1. Aqueous Equilibria Chapter 15 Applications of Aqueous Equilibria

2. Aqueous Equilibria The Common-Ion Effect Consider a solution of acetic acid: If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 − (aq)

3. 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.”

4. Aqueous Equilibria The Common-Ion Effect Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K a for HF is 6.8  10 −4. [H 3 O + ] [F − ] [HF] K a = = 6.8  10 -4

5. Aqueous Equilibria The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H 3 O + ] is not 0, but rather 0.10 M. [HF], M[H 3 O + ], M[F − ], M Initially0.200.100 Change−x−x+x+x+x+x At Equilibrium 0.20 − x  0.200.10 + x  0.10 x HF (aq) + H 2 O (l) H 3 O + (aq) + F − (aq)

6. Aqueous Equilibria The Common-Ion Effect = x 1.4  10 −3 = x (0.10) (x) (0.20) 6.8  10 −4 = (0.20) (6.8  10 −4 ) (0.10)

7. Aqueous Equilibria The Common-Ion Effect Therefore, [F − ] = x = 1.4  10 −3 [H 3 O + ] = 0.10 + x = 1.01 + 1.4  10 −3 = 0.10 M So,pH = −log (0.10) pH = 1.00

8. Aqueous Equilibria Acidic Solutions Containing Common Ions In section 14.5 we found that the equilibrium concentration of H + in a 1.0 M HF solution is 2.7  10 −2, & the % dissociation of HF is 2.7%. Calculate the [H + ] & the % dissociation of HF in a solution that is 1.0 M HF (K a for HF is 7.2  10 −4 ) & 1.0 M NaF.

9. Aqueous Equilibria Buffers: Solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when strong acid or base is added.

10. Aqueous Equilibria Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH − to make F − and water.

11. Aqueous Equilibria Buffers If acid is added, the F − reacts to form HF and water.

12. Aqueous Equilibria Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: [H 3 O + ] [A − ] [HA] K a = HA + H 2 OH 3 O + + A −

13. Aqueous Equilibria Buffer Calculations Rearranging slightly, this becomes [A − ] [HA] K a = [H 3 O + ] Taking the negative log of both side, we get [A − ] [HA] −log K a = −log [H 3 O + ] + − log pKapKa pH acid base

14. Aqueous Equilibria Buffer Calculations So pK a = pH − log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] This is the Henderson–Hasselbalch equation.

15. Aqueous Equilibria Henderson–Hasselbalch Equation 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? K a for lactic acid is 1.4  10 −4.

16. Aqueous Equilibria Henderson–Hasselbalch Equation pH = pK a + log [base] [acid] pH = −log (1.4  10 −4 ) + log (0.10) (0.12) pH = 3.85 + (−0.08) pH = 3.77

17. Aqueous Equilibria The pH of a Buffered Solution I A buffered solution contains 0.50 M acetic acid (HC 3 H 5 O 3 ) (K a for HC 3 H 5 O 3 is 1.8  10 −5 ) and 0.50 M sodium acetate. Calculate the pH of the solution.

18. Aqueous Equilibria pH Changes in Buffered Solutions Calculate the pH change that occurs when 0.010 mol of solid NaOH is added to 1L of the buffer in the previous probIem. Compare this change to the change in pH when 0.010 M NaOH is added to 1.0 L of water.

19. Aqueous Equilibria pH Changes in Buffered Solutions Calculate the pH change that occurs when 0.010 mol of solid NaOH is added to 1L of the buffer in the previous probIem. Compare this change to the change in pH when 0.010 M NaOH is added to 1.0 L of water.

20. Aqueous Equilibria The pH of a Buffered Solution II Calculate the pH of a solution containing 0.75 M lactic acid (K a = 1.4  10 −4 ) & 0.25 M sodium lactate. Lactic acid (HC 3 H 5 O 3 ) is a common constituent of biological systems. For example, it is found in milk & present in human muscle tissue during exertion.

21. Aqueous Equilibria The pH of a Buffered Solution III A buffered solution contains 0.25 M NH 3 (K b = 1.8  10 −5 ) & 0.40 M NH 4 Cl. Calculate the pH of this solution.

22. 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.

23. Aqueous Equilibria 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.

24. 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 the Henderson–Hasselbalch equation to determine the new pH of the solution.

25. Aqueous Equilibria Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added. Before the reaction, since mol HC 2 H 3 O 2 = mol C 2 H 3 O 2 − pH = pK a = −log (1.8  10 −5 ) = 4.74

26. Aqueous Equilibria Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC 2 H 3 O 2 (aq) + OH − (aq)  C 2 H 3 O 2 − (aq) + H 2 O (l) HC 2 H 3 O 2 C2H3O2−C2H3O2− OH − Before reaction0.300 mol 0.020 mol After reaction0.280 mol0.320 mol0.000 mol

27. Aqueous Equilibria Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + log (0.320) (0. 200) pH = 4.74 + 0.06 pH = 4.80

28. Aqueous Equilibria Adding a Strong Acid to a Buffered Solution Calculate the pH of this solution that results when 0.10 mol of gaseous HCl is added to 1.0 L of the buffered solution from the previous example.

29. Aqueous Equilibria Adding a Strong Acid to a Buffered Solution II Calculate the pH of this solution that results when 0.10 mol of gaseous HCl is added to 1.0 L of the following solutions (K a for acetic acid is 1.8  10 −5 ) : Solution A: 5.00 M HC 2 H 3 O 2 & 5.00 M NaC 2 H 3 O 2 Solution B: 0.050 M HC 2 H 3 O 2 & 0.050 M NaC 2 H 3 O 2

30. Aqueous Equilibria Preparing a Buffer A chemist needs a solution buffered at a pH of 4.30 & can choose from the following acids (& their sodium salts): a.Chloroacetic acid (K a = 1.35  10 −3 ) b.propanoic acid (K a = 1.3  10 −5 ) c.Benzoic acid (K a = 6.4  10 −5 ) d.Hypochlorous acid (K a = 3.5  10 −8 ) Calculate the ratio [HA]/[A - ] required for each system to yield a pH of 4.30. Which system works the best?

