Bronsted Lowry Acid and Base.pptx

Discussion and summary for acid and base under bronsted and lorys concept

1. 12-חומצות ובסיסים 1 Acid-Base Concepts: The Brønsted -Lowry Theory Arrhenius Acid : A substance that dissociates in water to produce hydrogen ions, H + Arrhenius Base : A substance that dissociates in water to produce hydroxide ions, OH – M + ( aq ) + OH – ( aq ) MOH( aq ) H + ( aq ) + A – ( aq ) HA( aq )

2. 12-חומצות ובסיסים 2 Acid-Base Concepts: The Brønsted -Lowry Theory Conjugate Acid-Base Pairs : Chemical species whose formulas differ only by one hydrogen ion, H + Brønsted-Lowry Acid : A substance that can transfer hydrogen ions, H + . In other words, a proton donor Brønsted-Lowry Base : A substance that can accept hydrogen ions, H + . In other words, a proton acceptor

3. 12-חומצות ובסיסים 3 Acid-Base Concepts: The Brønsted -Lowry Theory Acid-Dissociation Equilibrium

4. 12-חומצות ובסיסים 4 Acid-Base Concepts: The Brønsted -Lowry Theory Base-Dissociation Equilibrium

5. 12-חומצות ובסיסים 5 NH 3 (g) + H 2 O(ℓ) NH 4 + (aq) + OH - (aq) Base: H + acceptor Acid: H + donor Brønsted-Lowry Acid = Proton donor Brønsted-Lowry Base = Proton acceptor Works for non-aqueous solutions and explains why NH 3 is basic: Brønsted-Lowry Concept

6. 12-חומצות ובסיסים 6 Conjugate acid-base pairs are molecules or ions related by the loss/gain of one H + : Conjugate Acid Conjugate Base H 3 O + H 2 O CH 3 COOH CH 3 COO - NH 4 + NH 3 H 2 SO 4 HSO 4 - HSO 4 - SO 4 2- HCl Cl - donate H + accept H + NH 4 + and NH 2 - are not conjugate, conversion requires 2 H + . Conjugate Acid-Base Pairs

7. 12-חומצות ובסיסים 7 Identify the base conjugate to HF( aq ) and the acid conjugate to HCO 3 - ( aq ). HCO 3 - + H + H 2 CO 3 HF H + + F - ( CO 3 2- is the base conjugate to HCO 3 - ) Conjugate Acid Base: H + acceptor Conjugate Base Acid: H + donor Conjugate Acid-Base Pairs

8. 12-חומצות ובסיסים 8 Water acts as a base when an acid dissolves in water: HBr ( aq ) + H 2 O(ℓ) H 3 O + ( aq ) + Br - ( aq ) acid base acid base Water is amphiprotic - it can donate or accept a proton (act as acid or base). But water acts as an acid for some bases: H 2 O(ℓ) + NH 3 ( aq ) NH 4 + ( aq ) + OH - ( aq ) acid base acid base Water’s Role as Acid or Base

9. 12-חומצות ובסיסים 9 The most general acid-base definition: A Lewis acid accepts a pair of e - to form a bond. A Lewis base donates a pair of e - to form a bond. A B .. .. + A B Acid e - pair acceptor Base e - pair donor Coordinate covalent bond Coordinate covalent bond : a shared e - pair, with both e - donated by the same atom. Lewis Acids & Bases

10. 12-חומצות ובסיסים 10 Lewis acid Lewis base + H O H H H O H H Examples: + N H H H B F F F B F F F N H H H H + + H 2 O H 3 O + Lewis acid Lewis base Lewis Acids & Bases

11. 12-חומצות ובסיסים 11 Metal ions can act as Lewis acids. They have: Missing e - . Empty valence orbitals. Metal ion + Lewis base → complex ion Hydroxide ions are Lewis bases. O is e - rich (has large electronegativity): Can easily donate an e - pair to a bond. :O–H : : δ + δ - - Ag + ( aq ) + 2 :NH 3 ( aq ) [H 3 N:Ag:NH 3 ] + ( aq ) Positive Metal Ions as Lewis Acids

12. 12-חומצות ובסיסים 12 Many metal hydroxides are amphoteric – they react with acids and bases. Positive Metal Ions as Lewis Acids Al(OH) 3 (s) + OH - ( aq ) [Al(OH) 4 ] - ( aq ) Aluminum hydroxide can act as a Lewis acid: Al(OH) 3 (s) + 3 H 3 O + ( aq ) Al 3+ ( aq ) + 6 H 2 O(ℓ) or as a Br ø nsted-Lowry base:

13. 12-חומצות ובסיסים 13 δ - Non-metal oxides act as Lewis acids. O attracts e - from any multiple bond. Leaves the other non-metal e - deficient. Susceptible to attack by Lewis base: S O O = – δ + δ - S O O – = δ + δ - δ - O = C = O δ + δ - δ - Neutral Molecules as Lewis Acids O = C = O H O H O C = O O - -

