2. Aqueous solutions can be classified as acidic, basic, or neutral. This classification scheme is based on the quantities of 2 ions, hydronium ion, H 3 0 1+ and hydroxide ion, OH 1-. Where do these ions come from in solutions of pure water? Water molecules in motion will randomly collide with one another. When this happens occasionally a hydrogen nucleus from one molecule will be transferred from one molecule to the other. This can be illustrated.

3. Notice the nucleus of one hydrogen atom, a proton, was transferred, but the electron pair was left behind. This produces the H 3 0 1+ ion (hydronium) and the OH 1- ion (hydroxide)

4. the equilibrium expression is H 2 O + H 2 0 H 3 O 1+ (aq) + OH 1- (aq) Hydronium ion (H 3 O 1+ ) Hydroxide ion (OH 1- ) H 2 0 H 1+ (aq) + OH 1- (aq) Which is usually shortened to: water ionizing

5. The equilibrium constant for H 2 0 H 1+ (aq) + OH 1- (aq) is Ke = [H 1+ ][OH 1- ] [H 2 O] [H 2 O] in mol/L is mass of 1 L of water is 1000 g n = (1000g)/(18g/mol) =55.6 molL -1

6. Ke = [H 1+ ][OH 1- ] [H 2 O] To simplify this equation Here is a new constant [H 2 O]Ke = Kw the ion product constant of water K w = [H 1+ ][OH 1- ] = 1.0 x 10 -14 @ 25 o C at higher temperatures K w increases

7. K w = [H 1+ ][OH 1- ] = 1.0 x 10 -14 @ 25 o C if [H 1+ ]=[OH 1- ] then [H 1+ ]=[OH 1- ]= 1.0 x 10 -7 which is the situation for neutral water. A scale which simplifies this very small number is the pH (potency of H 1+ ) scale. It is based on powers of ten. Simply take the exponent from 10 -7 and then multiply it by -1. This produces the number 7.

8. In mathematical terms pOH = -log[OH 1- ] so if in an aqueous solution the [OH 1- ] = 2.4 x 10 -4, the pOH is 3.62 If the pOH = 4.56 then the [OH 1- ] = 2.8 x 10 -5 This is calculated using the equation [OH 1- ] = 10 -pOH, similarly [H 1+ ]= 10 -pH.

9. In mathematical terms pH = -log[H 1+ ] so if in an aqueous solution the [H 1+ ] = 2.4 x 10 -8, the pH is 7.62 Remember the whole number portion of a pH doesn’t count as a significant digit (SD), just like in the number 2.4 x 10 -8 the exponent -8 doesn’t count as a SD.

10. If the [H 1+ ][OH 1- ] = 1.0 x 10 -14 then taking the log of each side pH + pOH = 14 so if the pH of a solution is 2.3 the pOH is 11.7 This means for these 4 values [H 1+ ], [OH 1- ], pH, pOH if one of them is given the other 3 can be calculated. Here is an example

11. If the pH of a solution is 1.45 find the [H 1+ ], [OH 1- ], and the pOH. Remember here are the equations needed. 1.pH = -log[H 1+ ] 2.pOH = -log[OH 1- ] 3.[OH 1- ] = 10 -pOH 4.[H 1+ ] = 10 -pH. 5.[H 1+ ][OH 1- ] = 1.0 x 10 -14 6.pH + pOH = 14 If pH = 1.45 6. pOH = 14 - pH pOH = 12.55 3. [OH 1- ] = 10 -12.55 [OH 1- ] = 2.8 x 10 -13 M 4. [H 1+ ] = 10 -1.45 [H 1+ ] = 0.035 M

12. Take note the equations can be applied in varying orders and there are choices as to exactly which equations are used. These decisions are yours to make.

