1. CHAPTER 5 ELECTROLYTE EFFECTS AND EQUILIBRIUM: CALCULATIONS IN COMPLEX SYSTEMS Introduction to Analytical Chemistry 5-1

2. Copyright © 2011 Cengage Learning 5-2 5A Effects Of Electrolytes On Chemical Equilibria Concentration-based equilibrium constants are often indicated by adding a prime mark. Examples are K' w, K' sp, and K' a.

3. Copyright © 2011 Cengage Learning 5-3 Figure 5-1 Figure 5-1 Effect of electrolyte concentration on concentration-based equilibrium constants.

4. Copyright © 2011 Cengage Learning 5-4 5A Effects Of Electrolytes On Chemical Equilibria As the electrolyte concentration becomes very small, concentration-based equilibrium constants approach their thermodynamic values: K w, K sp, K a.

5. Copyright © 2011 Cengage Learning 5-5 5A-1 How Do Ionic Charges Affect Equilibria? The magnitude of the electrolyte effect is highly dependent on the charges of the participants in an equilibrium. When only neutral species are involved, the position of equilibrium is essentially independent of electrolyte concentration. With ionic participants, the magnitude of the electrolyte effect increases with charge.

6. Copyright © 2011 Cengage Learning 5-6 Figure 5-2 Figure 5-2 Effect of electrolyte concentration on the solubility of some salts.

7. Copyright © 2011 Cengage Learning 5-7 5A-2 What Is the Effect of Ionic Strength on Equilibria? Systematic studies have shown that the effect of added electrolyte on equilibria is independent of the chemical nature of the electrolyte but depends on a property of the solution called the ionic strength. This quantity is defined as where [A], [B], [C],... represent the molar species concentrations of ions A, B, C,... and Z A, Z B, Z C,... are their ionic charges.

8. Copyright © 2011 Cengage Learning 5-8 5A-2 What Is the Effect of Ionic Strength on Equilibria? For solutions with ionic strengths of 0.1 M or less, the electrolyte effect is independent of the kind of ions and dependent only on the ionic strength.

9. Copyright © 2011 Cengage Learning 5-9 Example 5-1 Calculate the ionic strength of (a) a 0.1 M solution of KNO₃ and (b) a 0.1 M solution of Na₂SO₄. 5-9

10. Copyright © 2011 Cengage Learning 5-10 5A-3 The Salt Effect The electrolyte effect, results from the electrostatic attractive and repulsive forces that exist between the ions of an electrolyte and the ions involved in an equilibrium.

11. Copyright © 2011 Cengage Learning 5-11 5B Activity And Activity Coefficients Chemists use the term activity, a, to account for the effects of electrolytes on chemical equilibria. The activity, or effective concentration, of species X depends on the ionic strength of the medium and is defined as where a X is the activity of X, [X] is its molar concentration, and γ X is a dimensionless quantity called the activity coefficient. (5-2)

12. Copyright © 2011 Cengage Learning 5-12 5B ACTIVITY AND ACTIVITY COEFFICIENTS The thermodynamic solubility product expression is defined by the equation Here, K'sp is the concentration solubility product constant and K sp is the thermo-dynamic equilibrium constant. The activity coefficients γ X and γ Y vary with ionic strength in such a way as to keep K sp numerically constant and independent of ionic strength (in contrast to the concentration constant K' sp ). (5-3) (5-4)

13. Copyright © 2011 Cengage Learning 5-13 5B-1 Properties of Activity Coefficients Activity coefficients have the following properties: 1. The activity coefficient of a species is a measure of the effectiveness with which that species influences an equilibrium in which it is a participant. At moderate ionic strengths, γ X < 1; as the solution approaches infinite dilution, however, and thus a X → [X] and K'sp → Ksp.

14. Copyright © 2011 Cengage Learning 5-14 5B-1 Properties of Activity Coefficients Activity coefficients have the following properties: 2. In solutions that are not too concentrated, the activity coefficient for a given species is independent of the nature of the electrolyte and dependent only on the ionic strength. 3. For a given ionic strength, the activity coefficient of an ion departs farther from unity as the charge carried by the species increases.

