibria pOH = –log(1.0  10  3 ) = 3.00 pH = 14.00 – pOH = 14.00 – 3.00 = 11.00 Solve In this case the volume in the titration flask is Hence, the concentration of OH – (aq) in the flask is and we have 50.0 mL + 51.0 mL = 101.0 mL = 0.1010 L

61. Aqueous Equilibria Sample Exercise 17.6 Calculating pH for a Strong Acid–Strong Base Titration Comment Note that the pH increased by only two pH units, from 1.00 (Figure 17.7) to 3.00, after the first 49.0 mL of NaOH solution was added, but jumped by eight pH units, from 3.00 to 11.00, as 2.0 mL of base solution was added near the equivalence point. Such a rapid rise in pH near the equivalence point is a characteristic of titrations involving strong acids and strong bases. Practice Exercise Calculate the pH when (a) 24.9 mL and (b) 25.1 mL of 0.100 M HNO 3 have been added to 25.0 mL of 0.100 M KOH solution. Answers: (a) 10.30, (b) 3.70

62. Aqueous Equilibria Titration of a Weak Acid with a Strong Base

63. Aqueous Equilibria Unlike in the previous case, the stronger conjugate base of the weak acid affects the pH when it is formed (hydrolysis)

64. Aqueous Equilibria At the equivalence point the pH is >7.

65. Aqueous Equilibria Unlike in the previous case, the stronger conjugate base of the weak acid affects the pH when it is formed (hydrolysis)

66. Aqueous Equilibria Titration of a Weak Acid with a Strong Base At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time. © 2012 Pearson Education, Inc.

67. Aqueous Equilibria Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle. © 2012 Pearson Education, Inc.

68. Aqueous Equilibria Sample Exercise 17.7 Calculating pH for a Weak Acid–Strong Base Titration Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH (K a = 1.8  10  5 ). Solution Analyze We are asked to calculate the pH before the equivalence point of the titration of a weak acid with a strong base. Plan We first must determine the number of moles of CH 3 COOH and CH 3 COO – present after the neutralization reaction. We then calculate pH using K a, [CH 3 COOH], and [CH 3 COO – ].

69. Aqueous Equilibria Solve Stoichiometry Calculation: The product of the volume and concentration of each solution gives the number of moles of each reactant present before the neutralization: The 4.50  10  3 of NaOH consumes 4.50  10  3 of CH 3 COOH: The total volume of the solution is 45.0 mL + 50.0 mL = 95.0 mL = 0.0950 L

70. Aqueous Equilibria The resulting molarities of CH 3 COOH and CH 3 COO – after the reaction are therefore Equilibrium Calculation: The equilibrium between CH 3 COOH and CH 3 COO – must obey the equilibrium-constant expression for CH 3 COOH: ReactionCH 3 COOH H + CH 3 COO - Initial0.053 M0 0.0474 Change- x+ x Equilibrium0.053 M -xx 0.0474 + x

71. Aqueous Equilibria Comment We could have solved for pH equally well using the Henderson–Hasselbalch equation – once the post neutralization concentrations have been calculated! Practice Exercise (a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (C 6 H 5 COOH, K a = 6.3  10  5. (b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH 3. Answers: (a) 4.20, (b) 9.26 Making similar assumptions as before: 0.053 M –x = 0.053 and 0.0474 + x = 0.0474

72. Aqueous Equilibria Sample Exercise 17.8 Calculating the pH at the 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. Solution Analyze We are asked to determine the pH at the equivalence point of the titration of a weak acid with a strong base. Because the neutralization of a weak acid produces its anion, which is a weak base, we expect the pH at the equivalence point to be greater than 7. (The conjugate base of the weak acid will hydrolyze the water) Plan The initial number of moles of acetic acid equals the number of moles of acetate ion at the equivalence point. We use the volume of the solution at the equivalence point to calculate the concentration of acetate ion. Because the acetate ion is a weak base, we can calculate the pH using K b and [CH 3 COO – ].