14. 12-חומצות ובסיסים 14 Acid Strength and Base Strength Weak Acid : An acid that is only partially dissociated in water and is thus a weak electrolyte

15. 12-חומצות ובסיסים 15 Acid Strength and Base Strength

16. 12-חומצות ובסיסים 16 Hydrated Protons and Hydronium Ions H + ( aq ) + A  ( aq ) HA( aq ) [H(H 2 O) n ] + For our purposes, H + is equivalent to H 3 O + . n = 4 H 9 O 4 + n = 1 H 3 O + n = 2 H 5 O 2 + n = 3 H 7 O 3 + Due to high reactivity of the hydrogen ion, it is actually hydrated by one or more water molecules.

17. 12-חומצות ובסיסים 17 Acid Strength and Base Strength H 3 O + ( aq ) + A  ( aq ) HA( aq ) + H 2 O( l ) Base Acid Base Acid Weaker acid + Weaker base Stronger acid + Stronger base With equal concentrations of reactants and products, what will be the direction of reaction?

18. 12-חומצות ובסיסים 18 Strong acids are better H + donors than weak acids Strong bases are better H + acceptors than weak bases Stronger acids have weaker conjugate bases. Weaker acids have stronger conjugate bases. Strong acid + H 2 O H 3 O + + conjugate base Fully ionized, reverse reaction essentially does not occur. The conjugate base is extremely weak . Weakly ionized, reverse reaction readily occurs. The conjugate base is relatively strong . Weak acid + H 2 O H 3 O + + conjugate base Relative Strength of Acids & Bases

19. 12-חומצות ובסיסים 19 Acid strength: HF > NH 4 + . (Base strength: NH 3 > F - ) HF has greater tendency to ionize than NH 4 + . Reactant Favored NH 4 + + F - NH 3 + HF Problem Is the following aqueous reaction product or reactant favored? Acid strength increasing Base strength increasing Conj acid. Conj. base H 2 SO 4 HSO 4 - HBr Br - HCl Cl - HF F - NH 4 + NH 3 OH - O 2- H 2 H - CH 4 CH 3 - Relative Strength of Acids & Bases

20. 12-חומצות ובסיסים 20 Factors That Affect Acid Strength Bond Polarity

21. 12-חומצות ובסיסים 21 Factors That Affect Acid Strength Bond Strength

22. 12-חומצות ובסיסים 22 Factors That Affect Acid Strength Oxoacids

23. 12-חומצות ובסיסים 23 Factors That Affect Acid Strength Oxoacids

24. 12-חומצות ובסיסים 24 Organic acids Contain the carboxylic acid functional group: Carboxylic Acids & Amines

25. 12-חומצות ובסיסים 25 O is very electronegative. O withdraws e - from the O–H bond (weakens O-H). Makes it easier for H + to leave. Once formed, the anion is stabilized by resonance: The carboxylic acid H is acidic (will ionize). All other protons are non acidic: CH 3 – C = O O - CH 3 – C = O - O + H 2 O + H 3 O + CH 3 – C = O OH CH 3 – C = O O - Carboxylic Acids & Amines

26. 12-חומצות ובסיסים 26 Strengths of Carboxylic Acids R-COOH Acid K a Acetic CH 3 COOH 1.8 x 10 -5 Hexanoic CH 3 CH 2 CH 2 CH 2 CH 2 COOH 1.4 x 10 -5 Chloroacetic CH 2 C l COOH 0.0013 Trichloroacetic CC l 3 COOH 0.3 Trifluoroacetic CF 3 COOH 0.6 All hydrocarbon R’s “pull” equally… ~equal strength. Add a halogen to R = more e - withdrawing = stronger acid

27. 12-חומצות ובסיסים 27 Amine groups are basic: R–NH 2 , R–NHR or R–NR 2 .. .. .. R = any hydrocarbon. Lone-pair on N accepts a proton (like NH 3 ). Carboxylic Acids & Amines

28. 12-חומצות ובסיסים 28 Name R glycine H alanine CH 3 valine (CH 3 ) 2 CH serine HOCH 2 … and others = − − H O R– C– C–O–H NH 2 Amino Acids & Zwitterions Amino acids contain acid and amine functional groups. Alpha carbon They are crystalline solids ( m.p . > 200 °C). Why? The solid is a zwitterion (has multiple charges) = − − H O R– C– C–O - NH 3 + The acid H + has transferred to the amine (base)

29. 12-חומצות ובסיסים 29 Amino Acids & Zwitterions Amino acid in solution have pH dependent structures. Alanine: = − − H O CH 3 – C– C–O - NH 3 + = − − H O CH 3 – C– C–O–H NH 3 + = − − H O CH 3 – C– C–O - NH 2 OH - H + base added acid added Positively charged Negatively charged Zwitterion (no net charge)