15. Copyright © 2011 Cengage Learning 5-15 5B-1 Properties of Activity Coefficients Activity coefficients have the following properties: 4. At any given ionic strength, the activity coefficients of ions of the same charge are approximately equal. 5. The activity coefficient of a given ion describes its effective behavior in all equilibria in which it participates.

16. Copyright © 2011 Cengage Learning 5-16 Figure 5-3 Figure 5-3 Effect of ionic strength on activity coefficients.

17. Copyright © 2011 Cengage Learning 5-17 5B-2 The Debye–Hückel Equation where γ X = activity coefficient of the species X Z X = charge on the species X μ = ionic strength of the solution α X = effective diameter of the hydrated ion X in nanometers (10 ¯⁹ m) (5-5)

18. Copyright © 2011 Cengage Learning 5-18 5B-2 The Debye–Hückel Equation When μ is less than 0.01, and Equation 5-5 becomes This equation is referred to as the Debye–Hückel limiting law (DHLL). Thus, in solutions of very low ionic strength, the DHLL can be used to calculate approximate activity coefficients.

19. Copyright © 2011 Cengage Learning 5-19 Table 5-1

20. Copyright © 2011 Cengage Learning 5-20 Example 5-4 Find the relative error introduced by neglecting activities in calculating the solubility of Ba(IO₃)₂ in a 0.033 M solution of Mg(IO₃)₂. The thermodynamic solubility product for Ba(IO₃)₂ is 1.57×10¯⁹(Appendix 1). At the outset, we write the solubility-product expression in terms of activities:

21. Copyright © 2011 Cengage Learning 5-21 Example 5-4 where and are the activities of barium and iodate ions. Replacing activities in this equation by activity coefficients and concentrations from Equation 5-2 yields

22. Copyright © 2011 Cengage Learning 5-22 Example 5-4 where and are the activity coefficients for the two ions. Rearranging this expression gives where K' sp is the concentration-based solubility product. (5-6)

23. Copyright © 2011 Cengage Learning 5-23 Example 5-4 The ionic strength of the solution is obtained by substituting into Equation 5-1:

24. Copyright © 2011 Cengage Learning 5-24 Example 5-4 In calculating μ, we have assumed that the Ba²⁺ and IO₃⁻ ions from the precipitate do not significantly affect the ionic strength of the solution. This simplification seems justified, considering the low solubility of barium iodate and the relatively high concentration of Mg(IO₃)₂. In situations where it is not possible to make such an assumption, the concentrations of the two ions can be approximated by a solubility calculation in which activities and concentrations are assumed to be identical (as in Examples 2-18 and 2-19). These concentrations can then be introduced to give a better value for μ.

25. Copyright © 2011 Cengage Learning 5-25 Example 5-4 Turning now to Table 5-1, we find that at an ionic strength of 0.1, If the calculated ionic strength did not match that of one of the columns in the table, and could be calculated from Equation 5-5.

26. Copyright © 2011 Cengage Learning 5-26 Example 5-4 Substituting into the thermodynamic solubility-product expression gives

27. Copyright © 2011 Cengage Learning 5-27 Example 5-4 Proceeding now as in earlier solubility calculations yields

28. Copyright © 2011 Cengage Learning 5-28 Example 5-4 If we neglect activities, the solubility is

29. Copyright © 2011 Cengage Learning 5-29 5B-4 Omitting Activity Coefficients in Equilibrium Calculations We shall ordinarily neglect activity coefficients and simply use molar concentrations in applications of the equilibrium law. This recourse simplifies the calculations and greatly decreases the amount of data needed.

30. Copyright © 2011 Cengage Learning 5-30 5C Equilibrium Calculations In Complex Systems: Solving Multiple-Equilibrium Problems By A Systematic Method Three types of algebraic equations are used in solving multiple-equilibrium problems: (1) equilibrium- constant expressions, (2) mass-balance equations, and (3) a single charge-balance equation. (5-9) (5-10) (5-8) (5-7)

31. Copyright © 2011 Cengage Learning 5-31 Example 5-6 Write mass-balance expressions for a 0.0100 M solution of HCl that is in equilibrium with an excess of solid BaSO₄.