13. If the [OH 1- ] of a solution is 9.4 x 10 -9 M find the [H 1+ ], pH, and the pOH. 1.pH = -log[H 1+ ] 2.pOH = -log[OH 1- ] 3.[OH 1- ] = 10 -pOH 4.[H 1+ ] = 10 -pH. 5.[H 1+ ][OH 1- ] = 1.0 x 10 -14 6.pH + pOH = 14 If [OH 1- ] = 9.4 x 10 -9 2. pOH = -log 9.4 x 10 -9 pOH = 8.03 6. pH = 14 - 8.03 = 5.97 4. [H 1+ ] = 10 -5.97 = 1.1 x 10 -6 M

14. If the [H 1+ ] of a solution is 6.2 x 10 -2 M find the [OH 1- ], pH, and the pOH. 1.pH = -log[H 1+ ] 2.pOH = -log[OH 1- ] 3.[OH 1- ] = 10 -pOH 4.[H 1+ ] = 10 -pH. 5.[H 1+ ][OH 1- ] = 1.0 x 10 -14 6.pH + pOH = 14 If [H 1+ ] = 6.2 x 10 -2 M 1. pH = -log 6.2 x 10 -2 pH = 1.21 6. pOH = 14 - 1.21 pOH = 12.79 3.[OH 1- ] = 10 -12.79 [OH 1- ] = 1.6 x 10 -13 M

15. Increasing basicity Increasing acidity -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 NeutralNeutral Since this is a logarithmic scale pH 9 is 10x’s more basic than pH 8, pH 12 is 1000 x’s more basic than pH 9 How much more acidic is pH 1 than pH 5? 10 000 x’s What pH is 1000x’s more acidic than pH 2? pH of -1

16. Conceptual Definitions of Acids and Bases Arrhenius’s Concept - acids are substances which react in water and produce hydronium ions. HCl (g) + H 2 0 -------> H 3 0 1+ (aq) + Cl 1- (aq) Bases are substances which react with water and produce hydroxide ions. NH 3 (g) + H 2 0 ------> NH 4 1+ (aq) + OH 1- (aq)

17. This concept has its limitations however. Can’t substances be classified as acids or bases without the involvement of water?

18. Bronstead’s Definition of Acids and Bases Acids are substances which donate protons and bases are substances which accept protons. In the examples above HCl(g) is an acid because it donates protons to H 2 O molecules and NH 3 is a base because it accepts protons from H 2 O molecules.

19. Strength of Acids and Bases is determined by the degree to which a substance produces ions in solution. A strong acid or base is a substance which completely ionizes. In other words if 100 molecules of a strong acid like HCl are placed in water all 100 of them will react with H 2 O producing 100 H 3 O 1+ ions and 100 Cl 1- ions. Weak acids and bases only partially ionize. Strong Acid - the reaction below goes to completion. HCl (g) + H 2 0 --------> H 3 0 1+ (aq) + Cl 1- (aq)

20. Weak Acid - the reaction occurs to a limited extent. In the example below if 100 acetic acid molecules are placed in water only a few of them will successfully react with water molecules producing hydronium ions. Most CH 3 COOH molecules remain intact. CH 3 COOH + H 2 0 H 3 0 1+ (aq) + CH 3 COO 1- (aq)

21. Strong Acids in order of decreasing strength are HClO 4, HI, HBr, H 2 SO 4, HCl, HNO 3 A table with the remaining moderate and weak acids can be found on page 803. Acid strength has to do with the ease with which an acid can lose a proton. If the binary acid strengths (HI, HBr, HCl) are compared it can be seen that HI is the strongest acid of this group because its iodide ion is the largest of the group so the force between the hydrogen ion and the iodide ion is the weakest so it loses its proton most easily.

22. I 1- Cl 1- Br 1- H 1+ Force is strongest since the ions are closest Force is weakest since the ions are furthest Remember the weaker the force the stronger the acid

23. Strong Bases include hydroxides of group 1A and Ca 2+, Ba 2+, and Sr 2+. A table with the remaining moderate and weak bases can be found on page 803. As with acids the weaker the bonds, the stronger the base since liberation of OH 1- ions is easiest when the bonds are weakest.

24. Polyprotic Acids donate protons in steps. For instance carbonic acid, H 2 CO 3 has two protons to donate and it does this in two steps: step 1 H 2 CO 3 + H 2 0HCO 3 1- + H 3 0 1+ step 2 HCO 3 1- + H 2 0CO 3 2- + H 3 0 1+ note: The arrows are constructed in this manner to show the reverse reaction has a greater tendency than the forward reaction.

25. Conjugate Acid - Base Pairs - When using the Bronsted concept for acids and bases it is convenient to consider all acid - base reactions as reversible equilibria. For instance when sulfurous acid, H 2 SO 3 reacts with water the following equilibrium is established: H 2 SO 3 + H 2 OH 3 0 1+ + HSO 3 1- conjugate pair acid baseacid base

26. In the forward direction the H 2 SO 3 is the proton donor so it’s the acid and the H 2 O is the proton acceptor so it’s the base. In the reverse direction the H 3 0 1+ is the proton donor so it’s the acid and the HSO 3 1- is the proton acceptor so it’s a base. H 2 SO 3 + H 2 OH 3 0 1+ + HSO 3 1- conjugate pair acid baseacid base

27. When looking at both forward and reverse reactions it is easy to pick out a pair of molecules which differ by a single proton (H atom without its electron). These pairs are called conjugate acid-base pairs. acid baseacid base H 2 SO 3 + H 2 OH 3 0 1+ + HSO 3 1- conjugate pair

28. Identify the acid base conjugate pairs in the equilibrium below: PO 4 3- (aq) + H 2 O (aq) HPO 4 2- (aq) + OH 1- (aq) acidbaseacidbase conjugate pair Show how HPO 4 -2 can act as both an acid and a base.

29. Amphoteric (Amphiprotic) Substances can behave as both acids or bases dependent on the circumstances. Water molecules, for instance, can sometimes accept protons and behave as bases or donate protons and behave as acids. HBr (g) + H 2 O H 3 0 1+ ( aq) + Br 1- (aq) base NH 3 (g) + H 2 O0H 1- (aq) + NH 4 1+ (aq) acid

30. Any substance which alters the pH of an aqueous solution is regarded as an acid or a base. Acids and Bases are classified into 2 categories, strong and weak. Strong Acids and Bases are completely ionized. This means in solution no molecules are present. All the molecules break apart (dissociate), to form ions. Weak acids and bases only partially dissociate and ionize and so are involved in equilibria.

31. H 1+ Cl 1- H 1+ Cl 1- H 1+ Cl 1- H 1+ Cl 1- H 1+ Cl 1- H 1+ Cl 1- H 1+ Cl 1- H 1+ Cl 1- H 1+ Cl 1- H 1+ Cl 1- H 1+ Cl 1- H 1+ F 1- H 1+ F 1- H 1+ F 1- H 1+ F 1- H 1+ F 1- H 1+ F 1- H 1+ F 1- H 1+ F 1- H 1+ F 1- H 1+ F 1- H 1+ F 1- H 1+ F 1- H 1+ F 1-

32. Is this a Strong or Weak Acid?

34. Find the pH and pOH of a 0.28 molL -1 solution of HClO 4. Since this is a strong acid the ionization is complete. This can be represented by … HClO 4 H 1+ (aq) + ClO 4 1- (aq) (molL -1 ) initial 0.280.0 Before any HClO 4 has dissolved -0.280.28 shift final 0.000.28

35. HClO 4 H 1+ (aq) + ClO 4 1- (aq) (molL -1 ) initial 0.280.0 -0.280.28 shift final 0.000.28 Remember pH = -log[H 1+ ] so pH = - log 0.28 pH = 0.55 pH + pOH = 14 so pOH = 14 - 0.55 pOH = 13.45

36. If the pOH of a solution of HCl is found to be 12.9 what is the concentration of this solution. Since this is a strong acid the ionization is complete. This can be represented by … HCl H 1+ (aq) + Cl 1- (aq) (molL -1 ) initial 0.080.0 -0.0790.079 shift final 00.079 In this case the pOH can be used to determine the final [H 1+ ], pH = 14 - 12.9, [H 1+ ] = 10 -1.1