73. Aqueous Equilibria Solve The number of moles of acetic acid in the initial solution is obtained from the volume and molarity of the solution: Moles = M  L = (0.100 mol)/L(0.0500 L) = 5.00  10  3 mol CH 3 COOH Hence, 5.00  10  3 mol of [CH 3 COO – ] is formed. It will take 50.0 mL of NaOH to reach the equivalence point (Figure 17.9). … since they are both 0.100 M solutions The volume of this salt solution at the equivalence point is the sum of the volumes of the acid and base, 50.0 mL + 50.0 mL = 100.0 mL = 0.1000 L. Thus, the concentration of CH 3 COO – is

74. Aqueous Equilibria The K b for CH 3 COO – can be calculated from the K a value of its conjugate acid, K b = K w /K a = (1.0  10  14 )/(1.8  10  5 ) = 5.6  10  10. Using the K b expression, we have Making the approximation that and then solving for x, we have x = [OH – ] = 5.3  10  6 M, which gives pOH = 5.28 pH = 8.72. Check The pH is above 7, as expected for the salt of a weak acid and strong base. Practice Exercise Calculate the pH at the equivalence point when (a) 40.0 mL of 0.025 M benzoic acid (C 6 H 5 COOH, K a = 6.3  10  5 ) is titrated with 0.050 M NaOH; (b) 40.0 mL of 0.100 M NH 3 is titrated with 0.100 M HCl. Answers: (a) 8.21, (b) 5.28 The CH 3 COO – ion is a weak base:

75. Aqueous Equilibria Titration of a Weak Base with a Strong Acid The pH at the equivalence point in these titrations is <7, so using phenolphthalein would not be a good idea. Methyl red is the indicator of choice. © 2012 Pearson Education, Inc.

76. Aqueous Equilibria pKa = pH at ½ equivalence point

77. Aqueous Equilibria For HC 2 H 3 O 2 : HC 2 H 3 O 2 + OH-  H 2 O + C 2 H 3 O 2 - 100 mL of 0.100 M = 0.010 moles of H + 50 mL of 0.100 M = 0.005 moles of OH + Only 0.005 Moles react; 0.005 moles HC 2 H 3 O 2 Left over 0.005 Moles OH - react; 0.005 Moles C 2 H 3 O 2 - produced pH depends on leftover HC 2 H 3 O 2 in water HC 2 H 3 O 2  H + + C 2 H 3 O 2 - Ka = [H + ][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] Ka = [H + ](.005 Mol/vol) (0.005 Mol/vol) pKa = pH Remaining solution Contains equal quantities of HC 2 H 3 O 2 and C 2 H 3 O 2 -

78. Aqueous Equilibria Titrations of Polyprotic Acids When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation. © 2012 Pearson Education, Inc. Ex: For H 2 CO 3

79. Aqueous Equilibria

80. Aqueous Equilibria 17.4 Solubility Products In a solution of a slightly soluble salt, an equilibrium will exist between the excess solid and the small concentration of dissolved ions. For example, Barium Sulfate; a mostly insoluble salt BaSO 4 (s)Ba 2+ (aq) + SO 4 2  (aq)

81. Aqueous Equilibria

82. Aqueous Equilibria Solubility Products The equilibrium constant expression for this equilibrium can be written: K sp = [Ba 2+ ] [SO 4 2  ] where the equilibrium constant, K sp, is called the solubility product constant. *Recall that solids (and liquids) are left out of the expression, so the Ksp only includes the products. *BaSO 4 (s)Ba 2+ (aq) + SO 4 2  (aq)

83. Aqueous Equilibria Sample Exercise 17.9 Writing Solubility-Product (K sp ) Expressions Solution Analyze We are asked to write an equilibrium-constant expression for the process by which CaF 2 dissolves in water. Plan We apply the general rules for writing an equilibrium-constant expression, excluding the solid reactant from the expression. We assume that the compound dissociates completely into its component ions: Solve The expression for K sp is K sp = [Ca 2+ ][F – ] 2 Write the expression for the solubility-product constant for CaF 2, Appendix D gives 3.9  10  11 for this K sp.

84. Aqueous Equilibria Practice Exercise Give the solubility-product-constant expressions and K sp values (from Appendix D) for (a) barium carbonate, (b) silver sulfate. Answers: (a) K sp = [Ba 2+ ][CO 3 2– ] = 5.0  10  9, (b) K sp = [Ag + ] 2 [SO 4 2– ] = 1.5  10  5

85. Aqueous Equilibria Solubility Products K sp is not the same as solubility. Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M). © 2012 Pearson Education, Inc.

