1. ACID-BASE EQUILIBRIUM ERT 207 ANALYTICAL CHEMISTRY SEMESTER 1, ACADEMIC SESSION 2015/16

2. Overview  ACID-BASE THEORIES  ACID DISSOCIATION CONSTANT  pH SCALE  METHODS OF MEASURING pH  POLYPROTIC ACIDS  WEAK BASES  BASE DISSOCIATION CONSTANT  RELATIONSHIP BETWEEN K w, K a AND K b 2 bblee@unimap

3. Overview  BEHAVIOR OF SALTS IN WATER  SALT SOLUTIONS  ACIDS-BASE REACTIONS  BUFFER SOLUTIONS  HENDERSON-HASSELBALCH EQUATION  PREPARING A BUFFER 3 bblee@unimap

4. ACID-BASE THEORIES bblee@unimap 4  Arrhenius (or Classical) Acid-Base Definition  An acid is a substance that contains hydrogen and dissociates in water to yield a hydronium ion : H 3 O +  A base is a substance that contains the hydroxyl group and dissociates in water to yield : OH –  Neutralization is the reaction of an H + (H 3 O + ) ion from the acid and the OH - ion from the base to form water, H 2 O. Arrhenius 1903 Nobel Prize

5. ACID-BASE THEORIES bblee@unimap 5 H 3 O + = H + (aq) = proton in water

6. ACID-BASE THEORIES bblee@unimap 6

7. ACID-BASE THEORIES bblee@unimap 7  The neutralization reaction is exothermic and releases approximately 56 kJ per mole of acid and base.  Brønsted-Lowry Acid-Base Definition  An acid is a proton donor, any species that donates an H + ion.  An acid must contain H in its formula; HNO 3 and H 2 PO 4 - are two examples, all Arrhenius acids are Brønsted-Lowry acids.

8. ACID-BASE THEORIES bblee@unimap 8  A base is a proton acceptor, any species that accepts an H + ion.  A base must contain a lone pair of electrons to bind the H + ion; a few examples are NH 3, CO 3 2-, F -, as well as OH -.  Brønsted-Lowry bases are not Arrhenius bases, but all Arrhenius bases contain the Brønsted-Lowry base OH -.  Therefore in the Brønsted-Lowry perspective, one species donates a proton and another species accepts it: an acid-base reaction is a proton transfer process.

9. ACID-BASE THEORIES bblee@unimap 9 Figure 1: Acid-Base Theories

10. ACID-BASE THEORIES bblee@unimap 10

11. ACID DISSOCIATION CONSTANT

12.  Strong Acids:100% dissociation  good H + donor  equilibrium lies far to right (HNO 3 )  generates weak base (NO 3 - )  Weak Acids:<100% dissociation  not-as-good H + donor  equilibrium lies far to left (CH 3 COOH)  generates strong base (CH 3 COO - )

13. ACID DISSOCIATION CONSTANT

16. Strength vs. K a ACID DISSOCIATION CONSTANT

17. bblee@unimap 17  [H 3 O + ] 1x10 0 to 1x10 -14 in water  [OH - ] 1x10 -14 to 1x10 0 in water [H 3 O + ] and [OH - ]:

18. ACID DISSOCIATION CONSTANT bblee@unimap 18  Finding [H 3 O + ] and [OH - ]:

19. pH SCALE bblee@unimap 19  pH:  pH is defined as the negative base-10 logarithm of the hydronium ion concentration. pH = –log [H 3 O + ]  The “p” in pH tells us to take the negative log of the quantity (in this case, hydronium ions).  Some similar examples are : pOH = –log [OH - ] pK w = –log K w

20. pH SCALE bblee@unimap 20

21. pH SCALE bblee@unimap 21  pH and pOH:  As [H 3 O + ] rises, [OH - ] falls  As pH falls, pOH rises

22. pH SCALE bblee@unimap 22

23. pH SCALE bblee@unimap 23  Some pH values:

24. METHODS OF MEASURING pH bblee@unimap 24 pH indicatorpH meter  It measures the voltage in the solution.

25. METHODS OF MEASURING pH bblee@unimap 25 Figure 1: pH indicators

26. METHODS OF MEASURING pH bblee@unimap 26  Relationship between K a and pK a :

27. METHODS OF MEASURING pH bblee@unimap 27 EXAMPLE 1: Calculating Ka from 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

28. METHODS OF MEASURING pH bblee@unimap 28 EXAMPLE 2: Calculating pH from K a  Calculate the pH of a 0.30 M solution of acetic acid, C 2 H 3 O 2 H, at 25°C.  K a for acetic acid at 25°C is 1.8  10 -5.