32. Copyright © 2011 Cengage Learning 5-32 Example 5-6 From our general knowledge of the behavior of aqueous solutions, we can write equations for three equilibria that must be present in this solution.

33. Copyright © 2011 Cengage Learning 5-33 Example 5-6 Because the only source for the two sulfate species is the dissolved BaSO₄, the barium ion concentration must equal the total concentration of sulfate- containing species, and a mass-balance equation can be written that expresses this equality.

34. Copyright © 2011 Cengage Learning 5-34 Example 5-6 Thus,

35. Copyright © 2011 Cengage Learning 5-35 Example 5-6 The hydronium ion concentration in this solution has two sources: one from the HCl and the other from the dissociation of the solvent. A second mass-balance expression is thus Since the only source of hydroxide is the dissociation of water, [OH¯] is equal to the hydronium ion concentration from the dissociation of water.

36. Copyright © 2011 Cengage Learning 5-36 5C-2 Charge-Balance Equation We know that electrolyte solutions are electrically neutral even though they may contain many millions of charged ions. no. mol/L positive charge = no. mol/L negative charge This equation represents the charge-balance condition and is called the charge-balance equation.

37. Copyright © 2011 Cengage Learning 5-37 Example 5-8 Write a charge-balance equation for the system in Example 5-7. 5-37

38. Copyright © 2011 Cengage Learning 5-38 5C-3 Steps for Solving Problems Involving Several Equilibria Step 1.  Write a set of balanced chemical equations for all pertinent equilibria. Step 2.  State in terms of equilibrium concentrations what quantity is being sought. Step 3.  Write equilibrium-constant expressions for all equilibria developed in step 1, and find numerical values for the constants in tables of equilibrium constants. Step 4.  Write mass-balance expressions for the system.

39. Copyright © 2011 Cengage Learning 5-39 5C-3 Steps for Solving Problems Involving Several Equilibria Step 5.  If possible, write a charge-balance expression for the system. Step 6.  Count the number of unknown concentrations in the equations developed in steps 3, 4, and 5, and compare this number with the number of independent equations. If the number of equations is equal to the number of unknowns, proceed to step 7. If the number is not, seek additional equations. If enough equations cannot be developed, try to eliminate unknowns by suitable approximations regarding the concentration of one or more of the unknowns. If such approximations cannot be found, the problem cannot be solved.

40. Copyright © 2011 Cengage Learning 5-40 5C-3 Steps for Solving Problems Involving Several Equilibria Step 7.  Make suitable approximations to simplify the algebra. Step 8.  Solve the algebraic equations for the equilibrium concentrations needed to give a provisional answer as defined in step 2. Step 9.  Check the validity of the approximations made in step 7 using the provisional concentrations computed in step 8.

41. Copyright © 2011 Cengage Learning 5-41 Figure 5-4 Figure 5-4 A systematic method for solving multiple equation problems.

42. Copyright © 2011 Cengage Learning 5-42 Example 5-10 Calculate the molar solubility of Mg(OH)₂ in water. Step 1. Pertinent Equilibria  Two equilibria that need to be considered are

43. Copyright © 2011 Cengage Learning 5-43 Example 5-10 Step 2. Definition of the Unknown  Since 1 mol of Mg²⁺ is formed for each mole of Mg(OH)₂ dissolved,

44. Copyright © 2011 Cengage Learning 5-44 Example 5-10 Step 3. Equilibrium-Constant Expressions (5-12) (5-11)

45. Copyright © 2011 Cengage Learning 5-45 Example 5-10 Step 4. Mass-Balance Expression  As shown by the two equilibrium equations, there are two sources of hydroxide ions: Mg(OH)₂ and H₂O. The hydroxide ion resulting from dissociation of Mg(OH)₂ is twice the magnesium ion concentration and that from the dissociation of water is equal to the hydronium ion concentration. Thus, (5-13)

46. Copyright © 2011 Cengage Learning 5-46 Example 5-10 Step 5. Charge-Balance Expression  Note that this equation is identical to Equation 5-13. Often a mass-balance and a charge-balance equation are the same.