30. 12-חומצות ובסיסים 30 Dissociation of Water H 3 O + ( aq ) + OH  ( aq ) 2 H 2 O( l ) K w = [H 3 O + ][OH  ] Ion-Product Constant for Water : Dissociation of Water : K w = (1.0 × 10  7 )(1.0 × 10  7 ) = 1.0 × 10  14 at 25 o C : [H 3 O + ] = [OH  ] = 1.0 × 10  7 M

31. 12-חומצות ובסיסים 31 Dissociation of Water K w = [H 3 O + ][OH  ] = 1.0 × 10  14 [OH  ] = [H 3 O + ] 1.0 × 10  14 [H 3 O + ] = [OH  ] 1.0 × 10  14

32. 12-חומצות ובסיסים 32 Dissociation of Water

33. 12-חומצות ובסיסים 33 Dissociation of Water Acidic: [H 3 O + ] > [OH  ] Neutral: [H 3 O + ] = [OH  ] Basic: [H 3 O + ] < [OH  ]

34. 12-חומצות ובסיסים 34 Example Calculate the hydronium and hydroxide ion concentrations at 25 ° C in a 6.0 M aqueous sodium hydroxide solution. K w = [H 3 O + ][OH - ] = 1.0 x 10 -14 NaOH( aq ) is strong (100% ionized) so [OH - ] = 6.0 M. [H 3 O + ](6.0) = 1.0 x 10 -14 [H 3 O + ] = 1.7 x 10 -15 M [OH - ] = 6.0 M Autoionization of Water

35. 12-חומצות ובסיסים 35 The pH Scale pH =  log[H 3 O + ] [H 3 O + ] = 10  pH Acidic: pH < 7 Neutral: pH = 7 Basic: pH > 7

36. 12-חומצות ובסיסים 36 The pH Scale The hydronium ion concentration for lemon juice is approximately 0.0025 M. What is the pH when [H 3 O + ] = 0.0025 M? pH =  log(0.0025) = 2.60 2 significant figures 2 decimal places

37. 12-חומצות ובסיסים 37 Base concentrations: pOH = − log 10 [OH - ] A neutral solution (25 °C) has: pOH = − log 10 [1.0 x 10 -7 ] = −(−7.00) = 7.00 Since K w = [ H 3 O + ][ OH - ] = 1.0 x 10 -14 − log( K W ) = − log[H 3 O + ] + ( − log[OH - ]) = − log(1.0 x 10 -14 ) (Valid in all aq. solns . at 25 °C: acidic, neutral or basic) p K w = pH + pOH = 14.00 The pOH scale

38. 12-חומצות ובסיסים 38 Solution A : pOH = −log[ OH - ] = 3.37 pH + pOH = pK w = 14.00 pH = 10.63 Solution B : pH = −log[ H 3 O + ] = 8.12 A has higher pH, B is more acidic Given two aqueous solutions (25 °C) . Solution A : [OH - ] = 4.3 x 10 -4 M, Solution B : [H 3 O + ] = 7.5 x 10 -9 M. Which has the higher pH? Which is more acidic? pH Calculations

39. 12-חומצות ובסיסים 39 H 3 O + concentrations can be measured with an: Electronic pH meter: fast and accurate. preferred method. Acid-base indicator: substance changes color over a small pH range. may have multiple colors (e.g. bromthymol blue). one “color” may be colorless (e.g. phenolphthalein). cheap and convenient. Measuring pH

40. 12-חומצות ובסיסים 40 Measuring pH Acid–Base Indicator : A substance that changes color in a specific pH range. Indicators exhibit pH-dependent color changes because they are weak acids and have different colors in their acid ( HIn ) and conjugate base (In  ) forms. H 3 O + ( aq ) + In  ( aq ) HIn ( aq ) + H 2 O( l ) Color B Color A

41. 12-חומצות ובסיסים 41 Equilibria in Solutions of Weak Acids Acid-Dissociation Constant : H 3 O + ( aq ) + A  ( aq ) HA( aq ) + H 2 O( l ) [H 3 O + ][A  ] [HA] K a =

42. 12-חומצות ובסיסים 42

43. 12-חומצות ובסיסים 43 Calculation of [H + ] in a Water Solution of a Weak Acid If Note x = [H + ] eq and a = [HB] o % ionization can be obtained by multiplying 100% If the % ionization is less than 5%, your assumption is valid. If not, solve for x using the quadratic equation

44. 12-חומצות ובסיסים 44 Equilibria in Solutions of Weak Acids The pH of 0.250 M HF is 2.036. What are the values of K a and p K a for hydrofluoric acid? H 3 O + ( aq ) + F  ( aq ) HF( aq ) + H 2 O( l ) 0.250 ≈0  x + x + x 0.250  x x x x = [H 3 O + ] = 10  2.036 = 0.00920 M p K a =  log( K a )

45. 12-חומצות ובסיסים 45 Equilibria in Solutions of Weak Acids (0.00920)(0.00920) 0.241 = 3.51 × 10  4 [HF] = 0.250  x = 0.250  0.00920 = 0.241 M [F  ] = [H 3 O + ] = 0.00920 M [H 3 O + ][F  ] [HF] K a = p K a =  log( K a ) =  log(3.51 × 10  4 ) = 3.455 [H 3 O + ][F  ] [HF] = K a =

46. 12-חומצות ובסיסים 46 Calculating Equilibrium Concentrations of Weak Acids Calculate the pH of a 0.10 M HCN solution. At 25 o C , K a = 4.9 × 10  10 . H 3 O + ( aq ) + CN  ( aq ) HCN( aq ) + H 2 O( l ) 0.10 ≈0  x + x + x 0.10  x x x [H 3 O + ][CN  ] [HCN] K a =

47. 12-חומצות ובסיסים 47 Calculating Equilibrium Concentrations of Weak Acids x 2 0.10 ≈ 4.9 × 10  10 = ( x )( x ) (0.10  x ) pH =  log([H 3 O + ]) =  log(7.0 × 10  6 ) = 5.15 x = [H 3 O + ] = 7.0 × 10  6 M

48. 12-חומצות ובסיסים 48 Percent Dissociation in Solutions of Weak Acids Percent dissociation = [HA] initial [HA] dissociated × 100%

49. 12-חומצות ובסיסים 49 Polyprotic Acids Some acids “release” multiple H + Each H + ionization has a different K a . 1 st proton is easiest to remove 2 nd is harder, etc. Acid( aq ) Name Acidic H H 2 S Hydrosulfuric Acid 2 H 3 PO 4 Phosphoric Acid 3 H 2 CO 3 Carbonic Acid 2 HOOC-COOH Oxalic Acid 2 C 3 H 5 (COOH) 3 Citric Acid 3

50. 12-חומצות ובסיסים 50 Polyprotic Acids

51. 12-חומצות ובסיסים 51 Phosphoric acid (H 3 PO 4 ) has three acidic protons: H 3 PO 4 (aq) + H 2 O(ℓ) H 3 O + (aq) + H 2 PO 4 - (aq) K a = 7.2 x 10 -3 H 2 PO 4 - (aq) + H 2 O(ℓ) H 3 O + (aq) + HPO 4 2- (aq) K a = 6.3 x 10 -8 HPO 4 2- (aq) + H 2 O(ℓ) H 3 O + (aq) + PO 4 3- (aq) K a = 4.6 x 10 -13 K a Values for Polyprotic Acids H + harder to remove

52. 12-חומצות ובסיסים 52 Equilibria in Solutions of Weak Bases NH 4 + ( aq ) + OH  ( aq ) NH 3 ( aq ) + H 2 O( l ) Base Acid Acid Base [NH 4 + ][OH  ] [NH 3 ] K b = BH + ( aq ) + OH  ( aq ) B( aq ) + H 2 O( l ) [BH + ][OH  ] [B] K b = Base-Dissociation Constant :

53. 12-חומצות ובסיסים 53 Equilibria in Solutions of Weak Bases

54. 12-חומצות ובסיסים 54 Equilibria in Solutions of Weak Bases Calculate the pH of a 0.40 M NH 3 solution. At 25 o C , K b = 1.8 × 10  5 . [NH 4 + ][OH  ] [NH 3 ] K b = NH 4 + ( aq ) + OH  ( aq ) NH 3 ( aq ) + H 2 O( l ) 0.40 ≈0  x + x + x 0.40  x x x

55. 12-חומצות ובסיסים 55 Equilibria in Solutions of Weak Bases x 2 0.40 ≈ 1.8 × 10  5 = ( x )( x ) (0.40  x ) pH =  log([H 3 O + ]) =  log(3.7 × 10  12 ) = 11.43 x = [OH  ] = 0.0027 M = 3.7 × 10  12 M 0.0027 1.0 × 10  14 [H 3 O + ] =

56. 12-חומצות ובסיסים 56 Relation Between K a and K b H 3 O + ( aq ) + OH  ( aq ) 2 H 2 O( l ) H 3 O + ( aq ) + NH 3 ( aq ) NH 4 + ( aq ) + H 2 O( l ) NH 4 + ( aq ) + OH  ( aq ) NH 3 ( aq ) + H 2 O( l ) [NH 4 + ][OH  ] [NH 3 ] [H 3 O + ][NH 3 ] [NH 4 + ] × K w K a K b = [H 3 O + ][OH  ] = K w = (5.6 × 10  10 )(1.8 × 10  5 ) = 1.0 × 10  14 K a × K b =

57. 12-חומצות ובסיסים 57 Relation Between K a and K b p K a + p K b = p K w = 14.00 K a × K b = K w K b = K a K w K a = K b K w conjugate acid-base pair

58. 12-חומצות ובסיסים 58 Acid–Base Properties of Salts Cations from strong bases : Alkali metal cations of group 1a (Li + , Na + , K + ) Alkaline earth metal cations of group 2a (Mg 2+ , Ca 2+ , Sr 2+ , Ba 2+ ), except for Be 2+ Anions from strong monoprotic acids : Cl  , Br  , I  , NO 3  , and ClO 4  The following ions do not react appreciably with water to produce either H 3 O + or OH  ions: Salts That Yield Neutral Solutions

59. 12-חומצות ובסיסים 59 Acid–Base Properties of Salts Ammonium ion (NH 4 + ) is the conjugate acid of the weak base ammonia (NH 3 ) while chloride ion ( Cl  ) is neither acidic nor basic. H 3 O + ( aq ) + NH 3 ( aq ) NH 4 + ( aq ) + H 2 O( l ) Salts such as NH 4 Cl that are derived from a weak base (NH 3 ) and a strong acid ( HCl ) yield acidic solutions. Salts That Yield Acidic Solutions

60. 12-חומצות ובסיסים 60 Acid–Base Properties of Salts Hydrated cations of small, highly charged metal ions, such as Al 3+ . Salts That Yield Acidic Solutions

61. 12-חומצות ובסיסים 61 Acid–Base Properties of Salts Salts That Yield Acidic Solutions

62. 12-חומצות ובסיסים 62 Acid–Base Properties of Salts Salts That Yield Basic Solutions Cyanide ion (CN  ) is the conjugate base of the weak acid hydrocyanic acid (HCN) while sodium ion (Na + ) is neither acidic nor basic. HCN( aq ) + OH  ( aq ) CN  ( aq ) + H 2 O( l ) Salts such as NaCN that are derived from a strong base ( NaOH ) and a weak acid (HCN) yield basic solutions.

63. 12-חומצות ובסיסים 63 Acid–Base Properties of Salts The pH of an ammonium carbonate solution, (NH 4 ) 2 CO 3 , depends on the relative acid strength of the cation and the relative base strength of the anion. Is it acidic or basic ? Salts That Contain Acidic Cations and Basic Anions

64. 12-חומצות ובסיסים 64 Acid–Base Properties of Salts Salts That Contain Acidic Cations and Basic Anions HCO 3  ( aq ) + OH  ( aq ) CO 3 2  ( aq ) + H 2 O( l ) K b H 3 O + ( aq ) + NH 3 ( aq ) NH 4 + ( aq ) + H 2 O( l ) K a (NH 4 ) 2 CO 3 : Three possibilities: K a > K b : The solution will contain an excess of H 3 O + ions (pH < 7). K a < K b : The solution will contain an excess of OH  ions (pH > 7). K a ≈ K b : The solution will contain approximately equal concentrations of H 3 O + and OH  ions (pH ≈ 7).

65. 12-חומצות ובסיסים 65 Acid–Base Properties of Salts = 1.8 × 10  4 5.6 × 10  11 1.0 × 10  14 K b for CO 3 2  = K a for HCO 3  K w = = 5.6 × 10  10 1.8 × 10  5 1.0 × 10  14 K a for NH 4 + = K b for NH 3 K w = Basic, K a < K b (NH 4 ) 2 CO 3 : Salts That Contain Acidic Cations and Basic Anions HCO 3  ( aq ) + OH  ( aq ) CO 3 2  ( aq ) + H 2 O( l ) K b H 3 O + ( aq ) + NH 3 ( aq ) NH 4 + ( aq ) + H 2 O( l ) K a

66. 12-חומצות ובסיסים 66 Acid–Base Properties of Salts

67. 12-חומצות ובסיסים 67 CO 3 2- (aq) + H 2 O(ℓ) HCO 3 - (aq) + OH - (aq) [ ] initial 1.50 0 0 [ ] change - x +x +x [ ] equil 1.50 - x x x K b = 2.1 x 10 -4 = = ≈ [HCO 3 - ][OH - ] [CO 3 2- ] x 2 (1.50 – x) x 2 1.50 x = 1.77 x 10 -2 pOH = −log(1.77 x 10 -2 ) = 1.75 pH = 14.00 - 1.75 = 12.25 What is the pH of a 1.50 M Na 2 CO 3 ( aq )? K b = 2.1 x 10 -4 pH of a Salt Solution

68. 12-חומצות ובסיסים 68 NH 4 + (aq) + H 2 O(ℓ) NH 3 (aq) + H 3 O + (aq) [ ] initial 0.132 0 0 [ ] change - x +x +x [ ] equil 0.132 - x x x x = 8.57 x 10 -6 pH = − log(8 .57 x 10 -6 ) = 5.07 K a = 5.6 x 10 -10 = = ≈ [NH 3 ][H 3 O + ] [NH 4 + ] x 2 (0.132 – x) x 2 0.132 0.132 M NH 4 Br? Acidic, basic or neutral? K a (NH 4 + ) = 5.6 x 10 -10 . pH of a Salt Solution Acidic !

69. 12-חומצות ובסיסים 69 The Common-Ion Effect H 3 O + ( aq ) + CH 3 CO 2 – ( aq ) CH 3 CO 2 H( aq ) + H 2 O( l ) Common-Ion Effect : The shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium

70. 12-חומצות ובסיסים 70 The Common-Ion Effect The pH of 0.10 M acetic acid is 2.89. Calculate the pH of a solution that is prepared by dissolved 0.10 mol of acetic acid and 0.10 mol sodium acetate in enough water to make 1.00 L of solution. H 3 O + ( aq ) + CH 3 CO 2 – ( aq ) CH 3 CO 2 H( aq ) + H 2 O( l ) 0.10 ≈0 0.10 – x + x + x 0.10 – x x 0.10 + x [H 3 O + ][CH 3 CO 2 – ] [CH 3 CO 2 H] K a =

71. 12-חומצות ובסיסים 71 The Common-Ion Effect x (0.10) 0.10 ≈ 1.8 × 10 –5 = ( x )(0.10 + x ) (0.10 – x ) pH = –log([H 3 O + ]) = –log(1.8 × 10 –5 ) = 4.74 x = [H 3 O + ] = 1.8 × 10 –5 M

72. 12-חומצות ובסיסים 72 The Common-Ion Effect

73. 12-חומצות ובסיסים 73 The Common-Ion Effect H 3 O + ( aq ) + CH 3 CO 2 – ( aq ) CH 3 CO 2 H( aq ) + H 2 O( l ) The addition of acetate ion to a solution of acetic acid suppresses the dissociation of the acid. The equilibrium shifts to the left. Le Châtelier’s Principle

74. 12-חומצות ובסיסים 74 The Common-Ion Effect

75. 12-חומצות ובסיסים 75 Buffer Solutions Buffer Solution : A solution that contains a weak acid and its conjugate base and resists drastic changes in pH CH 3 CO 2 H + CH 3 CO 2 – HF + F – NH 4 + + NH 3 H 2 PO 4 – + HPO 4 2– Weak acid + Conjugate base For example:

76. 12-חומצות ובסיסים 76 Buffer = chemical system that resists changes in pH. Example Add 0.010 mol of HCl or NaOH to: Buffer Solutions 1 L of solution pH Initial after HCl after NaOH Pure water 7.00 2.00 12.00 [C H 3 COOH] = 0.5 M + [CH 3 COONa] = 0.5 M 4.74 4.72 4.76 A buffered solution

77. 12-חומצות ובסיסים 77 Buffers must contain: A weak acid to react with any added base. A weak base to react with any added acid. These components must not react with each other. Buffers are made with ~equal quantities of a conjugate acid-base pair. (e.g. CH 3 COOH + CH 3 COONa) Buffer Action

78. 12-חומצות ובסיסים 78 Added OH - removed by the acid: CH 3 COOH + OH - CH 3 COO - + H 2 O Added H 3 O + removed by the conjugate base: CH 3 COO - + H 3 O + CH 3 COOH + H 2 O Buffer Action

79. 12-חומצות ובסיסים 79 The equilibrium maintains the acid/base ratio. CH 3 COOH + H 2 O CH 3 COO - + H 3 O + pH remains stable. Buffer Action

80. 12-חומצות ובסיסים 80 Blood pH Blood is buffered at pH = 7.40 ± 0.05. pH too low: acidosis. pH too high: alkalosis. CO 2 generates the most important blood buffer. CO 2 (aq) + H 2 O(ℓ) H 2 CO 3 (aq) H 2 CO 3 (aq) + H 2 O(ℓ) H 3 O + (aq) + HCO 3 - (aq) Buffer Action

81. 12-חומצות ובסיסים 81 Henderson-Hasselbalch equation Depends on the [acid]/[base] – not absolute amounts. [A - ] [HA] pH = p K a + log With p K a = -log K a Note: pH = p K a when [HA] = [A - ] The pH of Buffer Solutions [H 3 O + ] = K a [HA] [A - ] [HA] [A - ] log [H 3 O + ] = log K a + log [BH + ] [B] pOH = p K b + log B: weak base

82. 12-חומצות ובסיסים 82 What is the pH of a 0.050 M (monoprotic) pyruvic acid + 0.060 M sodium pyruvate buffer? K a = 3.2 x 10 -3 . pH = – log(3.2 x 10 -3 ) + log(0.060/0.050) = 2.49 + 0.08 = 2.57 The pH of Buffer Solutions What is the HPO 4 2- /H 2 PO 4 - ratio in blood at pH =7.40. K a (H 2 PO 4 - ) = K a,2 (H 3 PO 4 ) = 6.2 x 10 -8 . 7.40 = −log(6.2 x 10 -8 ) + log([HPO 4 2- ]/[H 2 PO 4 - ]) log([HPO 4 2- ]/[H 2 PO 4 - ]) = 0.192 [HPO 4 2- ]/[H 2 PO 4 - ] = 1.5