37. Find the pH and pOH of a 0.12 molL -1 solution of Sr(OH) 2. Since this is a strong base the ionization is complete. This can be represented by … Sr(OH) 2 Sr 2+ (aq) + 2 OH 1- (aq) (molL -1 ) initial 0.120.0 -0.120.120.24 shift final 0.000.120.24 Remember pOH = -log[OH 1- ] so pOH = - log 0.24 = 0.62 pH = 14 - 0.62 = 13.38

38. Find the pH and pOH of a 1.68 molL -1 solution of HCN (Ka = 6.2 x 10 -10. Since this is a weak acid the ionization is incomplete. This can be represented by … HCNH 1+ (aq) + CN 1- (aq) (molL -1 ) initial 1.680.0 Before any HCN has dissociated - x x x shift final 1.68 - x x x

39. To determine the value of x the Ke for this reaction must be known. It is found in tables called Ka (equilibrium constants for acids). For HCN Ka = 6.2 x 10 -10. HCNH 1+ (aq) + CN 1- (aq) Ka = [H 1+ ][ CN 1- ] [ HCN ] = [x] 2 [1.68-x] Since the amount of acid which ionizes is very small this x is negligible 6.2 x 10 -10 = x 2 1.68 x = 3.2 x 10 -5, pH = 4.49, pOH = 9.51

40. What is the percentage ionization of the 1.68 mol/L HCN solution?

41. If an unknown 0.34 mol/L monoprotic acid is 1.3 x 10 -5 % ionized what is its Ka?

42. If the pH of a solution of HOCl is found to be 4.9 what is the concentration of this solution. Ka = 3.0 x 10 -8 Since this is a weak acid the ionization is incomplete. This can be represented by … HOCl H 1+ (aq) + OCl 1- (aq) (molL -1 ) initial x0.0 1.26 x 10 -5 shift final x - 1.26 x 10 -5 1.26 x 10 -5 In this case the pH can be used to determine the final [H 1+ ], [H 1+ ] = 10 -4.9 = 1.26 x 10 -5. -1.26 x 10 -5

43. Ka = [H 1+ ][ OCl 1- ] [ HOCl ] = [1.26 x 10 -5 ] 2 [x - 1.26 x 10 -5 ] 3.0 x 10 -8 = 1.58 x 10 -10 x HOCl H 1+ (aq) + OCl 1- (aq) (molL -1 ) initial x0.0 1.26 x 10 -5 shift final x - 1.26 x 10 -5 1.26 x 10 -5 -1.26 x 10 -5 This amount is negligible x = 5 x 10 -3 molL -1

44. Strong acids totally ionize because the force of attraction between the two ions is relatively weak and water molecules pulling on them can separate them. H 1+ NO 3 1 H 1+ This means NO 3 1- has a weak pull on H 1+ ions

45. Conversely weak acids have negative ions with relatively strong attractions for H 1+ ions NO 2 1- H 1+ These ions cannot be easily ripped apart Water molecules can’t do it If a salt like NaNO 2 is added to water it is easily pulled apart since sodium ions are quite large compared to H 1+ ions making the attraction between Na 1+ and NO 2 1- fairly weak.

46. Na 1+ NO 2 1- Na 1+ NO 2 1 H 1+ O 2- NO 2 1 H 1+ O 2- NO 2 1 H 1+ O 2- NO 2 1 H 1+ O 2- NO 2 1- H 1+ O 2- All negative ions except those found in strong acids, Cl 1-, Br 1-, I 1-, NO 3 1-, ClO 4 1-, HSO 4 1-, are capable of creating OH 1- ions, this means they act like bases because they raise pH

