1. Chapter 17 Acid-Base & Solubility Equilibria 1

2. 17.1The Common Ion Effect 17.2Buffer Solutions 17.3Acid-Base Titrations (omitted) 17.4Solubility Equilibria 17.5Factors Affecting Solubility 17.6Separation of Ions Using Differences in Solubility (omitted)

3. The Effect of Common Ion  A phenomenon known as the common ion effect states that: When a compound containing an ion in common with an already dissolved substance (a weak electrolyte) is added to an aqueous solution at equilibrium, the equilibrium shifts to the left. The ionization of the weak electrolyte is being suppressed by adding the common ion to the solution. 3 17.1

4. The Common Ion Effect  A 1.0 L of 0.10 M solution of CH 3 COOH. Adding 0.050 mol of CH 3 COONa: The result is that fewer H + ions present with a higher pH. 4 17.1 CH 3 COO ‒ is a common ion Le Châtelier's principle

5. Equilibrium Calculation Involving Common Ion Effect  Before adding the CH 3 COO ‒ ions: 5 (M)(M)CH 3 COOHH+H+ CH 3 COO ‒ Initial conc.0.1000 Change in conc.- x+ x Equilibrium conc.0.10 - xxx x = 1.34  10 -3 M K a < 10 -3 [H + ] = 1.34  10 -3 M pH = -log (1.34  10 -3 ) = 2.87 17.1 1.0 L of 0.10 M

6. Equilibrium Calculation Involving Common Ion Effect  After adding the CH 3 COO ‒ (0.05 M) ions: 6 (M)(M)CH 3 COOHH+H+ CH 3 COO ‒ Initial conc.0.1000.050 Change in conc.- x+ x Equilibrium conc.0.10 - xx0.050 + x x = 3.6  10 -5 M [H + ] = x = 3.6  10 -5 M pH = -log (3.6  10 -5 ) = 4.44 17.1 Method II CH 3 COONa

7. The Common Ion Effect  Exercise: Which of the following when dissolved in aqueous NH 3 solution is (are) going to decrease the dissociation of NH 3 ? NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) (a) Ca(OH) 2 (b) HNO 2 (c) CH 3 COONa (d) NH 4 NO 3 (e) NH 3 7 17.1

8. Buffer Solutions  A buffer solution is the one that resists the change in its pH when small amounts of either H + or OH ‒ ions are added. Buffers are useful applications of the common ion effect.  Buffer solutions are important for: o Biological systems. (some enzymes can only function at a specific pH, pH of blood is always about 7.4, gastric “stomach” juices maintain a pH of about 1.5) o Chemical applications. (fermentation processes, dyes used in coloring fabrics, calibration for pH meters). 8 17.2

9. Buffer Solutions  A buffer solution can be: o a solution containing a weak acid and its conjugate base, or CH 3 COOH (aq)CH 3 COO ‒ (aq) + H + (aq) It is known as an acidic buffer solution and it maintains a pH value that is less than 7. o a solution containing a weak base and its conjugate acid. NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH ‒ (aq) It is known as a basic buffer solution and it maintains a pH value that is greater than 7. 9 17.2 weak acidconjugate base weak baseconjugate acid

10. Henderson-Hasselbalch Equation  For any buffer solution when a valid approximation is applied, its equilibrium expression is: K a = [H + ][A ‒ ] [HA] HA is the weak acid A ‒ is the conjugate base [H + ] = K a [HA] [A ‒ ] ‒ log [H + ] = ‒ log K a + log [HA] [A ‒ ] pH = pK a + log [weak acid] [conjugate base] From H-H equation, when the concentrations of the weak acid and its conjugate base in a buffer are equal, its pH = its pK a. The slight change in the pH of the buffer is due to change in the concentrations of the weak acid and its conjugate base when small amounts of either H + or OH ‒ ions are added to the buffer. How to calculate pH of buffer?