86. Aqueous Equilibria Sample Exercise 17.10 Calculating K sp from Solubility Solid silver chromate is added to pure water at 25  C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3  10  4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving Ag + or CrO 4 2– ions in the solution, calculate K sp for this compound.

87. Aqueous Equilibria Solution Analyze We are given the equilibrium concentration of Ag + in a saturated solution of Ag 2 CrO 4 and asked to determine the value of K sp for Ag 2 CrO 4. Plan The equilibrium equation and the expression for K sp are To calculate K sp, we need the equilibrium concentrations of Ag + and CrO 4 2–.We know that at equilibrium [Ag + ] = 1.3  10  4 M. All the Ag + and CrO 4 2– ions in the solution come from the Ag 2 CrO 4 that dissolves. Thus, we can use [Ag + ] to calculate [CrO 4 2– ].

88. Aqueous Equilibria Solve From the chemical formula of silver chromate, we know that there must be 2 Ag + ions in solution for each CrO 4 2– ion in solution. Consequently, the concentration of CrO 4 2– is half the concentration of Ag + : and K sp is K sp = [Ag + ] 2 [CrO 4 2– ] = (1.3  10  4 ) 2 (6.5  10  5 ) = 1.1  10  12 Check We obtain a small value, as expected for a slightly soluble salt. Furthermore, the calculated value agrees well with the one given in Appendix D, 1.2  10  12

89. Aqueous Equilibria Practice Exercise A saturated solution of Mg(OH) 2 in contact with undissolved Mg(OH) 2 (s) is prepared at 25  C. The pH of the solution is found to be 10.17. Assuming that Mg(OH) 2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg 2+ or OH – ions, calculate K sp for this compound. Answer: 1.6  10  12

90. Aqueous Equilibria Sample Exercise 17.11 Calculating Solubility from K sp The K sp for CaF 2 is 3.9  10  11 at 25  C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter.

91. Aqueous Equilibria Solution Analyze We are given K sp for CaF 2 and asked to determine solubility. Recall that the solubility of a substance is the quantity that can dissolve in solvent, whereas the solubility-product constant, K sp, is an equilibrium constant. Plan To go from K sp to solubility, we follow the steps indicated by the red arrows in Figure 17.16. We first write the chemical equation for the dissolution and set up a table of initial and equilibrium concentrations. We then use the equilibrium- constant expression. In this case we know K sp, and so we solve for the concentrations of the ions in solution. Once we know these concentrations, we use the formula weight to determine solubility in g/L.

92. Aqueous Equilibria Solve Assume that initially no salt has dissolved, and then allow x mol L of CaF 2 to dissociate completely when equilibrium is achieved: That makes the concentration of Ca 2+ ions be X, and F - 2X

93. Aqueous Equilibria The stoichiometry of the equilibrium dictates that 2x mol L - of F – are produced for each x mol L - of CaF 2 that dissolve. We now use the expression for K sp and substitute, and simplify the equilibrium concentrations to solve for the value of x:

94. Aqueous Equilibria (Remember that. To calculate the cube root of a number, use the y x function on your calculator, with ) Thus, the molar solubility of CaF 2 is 2.1  10  4 mol L. The mass of CaF 2 that dissolves in water to form 1 L of solution is

95. Aqueous Equilibria Check We expect a small number for the solubility of a slightly soluble salt. If we reverse the calculation, we should be able to recalculate the solubility product: K sp = (2.1  10  4 )(4.2  10  4 ) 2 = 3.7  10  11 close to the value given in the problem statement, 3.9  10  11. Comment Because F – is the anion of a weak acid, you might expect hydrolysis of the ion to affect the solubility of CaF 2. The basicity of F – is so small ( K b = 1.5  10  11 ), however, that the hydrolysis occurs to only a slight extent and does not significantly influence the solubility. The reported solubility is 0.017 g/L at 25  C, in good agreement with our calculation. Practice Exercise The K sp for LaF 3 is 2  10  19. What is the solubility of LaF 3 in water in moles per liter? Answer: 9  10  6 mol/L

96. Aqueous Equilibria 17.5 Factors Affecting Solubility The Common-Ion Effect –If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease: BaSO 4 (s)Ba 2+ (aq) + SO 4 2  (aq) Example: adding sulfuric acid H 2 SO 4 (aq) to a solution of Barium sulfate, will increase the concentration of SO 4 2- ions. The equilibrium above shifts left to remove the excess sulfate and BaSO 4 precipitates out.