29. POLYPROTIC ACIDS bblee@unimap 29  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.

30. POLYPROTIC ACIDS bblee@unimap 30

31. WEAK BASES bblee@unimap 31  Strength of Bases:  Strong:  100% dissociation  OH - supplied to solution  NaOH (s)  Na + (aq) + OH - (aq)  Weak:  <100% dissociation  OH - by reaction with water  CH 3 NH 2(aq) + H 2 O (l) CH 3 NH 2(aq) + OH - (aq)

32. WEAK BASES bblee@unimap 32  Sustainable  Sustainability.

33. BASE DISSOCIATION CONSTANT bblee@unimap 33  Bases react with water to produce hydroxide ion.  Sustainability.

34. BASE DISSOCIATION CONSTANT bblee@unimap 34  K b can be used to find [OH – ] and, through it, pH.

35. BASE DISSOCIATION CONSTANT bblee@unimap 35

36. BASE DISSOCIATION CONSTANT bblee@unimap 36  EXAMPLE 3: Calculating pH of Basic Solutions  What is the pH of a 0.15 M solution of NH 3 ? [NH 4 + ] [OH − ] [NH 3 ] K b = = 1.8  10 -5

37. RELATIONSHIP BETWEEN K w, K a AND K b bblee@unimap 37

38. BEHAVIOR OF SALTS IN WATER bblee@unimap 38

39. BEHAVIOR OF SALTS IN WATER bblee@unimap 39  Properties of salt solutions:  The spectator ions in acid-base reactions form salts. Salts completely ionize in their aqueous solutions.  A salt NH 4 A, the ionizes NH 4 A ↔ NH 4 + + A -  The ions of salts interact with water (called hydration), A - + H 2 O ↔ HA + OH -

40. BEHAVIOR OF SALTS IN WATER bblee@unimap 40  If the anions are stronger base than H 2 O, the solution is basic. NH 4 + + H 2 O = NH 3 + H 3 O +  If the cations are stronger acid than H 2 O, the solution is acidic.  These competitive reactions make the solution acidic or basic depending on the strength of the acids and bases.

41. SALT SOLUTIONS bblee@unimap 41  Neutral Salt Solutions:  Salts consisting of the Anion of a Strong Acid and the Cation of a Strong Base yield a neutral solutions because the ions do not react with water.  Nitrate (NO 3 - ) is a weaker base than water (H 2 O) ; reaction goes to completion as the NO 3 - becomes fully hydrated; does not react with water

42. SALT SOLUTIONS bblee@unimap 42  Na + & NO 3 - do not react with water, leaving just the “autoionization” of water, i.e., a neutral solution.  Salts that produce Acidic Solutions:  A salt consisting of the anion of a strong acid and the cation of a weak base yields an acidic solution.  The cation acts as a weak acid

43. SALT SOLUTIONS bblee@unimap 43  The anion does not react with water.  In a solution of NH 4 Cl, the NH 4 + ion that forms from the weak base, NH 3, is a weak acid.  The Chloride ion, the anion from a strong acid does not react with water

44. SALT SOLUTIONS bblee@unimap 44  Salts that produce Basic Solutions:  A salt consisting of the anion of a weak aid and the cation of a strong base yields a basic solution.  The anion acts as a weak base.  The cation does not react with water.  The anion of the weak acid accepts a proton from water to yield OH - ion, producing a “Basic” solution.

