1. ACIDS AND BASES ACID – A compound that produces hydrogen ions in a water solution HCl (g) → H + (aq) + Cl - (aq) BASE – A compound that produces hydroxide ions in a water solution NaOH (s) → Na + (aq) + OH - (aq) 1884 SVANTE ARRHENIUS Proposed the first definitions of acids and bases 1E-1 (of 25)

2. FORMATION OF ACIDS AND BASES Bases are produced from the reaction of metal oxides with water Na 2 O (s) + H 2 O (l) → 2NaOH (aq) Acids are produced from the reaction of nonmetal oxides with water CO 2 (g) + H 2 O (l) → H 2 CO 3 (aq) 1E-2 (of 25)

3. Na 2 OMgOAl 2 O 3 SiO 2 P 2 O 5 SO 3 Cl 2 O 7 NaOHMg(OH) 2 Al(OH) 3 Si(OH) 4 PO(OH) 3 SO 2 (OH) 2 ClO 3 (OH) H 4 SiO 4 H 3 PO 4 H 2 SO 4 HClO 4 AMPHOTERIC – A substance that can act as an acid or a base base base or acid acid strongest base strongest acid Products of elemental oxides with water: 1E-3 (of 25)

4. base Products of elemental oxides with water: 1+2+ 3+ 4+5+6+7+ The higher the element’s oxidation number, the more acidic its hydroxide Cr(OH) 2 Cr(OH) 3 Cr(OH) 6 amphoteric acid 1E-4 (of 25) Na 2 OMgOAl 2 O 3 SiO 2 P 2 O 5 SO 3 Cl 2 O 7 NaOHMg(OH) 2 Al(OH) 3 Si(OH) 4 PO(OH) 3 SO 2 (OH) 2 ClO 3 (OH) H 4 SiO 4 H 3 PO 4 H 2 SO 4 HClO 4 base base or acid acid

5. 1923 Expanded the definitions of acids and bases THOMAS LOWRY 1E-5 (of 25)

6. ACID – A hydrogen ion (or proton) donor BASE – A hydrogen ion (or proton) acceptor 1923 Expanded the definitions of acids and bases JOHANNES BRØNSTED 1E-6 (of 25)

7. HCl + H 2 O → 1E-7 (of 25)

8. HCl + H 2 O →Cl - + H 3 O + acidbase HYDRONIUM ION – H 3 O +, formed when a hydrogen ion attaches to a water +- 1E-8 (of 25)

9. NH 3 + H 2 O → 1E-9 (of 25)

10. NH 4 + + OH - NH 3 + H 2 O → baseacid Water is amphoteric +- 1E-10 (of 25)

11. Acids turn into bases, and bases turn into acids HCl + H 2 O Cl - + H 3 O + acidbase → 1E-11 (of 25)

12. conjugate base of HCl conjugate acid of H 2 O Acids turn into bases, and bases turn into acids HCl + H 2 O Cl - + H 3 O + acidbase ← NH 3 + H 2 O → NH 4 + + OH - baseacid conjugate acid of NH 3 conjugate base of H 2 O 1E-12 (of 25)

13. conjugate base of NH 4 + conjugate acid of NO 2 - NH 4 + + NO 2 - → NH 3 + HNO 2 acidbase ← The favored reaction direction turns strong acids and bases into weak acids and bases HNO 2 is a stronger acid than NH 4 + NH 3 is a stronger base than NO 2 - 1E-13 (of 25)

14. AUTOIONIZATION OF WATER Water ionizes itself to a small extent 1E-14 (of 25)

15. AUTOIONIZATION OF WATER Water ionizes itself to a small extent 2H 2 O (l) → H 3 O + (aq) + OH - (aq) - + Makes solutions acidicMakes solutions basic Equal amounts of H 3 O + and OH - make a solution NEUTRAL  pure water is neutral 1E-15 (of 25)