97. Aqueous Equilibria Notice: as concentrated acid H 2 SO 4 is added, white BaSO 4 precipitates BaSO 4 (s)  Ba 2+ + SO 4 2-

98. Aqueous Equilibria Sample Exercise 17.12 Calculating the Effect of a Common Ion on Solubility Calculate the molar solubility of CaF 2 at 25  C in a solution that is (a) 0.010 M in Ca(NO 3 ) 2, (b) 0.010 M in NaF. Solution Analyze We are asked to determine the solubility of CaF 2 in the presence of two strong electrolytes, each containing an ion common to CaF 2. In (a) the common ion is Ca 2+, and NO 3 – is a spectator ion. In (b) the common ion is F –, and Na + is a spectator ion.

99. Aqueous Equilibria Plan Because the slightly soluble compound is CaF 2, we need to use K sp for this compound, which Appendix D gives as 3.9  10  11. The value of K sp is unchanged by the presence of additional solutes. Because of the common-ion effect, however, the solubility of the salt decreases in the presence of common ions. We use our standard equilibrium techniques of starting with the equation for CaF 2 dissolution, setting up a table of initial and equilibrium concentrations, and using the K sp expression to determine the concentration of the ion that comes only from CaF 2.

100. Aqueous Equilibria Solve (a) The initial concentration of Ca 2+ is 0.010 M because of the dissolved Ca(NO 3 ) 2 : Substituting into the solubility-product expression gives K sp = 3.9  10  11 = [Ca 2+ ][F – ] 2 = (0.010 + x)(2x) 2

101. Aqueous Equilibria This would be a messy problem to solve exactly, but fortunately it is possible to simplify matters. Even without the common-ion effect, the solubility of CaF 2 is very small (2.1  10  4 M). Thus, we assume that the 0.010 M concentration of Ca 2+ from Ca(NO 3 ) 2 is very much greater than the small additional concentration resulting from the solubility of CaF 2 ; that is, x is much smaller than 0.010 M, and. We then have

102. Aqueous Equilibria [Ca 2+ ] = x and [F – ] = 0.010 + 2x (b) The common ion is F –, and at equilibrium we have Assuming that 2x is much smaller than 0.010 M (that is, ), we have Thus, 3.9  10  7 mol of solid CaF 2 should dissolve per liter of 0.010 M NaF solution.

103. Aqueous Equilibria Comment The molar solubility of CaF 2 in water is 2.1  10  4 M (Sample Exercise 17.11). By comparison, our calculations here give a CaF 2 solubility of 3.1  10  5 M in the presence of 0.010 M Ca 2+ and 3.9  10  7 M in the presence of 0.010 M F – ion. Thus, the addition of either Ca 2+ or F – to a solution of CaF 2 decreases the solubility. However, the effect of F – on the solubility is more pronounced than that of Ca 2+ because [F – ] appears to the second power in the K sp expression for CaF 2, whereas [Ca 2+ ] appears to the first power.

104. Aqueous Equilibria Practice Exercise For manganese(II) hydroxide,Mn(OH) 2, K sp = 1.6  10  13. Calculate the molar solubility of Mn(OH) 2 in a solution that contains 0.020 M NaOH. Answer: Continued Sample Exercise 17.12 Calculating the Effect of a Common Ion on Solubility 4.0  10  10 M

105. Aqueous Equilibria Factors Affecting Solubility pH –If a substance has a basic anion, it will be more soluble in an acidic solution. –Substances with acidic cations are more soluble in basic solutions. © 2012 Pearson Education, Inc.