45. SALT SOLUTIONS bblee@unimap 45  Salts of Weakly Acidic Cations and Weakly Basic Anions:  Overall acidity of solution depends on relative acid strength (K a ) or base strength (K b ) of the separated ions.  Eg. NH 4 CN - Acidic or Basic?  Write equations for any reactions that occur between the separated ions and water

46. SALT SOLUTIONS bblee@unimap 46  Compare K a of NH 4 + & K b of CN -  Magnitude of K b (K b > K a ) = (1.6 x 10 -5 / 5.7 x 10 -10 = 3 x 10 4 ), K b of CN - >> K a of NH 4 + (Solution is Basic)  Acceptance of proton from H 2 O by CN - proceeds much further than the donation of a proton to H 2 O by NH 4 +.

47. SALTS OF WEAKLY ACIDIC CATIONS AND WEAKLY BASIC ANIONS bblee@unimap 47

48. ACIDS-BASE REACTIONS bblee@unimap 48 four  There are four classifications or types of reactions: i. strong acid with strong base, ii. strong acid with weak base, iii. weak acid with strong base, and iv. weak acid with weak base. NOTE:  For all four reaction types the limiting reactant problem is carried out first.  Once this is accomplished, one must determine which reactants and products remain and write an appropriate equilibrium equation for the remaining mixture.

49. ACIDS-BASE REACTIONS bblee@unimap 49 (i) STRONG ACID WITH STRONG BASE:  The net reaction is:  The product, water, is neutral. (ii) STRONG ACID WITH WEAK BASE:  The net reaction is:  The product is HB + and the solution is acidic.

50. ACIDS-BASE REACTIONS bblee@unimap 50 (iii) WEAK ACID WITH STRONG BASE:  The net reaction is:  The product is A - and the solution is basic. (iv) WEAK ACID WITH WEAK BASE:  The net reaction is:  Notice that K net may even be less than one.  This will occur when K a HB + > K a HA.

51. ACIDS-BASE REACTIONS bblee@unimap 51 Example 4:  You titrate 100 mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point (mol HBz = mol NaOH).  What is the pH of (a) the final solution (b) half way point? Note: Note: HBz and NaOH are used up! HBz + NaOH ---> Na + + Bz - + H 2 O K a = 6.3 × 10 -5 K b = 1.6 × 10 -10 C 6 H 5 CO 2 H = HBz Benzoate ion = Bz -

52. BUFFER SOLUTIONS bblee@unimap 52 HCl is added to pure water. HCl is added to a solution of a weak acid H 2 PO 4 - and its conjugate base HPO 4 2-.

53. BUFFER SOLUTIONS bblee@unimap 53  The function of a buffer is to resist changes in the pH of a solution.  Buffers are just a special case of the common ion effect.  Buffer Composition Weak Acid+Conj. Base HC 2 H 3 O 2 +C 2 H 3 O 2 - H 2 PO 4 - +HPO 4 2- Weak Base+Conj. Acid NH 3 +NH 4 +

54. BUFFER SOLUTIONS bblee@unimap 54  Consider HOAc/OAc - to see how buffers work.  ACID USES UP ADDED OH -.  We know that OAc - + H 2 O HOAc + OH - has K b = 5.6 x 10 -10  Therefore, the reverse reaction of the WEAK ACID with added OH - has K reverse = 1/ K b = 1.8 x 10 9  K reverse is VERY LARGE, so HOAc completely uses up the OH - !!!! Acetic acid (HOAc) & a salt of the acetate ion (OAc)

55. BUFFER SOLUTIONS bblee@unimap 55  Consider HOAc/OAc - to see how buffers work.  CONJUGATE BASE USES UP ADDED H + HOAc + H 2 O OAc - + H 3 O + has K a = 1.8 x 10 -5.  Therefore, the reverse reaction of the WEAK BASE with added H + has K reverse = 1/ K a = 5.6 x 10 4  K reverse is VERY LARGE, so OAc - completely uses up the H + !