47. Copyright © 2011 Cengage Learning 5-47 Example 5-10 Step 6. Number of Independent Equations and Unknowns  We have developed three independent algebraic equations (Equations 5-11, 5-12, and 5-13) and have three unknowns ([Mg²⁺], [OH¯], and [H₃O⁺]).Therefore, the problem can be solved rigorously.

48. Copyright © 2011 Cengage Learning 5-48 Example 5-10 Step 7. Approximations  We can make approximations only in Equation 5-13. Since the solubility-product constant for Mg(OH)₂ is relatively large, the solution will be somewhat basic. Therefore, it is reasonable to assume that [H₃O⁺]<<[OH¯]. Equation 5-13 then simplifies to (5-14)

49. Copyright © 2011 Cengage Learning 5-49 Example 5-10 Step 8. Solution to Equations  Substitution of Equation 5-14 into Equation 5-11 gives

50. Copyright © 2011 Cengage Learning 5-50 Example 5-10 Step 9. Check of Assumptions  Substitution into Equation 5-14 yields  Thus, our assumption that 4.1 × 10¯¹¹<< 1.6 × 10¯⁴ is certainly valid.

51. Copyright © 2011 Cengage Learning 5-51 Example 5-11 Calculate the molar solubility of calcium oxalate in a solution that has been buffered so that its pH is constant and equal to 4.00. Step 1. Pertinent Equilibria (5-15)

52. Copyright © 2011 Cengage Learning 5-52 Example 5-11  Oxalate ions react with water to form HC₂O₄¯and H₂C₂O₄. Thus, there are three other equilibria present in this solution: (5-17) (5-16)

53. Copyright © 2011 Cengage Learning 5-53 Example 5-11 Step 2. Definition of the Unknown  Calcium oxalate is a strong electrolyte so that its molar analytical concentration is equal to the equilibrium calcium ion concentration. That is, (5-18)

54. Copyright © 2011 Cengage Learning 5-54 Example 5-11 Step 3. Equilibrium-Constant Expressions (5-20) (5-19) (5-21)

55. Copyright © 2011 Cengage Learning 5-55 Example 5-11 Step 4. Mass-Balance Expressions  Because CaC₂O₄ is the only source of Ca²⁺ and the three oxalate species,  Moreover, the problem states that the pH is 4.00. Thus, (5-22)

56. Copyright © 2011 Cengage Learning 5-56 Example 5-11 Step 5. Charge-Balance Expressions  A buffer is required to maintain the pH at 4.00. The buffer most likely consists of some weak acid HA and its conjugate base A¯ (Section 7C-1). The identity of the three species and their concentrations have not been specified, however, so we do not have enough information to write a charge-balance equation.

57. Copyright © 2011 Cengage Learning 5-57 Example 5-11 Step 6. Number of Independent Equations and Unknowns  We have four unknowns([Ca²⁺], [C₂O₄ 2 ¯], [HC₂O₄¯],and [H₂C₂O₄]) as well as four independent algebraic relationships (Equations 5-19, 5-20, 5-21, and 5-22). Therefore, an exact solution can be obtained, and the problem becomes one of algebra. Step 7. Approximations  An exact solution is so readily obtained in this case that we will not bother with approximations.

58. Copyright © 2011 Cengage Learning 5-58 Example 5-11 Step 8. Solution to Equations  A convenient way to solve the problem is to substitute Equations 5-20 and 5-21 into Equation 5-22 in such a way as to develop a relationship between [Ca²⁺], [C₂O₄ 2 ¯], and [H₃O⁺]. Thus, we rearrange Equation 5-21 to give

59. Copyright © 2011 Cengage Learning 5-59 Example 5-11  Substituting numerical values for [H₃O⁺] and K₂ gives  Substituting this relationship into Equation 5-20 and rearranging gives