83. 12-חומצות ובסיסים 83 pH weak acid weak base K a (weak acid) p K a 4 lactic acid lactate ion 1.4 x 10 -4 3.85 5 acetic acid acetate ion 1.8 x 10 -5 4.74 6 carbonic acid hydrogen carbonate ion 4.3 x 10 -7 6.37 7 dihydrogen phosphate hydrogen phosphate ion 6.3 x 10 -8 7.20 8 hypochlorous acid hypochlorite ion 6.8 x 10 -8 7.17 9 ammonium ion ammonia 5.6 x 10 -10 9.25 10 hydrogen carbonate carbonate ion 4.7 x 10 -11 10.33 Common Buffers Useful buffer range: pH = p K a ±1 (10:1 or 1:10 ratio of [A - ]/[HA ]). Buffer Capacity : The amount of acid (or base) that can be added without large pH changes.

84. 12-חומצות ובסיסים 84 1.0 L of buffer is prepared with [NaH 2 PO 4 ] = 0.40 M and [Na 2 HPO 4 ] = 0.25 M. Calculate the pH of: (a) the buffer (b) after 0.10 mol of NaOH is added. Addition of Acid or Base to a Buffer [A - ] [HA] pH = p K a + log p K a = - log(6.3 x 10 -8 ) = 7.20 (a) No base added: 0.25 0.40 pH = 7.20 + log = 7.00 K a (H 2 PO 4 - ) = 6.3 x 10 -8

85. 12-חומצות ובסיסים 85 1.00 L Buffer: [NaH 2 PO 4 ]= 0.40 M ; [Na 2 HPO 4 ] = 0.25 M. (b) calculate pH after 0.10 mol of NaOH is added. K a (H 2 PO 4 - ) = 6.3 x 10 -8 (b) 0.10 mol of NaOH, converts conj. acid to base: H 2 PO 4 - + OH - → HPO 4 2- + H 2 O [A - ] [HA] pH = p K a + log 0.35 0.30 pH = 7.20 + log = 7.27 n initial 0.40 0.25 n added 0.10 n after (0.40 – 0.10) 0 (0.25 + 0.10) Use n directly. [ ] = n/ V and V is the same for both (cancels) Addition of Acid or Base to a Buffer

86. 12-חומצות ובסיסים 86 Standard solution ( titrant ) is added from a buret. Equivalence occurs when a stoichiometric amount of titrant has been added. Acid-Base Titrations

87. 12-חומצות ובסיסים 87 Acid-Base Titrations An indicator is used to find the end point . Color change observed. End point ≠ equivalence point (should be close...). At the equivalence point: n titrant = n analyte n titrant = V titrant [titrant] n analyte = V analyte [analyte]

88. 12-חומצות ובסיסים 88 The acid-Base i ndicator is a weak acid that changes color with changes in pH. Detection of the Equivalence Point [H 3 O + ][In - ] [ HIn ] K a = Observed color will vary (depends on the relative amounts of HIn and In - in solution). HIn(aq) + H 2 O(ℓ) H 3 O + (aq) + In - (aq) color in acid color in base

89. 12-חומצות ובסיסים 89 The acidic form dominates when the [HIn] >> [In - ] Detection of the Equivalence Point -p K a = -pH - 1 pH = p K a -1 The b asic color dominates when [In - ] >> [ HIn ] -p K a = 1 - pH pH = p K a +1 [H 3 O + ][In - ] [ HIn ] K a = [H 3 O + ] 10 = [ HIn ] [In - ] = 10 If [H 3 O + ][In - ] [ HIn ] K a = = 10 [H 3 O + ] [In - ] [ HIn ] = 10 If

90. 12-חומצות ובסיסים 90 Detection of the Equivalence Point

91. 12-חומצות ובסיסים 91 Acid-base titration curve = plot of pH vs V titrant added. Titrate 50.0 mL of 0.100 M HCl with 0.100 M NaOH n acid = 0.0500 L = 5.00 x 10 -3 mol 0.100 mol L Initial pH = -log(0.100) = 1.000 (fully ionized acid) Titration of Strong Acid with Strong Base

92. 12-חומצות ובסיסים 92 After 40.0 mL of base added (before equivalence) Base removes H 3 O + : H 3 O + + OH -  2 H 2 O [H 3 O + ] = n acid – n base added V acid (ℓ ) + V base (ℓ ) added [H 3 O + ] = (5.00 x 10 -3 – 4.00 x 10 -3 ) mol (0.0500 + 0.0400) L = 0.0111 M pH = -log(0.0111) = 1.95 At Equivalence (50.0 mL added) All acid and base react. Neutral salt. pH = 7.00. Titration of Strong Acid with Strong Base Acid concentration becomes:

93. 12-חומצות ובסיסים 93 After Equivalence (50.2 mL added) All acid consumed. n OH - added = 0.0502 L = 5.02 x 10 -3 mol 0.100 mol L original n acid = 0.0500 L = 5.00 x 10 -3 mol 0.100 mol L V total = (0.0500 + 0.0502) L = 0.1002 L pOH = - log(0.02 x10 -3 / 0.1002) = 3.70 pH = 14.00 – 3.70 = 10.30 n OH - remaining = 0.02 x 10 -3 mol Titration of Strong Acid with Strong Base

94. 12-חומצות ובסיסים 94 V titrant /mL V excess /mL V total /mL [OH - ]/mol L -1 pH 50.0 0 100.0 0 7.00 50.1 0.1 100.1 0.0001 10.00 50.2 0.2 100.2 0.0002 10.30 51 1 101.0 0.0010 11.00 55 5 105.0 0.0048 11.67 60 10 110.0 0.0091 11.96 70 20 120.0 0.0167 12.22 80 30 130.0 0.0200 12.30 0.1 mL of base increased the pH by 3 units! Titration of Strong Acid with Strong Base

95. 12-חומצות ובסיסים 95 Titration of Strong Acid with Strong Base

96. 12-חומצות ובסיסים 96 More complicated. Weak acid is in equilibrium with its conjugate base. HA + H 2 O H 3 O + + A - Example Titrate 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH. What is the pH after the following titrant additions: 0 mL, 40.0 mL, 50.0 mL, and 50.2 mL? x 2 (0.100) ≈ [H 3 O + ][A - ] [HA] K a = 1.8 x 10 -5 = x = 0.0013 pH = -log(0.0013) = 2.88 mL added Titration of Weak Acid with Strong Base

97. 12-חומצות ובסיסים 97 HA(aq) + OH - (aq) A - (aq) + H 2 O(ℓ) 40.0 mL added n initial 0.00500 n added 0.00400 n left 0.00100 0.00400 Each OH - removes 1 HA… n left = 0.00500 - 0.00400 …and makes 1 A - 40 mL of 0.100 M 50 mL of 0.100 M V total = (50.0 + 40.0) mL = 90.0 mL = 0.0900 L Titration of Weak Acid with Strong Base

98. 12-חומצות ובסיסים 98 40.0 mL added (continued) (0.00400/0.0900) (0.00100/0.0900) pH = -log(1.8 x 10 -5 ) + log 0.0400 0.0100 pH = 4.74 + log = 5.35 Henderson-Hasselbalch: [A - ] = n A- / V [HA] = n HA / V Note: V cancels (could be omitted) Titration of Weak Acid with Strong Base

99. 12-חומצות ובסיסים 99 HA(aq) + OH - (aq) A - (aq) + H 2 O(ℓ) (c) 50.0 mL added Equivalence. All HA is converted to A - . V total = 100.0 mL, so [A - ] = 0.00500 mol/0.100 L = 0.0500 M n initial 0.00500 n added 0.00500 n left 0 0.00500 Titration of Weak Acid with Strong Base A - is basic!

100. 12-חומצות ובסיסים 100 Use K b to solve for the [OH - ] generated by [A - ]: K b = 5.6 x 10 -10 ≈ x 2 0.0500 pOH = 5.28 pH = 14 - pOH = 8.72 [OH - ][HA] [A - ] K b = A - + H 2 O HA + OH - (0.0500 - x) x x K b = K w / K a = 5.6 x 10 -10 x = 5.3 x 10 -6 M Titration of Weak Acid with Strong Base

101. 12-חומצות ובסיסים 101 (c) 50.2 mL added 0.2 mL of “extra” OH - dominates pH. Ignore any contribution from A - . [OH - ] = (0.0002 L x 0.100 mol/L) / 0.1002 L = 2.0 x 10 -4 M pOH = 3.7 pH = 14.0 – 3.7 = 10.3 Titration of Weak Acid with Strong Base

102. 12-חומצות ובסיסים 102 pH = p K a at 50% titration (25 mL) p K a (acetic acid) = 4.74. Short vertical section compared to strong acid/strong base. Titrate of 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH. Titration of Weak Acid with Strong Base

103. 12-חומצות ובסיסים 103 pH = p K a at 50% titration Weaker acid = shorter vertical section Titration of Weak Acid with Strong Base

104. 12-חומצות ובסיסים 104 50.0 mL of 0.100 M ammonia titrated with 0.100 M HCl. pH = p K a at 50% to equivalence (p K a = 9.25). Titration of Weak Base with Strong Acid

105. 12-חומצות ובסיסים 105 Titration of Polyprotic Acids & Bases Maleic acid: Sodium carbonate HOOC–CH=CH–COOH Na + ( aq ) + CO 3 2- ( aq ) (diprotic acid) (basic anion)