1. AP Chapter 17 Additional Aspects of Equilibrium Acid Base and Solubility Equilibria HW:7, 15, 29, 39, 43. 67

2. 17.1 - 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. 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.” AKA – adding a common ion decreases the proportion of ionized substance

4. Example 17.1: What is the pH of a solution made by adding 0.30 mole of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution? Ka = 1.8 x 10 -5 (can use assumption) pH = 1.8 x 10 -5

5. 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.

6. 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)

7. 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)

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

9. 17.2 – Buffered Solutions / Henderson – Hasselbach Equation Option for Acid – Base common ion / buffer calculations Must have both of an acid / base pair – Example – Acid with its conjugate base – Example – Base with its conjugate acid

10. Henderson - Hasselbach 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 −

11. Henderson - Hasselbach 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

12. So pK a = pH − log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] This is the Henderson–Hasselbalch equation. Henderson - Hasselbach

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

14. 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.

15. 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

16. 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. Could have done this using Ka expression:

17. 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.

18. Buffers If acid is added to the HF in NaF, the F − reacts to form HF and water.

19. Buffers Calculate the pH of a 1.0 M acetic acid with 1.0 M sodium acetate solution.

20. 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.

21. 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.

22. 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. 3.Be careful – if the strong acid or base is left over, use that for the calculation

23. 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. Ka of Acetic Acid is 1.8 x 10 -5. Calculate the pH of the original solution. Calculate the pH of this solution after 0.020 mol of NaOH is added.

24. Calculating pH Changes in Buffers Before the reaction, since mol HC 2 H 3 O 2 = mol C 2 H 3 O 2 − pH = pK a pH= −log (1.8  10 −5 ) = 4.74 pH = pK a + log [base] [acid] 1 0

25. 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

26. Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + log (0.320) (0. 280) pH = 4.74 + 0.06 pH = 4.80

27. 17.3 - Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base).

28. 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.

29. 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.

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

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

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

33. Calculation of SA + SB:

35. 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.

36. 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.

37. 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.

38. Calculation – Weak Acid + Strong Base

40. 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.

41. On Titration Curves -Greatest pH change occurs at the equivalence point -Pick an indicator that changes at that pH At ½ Veq… pH = pK a + log [base] [acid]

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

43. Indicators Endpoint vs. Equivalence Point Look at color change as well.

44. Indicators pH RangeAcidBase Thymol Blue1.2-2.8redyellow Methyl yellow2.9-4.0redyellow Methyl orange3.1-4.4redorange Bromphenol blue3.0-4.6yellowblue-violet Bromcresol green4.0-5.6yellowblue Methyl red4.4-6.2redyellow Bromcresol purple5.2-6.8yellowpurple Bromphenol blue6.2-7.6yellowblue Phenol red6.4-8.0yellowred Cresol red7.2-8.8yellowred Thymol blue8.0-9.6yellowblue Phenolphthalein8.0-10.0colorlessred Thymolphthalein9.4-10.6colorlessblue Color of indicator when BELOW LOWER pH Range value Color of indicator when ABOVE HIGHER pH Range value Example: Thymol Blue: When pH less than 1.2 When pH greater than 2.8 When pH between 1.2 – 2.8 Red Yellow Orange

45. 17.4 - Solubility Equilibrium Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s) Ba 2+ (aq) + SO 4 2− (aq)

46. 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.

47. 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).

48. 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)

49. 17.5 - 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.

50. 17.5 – Factors Affecting Solubility Formation of Complex Ions Complex Ions –Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.

51. Complex Ions –The formation of these complex ions increases the solubility of these salts.

52. 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 will dissolve until Q = K sp. – If Q > K sp, the salt will precipitate until Q = K sp.

53. Example Problems

54. Predicting Precipitate Reactions

55. 17.7 – Qualitative Analysis Use differences in solubilities of salts to separate ions in a mixture.