11. A solution is only a buffer if it has the capacity to resist pH change when either an acid or a base is added. If the concentrations of a weak acid and conjugate base differ by more than a factor of 10, the solution does not have this capacity. Requirement for a solution to be buffer?

12. 17.2 Therefore, we consider a solution a buffer, and can use to calculate its pH, only if the following condition is met: Consequently, the log term can only have values from – 1 to 1, and the pH of a buffer cannot be more than one pH unit different from the pK a of the weak acid it contains. This is known as the range of the buffer, where pH = pK a  1.

13. Determine the pH at 25°C of a solution prepared by adding 0.050 mole of sodium acetate to 1.0 L of 0.10 M acetic acid. K a = 1.8 x 10 -5 for acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution.) The solution will work as buffer or not? Exercise:

14. pH = pK a + log [weak acid] [conjugate base] pK a = ‒ log K a = ‒ log (1.8 x 10 -5 ) = 4.74 Log [HA] [A ‒ ] = [0.10] [.050] Log = Log 0.5 = -0.3 pH = 4.74 – 0.3 = 4.44

15. What happens when acid or base is added into a buffer solution? (1.00 M of acetic acid and 1.00 M of sodium acetate) 1.00 mol 1.10 mol

16. Consider what happens to the pH when we add 0.10 mole of HCl to (a)Pure water (b)The buffer solution prepared by adding 1.00 mole of sodium acetate to 1.0 L of 1.00 M acetic acid. (K a = 1.8 x 10 -5 for acetic acid or pK a = 4.74). (Assume that the addition of HCl causes no change in the volume of the solution.)

17. pH = pK a + log [weak acid] [conjugate base] pK a = ‒ log K a = ‒ log (1.8 x 10 -5 ) = 4.74 log [WA] [CB] = [1.10] [0.90] log = log 0.82 = - 0.08 pH = 4.74 – 0.08 = 4.66 A change of only 0.08 pH units. How about the pH of pure water? pH = 1 Since we are adding 0.1 M strong acid (HCl); [acid] = [1 M + 0.1 M] = [1.1 M], [conjugate base] = [1 M -.01 M] = [0.9 M] OR

18. Exercise: Starting with 1.00 L of a buffer that is 1.00 M in acetic acid and 1.00 M in sodium acetate, calculate the pH after the addition of 0.100 mole of NaOH. (Assume that the addition does not change the volume of the solution.) (K a = 1.8 x 10 -5 for acetic acid or pK a = 4.74)

19. pK a = ‒ log K a = ‒ log (1.8 x 10 -5 ) = 4.74 log [WA] [CB] = [0.90] [1.10] log = log 1.22 = + 0.08 pH = 4.74 + 0.08 = 4.82 A change of only 0.08 pH units. How about the pH of pure water? 1.00 mol 1.10 mol Since we are adding 0.1 M strong base (NaOH); [acid] = [1 M - 0.1 M] = [0.9 M], [conjugate base] = [1 M +.01 M] = [1.1 M] OR pH = 13

20. Exercise: Calculate the pH of 2.0 L of a buffer that is 1.0 mole in both acetic acid and sodium acetate after adding 0.15 mol of Ca(OH) 2. (pK a = 4.74 for acetic acid)

21. pH = pK a + log [CH 3 COOH] [CH 3 COO ‒ ] = 4.74 + log (0.2) (0.8) = 4.74 + 0.6 = 5.34 Concentration = L Mole 2 L 1.0 = = 0.5 M

22. 17.2 22 1.Choose a weak acid whose pK a is close to the desired pH. How to Prepare a Buffer Solution with a Specific pH

23. 17.2 23 2.Substitute the pH and pK a values into the following equation to obtain the necessary ratio of [conjugate base]/[weak acid].

24. 17.3 24 Exercise: How you would prepare a buffer with a pH of 9.50.?

25. 17.3 25 One way to achieve this would be to dissolve 0.41 mol of C 6 H 5 ONa and 1.00 mol of C 6 H 5 OH in 1 L of water.

26. Exercise: Which of the following compounds, when added to an aqueous solution of HF, would cause an increase in the pH? Decrease in the pH? Which would not have effect on pH? 1- NaF 2- SnF 2 3- HCl 4- NaCl 5- NaOH

27. Solubility Product Expression and K sp Calculations Involving K sp and Solubility Predicting Precipitation Reactions

28. 17.4 Solubility Product Expression and K sp 28 The solubility of ionic compounds is important in industry, medicine, and everyday life. Consider – soluble, less soluble, insoluble Equilibrium constant is represented by K sp - called the solubility product constant. Solubility, solute, saturated solution, precipitate??

29. All above compounds are very slightly soluble in water. None of them are soluble in water. The smaller the K sp value, the less soluble the compound. This is valid for compounds of similar formulas, such as comparing AgCl with CuBr, and CaF 2 with Fe(OH) 2.

30. 17.4 Solubility Product Expression and K sp 30 K sp is equal to the concentrations of products over the concentrations of reactants, each raised to its coefficient from the balanced chemical equation. The smaller the K sp, the less soluble the compound.

31. 17.4 Calculations Involving K sp and Solubility 31 There are two ways to express the solubility of a substance: 1.The molar solubility: number of moles of solute in 1 L of a saturated solution (mol/L). 2.The solubility: number of grams of solute in 1 L of a saturated solution (g/L). Both of these refer to concentrations at a particular temperature (usually 25°C). How to convert solubility into molar solubility or vice-versa?

32. 17.4 Calculations Involving K sp and Solubility 32 To calculate molar solubility from K sp : The procedure is essentially identical to the procedure for solving weak acid or weak base equilibrium problems 1.Construct an equilibrium or ICE table. 2.Fill in what we know. 3.Figure out what we don’t know.

33. Exercise: Calculate the solubility of AgBr? 33 The K sp of silver bromide (AgBr) is 7.7 × 10 ‒ 13. Let s be the molar solubility (in mol/L) of AgBr. At equilibrium: [Ag + ] = [Br ‒ ] = s

34. 17.4 Calculations Involving K sp and Solubility 34 The equilibrium expression is Therefore, Is this solubility or molar solubility?

35. Calculations Involving K sp and Solubility 35 The molar solubility of AgBr is 8.8 × 10 ‒ 7 M. Express this solubility in g/L by multiplying the molar solubility by the molar mass of AgBr:

36. 17.7 36 Calculate the solubility of copper(II) hydroxide [Cu(OH) 2 ] in g/L. (K sp for Cu(OH) 2 = 2.2 × 10 ‒20 ) and the molar mass of Cu(OH) 2 is 97.57 g/mol. The equation for the dissociation of Cu(OH) 2 is K sp = [Cu 2+ ][OH ‒ ] 2 Exercise:

37. 17.7 37 Therefore, Remember to raise an entire term to the appropriate power! K sp = [Cu 2+ ][OH ‒ ] 2

38. 17.7 38 The molar solubility of Cu(OH) 2 is 1.8 × 10 ‒ 7 M. Multiplying by its molar mass gives:

39. 17.8 39 The solubility of calcium sulfate (CaSO 4 ) is measured experimentally and found to be 0.67 g/L. Calculate the value of K sp for calcium sulfate. The molar mass of CaSO 4 is 136.2 g/mol. The molar solubility of CaSO 4 is Exercise:

40. 17.8 40 Think about it – what about K sp of Mg(OH) 2 ?

41. Predicting Precipitation Reactions  To predict whether a precipitation will form or not, we calculate the reaction quotient (Q) for the possible precipitation for the initial state of mixing two solutions. o If Q < K sp, no precipitation is going to form. o If Q > K sp, precipitation is going to form. o If Q = K sp,  At the first stage, you should be able to determine which compound is soluble in water and which is very slightly soluble in water. You calculate Q for the latter compound and then compare it with the listed K sp values. 41 Q = [Ag + ] [Cl ‒ ] the solution is saturated.

42. Exercise: Predict whether a precipitate will form when 0.001 M Ag 2 SO 4 is prepared? Ag 2 SO 4, K sp = 1.5 × 10 ‒ 5

43. Exercise: K sp values of Ag 2 SO 4, K sp = 1.5 × 10 ‒ 5 Predict whether a precipitate will form when 0.002 M AgNO 3 is added to 0.100 M K 2 SO 4 ? (Assume that the addition does not change the volume of the solution.)

44. 17.9 44 Predict whether a precipitate will form when 250 mL of 0.0040 M BaCl 2 is added to 650 mL of 0.0080 M K 2 SO 4 : The K sp values of BaSO 4, K sp = 1.1 × 10 ‒ 10 Exercise:

45. 17.9 45 Solution Concentrations of the constituent ions of BaSO 4 are: Q = [Ba 2+ ][SO 4 2 ‒ ] = (0.0011)(0.0058) = 6.4 × 10 ‒ 6 Q > K sp (1.1 × 10 ‒ 10 ) BaSO 4 will precipitate.

46. Factors Affecting Solubility  There are some factors that affect solubility. We will be discussing: o The common ion effect. o The pH. o Complex ion formation 46 17.5

47. Any factor that will shift the equilibrium left (reverse) will decrease the solubility. Any factor that will shift the equilibrium right (forward) will increase the solubility.

48. Factors Affecting Solubility  The common ion effect. Consider dissolving AgCl salt in pure water to get a saturated aqueous solution of AgCl. The solubility of AgCl in water at 25  C is 1.3  10 -5 M. 48 17.5 The solubility of AgCl in water can be calculated from its solubility product constant (K sp = 1.6  10 -10 ) 1.0 L Pure water AgCl salt So how about if we dissolve AgCl in solution other than water? AgCl (s) Ag + Cl ‒

49. Factors Affecting Solubility  The common ion effect. The common ion effect is an example of Le Châtelier’s principle. The presence of a second salt (normally very soluble in water) that produces an ion common to a solubility equilibrium will reduce solubility. AgCl (s) Ag + (aq) + Cl ‒ (aq) 49 17.5 AgNO 3 solutionNaCl solution What about NaNO 3 ?

50. Factors Affecting Solubility  Example: Calculate the molar solubility of BaSO 4 in 0.0010 M Na 2 SO 4. BaSO 4 (s) Ba 2+ (aq) + SO 4 2‒ (aq) K sp of BaSO 4 = 1.1  10 -10 50 17.5 (M)(M)BaSO 4 (s)Ba 2+ (aq)SO 4 2‒ (aq) Initial conc.0 1  10 -3 Change in conc.+ s Equilibrium conc.s 1  10 -3 + s K sp = 1.1  10 -10 = [Ba 2+ ][SO 4 2 ‒ ] = (s)(1  10 -3 + s) 1.1  10 -10 ≈ (s)(1  10 -3 ) s = 1  10 -7 M For comparison, the solubility in pure water is: s = (1.1  10 -10 ) 1/2 = 1.0  10 -5 M (less soluble) (soluble)

51. In general, the common ion effect decreases the solubility of a substance.

52. Factors Affecting Solubility  The pH HF, CH 3 COOH etc. Mg(OH) 2, HCl, HBr, HNO 3, H 2 SO 4 NaOH 52 17.5

53. Exercise: Which of the following compounds will be more soluble in acidic solution than in water: (a) CuS, (b) AgCl, (c) PbSO 4 ? Think about it: How solubility will change under basic solution? Exercise: Which of the following compounds will be more soluble in acidic solution than in water: (a) Ca(OH) 2, (b) PbBr 2 ? Think about it: How solubility will change under basic solution?