16. The ionization of water is an endothermic process 2H 2 O (l) ⇆ H 3 O + (aq) + OH - (aq) Write the equilibrium constant expression for the autoionization of water K eq = [H 3 O + ][OH - ] ION-PRODUCT CONSTANT FOR WATER (K w ) – The equilibrium constant for the ionization of water 20 25 30 0.68 x 10 -14 1.00 x 10 -14 1.47 x 10 -14 Temp. (°C)K w energy + 1E-16 (of 25)

17. Find [H 3 O + ] and [OH - ] in pure water at 25°C. xx 2H 2 O (l) ⇆ H 3 O + (aq) + OH - (aq) Initial M’s Change in M’s Equilibrium M’s 00 + x K w = [H 3 O + ][OH - ] 1.00 x 10 -14 M 2 = x 2 1.00 x 10 -7 M = x= [H 3 O + ] = [OH - ] 1E-17 (of 25)

18. THE pH SCALE pH – The negative logarithm of [H 3 O + ] of a solution The common logarithm of a number is: the exponent to which 10 must be raised to equal the number 1000.010.001 100 = 10 2 0.01 = 10 -2 0.001 = 10 -3 0.002 0.002 = 10 -2.699 log 100 = 2log 0.01 = -2log 0.001 = -3log 0.002 = -2.699 pOH – The negative logarithm of [OH - ] of a solution 1E-18 (of 25)

19. Calculate the pH of orange juice if its [H 3 O + ] = 2.5 x 10 -4 M. = -log(2.5 x 10 -4 M) pH= -log[H 3 O + ] = -(0.40 + -4.000000…) 1E-19 (of 25) = 3.60 For logarithmic numbers, only the digits after the decimal point are significant figures, not the digits before = 3.6incorrect number of significant figures = -[log(2.5) + log(10 -4 )]

20. Calculate the pH and pOH of pure water at 25ºC. = -log(1.00 x 10 -7 M) pH= -log[H 3 O + ] For pure water: [H 3 O + ] = 1.00 x 10 -7 M = 7.000 = -log(1.00 x 10 -7 M) pOH= -log[OH - ] [OH - ] = 1.00 x 10 -7 M = 7.000 1E-20 (of 25)

21. In any water solution: K w =[H 3 O + ][OH - ] K w =[OH - ] ________ [H 3 O + ] at 25°C 1.00 x 10 -14 =[OH - ] _____________ [H 3 O + ] 1E-21 (of 25) For orange juice: [H 3 O + ] = 2.5 x 10 -4 M 1.00 x 10 -14 M 2 = [OH - ] _________________ 2.5 x 10 -4 M = 4.0 x 10 -11 M

22. pH 7 = Neutral < 7 is Acidic > 7 is Basic 78910111213146503421 [H 3 O + ] = 10 -7 M [OH - ] = 10 -7 M [H 3 O + ] = 10 -3 M [OH - ] = 10 -11 M [H 3 O + ] = 10 -12 M [OH - ] = 10 -2 M 15 [H 3 O + ] = 10 1 M [OH - ] = 10 -15 M 1E-22 (of 25)

23. In any water solution: K w =[H 3 O + ][OH - ] log K w =log ([H 3 O + ][OH - ]) -log K w =-log ([H 3 O + ][OH - ]) -log K w =- log[H 3 O + ] - log[OH - ] pK w =pH + pOH at 25°C, -log(1.00 x 10 -14 ) = 14.000 ∴ 14.000= pH + pOH 1E-23 (of 25)

24. Calculate the pH and pOH of soda if its [H 3 O + ] = 1.6 x 10 -3 M. = -log(1.6 x 10 -3 M) pH= -log[H 3 O + ] [H 3 O + ] = 1.6 x 10 -3 M = 2.80 = 14.000 - 2.80 pOH= pK w - pH pH + pOH = pK w = 11.20 1E-24 (of 25)

25. Calculate the [H 3 O + ] of blood, which has a pH of 7.4. = 0.0000000398 M pH= -log[H 3 O + ] -pH= log[H 3 O + ] antilog (-pH)= [H 3 O + ] antilog (-7.4)= [H 3 O + ] For logarithmic numbers, only the digits after the decimal point are significant figures, not the digits before = 3.98 x 10 -8 M = 4 x 10 -8 M 1E-25 (of 25)

