1. Chemistry 100 Acids and Bases

2. The Brønsted Definitions Brønsted Acid  proton donor Brønsted Base  proton acceptor Conjugate acid - base pair  an acid and its conjugate base or a base and its conjugate acid

3. Example Acid-Base Reactions Look at acetic acid dissociating CH 3 COOH(aq) ⇌ CH 3 COO - (aq) + H + (aq)  Brønsted acid Conjugate base Look at NH 3 (aq) in water NH 3 (aq) + H 2 O(l) ⇌ NH 4 + (aq) + OH - (aq)   Brønsted baseconjugate acid

4. Representing Protons in Aqueous Solution CH 3 COOH(aq) ⇌ CH 3 COO - (aq) + H + (aq) CH 3 COOH(aq) + H 2 O(l) ⇌ CH 3 COO - (aq) + H 3 O + (aq) HCl (aq)  Cl - (aq) + H + (aq) HCl(aq) + H 2 O(l)  Cl - (aq) + H 3 O + (aq)

5. Representing Protons Both representations of the proton are equivalent H 5 O 2 + (aq), H 7 O 3 + (aq), H 9 O 4 + (aq) have been observed We will use either H + (aq) or H 3 O + (aq)

6. What is H + (aq)? H3O+H3O+ H5O2+H5O2+ H+H+ H9O4+H9O4+

7. The Hydroxide Bases KOH, RbOH, NaOH, are not strictly Brønsted Bases since none of these substances accepts a proton KOH(aq)  K + (aq) + OH - (aq) NaOH(aq)  Na + (aq) + OH - (aq) OH - (aq) + H 3 O + (aq) ⇌ 2 H 2 O(l)

8. The Autoionization of Water Water autoionizes (self-dissociates) to a small extent 2H 2 O(l) ⇌ H 3 O + (aq) + OH - (aq) H 2 O(l) ⇌ H + (aq) + OH - (aq) These are both equivalent definitions of the autoionization reaction. Water is acting as a base and an acid in the above reaction  water is amphoteric.

9. The Autoionization Equilibrium ◦ from the preceding chapter – but we know [H 2 O] is constant

10. The Defination of K w K eq [H 2 O] = K w = [H + ][OH - ] Ion product constant for water, K w, is the product of the molar concentrations of H + and OH - ions in pure water at a temperature of 298.15 K K w = [H + ][OH - ] = 1.0x10 -14 at 298.2 K

11. The Definition of an Acidic Solution We define an acidic solution as one where the [H + ] in the solution is greater than the the [H + ] in pure water acidic solution  [H + ] > 1.0 x 10 -7 mole/L at 298.2 K

12. The Definition of a Basic Solution Basic solutions are those where the [H + ] in the solution is less than its concentration in pure water at 298.2 K. Basic solution  [H + ] < 1.0 x 10 -7 mole/L An alternative definition of a a basic solution is as follows Basic solution  [OH - ] > 1.0 x 10 -7 mole/L

13. The Definition of a Neutral Solution A neutral solution is defined as one where the [H + ] in the solution is equal to the hydrogen ion concentration in pure water Neutral solution  [H + ] = [OH - ] = 1.0 x 10 -7 mole/L at 298.2 K!

14. The Dependence of K w on Temperature In our definitions of an acidic, basic, and a neutral solutions, we had explicitly stated the temperature as 298.2 K. Why? K w is temperature dependent How will that affect our definition of an acidic, basic, or a neutral solution?

15. Neutrality at Body Temperature At T = 310.15 K (physiological temperature)  K w = 2.4 x 10 -14 A neutral solution has [H + ] = [OH - ] = (K w )½ At 310.15 K, a neutral solution is one where [H + ] = [OH - ] = 1.5 x 10 -7 M (UNLESS OTHERWISE INDICATED, ALL CALCULATIONS WILL BE AT 298.15 K)

16. The pH scale Sørenson - 1909 pH = -log [H+] Solution Type [H+] / MpH range neutral solutions [H + ] = [OH - ] = 1.0x10 -7 pH = 7.00 basic solutions [H+] <1.0x10 -7 pH > 7.00 acid solutions [H+] >1.0x10 -7 pH < 7.00

17. The Relationship between pH and pOH pH  -log [H + ] pOH  -log [OH - ] From the K w expression K w = [H + ][OH - ] = 1.0x10 -14 at 298.2 K -log (1 x 10 -14 ) = -log [H + ] -log [OH - ] 14.00 = pH + pOH

18. Acid Strength and % Dissociation CH 3 COOH(aq) ⇌ CH 3 COO - (aq) + H + (aq) HCOOH(aq) ⇌ HCOO - (aq) + H + (aq) both weak acids < 5% ionized Other examples of weak acids  HF, HNO 2, HCN

19. Acid Strength The strength of an acid is directly dependent on its % dissociation (  value) For an acid  n = the number of groups that donate a proton

20. Base Strength and % Dissociation Strong Bases  also 100% ionized in water NaOH(aq)  Na + (aq) + OH - (aq) Ba(OH) 2 (aq)  Ba 2+ (aq) + 2OH - (aq) Some bases are weak bases; they don’t ionize completely. NH 3 (aq) + H 2 O(l) ⇌ NH 4 + (aq) + OH - (aq) < 5% ionized in aqueous solution

21. Base Strength The strength of a base is also directly dependent on its % dissociation (  value) For a base  [base] o  the original concentration of base  m = the number of basic groups in the molecule

22. Conjugate Acid-Base Strengths CH 3 COOH (aq) ⇌ CH 3 COO - (aq) + H + (aq) Note that the conjugate base of acetic acid is a reasonable proton acceptor CH 3 COO - (aq) + H 2 O (l) ⇌ CH 3 COOH (aq) + OH - (aq)

23. Other Examples HNO 3 (aq)  H + (aq) + NO 3 - (aq)  conjugate base (very weak) HCOOH (aq) ⇄ HCOO - (aq) + H + (aq)  conjugate base is relatively strong NH 3 (aq) + H 2 O(l) ⇄ NH 4 + (aq) + OH - (aq)  relatively strong conjugate acid

24. HCl (aq)  Cl - (aq) + H + (aq) The conjugate base, Cl - ion, is extremely weak Cl - (aq) + H 2 O (l)  HCl (aq) + OH - (aq)

25. S 2- (aq) + H 2 O (l)  HS - (aq) + OH - (aq) The conjugate acid, the HS - ion, is extremely weak HS - (aq) ⇌ H + (aq) + S 2- (aq) The equilibrium lies very far to the left for this reaction

26. The greater the acid strength (large K a ), the weaker the conjugate base of that acid The weaker the acid (smaller K a ), the stronger its conjugate base If the base strength is high (K b is large), its conjugate acid is very weak The weaker the base (small K b value), the stronger the conjugate acid of the base

27. Calculating the pH of Solution of Strong Acids For the dissolution of HCl, HI, or any of the other seven strong acids in water HCl (aq)  H + (aq) + Cl - (aq) HI (aq)  H + (aq) + I - (aq) The pH of these solutions is obtained from the molarity of the dissolved acid pH  -log [H + ] = -log[HCl]

28. Calculating the pH of Solution of Strong Bases For the dissolution of NaOH, Ba(OH) 2, or any of the other strong bases in water NaOH (aq)  Na + (aq) + OH - (aq) Ba(OH) 2 (aq)  Ba 2+ (aq) + 2OH - (aq)

29. The pH of these solutions is obtained by first calculating the pOH from the molarity of the dissolved base pOH  -log [OH - ] = -log[NaOH] pOH  -log [OH - ] = -log{2 [Ba(OH) 2 ]} pH = 14.00 - pOH

30. The Seven Strong Acids chloric acidHClO 3 hydrobromic acidHBr hydrochloric acidHCl hydroiodic acidHI nitric acidHNO 3 perchloric acidHClO 4 sulphuric acidH 2 SO 4 What about the relative strength of the strong acids?

31. The Leveling Effect H + (aq) (or H 3 O + (aq)) is the strongest acid that can exist in aqueous solution. Any acid stronger than H + (aq) reacts with water completely to produce H + (aq) and the weak conjugate base.

32. HNO 3 (aq) is a stronger acid than H + (aq) (H 3 O + )  reacts with water completely to form H + (aq) HNO 3 (aq)  H + (aq) + NO 3 - (aq) Acids weaker than H + (aq) have the equilibrium lying primarily to the left. HNO 2 (aq) ⇌ H + (aq) + NO 2 - (aq)

33. The OH - ion is the strongest base that can exist in aqueous solution. Bases stronger than OH - (aq) react with water to produce the hydroxide ion (OH - ).

34. NH 2 - (the amide ion) is an extremely strong base (much stronger than OH - ). Therefore, NaNH 2 (aq) + H 2 O (l)  NH 3 (aq) + NaOH(aq) NH 2 - cannot exist in aqueous solution. NH 3 is a much weaker base than OH -. Therefore, when it reacts with water, the equilibrium favours the reactants NH 3 (aq) + H 2 O (l) ⇌ NH 4 + (aq) + OH - (aq)

35. The Leveling Effect Defined Any acid that is stronger than H + (aq) means that we have 100% ionisation of the acid. For acids like HCl(aq), HClO 4 (aq), HNO 3 (aq), the appearance is one of equal acid strength. Water is said to have a levelling effect on the acid strength

36. Equilibria in Aqueous Solutions of Weak Acids/ Weak Bases By definition, a weak acid or a weak base does not ionize completely in water (  <<100%). How would we calculate the pH of a solution of a weak acid or a weak base in water?

37. The K a Value To obtain the pH of a weak acid solution, we must apply the principles of chemical equilibrium Define the acid dissociation constant K a For a general weak acid reaction HA (aq) ⇌ H + (aq) + A - (aq)

38. Weak Acid/Bases and pH For a solution of hydrofluoric acid in water HF (aq) ⇌ H + (aq) + F - (aq)

39. Equilibria of Weak Bases in Water To calculate the percentage dissociation of a weak base in water (and the pH of the solutions) CH 3 NH 2 (aq) + H 2 O ⇌ CH 3 NH 3 + (aq) + OH - (aq) We approach the problem as in the case of the weak acid above, i.e., from the chemical equilibrium viewpoint.