106. Aqueous Equilibria Sample Exercise 17.13 Predicting the Effect of Acid on Solubility Which of these substances are more soluble in acidic solution than in basic solution: (a) Ni(OH) 2 (s), (b) CaCO 3 (s), (c) BaF 2 (s), (d) AgCl(s)? Solution Analyze The problem lists four sparingly soluble salts, and we are asked to determine which are more soluble at low pH than at high pH. Plan Ionic compounds that dissociate to produce a basic anion are more soluble in acid solution. Solve (a) Ni(OH) 2 (s) is more soluble in acidic solution because of the basicity of OH – ; the H + reacts with the OH – ion, forming water:

107. Aqueous Equilibria Solution (b) CaCO 3 (s) ; Similarly, CaCO 3 (s) dissolves in acid solutions because CO 3 2– is a basic anion: The reaction between CO 3 2– and H + occurs in steps, with HCO 3 – forming first and H 2 CO 3 forming in appreciable amounts only when [H + ] is sufficiently high. What would be observed as the CaCO 3 dissolves? Bubbles! …of CO 2

108. Aqueous Equilibria Continued (c) BaF 2 (s)?, The solubility of BaF 2 is enhanced by lowering the pH because F – is a basic anion: (d) AgCl(s)? The solubility of AgCl is unaffected by changes in pH because Cl – is the anion of a strong acid and therefore has negligible basicity. Sample Exercise 17.13 Predicting the Effect of Acid on Solubility

109. Aqueous Equilibria Practice Exercise Write the net ionic equation for the reaction between an acid and (a) CuS, (b) Cu(N 3 ) 2. Answers: (a) (b)

110. Aqueous Equilibria Factors Affecting Solubility Complex Ions –Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent. © 2012 Pearson Education, Inc.

111. Aqueous Equilibria Factors Affecting Solubility Complex Ions –The formation of these complex ions increases the solubility of these salts.

112. Aqueous Equilibria Sample Exercise 17.14 Evaluating an Equilibrium Involving a Complex Ion Calculate the concentration of Ag + present in solution at equilibrium when concentrated ammonia is added to a 0.010 M solution of AgNO 3 to give an equilibrium concentration of [NH 3 ] = 0.20 M. Neglect the small volume change that occurs when NH 3 is added. Solution Analyze Addition of NH 3 (aq) to Ag + (aq) forms Ag(NH 3 ) 2 +, as shown in Equation 17.22. We are asked to determine what concentration of Ag + (aq) remains uncombined when the NH 3 concentration is brought to 0.20 M in a solution originally 0.010 M in AgNO 3.

113. Aqueous Equilibria Plan We assume that the AgNO 3 is completely dissociated, giving 0.010 M Ag +. Because K f for the formation of Ag(NH 3 ) 2 + is quite large, we assume that essentially all the Ag + is converted to Ag(NH 3 ) 2 + and approach the problem as though we are concerned with the dissociation of Ag(NH 3 ) 2 + rather than its formation. To facilitate this approach, we need to reverse Equation 17.22 and make the corresponding change to the equilibrium constant:

114. Aqueous Equilibria Solve If [Ag + ] is 0.010 M initially, [Ag(NH 3 ) 2 + ] will be 0.010 M following addition of the NH 3.We construct a table to solve this equilibrium problem. Note that the NH 3 concentration given in the problem is an equilibrium concentration rather than an initial concentration. Because [Ag + ] is very small, we can ignore x, so that. Substituting these values into the equilibrium-constant expression for the dissociation of Ag(NH 3 ) 2 +, we obtain Formation of the Ag(NH 3 ) 2 + complex drastically reduces the concentration of free Ag + ion in solution.

115. Aqueous Equilibria Practice Exercise Calculate [Cr 3+ ] in equilibrium with Cr(OH) 4  when 0.010 mol of Cr(NO 3 ) 3 is dissolved in 1 L of solution buffered at pH 10.0. Answer: 1  10  16 M

116. Aqueous Equilibria Factors Affecting Solubility Amphoterism –Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. –Examples of such cations are Al 3+, Zn 2+, and Sn 2+.

117. Aqueous Equilibria 17.6 Will a Precipitate Form? In a solution, –If Q = K sp, the system is at equilibrium and the solution is saturated. –If Q < K sp, more solid can dissolve until Q = K sp. –If Q > K sp, the salt will precipitate until Q = K sp. © 2012 Pearson Education, Inc.