60. Copyright © 2011 Cengage Learning 5-60 Example 5-11  Substituting numerical values for [H₃O⁺] and K₁ yields

61. Copyright © 2011 Cengage Learning 5-61 Example 5-11  Substituting these expressions for [HC₂O₄¯ ]and [H₂C₂O₄] into Equation 5-22 gives or

62. Copyright © 2011 Cengage Learning 5-62 Example 5-11  Substituting into Equation 5-19 gives

63. Copyright © 2011 Cengage Learning 5-63 5D-3 The Solubility of Precipitates in the Presence of Complexing Agents

64. Copyright © 2011 Cengage Learning 5-64 Complex Formation with a Common Ion

65. Copyright © 2011 Cengage Learning 5-65 Figure 5-5 Figure 5-5 Solubility of silver chloride in potassium chloride solutions. The dashed curve is calculated from K sp ; the solid curve is plotted from experimental data of A. Pinkus and A. M. Timmermans, Bull. Soc. Belges, 1937, 46, 46–73.

66. Copyright © 2011 Cengage Learning 5-66 5D-3 The Solubility of Precipitates in the Presence of Complexing Agents In gravimetric procedures, a small excess of precipitating agent minimizes solubility losses, but a large excess often causes increased losses due to complex formation.

67. Copyright © 2011 Cengage Learning 5-67 5E Separating Ions By pH Control: Sulfide Separations Several precipitating agents permit separation of ions based on solubility differences.

68. Copyright © 2011 Cengage Learning 5-68 5E Separating Ions By pH Control: Sulfide Separations solubility =[M²⁺] 0.1=[S²⁻]+[HS⁻]+[H₂S] assuming that ([S²⁻]+[HS⁻])<<[H₂S] [H₂S] ≈ 0.10 mol/L

69. Copyright © 2011 Cengage Learning 5-69 5E Separating Ions By pH Control: Sulfide Separations (5-29)

70. Copyright © 2011 Cengage Learning 5-70 Figure 5-6 Figure 5-6 Sulfide ion concentration as a function of pH in a saturated H 2 S solution.

71. Copyright © 2011 Cengage Learning 5-71 Example 5-12 Cadmium sulfide is less soluble than thallium(I) sulfide. Find the conditions under which Cd²⁺and Tl⁺can, in theory, be separated quantitatively with H₂S from a solution that is 0.1 M in each cation. The constants for the two solubility equilibria are

72. Copyright © 2011 Cengage Learning 5-72 Example 5-12 Since CdS precipitates at a lower [S²¯] than does Tl₂S, we first compute the sulfide ion concentration necessary for quantitative removal of Cd²⁺ from solution. To make such a calculation, we must first specify what constitutes a quantitative removal. The decision is arbitrary and depends on the purpose of the separation. 5-72

73. Copyright © 2011 Cengage Learning 5-73 Example 5-12 In this example, we shall consider a separation to be quantitative when all but 1 part in 1000 of the Cd²⁺has been removed; that is, the concentration of the cation has been lowered to 1.00×10¯⁴M. Substituting this value into the solubility-product expression gives

74. Copyright © 2011 Cengage Learning 5-74 Example 5-12 Thus, if we maintain the sulfide concentration at this level or greater, we may assume that quantitative removal of the cadmium will take place. Next, we compute the [S²⁻] needed to initiate precipitation of Tl₂S from a 0.1 M solution. Precipitation will begin when the solubility product is just exceeded. Since the solution is 0.1 M in Tl⁺,

75. Copyright © 2011 Cengage Learning 5-75 Example 5-12 These two calculations show that quantitative precipitation of Cd²⁺takes place if [S²⁻] is made greater than 1×10⁻²³. No precipitation of Tl occurs, however, until [S²⁻] becomes greater than 6×10⁻²⁰ M. Substituting these two values for [S²⁻] into Equation 5-29 permits calculation of the [H₃O⁺] range required for the separation.

76. Copyright © 2011 Cengage Learning 5-76 Example 5-12 By maintaining [H₃O⁺] between approximately 0.045 and 3.5 M, we can in theory separate CdS quantitatively from Tl₂S.

77. Copyright © 2011 Cengage Learning 5-77 THE END