40. The K b Value Define the base dissociation constant K b For a general weak base reaction with water B (aq) + H 2 O (l) ⇌ B + (aq) + OH - (aq) For the above system

41. Diprotic/Polyprotic Acids Look at the following system. H 2 C 2 O 4 (aq) ⇌ HC 2 O 4 - (aq) + H + (aq) K a1 HC 2 O 4 - (aq) ⇌ C 2 O 4 2- (aq) + H + (aq)K a2 For the dissociation of diprotic and polyprotic acids, the magnitudes of the dissociation constants decrease in the direction K a1 > K a2 > K a3 etc.

42. Example For oxalic acid in water, K a1 = 6.5 x 10 -2 K a2 = 6.1 x 10 -5 Since K a1 >> K a2, the [H + ] (and the pH) in the solution is due primarily to the first dissociation ONLY.

43. Obtaining the Relationship between K a and K b We have already seen that there is a relationship between the strength of an acid and the ability of its conjugate base to hydrolyse. HCOOH (aq) ⇌ HCOO - (aq) + H + (aq) K a (HCOOH) = 1.8 x 10 -4 Examine the reverse reaction, the hydrolysis (reaction of the substance with water) of HCOO - HCOO - (aq) + H 2 O (l) ⇌ HCOOH (aq) + OH - (aq)

44. Obtaining the K b of the Conjugate Base HCOOH (aq) ⇌ HCOO - (aq) + H + (aq) HCOO - (aq) + H 2 O (l) ⇌ HCOOH (aq) + OH - (aq) K eq (1) = K a (HCOOH) K eq (2) = K b (HCOO-) Add the two reactions together HCOOH (aq)  HCOO - (aq) + H + (aq) HCOO - (aq) + H 2 O (l) ⇌ HCOOH (aq) + OH - (aq)

45. We are left with the overall reaction H 2 O (l) ⇌ H + (aq) + OH - (aq) K w = [H + ][ OH - ] From our rules for the equilibria of multiple reactions. K w = K (1) K (2) K w = K b K a

46. Variation of Conjugate base Strength with K a HCOOH (aq) ⇌ HCOO - (aq) + H + (aq) K a (HCOOH) = 1.8 x 10 -4 K b (HCOO - ) = 5.6 x 10 -11 CH 3 COOH (aq) ⇌ CH 3 COO - (aq) + H + (aq) K a (CH 3 COOH) = 1.8 x 10 -5 K b (CH 3 COO - ) = 5.6 x 10 -10

47. Salts of Conjugate Bases Look at the dissolution of CH 3 COONa in water. CH 3 COONa (aq)  Na + (aq) + CH 3 COO - (aq) But we know that the acetate ion, CH 3 COO - (aq), will hydrolyze in aqueous solution according to the following reaction. CH 3 COO - (aq) + H 2 O (l) ⇌ CH 3 COOH (aq) + OH - (aq)

48. Hydrolysis reaction produces OH - Dissolving a salt of a strong base/ weak acid in water  basic solution.

49. Salts of Conjugate Acids Look at the dissolution of NH 4 Cl in water NH 4 Cl (aq)  NH 4 + (aq) + Cl - (aq) But we know that the ammonium ion, NH 4 + (aq), will donate a proton aqueous solution NH 4 + (aq) ⇌ NH 3 (aq) + H + (aq)

50. Hydrolysis reaction produces H + (aq) Dissolving a salt of a strong acid/ weak base in water  acidic solution.

51. Both the cation and anion Hydrolyse What about salts in which both the cation and anion hydrolyze? The pH of the solution will depend on the magnitude of the K a and the K b values of the respective acidic and basic ions.

53. Salts of Weak Acids/Strong Bases How do we prepare a solution of HCOONa? Titration of HCOOH with NaOH according to the following reaction Dissolution of the salt of a weak acid/strong base produces a basic solution (pH > 7.00). HCOOH (aq) + NaOH (aq)  HCOONa (aq) + H 2 O (l)    Weak Acid Strong Base Basic Salt

54. The Strong Acid/Weak Base Case How do we prepare a solution of NH 4 Cl? Titration of HCl with NH 3 according to the following reaction Dissolution of the salt of a strong acid/weak base produces a acidic solution (pH < 7.00). HCl (aq) + NH 3 (aq)  NH 4 Cl (aq)    Strong Weak Acidic Acid Base Salt

55. The Weak Acid/Weak Base Case What would be the pH of a solution of CH 3 COONH 4 ? Look at the following reactions NH 4 + (aq) ⇌ NH 3 (aq) + H + (aq) K = K a (NH 4 + ) CH 3 COO - (aq) + H 2 O (l) ⇌ CH 3 COOH (aq) + OH - (aq) K = K b (CH 3 COO - )

56. The pH of a solution the salt of a weak acid/weak base depends on the magnitudes of the equilibrium constants.