118. Aqueous Equilibria Sample Exercise 17.15 Predicting Whether a Precipitate Forms Does a precipitate form when 0.10 L of 8.0  10  3 M Pb(NO 3 ) 2 is added to 0.40 L of 5.0  10  3 M Na 2 SO 4 ? Solution Analyze The problem asks us to determine whether a precipitate forms when two salt solutions are combined. Plan We should determine the concentrations of all ions just after the solutions are mixed and compare the value of Q with K sp for any potentially insoluble product. The possible metathesis products are PbSO 4 and NaNO 3. Like all sodium salts is soluble, but PbSO 4 has a K sp of 6.3  10  7 (Appendix D) and will precipitate if the Pb +2 and SO 4 2  concentrations are high enough for Q to exceed K sp.

119. Aqueous Equilibria Solve When the two solutions are mixed, the volume is 0.10 L + 0.40 L = 0.50 L. The number of moles of Pb 2+ in 0.10 L of 8.0  10  3 M Pb(NO 3 ) 2 is The concentration of Pb 2+ in the 0.50-L mixture is therefore The number of moles of SO 4 2  in 0.40 L of 5.0  10  3 M Na 2 SO 4 is Therefore and Q = [Pb 2+ ][SO 4 2  ] = (1.6  10  3 )(4.0  10  3 ) = 6.4  10  6 Because Q > K sp (6.3  10  7 ), PbSO 4 precipitates.

120. Aqueous Equilibria Continued Practice Exercise Does a precipitate form when 0.050 L of 2.0  10  2 M NaF is mixed with 0.010 L of 1.0  10  2 M Ca(NO 3 ) 2 ? Answer: Sample Exercise 17.15 Predicting Whether a Precipitate Forms Yes, CaF 2 precipitates because Q = 4.6  10  8 is larger than K sp = 3.9  10  11

121. Aqueous Equilibria Sample Exercise 17.16 Calculating Ion Concentrations for Precipitation A solution contains 1.0  10  2 M Ag + and 2.0  10  2 M Pb 2+. When Cl  is added, both AgCl (K sp = 1.8  10  10 ) and PbCl 2 (K sp = 1.7  10  5 M) can precipitate. What concentration of Cl  is necessary to begin the precipitation of each salt? Which salt precipitates first? Solution Plan We are given K sp values for the two precipitates. Using these and the metal ion concentrations, we can calculate what Cl  concentration is necessary to precipitate each salt. The salt requiring the lower Cl  ion concentration precipitates first. Solve For AgCl we have K sp = [Ag + ][Cl  ] = 1.8  10  10 Because [Ag + ] = 1.0  10  2 M, the greatest concentration of Cl  that can be present without causing precipitation of AgCl can be calculated from the K sp expression: Any Cl  in excess of this very small concentration will cause AgCl to precipitate from solution.

122. Aqueous Equilibria Proceeding similarly for PbCl 2, we have Thus, a concentration of Cl  in excess of causes PbCl 2 to precipitate. Comparing the concentration required to precipitate each salt, we see that as Cl  is added, AgCl precipitates first because it requires a much smaller concentration of Cl . Thus, Ag + can be separated from by Pb 2+ slowly adding Cl  so that the chloride ion concentration remains between 1.8  10  10 M and 2.9  10  2 M. Comment Precipitation of AgCl will keep the Cl  concentration low until the number of moles of Cl  added exceeds the number of moles of Ag + in the solution. Once past this point, [Cl  ] rises sharply and PbCl 2 will soon begin to precipitate.

123. Aqueous Equilibria Continued Practice Exercise A solution consists of 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? [K sp = 1.8  10  11 for Mg(OH) 2, and K sp = 4.8  10  20 for Cu(OH) 2.] Answer: Sample Exercise 17.16 Calculating Ion Concentrations for Precipitation Cu(OH) 2 precipitates first, beginning when [OH  ] > 9.8  10  10 M. Mg(OH) 2 begins to precipitate when [OH  ] > 1.9  10  5 M.

124. Aqueous Equilibria 17.7 Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture. © 2012 Pearson Education, Inc.