54. Factors Affecting Solubility  Complex Ion Formation. A Complex ion is an ion that involves a central metal cation (mostly are transition metal ions) bonded to one or more ions or molecules ( ligands). 54 17.5 Tetra-ammine- copper(II) cation, Cu(NH 3 ) 4 2+, is one example of complex ions. Complex ions exhibit colors when transition metal ions are contained at the central position. Co(H 2 O) 6 2+ CoCl 4 2 

55. Factors Affecting Solubility  Complex Ion Formation. 55 17.5 K f = Consider adding aqueous ammonia to a saturated AgCl solution. AgCl in water AgCl dissolves, and Ag + ions form Ag(NH 3 ) 2 + complex NH 3 (aq)

56. Factors Affecting Solubility  Complex Ion Formation. We measure the tendency of a metal ion to form a complex ion using the formation constant, K f, (or stability constant). 56 17.5

57. Factors Affecting Solubility  Complex Ion Formation. Let’s write equilibrium equations for the previous experiment. AgCl (s) Ag + (aq) + Cl ‒ (aq) Ag + (aq) + 2NH 3 (aq) Ag(NH 3 ) 2 + (aq) AgCl (s) + 2NH 3 (aq) Ag(NH 3 ) 2 + (aq) + Cl ‒ (aq) K” = K sp  K f = 1.6  10 -10  1.5  10 7 = 2.4  10 -3 K” >> K sp In general, the formation of complex ions increases the solubility of a substance. We can also apply Le Châtelier’s principle to understand! 57 17.5 K sp = 1.6  10 -10 K f = 1.5  10 7 K ” = ?

58. Exercise: Determine whether the solubility of Cu(OH) 2 will increase or decrease when mixed with NaCN? K sp of Cu(OH) 2 = 1.6  10 -19 K f Cu(CN) 4 2- = 1  10 25

59. 17.5Factors Affecting Solubility Complex Ion Formation 59 Solving an equilibrium problem involving complex ion formation is complicated by the magnitude of K f and stoichiometry. Consider the following reaction Determine the molar conc. of free Cu 2+ ion in solution when 0.10 mole of Cu(NO 3 ) 2 is dissolved in a L of 3.0 M NH 3 to form Cu(NH 3 ) 4 2+. Kf for Cu(NH 3 ) 4 2+ = 5.0  10 14 We cannot neglect x, and have to raise [NH 3 ]to the fourth power. This equation is not easily solved. Another approach is needed.

60. 17.5Factors Affecting Solubility Complex Ion Formation 60 1.Assume that all the copper(II) ion is consumed to form the complex ion (since K f is so large.) 2.Consider the equilibrium in terms of the reverse reaction; the dissociation of Cu(NH 3 ) 4 2+, for which the equilibrium constant is the reciprocal of K f.

61. 17.5Factors Affecting Solubility Complex Ion Formation 61 Because this K is so small, we can expect x to be insignificant. Note that [NH 3 ], which had been 3.0 M, has been diminished by 4 × 0.10 M due to the amount required to complex 0.10 mole of copper(II) ion.

62. 17.5Factors Affecting Solubility Complex Ion Formation 62 Neglecting x, x = 4.4 × 10 ‒ 18 M Because K f is so large, the amount of copper that remains uncomplexed is extremely small.

63. 17.12 63 In the presence of aqueous cyanide, cadmium(II) forms the complex ion Cd(CN) 4 2 ‒. Determine the molar concentration of free (uncomplexed) cadmium(II) ion in solution when 0.20 mole of Cd(NO 3 ) 2 is dissolved in a liter of 2.0 M sodium cyanide (NaCN). Setup K f for the complex ion Cd(CN) 4 2 ‒ is 7.1 × 10 16. The reverse process has an equilibrium constant of 1/K f = 1.4 × 10 ‒ 17

64. 17.12 64 The equilibrium expression for the dissociation is Stoichiometry indicates that four CN ‒ ions are required to react with one Cd 2+ ion. Therefore, [CN ‒ ] will be [2.0 M ‒ 4(0.20 M)] = 1.2 M. Solution

65. 17.12 65 Neglecting x, x = 1.4 × 10 ‒ 18 M Don’t forget to adjust the concentration of the complexing agent before entering it in the equilibrium table.