1. Acid-Base Equilibria

2. Some Definitions Arrhenius – An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions. – A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions.

3. Some Definitions Brønsted-Lowry – An acid is a proton donor. – A base is a proton acceptor. A Brønsted-Lowry acid… …must have a removable (acidic) proton. A Brønsted-Lowry base… …must have a pair of nonbonding electrons.

4. If it can be either… …it is amphiprotic. HCO 3 - HSO 4 - H 2 O

5. What Happens When an Acid Dissolves in Water? Water acts as a Brønsted-Lowry base and abstracts a proton (H + ) from the acid. As a result, the conjugate base of the acid and a hydronium ion are formed.

6. Conjugate Acids and Bases The term conjugate comes from the Latin word “conjugare,” meaning “to join together.” Reactions between acids and bases always yield their conjugate bases and acids.

7. Acid and Base Strength Strong acids are completely dissociated in water. – Their conjugate bases are quite weak. Weak acids only dissociate partially in water. – Their conjugate bases are weak bases.

8. Acid and Base Strength Substances with negligible acidity do not dissociate in water. – Their conjugate bases are exceedingly strong. Example: CH 4 contains hydrogen but does not demonstrate any acidic behavior in water. Its conjugate base (CH 3 - ) is a strong base.

9. Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) H 2 O is a much stronger base than Cl -, so the equilibrium lies so far to the right that K is not measured (K>>1).

10. Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. CH 3 CO 2 H (aq) + H 2 O (l)H 3 O + (aq) + CH 3 CO 2 - (aq) Acetate is a stronger base than H 2 O, so the equilibrium favors the left side (K<1).

11. Autoionization of Water As we have seen, water is amphoteric. In pure water, a few molecules act as bases and a few act as acids. This is referred to as autoionization. H 2 O (l) + H 2 O (l)H 3 O + (aq) + OH - (aq)

12. Ion-Product Constant The equilibrium expression for this process is K c = [H 3 O + ] [OH - ] This special equilibrium constant is referred to as the ion-product constant for water, K w. At 25  C, K w = 1.0  10 -14

13. pH pH is defined as the negative logarithm in base 10 of the concentration of hydronium ion. pH = -log [H 3 O + ] or pH = -log [H + ]

14. pH In pure water, K w = [H 3 O + ] [OH - ] = 1.0  10 -14 Since in pure water [H 3 O + ] = [OH - ], [H 3 O + ] = 1.0  10 -14 = 1.0  10 -7

15. pH Therefore, in pure water, pH = -log (1.0  10 -7 ) = 7.00 An acid has a higher [H 3 O + ] than pure water, so its pH is <7. A base has a lower [H 3 O + ] than pure water, so its pH is >7.

16. pH These are the pH values for several common substances.

17. Other “p” Scales The “p” in pH tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions). Some similar examples are – pOH: -log [OH - ] – pK w : -log K w

18. Watch This! Because [H 3 O + ] [OH - ] = K w = 1.0  10 -14, we know that -log [H 3 O + ] + -log [OH - ] = -log K w = 14.00 or, in other words, pH + pOH = pK w = 14.00

19. How Do We Measure pH? – Litmus paper “Red” paper turns blue above ~pH = 8 “Blue” paper turns red below ~pH = 5 – Or an indicator. For less accurate measurements, one can use

20. How Do We Measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

21. Strong Acids The seven strong acids are HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 3, and HClO 4. These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. For the monoprotic strong acids, [H 3 O + ] = [acid].

22. Strong Bases Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca 2+, Sr 2+, and Ba 2+ ). Again, these substances dissociate completely in aqueous solution.