27. IONIZATION OF ACIDS HF (aq) + H 2 O(l) ⇆ H 3 O + (aq) + F - (aq) Write the K eq expression for the ionization of hydrofluoric acid = [H 3 O + ][F - ] ____________ [HF] ACID IONIZATION CONSTANT (K a ) – The equilibrium constant for an acid ionizing K eq 1F-1 (of 19)

28. IONIZATION OF ACIDS HF (aq) + H 2 O(l) ⇆ H 3 O + (aq) + F - (aq) Write the K eq expression for the ionization of hydrofluoric acid = [H 3 O + ][F - ] ____________ [HF] ACID IONIZATION CONSTANT (K a ) – The equilibrium constant for an acid ionizing Ka Ka 1F-2 (of 19)

29. Acids are a) strong if every acid molecule gives up a hydrogen ion HCl, HBr, HI, any acid with 2 or more O’s than H’s K a = large b)weak if less than every acid molecule gives up a hydrogen ion all other acids K a = small 1F-3 (of 19) ()()

30. IONIZATION OF BASES NH 3 (aq) + H 2 O(l) ⇆ NH 4 + (aq) + OH - (aq) Write the K eq expression for the ionization of ammonia = [NH 4 + ][OH - ] ______________ [NH 3 ] BASE DISSOCIATION CONSTANT (K b ) – The equilibrium constant for a base ionizing K eq 1F-4 (of 19)

31. IONIZATION OF BASES NH 3 (aq) + H 2 O(l) ⇆ NH 4 + (aq) + OH - (aq) Write the K eq expression for the ionization of ammonia = [NH 4 + ][OH - ] ______________ [NH 3 ] BASE DISSOCIATION CONSTANT (K b ) – The equilibrium constant for a base ionizing Kb Kb 1F-5 (of 19)

32. Bases are a) strong if every molecule/ion accepts a hydrogen ion alkali metal hydroxides, dilute Ca(OH) 2, Sr(OH) 2, Ba(OH) 2 solutions K b = large b)weak if less than every molecule/ion accepts a hydrogen ion all other bases, including NH 3 K b = small 1F-6 (of 19)

33. Calculate the pH of 0.010 M nitric acid. 1F-7 (of 19) CALCULATING THE pH OF STRONG ACID OR BASE SOLUTIONS Assume complete ionization or dissociation 0.010 HNO 3 (aq) + H 2 O (l) → H 3 O + (aq) + NO 3 - (aq) Initial M’s Change in M’s Final M’s 0.010 00 - x+ x 0 = -log(0.010 M) pH= -log[H 3 O + ] = 2.00 - 0.010+ 0.010

34. Calculate the pH of 0.0010 M sodium hydroxide. 1F-8 (of 19) 0.0010 NaOH (aq) → Na + (aq) + OH - (aq) Initial M’s Change in M’s Final M’s 0.0010 00 - x+ x 0 = -log(0.0010 M) pOH= -log[OH - ] = 3.00= 14.000 - 3.00 pH= pK w - pOH pH + pOH = pK w = 11.00 - 0.0010+ 0.0010

35. the K a for HNO 2 can be looked up Calculate the pH of 0.010 M nitrous acid. 1F-9 (of 19) CALCULATING THE pH OF WEAK ACID OR BASE SOLUTIONS Assume INCOMPLETE ionization or dissociation, reaching a state of equilibrium xx HNO 2 (aq) + H 2 O (l) ⇆ H 3 O + (aq) + NO 2 - (aq) Initial M’s Change in M’s Equilibrium M’s 0.010 00 - x + x 0.010 - x : 4.0 x 10 -4 K a = [H 3 O + ][NO 2 - ] ________________ [HNO 2 ] K a = x 2 _____________ (0.010 – x)

36. 4.0 x 10 -4 = x 2 ______________ (0.010 – x) If the K a is < 10 -2 the acid does not ionize much, so you may assume that x is very small, and ignore it when it is subtracted from the initial molarity 4.0 x 10 -4 = x 2 _______ 0.010 2.00 x 10 -3 = x For x values that are >1% of the original acid molarity, they should be put back in place of x in the denominator of the K a expression, and solved again 1F-10 (of 19)

37. 4.0 x 10 -4 = x 2 _________________________ (0.010 – 2.00 x 10 -3 ) 1.79 x 10 -3 = x Repeat until the answer (to 2 sig fig’s) is the same twice in a row 4.0 x 10 -4 = x 2 _________________________ (0.010 – 1.79 x 10 -3 ) 1.81 x 10 -3 = x = -log(1.81 x 10 -3 M)pH= -log[H 3 O + ]= 2.74 = [H 3 O + ] 1F-11 (of 19)

38. Calculate the pH of 0.20 M acetic acid if its K a = 1.8 x 10 -5. xx HC 2 H 3 O 2 (aq) + H 2 O (l) ⇆ H 3 O + (aq) + C 2 H 3 O 2 - (aq) Initial M’s Change in M’s Equilibrium M’s 0.20 00 - x + x 0.20 - x K a = [H 3 O + ][C 2 H 3 O 2 - ] ____________________ [HC 2 H 3 O 2 ] 1.8 x 10 -5 = x 2 ____________ (0.20 – x) 1.8 x 10 -5 = x 2 ______ 0.20 1.90 x 10 -3 = x 1F-12 (of 19)

39. 1.8 x 10 -5 = x 2 ________________________ (0.20 – 1.90 x 10 -3 ) 1.89 x 10 -3 = x = -log(1.89 x 10 -3 M)pH = -log[H 3 O + ]= 2.72 Calculate the pH of 0.20 M acetic acid if its K a = 1.8 x 10 -5. = [H 3 O + ] Calculate the percent ionization of acetic acid. 1.89 x 10 -3 M _________________ 0.20 M x 100 = 0.95% 1F-13 (of 19)

40. Calculate the pH of a 0.10 M ammonia solution if its K b = 1.8 x 10 -5. xx NH 3 (aq) + H 2 O (l) ⇆ NH 4 + (aq) + OH - (aq) Initial M’s Change in M’s Equilibrium M’s 0.10 00 - x + x 0.10 - x K b = [NH 4 + ][OH - ] ______________ [NH 3 ] 1.8 x 10 -5 = x 2 ____________ (0.10 – x) 1.8 x 10 -5 = x 2 ______ 0.10 1.34 x 10 -3 = x 1F-14 (of 19)

41. 1.8 x 10 -5 = x 2 ________________________ (0.10 – 1.34 x 10 -3 ) 1.33 x 10 -3 = x = -log(1.33 x 10 -3 M)pOH= -log[OH - ]= 2.88 Calculate the pH of a 0.10 M ammonia solution if its K b = 1.8 x 10 -5. = [OH - ] = 14.000 - 2.88 pH= pK w - pOH pH + pOH = pK w = 11.12 1F-15 (of 19)

42. A 0.500 M solution of the weak acid HA has a pH 2.010. Calculate its K a. xx HA (aq) + H 2 O (l) ⇆ H 3 O + (aq) + A - (aq) Initial M’s Change in M’s Equilibrium M’s 0.500 00 - x + x 0.500 - x K a = [H 3 O + ][A - ] _____________ [HA] K a = x 2 ______________ (0.500 – x) x = [H 3 O + ] = antilog (-2.010) = antilog (-pH) = 0.009772 M 1F-16 (of 19) CALCULATING THE K a OF A WEAK ACID FROM pH

43. A 0.500 M solution of the weak acid HA has a pH 2.010. Calculate its K a. K a = (0.009772 M) 2 _____________________________ (0.500 M - 0.009772 M) = 1.95 x 10 -4 What is the pK a of HA? = -log(1.947 x 10 -4 ) pK a = -log K a = 3.710 The smaller the K a (or the bigger the pK a ) the weaker the acid 1F-17 (of 19) CALCULATING THE K a OF A WEAK ACID FROM pH