47. The equilibrium for NO 2 1- where it alters pH is NO 2 1- (aq) + H 2 0 HNO 2 (aq) + 0H 1- (aq) Show how K 2 CO 3 alters pH CO 3 2- (aq) + H 2 0 HCO 3 1- (aq) +0H 1- (aq) Show how NaF alters pH F 1- (aq) + H 2 0 HF (aq) +0H 1- (aq) Show how NaCl alters pH it doesn’t, Cl 1- has too weak a pull on H 1+ remember, Cl 1- is part of the strong acid HCl

48. What is the pH of a 0.24 molL -1 solution of potassium sulfite Kb of sulfite ion is 1.5 x 10 -7 ? When potassium sulfite is placed in water it dissociates into ions. This can be shown by: K 2 SO 3 (s) 2K 1+ (aq) + SO 3 2- (aq) Since SO 3 2- is not part of a strong acid it is capable of pulling apart (hydrolyzing) water molecules. This can be shown by: SO 3 2- (aq) + H 2 0HSO 3 1- (aq) + 0H 1- molL -1 initial 0.240.00 shift -x+x @E 0.24-x+x

49. SO 3 2- (aq) + H 2 0HSO 3 1- (aq) + 0H 1- molL -1 initial 0.240.00 shift -x+x @E 0.24-x+x The Kb of SO 3 2- is 1.5 x 10 -7. So to find x 1.5 x 10 -7 = x2x2 0.24 -x X is so small compared to 0.24 it can be regarded as negligible x =1.9 x 10 -4 = [OH 1- ] pOH = -log (1.9 x 10 -4 ) pOH = 3.72 pH = 14 - 3.72 = 10.28

50. Positive ions like NH 4 1+, can donate H 1+ ions to water molecules. This increases the amount of hydronium (H 3 0 1+ ) creating an acidic affect. This can be illustrated by the following:

51. H2OH2O NH 4 1+ H 3 O 1+ NH 3 NH 4 1+ + H 2 ONH 3 + H 3 O 1+

52. Multivalent metal ions with a large charge density can also donate H 1+. They do this by attracting water molecules. Once the water molecules are attached to these highly charged ions the electron pairs shared by the hydrogen and oxygen atoms are pulled toward the oxygen end of the water molecule making the H 1+ ions vulnerable to removal by other colliding water molecules. This can be illustrated by the following:

53. H HO Al 3+ H HO H HO H HO H HO H HO H HO H HO O H H H [Al (H 2 O) 6 ] 3+ + H 2 O [Al(H 2 O) 5 OH] 2+ +H 3 O 1+ Notice the electron pairs moving closer to the Al 3+ Only small trivalent ions and Be 2+ act in this way H HHO

54. If the pH of a solution of NH 4 Cl is found to be 5.23 what is the concentration of this solution Ka NH 4 1+ = 5.7 x 10 -10 ? Since the pH is below 7 this substance is acting like an acid. It is not among the strong acids so it must be a weak acid. Since it has no H 1+ to donate it must be acting like an acid by hydrolyzing water. Positive ions can behave this way. When NH 4 Cl is dissolved it dissociates into ions. This is shown by NH 4 Cl(s)NH 4 1+ (aq) + Cl 1- (aq) The NH 4 1+ acts like a weak acid so it must ….

55. (molL -1 ) initial x0.0 - 5.9 x 10 -6 5.9 x 10 -6 shift final x - 5.9 x 10 -6 5.9 x 10 -6 In this case the pH can be used to determine the final [H 1+ ], pH = 5.23, [H 1+ ] = 10 -5.23 = 5.9 x 10 -6 NH 4 1+ + H 2 ONH 3 + H 3 O 1+ Ka = [H 1+ ][ NH 3 ] [ NH 4 1+ ] = [5.9 x 10 -6 ] 2 [x - 5.9 x 10 -6 ] 5.7 x 10 -10 = 3.47 x 10 -11 x This amount is negligible x = 0.061 molL -1

56. There are experimental situations where real life situations are simulated. Living organisms have fluids which are maintained within very narrow pH ranges so these simulated experiments require solutions in which pH changes are minimal despite the introduction of acids and bases. Mixtures which resist changes in pH when acids or bases are added are called buffers. Buffers are equilibrium mixtures which shift when acids or bases are added.