23. Weak Acids For a generalized acid dissociation, the equilibrium expression would be This equilibrium constant is called the acid- dissociation constant, K a. HA (aq) + H 2 O (l) A - (aq) + H 3 O + (aq) [H 3 O + ] [A - ] [HA] K c = Dissociation Constants

24. The greater the value of K a, the stronger is the acid.

25. Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25  C is 2.38. Calculate K a for formic acid at this temperature. We know that HCOOH + H 2 O H 3 O + + HCOO - [H 3 O + ] [HCOO - ] [HCOOH] K a =

26. Calculating K a from the pH To calculate K a, we need the equilibrium concentrations of all three things. We can find [H 3 O + ], which is the same as [HCOO - ], from the pH.

27. pH = -log [H 3 O + ] 2.38 = -log [H 3 O + ] -2.38 = log [H 3 O + ] 4.2  10 -3 = [H 3 O + ] = [HCOO - ] [4.2  10 -3 ] [0.10] K a = = 1.8  10 -4

28. Calculating Percent Ionization Percent Ionization =  100 In this example [H 3 O + ] eq = 4.2  10 -3 M [HCOOH] initial = 0.10 M [H 3 O + ] eq [HA] initial Percent Ionization =  100 4.2  10 -3 0.10 = 4.2%

29. Calculating pH from K a Calculate the pH of a 0.30 M solution of acetic acid, HC 2 H 3 O 2, at 25  C. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) K a for acetic acid at 25  C is 1.8  10 -5.

30. The equilibrium constant expression is [H 3 O + ] [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] K a = (x) 2 (0.30) 1.8  10 -5 = 2.3  10 -3 = x pH = -log [H 3 O + ] pH = -log (2.3  10 -3 ) pH = 2.64

31. Polyprotic Acids… …have more than one acidic proton If the difference between the K a for the first dissociation and subsequent K a values is 10 3 or more, the pH generally depends only on the first dissociation.

32. Weak Bases Bases react with water to produce hydroxide ion.

33. Weak Bases The equilibrium constant expression for this reaction is [HB] [OH - ] [B - ] K b = where K b is the base-dissociation constant.

34. Weak Bases K b can be used to find [OH - ] and, through it, pH.

35. pH of Basic Solutions What is the pH of a 0.15 M solution of NH 3 ? NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) [NH 4 + ] [OH - ] [NH 3 ] K b = = 1.8  10 -5

36. [NH 4 + ] [OH - ] [NH 3 ] K b = = 1.8  10 -5 (x) 2 (0.15) 1.8  10 -5 = 1.6  10 -3 = x 2 Therefore, [OH - ] = 1.6  10 -3 M pOH = -log (1.6  10 -3 ) pOH = 2.80 pH = 14.00 - 2.80 pH = 11.20

37. K a and K b K a and K b are related in this way: K a  K b = K w Therefore, if you know one of them, you can calculate the other.

38. Reactions of Anions with Water Anions are bases. As such, they can react with water in a hydrolysis reaction to form OH - and the conjugate acid: X - (aq) + H 2 O (l)HX (aq) + OH - (aq)

39. Reactions of Cations with Water Cations with acidic protons (like NH 4 + ) will lower the pH of a solution. Most metal cations that are hydrated in solution also lower the pH of the solution.

40. Reactions of Cations with Water Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water. This makes the O-H bond more polar and the water more acidic. Greater charge and smaller size make a cation more acidic.

41. 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. CH 3 COOH(aq) + H 2 O(l)H 3 O + (aq) + CH 3 COO − (aq)

42. 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

43. The Common-Ion Effect HF(aq) + H 2 O(l)H 3 O + (aq) + F − (aq) 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

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

45. 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.

46. 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.

47. Buffers Similarly, if acid is added, the F − reacts with it to form HF and water.

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

49. 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

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

51. What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate? K a for lactic acid is 1.4  10 −4. 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

52. 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.

53. 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.

54. 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.

55. 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 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)

56. HC 2 H 3 O 2 C2H3O2−C2H3O2− OH − Before reaction0.300 mol 0.020 mol After reaction0.280 mol0.320 mol0.000 mol Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + log (0.320) (0.280) pH pH = 4.80 pH = 4.74 + 0.06

57. 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) 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.

58. 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).