56. BUFFER SOLUTIONS bblee@unimap 56 Example 5  What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x 10 -5 K a = 1.8 x 10 -5

57. BUFFER SOLUTIONS  Notice that the expression for calculating the H + concentration of the buffer is  This leads to a general equation for finding the H + or OH - concentration of a buffer.  Notice that the H + or OH - concentrations depend on K and the ratio of acid and base concentrations. bblee@unimap 57 [H 3 O + ] = Orig. conc. of HOAc Orig. conc. of OAc K a [OH  ]  [Base] [Conj. acid] K b [H 3 O  ]  [Acid] [Conj. base] K a

58. HENDERSON-HASSELBALCH EQUATION bblee@unimap 58  Take the negative log of both sides of this equation: or  This is called the Henderson-Hasselbalch equation. [H 3 O + ] = [Acid] [Conj. base] K a pH = pK a + log [Conj. base] [Acid] pH = pK a - log [Acid] [Conj. base]

59. HENDERSON-HASSELBALCH EQUATION bblee@unimap 59  This shows that the pH is determined largely by the pK a of the acid and then adjusted by the ratio of acid and conjugate base.  Sustainability. pH  pK a + log [Conj. base] [Acid]

60. HENDERSON-HASSELBALCH EQUATION bblee@unimap 60 EXAMPLE 6  What is the new pH when 1.00 mL of 1.00 M HCl is added to: a) 1.00 L of pure water (before HCl, pH = 7.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)

61. PREPARING A BUFFER bblee@unimap 61  You want to buffer a solution at pH = 4.30.  This means [H 3 O + ] = 10 -pH = 5.0 x 10 -5 M  It is best to choose an acid such that [H 3 O + ] is about equal to K a (or pH ­ pK a ).  You get the exact [H 3 O + ] by adjusting the ratio of acid to conjugate base.  Buffer is prepared from:  HCO 3 - weak acid  CO 3 2- conjugate base

62. PREPARING A BUFFER bblee@unimap 62  You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x 10 -5 M POSSIBLE ACIDSK a HSO 4 - / SO 4 2- 1.2 x 10 -2 HOAc / OAc - 1.8 x 10 -5 HCN / CN - 4.0 x 10 -10  Best choice is acetic acid / acetate.

63. PREPARING A BUFFER bblee@unimap 63  You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x 10 -5 M  Solve for [HOAc]/[OAc - ] ratio = 2.78/1  Therefore, if you use 0.100 mol of NaOAc and 0.278 mol of HOAc, you will have pH = 4.30. [H 3 O  ]  5.0 x 10 -5 = [HOAc] [OAc - ] (1.8 x 10 -5 )

64. PREPARING A BUFFER bblee@unimap 64  The concentration of the acid and conjugate base are not important.  It is the RATIO OF THE NUMBER OF MOLES of each.  This simplifying approximation will be correct for all buffers with 3<pH<11, since the [H] + will be small compared to the acid and conjugate base.

65. PREPARING A BUFFER bblee@unimap 65  Preparing Buffers 1) Solid/Solid: mix two solids. 2) Solid/Solution: mix one solid and one solution. 3) Solution/Solution: mix two solutions. 4) Neutralization: Mix weak acid with strong base or weak base with strong acid.

66. PREPARING A BUFFER bblee@unimap 66 Example 7:  Preparing buffer: solid/solid  Calculate the pH of a solution made by mixing 1.5 moles of phthalic acid and 1.2 moles of sodium hydrogen phthalate in 500. mL of solution.  It is given that K a = 3.0 x 10 -4

67. PREPARING A BUFFER bblee@unimap 67 Example 8:  Preparing buffer: Solid/solution  How many moles of sodium acetate must be added to 500. mL of 0.25 M acetic acid to produce a solution with a pH of 5.50?