44. A 0.500 M solution of the weak acid HX is 3.15% ionized. Calculate its K a. xx HX (aq) + H 2 O (l) ⇆ H 3 O + (aq) + X - (aq) Initial M’s Change in M’s Equilibrium M’s 0.500 00 - x + x 0.500 - x K a = [H 3 O + ][X - ] _____________ [HX] K a = x 2 ______________ (0.500 – x) 3.15 = x (100) _______ 0.500 0.01575 = x 1F-18 (of 19) CALCULATING THE K a OF A WEAK ACID FROM PERCENT IONIZATION

45. A 0.500 M solution of the weak acid HX is 3.15% ionized. Calculate its K a. K a = (0.01575 M) 2 _____________________________ (0.500 M - 0.01575 M) = 5.12 x 10 -4 What is the pK a of HX? = -log(5.123 x 10 -4 ) pK a = -log K a = 3.291 1F-19 (of 19) CALCULATING THE K a OF A WEAK ACID FROM PERCENT IONIZATION

47. 1G-1 (of 13) POLYPROTIC ACIDS Polyprotic acids have K a values for the ionization of each H + H 2 CO 3 (aq) + H 2 O (l) ⇆ H 3 O + (aq) + HCO 3 - (aq) K a1 =[H 3 O + ][HCO 3 - ] __________________ [H 2 CO 3 ] K a1 =4.3 x 10 -7 HCO 3 - (aq) + H 2 O (l) ⇆ H 3 O + (aq) + CO 3 2- (aq) K a2 =[H 3 O + ][CO 3 2- ] _________________ [HCO 3 - ] K a2 =5.6 x 10 -11 Successive H + ’s are harder to remove ∴ H 2 CO 3 is a stronger acid than HCO 3 -

48. H 2 CO 3 (aq) + H 2 O (l) ⇆ H 3 O + (aq) + HCO 3 - (aq) K a1 =4.3 x 10 -7 HCO 3 - (aq) + H 2 O (l) ⇆ H 3 O + (aq) + CO 3 2- (aq) K a2 =5.6 x 10 -11 H 2 CO 3 (aq) + 2H 2 O (l) ⇆ 2H 3 O + (aq) + CO 3 2- (aq) K a1,2 =? K a1,2 = (4.3 x 10 -7 )(5.6 x 10 -11 )= 2.4 x 10 -17 1G-2 (of 13) [H 3 O + ][HCO 3 - ] __________________ [H 2 CO 3 ] x [H 3 O + ][CO 3 2- ] _________________ [HCO 3 - ] [H 3 O + ] 2 [CO 3 2- ] __________________ [H 2 CO 3 ] = K a1 K a2 = K a1,2

49. Find the concentrations of each species in a 0.10 M H 2 CO 3 solution. xx H 2 CO 3 (aq) + H 2 O (l) ⇆ H 3 O + (aq) + HCO 3 - (aq) Initial M’s Change in M’s Equilibrium M’s 0.10 00 - x + x 0.10 - x K a1 = [H 3 O + ][HCO 3 - ] _________________ [H 2 CO 3 ] 4.3 x 10 -7 = x 2 ____________ (0.10 – x) x = 2.07 x 10 -4 [H 2 CO 3 ]=0.10 M – 0.000207 M= 0.10 M [H 3 O + ]= = 0.00021 M [HCO 3 - ]== 0.00021 M 1G-3 (of 13)

50. Find the concentrations of each species in a 0.10 M H 2 CO 3 solution. 0.000207 + yy HCO 3 - (aq) + H 2 O (l) ⇆ H 3 O + (aq) + CO 3 2- (aq) Initial M’s Change in M’s Equilibrium M’s 0.000207 0 - y + y 0.000207 - y K a2 = [H 3 O + ][CO 3 2- ] _________________ [HCO 3 - ] 5.6 x 10 -11 = (0.000207 + y) y __________________ (0.000207 – y) y = 5.6 x 10 -11 [H 2 CO 3 ]== 0.10 M [H 3 O + ]= 0.000207 + 5.6 x 10 -11 = 0.00021 M [HCO 3 - ]=0.000207 – 5.6 x 10 -11 = 0.00021 M [CO 3 2- ]= = 5.6 x 10 -11 M 1G-4 (of 13)

51. pH OF SALT SOLUTIONS Acid HNO 3 HNO 2 Strong acids have non conjugate bases Weak acids have weak conjugate bases Strong Weak Conjugate Base NO 3 - NO 2 - Non Weak Base NH 3 NaOH NaOH(H 2 O) 5 Strong bases have non conjugate acids Weak bases have weak conjugate acids Weak Strong Conjugate Acid NH 4 + Na + Na(H 2 O) 6 + Weak Non 1G-5 (of 13)

52. from LiOH which is a strong base ∴ Li + is a non acid from H 3 PO 4 which is a weak acid ∴ PO 4 3- is a weak base Li 3 PO 4 ∴ pH > 7 1G-6 (of 13) Li + PO 4 3- PO 4 3- (aq) + H 2 O (l) ⇆ HPO 4 - (aq) + OH - (aq) HYDROLYSIS – The reaction of a dissolved ion with water

53. from NH 3 which is a weak base ∴ NH 4 + is a weak acid from HCl which is a strong acid ∴ Cl - is a non base NH 4 Cl ∴ pH < 7 1G-7 (of 13) NH 4 + Cl - NH 4 + (aq) + H 2 O (l) ⇆ NH 3 (aq) + H 3 O + (aq)

54. from KOH which is a strong base ∴ K + is a non acid from HNO 3 Which is a strong acid ∴ NO 3 - is a non base KNO 3 ∴ pH = 7 1G-8 (of 13) K+K+ NO 3 -

55. Find the pH of a 0.10 M potassium acetate solution x x C 2 H 3 O 2 - (aq) + H 2 O (l) ↔ HC 2 H 3 O 2 (aq) + OH - (aq) Initial M’s Change in M’s Equilibrium M’s 0.10 0 0 - x + x 0.10 - x K b = [HC 2 H 3 O 2 ][OH - ] ___________________ [C 2 H 3 O 2 - ] 5.6 x 10 -10 = x 2 ____________ (0.10 – x) x = 7.48 x 10 -6 Potassium ion is a non acid; acetate ion is a weak base Need the K b for acetate: 5.6 x 10 -10 1G-9 (of 13) CALCULATING THE pH OF A SALT SOLUTION

56. 7.48 x 10 -6 = x = -log(7.48 x 10 -6 M)pOH= -log[OH - ]= 5.13 = [OH - ] = 14.000 – 5.13pH= pK w - pOH pH + pOH = pK w = 8.87 Find the pH of a 0.10 M potassium acetate solution 1G-10 (of 13) CALCULATING THE pH OF A SALT SOLUTION

57. 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 C 2 H 3 O 2 - (aq) + H 2 O (l) ⇆ HC 2 H 3 O 2 (aq) + OH - (aq) K b for acetate 2H 2 O (l) ⇆ H 3 O + (aq) + OH - (aq) KwKw K a K b = K w 1G-11 (of 13)

58. Find the pH of a 0.25 M ammonium chloride solution if the K b for ammonia is 1.8 x 10 -5. Ammonium ion is a weak acid; chloride ion is a non base Need the K a for ammonium Have the K b for ammonia : 1.8 x 10 -5 K a = K w ____ K b K a K b = K w = 1.00 x 10 -14 _______________ 1.8 x 10 -5 = 5.56 x 10 -10 1G-12 (of 13)

59. Find the pH of a 0.25 M ammonium chloride solution if the K b for ammonia is 1.8 x 10 -5. xx NH 4 + (aq) + H 2 O (l) ⇆ H 3 O + (aq) + NH 3 (aq) Initial M’s Change in M’s Equilibrium M’s 0.25 00 - x + x 0.25 - x K a = [H 3 O + ][NH 3 ] ______________ [NH 4 + ] 5.56 x 10 -10 = x 2 ____________ (0.25 – x) = -log(1.18 x 10 -5 M) pH = -log[H 3 O + ]= 4.93 x = 1.18 x 10 -5 M 1G-13 (of 13)