57. Buffers contain 1.a weak acid and a soluble salt containing the conjugate base of the weak acid OR 2.A weak base and a soluble salt containing the conjugate acid of the weak base

58. If the weak acid is HF then a soluble salt containing its conjugate base partner is NaF. Remember that conjugate acid - base partnerships are different by a single H, bases gain H’s, acids lose H’s. If the weak acid is HC 2 H 3 O 2 then a soluble salt containing its conjugate base partner is KC 2 H 3 O 2. If the weak base is NH 3, then a soluble salt containing its conjugate acid partner is NH 4 Cl.

59. Let’s now look at how buffers work. If the buffer is HF and NaF the equilibrium mixture is HFHFH 1+ + F 1- If the buffer is HC 2 H 3 O 2 and NaC 2 H 3 O 2 the equilibrium mixture is HC2H3O2HC2H3O2 H 1+ + C 2 H 3 O 2 1- If the buffer is H 2 CO 3 and NaHCO 3 the equilibrium mixture is H 2 CO 3 H 1+ + HCO 3 1-

60. H 2 CO 3 H 1+ + HCO 3 1- Why do buffers require a salt? If you can recall the definition of a weak acid they only ionize a small amount, so there is only a small amount of HCO 3 1- present at equilibrium. If the equilibrium is going to resist a change in pH when acids or bases are added there must be sufficient quantities of substances on both sides of this equilibrium. When acids are added (a source of H 1+, the equilibrium shifts

61. H 2 CO 3 H 1+ + HCO 3 1- to get rid of it. If a base is added, the OH 1- produced reacts with the H 1+ forming water. This means the equilibrium shifts to replace the H 1+ removed by the addition of the OH 1-. A buffer’s capacity to resist large changes in pH is determined by the quantity of weak acid and conjugate base present. When either of these quantities is used up the solution can no longer act as a buffer.

62. Buffer Problems 1.1.0 L of a buffer is prepared by mixing acetic acid and sodium acetate. If both of these substances are 1.0 molL -1 at equilibrium find the pH if the Ka for acetic acid is 1.8 x 10 -5 ? Equilibrium is HC2H3O2HC2H3O2 H 1+ + C 2 H 3 O 2 1- @E1.0 x Ke = [H 1+ ] [C 2 H 3 O 2 1- ] [HC 2 H 3 O 2 ] 1.8 x 10 -5 = [x] [1.0] [1.0] pH = -log 1.8 x 10 -5 pH = 4.74

63. 1.If 1.0 mL of 2.0 M HCl is now added to this buffer what is the new pH? (assume no volume change of 1.0L.) HC2H3O2HC2H3O2 H 1+ + C 2 H 3 O 2 1- @E1.0 1.8 x 10 -5 upset 2.0 M x 0.0010L=0.0020 shift+0.0020 -0.0020 @E1.002 0.998x Assume all the HCl is consumed by the equilibrium shifting right to left

64. 1.If 5.0 mL of 2.0 M NaOH is now added to this buffer what is the new pH? (assume no volume change.) HC2H3O2HC2H3O2 H 1+ + C 2 H 3 O 2 1- @E1.0 1.8 x 10 -5 upset 2.0 M x 0.0050 L= - 0.010 shift-0.010 +0.0100 +0.010 @E0.99 1.010x Assume all the OH 1- is consumed by the H 1+ equilibrium shifts left to right to replace the H 1+

65. HC2H3O2HC2H3O2 H 1+ + C 2 H 3 O 2 1- @E0.99 1.01x Ke = [H 1+ ] [C 2 H 3 O 2 1- ] [HC 2 H 3 O 2 ] 1.8 x 10 -5 = [x] [1.01] [0.99] x = [1.8 x 10 -5 ] [1.01] [0.99] pH = -log x = 4.75