68. EXAMPLE 1 To calculate K a, we need all equilibrium concentrations. We can find [H 3 O + ], which is the same as [HCOO − ], from the pH. pH = –log [H 3 O + ] – 2.38 = log [H 3 O + ] 10 -2.38 = 10 log [H 3 O + ] = [H 3 O + ] 4.2  10 -3 = [H 3 O + ] = [HCOO – ] bblee@unimap 68

69. EXAMPLE 1 In table form: bblee@unimap 69

70. EXAMPLE 2 The equilibrium constant expression is: Use the ICE (Initial Change and Equilibrium) table: bblee@unimap 70

71. EXAMPLE 2 Simplify: x is relatively same, bblee@unimap 71

72. EXAMPLE 2 bblee@unimap 72 (1.8  10 -5 ) (0.30) = x 2 5.4  10 -6 = x 2 2.3  10 -3 = x pH = –log [H 3 O + ] pH = – log (2.3  10 −3 ) pH = 2.64

73. EXAMPLE 3 Tabulate the data. Simplify: bblee@unimap 73

74. EXAMPLE 3 Therefore, [OH – ] = 1.6  10 -3 M pOH = –log (1.6  10 -3 ) pOH = 2.80 pH = 14.00 – 2.80 pH = 11.20 bblee@unimap 74 (1.8  10 -5 ) (0.15) = x 2 2.7  10 -6 = x 2 1.6  10 -3 = x (x) 2 (0.15) 1.8  10 -5 =

75. EXAMPLE 4 (a) The product of the titration of benzoic acid, the benzoate ion, Bz -, is the conjugate base of a weak acid.  The final solution is basic. bblee@unimap 75 + + K b = 1.6 x 10 -10

76. EXAMPLE 4 (a) This is a two-step problem: 1 st :stoichiometry of acid-base reaction 2 nd :equilibrium calculation 1. Calculate moles of NaOH required. (0.100L HBz)(0.025M) = 0.0025mol HBz 0.0025 mol NaOH This requires 0.0025 mol NaOH 2. Calculate volume of NaOH required. 0.0025 mol (1 L / 0.100 mol) = 0.025 L 25 mL of NaOH required bblee@unimap 76

77. EXAMPLE 4 (a) 3.Moles of Bz - produced = moles HBz = 0.0025 mol Bz - 4.Calculate concentration of Bz -.  There are 0.0025 mol of Bz - in a TOTAL SOLUTION VOLUME of 125 mL [Bz - ] = 0.0025 mol / 0.125 L = 0.020 M bblee@unimap 77

78. EXAMPLE 4 (a) Solving: x = [OH - ] = 1.8 x 10 -6, pOH = 5.75, pH = 8.25 bblee@unimap 78 K b  1.6 x 10 -10 = x 2 0.020- x

79. EXAMPLE 4 (b) [H 3 O + ] = { [HBz] / [Bz - ] } K a At the half-way point, [HBz] = [Bz - ], so [H 3 O + ] = K a = 6.3 x 10 -5 pH = 4.20 bblee@unimap 79

80. EXAMPLE 5 Assuming that x << 0.700 and 0.600, we have Assuming that x << 0.700 and 0.600, we have [H 3 O + ] = 2.1 x 10 -5 and pH = 4.68 bblee@unimap 80

81. EXAMPLE 6 (a) Calculate [HCl] after adding 1.00 mL of HCl to 1.00 L of water M 1 V 1 = M 2 V 2 M 2 = 1.00 x 10 -3 M pH = 3.00 (b) Step 1 — do the stoichiometry H 3 O + (from HCl) + OAc - (from buffer) ---> HOAc (from buffer)  The reaction occurs completely because K is very large. bblee@unimap 81

82. EXAMPLE 6 Step 2—Equilibrium. bblee@unimap 82

83. EXAMPLE 6 Because [H 3 O + ] = 2.1 x 10 -5 M BEFORE adding HCl, we again neglect x relative to 0.701 and 0.599. [H 3 O + ] = 2.1 x 10 -5 M ------> pH = 4.68  The pH has not changed significantly upon adding HCl to the buffer! bblee@unimap 83 [H 3 O  ]  [HOAc] [OAc - ] K a  0.701 0.599 (1.8 x 10 -5 )

84. EXAMPLE 7 bblee@unimap 84 Conjugates do not react!!

85. EXAMPLE 8 Let X = moles NaC 2 H 3 O 2, bblee@unimap 85 Conjugates do not react!!