31. Aqueous Equilibria Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base).

32. 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.

33. 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.

34. Aqueous Equilibria Titration of a Strong Acid with a Strong Base Just before and after the equivalence point, the pH increases rapidly.

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

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

37. Aqueous Equilibria The titration of a strong base with a strong acid Calculate the pH at the following points of a titration of 50.0 ml of 0.200 M NaOH with 0.100 M HNO 3 a.No HNO 3 added b.20.0 ml HNO 3 added c.50.0 ml HNO 3 added

38. Aqueous Equilibria The titration of a strong base with a strong acid Calculate the pH at the following points of a titration of 50.0 ml of 0.200 M NaOH with 0.100 M HNO 3 c.100.0 ml HNO 3 added d.150.0 ml HNO 3 added

39. Aqueous Equilibria Titration of a Weak Acid with a Strong Base Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. The pH at the equivalence point will be >7. Phenolphthalein is commonly used as an indicator in these titrations.

40. 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.

41. 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.

42. Aqueous Equilibria Titration of a Weak Base with a Strong Acid The pH at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice.

43. Aqueous Equilibria The titration of a weak acid with a strong base Hydrogen cyanide gas (HCN), a powerful respiratory inihbitor, is highly toxic. It is a very weak acid (Ka = 6.2 x 10 -10 ) when dissolved in water. If a 50.0 ml sample of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH of the solution. a.After 8.00 ml NaOH has been added

44. Aqueous Equilibria The titration of a weak acid with a strong base b. At the halfway point of the titration c. At the equivalence point of the titration.

45. Aqueous Equilibria The titration of a weak base with a strong acid If a 100.0 ml sample of 0.050 M NH 3 is titrated with 0.100 M HCl, calculate the pH of the solution. a.Before the addition of any HCl b.After 10.0 ml NaOH has been added

46. Aqueous Equilibria The titration of a weak base with a strong acid b. At the halfway point of the titration c. At the equivalence point of the titration.

47. Aqueous Equilibria Calculating K a A chemist has synthesized a monoprotic weak acid & wants to determine its K a value. To do so, the chemist dissolves 2.00 mmol of the solid acid in 100.0 ml water & titrates the resulting solution with 0.0500 M NaOH. After 20.0 ml NaOH has been added, the pH is 6.00. What is the K a value for the acid.

48. Aqueous Equilibria Titrations of Polyprotic Acids In these cases there is an equivalence point for each dissociation.

49. Aqueous Equilibria Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s) Ba 2+ (aq) + SO 4 2− (aq)

50. Aqueous Equilibria Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2− ] where the equilibrium constant, K sp, is called the solubility product. Solubility is often expressed as grams of solute per liter of of solution (g/L). Molar Solubility is the number of moles per liter that will dissolve (mol/L).

51. 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).

52. Aqueous Equilibria Calculating K sp from Solubility Solid silver chromate is added to pure water at 25 0 C, & some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissloved Ag 2 CrO 4(s) & the solution. Analysis of the equilibrated solution shows that the silver ion concentration is 1.3x10 -4 M. Assuming that Ag 2 CrO 4 dissociates completely in water & that there are no other important equilibria involving Ag + or CrO 4 - ions in the solution, calculate K sp for this compound.

53. Aqueous Equilibria Calculating Solubility from K sp The K sp for CaF 2 is 3.9x10 -11 at 25 0 C. Assuming that CaF 2 dissociates completely upon dissolving & that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter.

54. Aqueous Equilibria 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)

55. Aqueous Equilibria Calculating the Effect of a Common Ion on Solubility The K sp for CaF 2 is 3.9x10 -11 at 25 0 C. Calculate the molar solubility of CaF 2 in a solution that is a.0.010 M in Ca(NO 3 ) 2 b.0.010 M in NaF

56. 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.

57. Aqueous Equilibria Predicting the Effect of Acid on Solubility Which of the following substances will be more soluble in acidic solution than in a basic solution a.Ni(OH) 2(s) b.CaCO 3(s) c.BaF 2(s) d.AgCl (s)

58. Aqueous Equilibria Factors Affecting Solubility Complex Ions  Metal ions can act as Lewis acids and form complex ions with Lewis bases (Ligands) in the solvent.

59. Aqueous Equilibria Factors Affecting Solubility Complex Ions  The formation of these complex ions increases the solubility of these salts.  Coordination number The number of ligands attached to the metal ion Most common coordination numbers are 6, 4, & 2 In general – double the charge of the metal ion to get the ligand number

60. 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+.

61. Aqueous Equilibria 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 will dissolve until Q = K sp.  If Q > K sp, the salt will precipitate until Q = K sp.

62. Aqueous Equilibria Predicting Whether a Precipitate Will Form Will a precipitate form when 0.10 L of 8.0x10 -3 M Pb(NO 3 ) 2 is added to 0.40 L of 5.0x10 -3 M Na 2 SO 4 ?

63. Aqueous Equilibria Predicting Whether a Precipitate Will Form A solution contains 1.0x10 -2 M Ag + & 2.0x10 -2 M Pb 2+. When Cl - is added to the solution, both AgCl (K sp = 1.8x10 -10 ) & PbCl 2 (K sp = 1.7x10 -5 ) precipitate from the solution. What concentration of Cl - is necessary to begin precipitation of each salt? Which salt precipitates first?

64. Aqueous Equilibria Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture.