Ch14.1 Acids and Bases Acids – taste sour Arrhenius Theory of Acids – they produce hydrogen ions in aqueous solutions. Bases – Taste bitter, feel slippery.

Ch14.1 Acids and Bases Acids – taste sour Arrhenius Theory of Acids – they produce hydrogen ions in aqueous solutions. Bases – Taste bitter, feel slippery Arrhenius Theory of bases – they produce hydroxide ions in aqueous solutions.

Ch14.1 Acids and Bases Acids – taste sour Arrhenius Theory of Acids – they produce hydrogen ions in aqueous solutions. Bronsted-Lowry model – acid is a proton donor Bases – Taste bitter, feel slippery Arrhenius Theory of bases – they produce hydroxide ions in aqueous solutions. Bronsted-Lowry model – base is a proton acceptor Ex1) Id each BL acid and base: HCl + H 2 O  H 3 O + + Cl -

Ex1) Id each BL acid and base: HCl + H(OH)  H 3 O + + Cl - Acid Base Conjugate Conjugate AcidBase Acid Dissociation Constant K a =

Ex2) Write the simple dissociation for each acid: a. Acetic acid, HC 2 H 3 O 2 b. Ammonium ion, NH 4 + c. Anilinium ion, C 6 H 5 NH 3 + d. hydrated aluminum(III) ion, [Al(H 2 O) 6 ] +3

Ch14.2 Acid Strength HA + H 2 O H 3 O + + A - Acid Strength: StrongWeak 1. Look at K a largesmall 2. Position of far to rightfar to left dissociation 3. Conc of H + [H + ] ≈ [HA] [H + ] << [HA] compared to [HA] 4. Strength of [A - ] is weaker [A - ] is stronger conj basebase than H 2 O base than H 2 O Compared to H 2 O

Ch14.2 Acid Strength HA + H 2 O H 3 O + + A - Common strong acids: 1. Hydrochloric acid, HCl (monoprotic) 2. Nitric acid, HNO 3 3. Perchloric acid, HClO 4 4. Sulfuric acid, H 2 SO 4 (diprotic – 2 ionizable hydrogens) 1 st H is nearly 100% ionizable. 2 nd H has to be removed from the HSO 4 -. 1. Most acids are oxyacids – acidic proton attached to an oxygen. (Sometimes refered to as ternary acids.) HCl is a binary acid. 2. Organic acids – have carbon as the backbone, called carboxyl acids. Ex: acetic acid

Ex3) HCl + H 2 O H 3 O + + Cl - K a =

Ex4) Determine the B-L acids and bases and conjugates: HCl + HOH H 3 O + + Cl - NH 3 + HOH NH 4 + + OH -

Ex4) Determine the B-L acids and bases and conjugates: HCl + HOH H 3 O + + Cl - Acid BaseCA CB NH 3 + HOH NH 4 + + OH - Base AcidCA CB Substances, like water, are amphoteric – can act as acid or base. Water self-ionizes: H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq)

Water self-ionizes: H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) K =

Water self-ionizes: H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) K w = [H + ][OH - ] = 1x10 -14 No matter what an aqueous solution contains, K w always equals 1x10 -14 (at 25 o C, of course)

Ex6) Calc [H + ] and [OH - ] for each a. [OH - ] = 1x10 -5 b. [OH - ] = 1x10 -7 c. [H + ] = 10M

Ex7) At 60 o C, K w = 1x10 -13. Use LeChatlier’s principle to predict if its exothermic or endothermic. 2H 2 O (l) H 3 O + (aq) + OH - (aq) and calc [H + ] and [OH - ] Ch14 HW#1 p712 28,29,33,35,37,39

28. Write the dissociation reaction and the corresponding K a equilibrium expression for each of the following acids in water. a.) HC 2 H 3 O 2 b.) Co(H 2 O) 6 3+ c.) CH 3 NH 3 + How do you break this up? + C – H

Ch14 HW#1 p712 28,29,33,35,37,39 28. Write the dissociation reaction and the corresponding K a equilibrium expression for each of the following acids in water. a.) HC 2 H 3 O 2  H + + C 2 H 3 O 2 - b.) Co(H 2 O) 6 3+  Co +3 + 6H 2 O c.) CH 3 NH 3 +  How do you break this up? + C – H

Ch14 HW#1 p712 28,29,33,35,37,39 28. Write the dissociation reaction and the corresponding K a equilibrium expression for each of the following acids in water. a.) HC 2 H 3 O 2  H + + C 2 H 3 O 2 - b.) Co(H 2 O) 6 3+  Co +3 + 6H 2 O c.) CH 3 NH 3 +  CH 3 NH 2 + H + How do you break this up? + C – N

29. For each of the following aqueous reactions, identify the acid, the base, the conjugate base, and the conjugate acid. a. HF + H 2 O  F - + H 3 O + b. H 2 SO 4 + H 2 O  H 3 O + + HSO 4 - c. HSO 4 - + H 2 O  SO 4 2- + H 3 O +

29. For each of the following aqueous reactions, identify the acid, the base, the conjugate base, and the conjugate acid. a. HF + H 2 O  F - + H 3 O + Acid BaseCB CA b. H 2 SO 4 + H 2 O  H 3 O + + HSO 4 - Acid Base CA CB c. HSO 4 - + H 2 O  SO 4 2- + H 3 O + Base Acid CB CA

33. Use Table 14.2 to order the following from strongest to weakest acid. H 2 O, HNO 3, HOCl, NH 4 +

33. Use Table 14.2 to order the following from strongest to weakest acid. H 2 O, HNO 3, HOCl, NH 4 + HNO 3 : > 1 HOCl: 3.5x10 -8 NH 4 + : 5.6x10 -10 H 2 O: 1x10 -14

35. Use Table 14.2 a. Which is the stronger acid, HCl or H 2 O? b. Which is the stronger acid, H 2 O or HNO 2 ? c. Which is the stronger acid, HCN or HOC 6 H 5 ?

35. Use Table 14.2 a. Which is the stronger acid, HCl or H 2 O? HCl b. Which is the stronger acid, H 2 O or HNO 2 ? HNO 2 c. Which is the stronger acid, HCN or HOC 6 H 5 ? HCN (barely)

37. Calculate the [OH - ] of each of the following solutions at 25˚C. Identify each solution as neutral, acidic, or basic. [H + ]. [OH - ] = 1x10 -14 a. [H + ] = 1.0х10 -7 M [OH - ] = b. [H + ] = 6.7х10 -4 M[OH - ] = c. [H + ] = 1.9х10 -11 M[OH - ] = d. [H + ] = 2.3 M[OH - ] =

37. Calculate the [OH - ] of each of the following solutions at 25˚C. Identify each solution as neutral, acidic, or basic. [H + ]. [OH - ] = 1x10 -14 a. [H + ] = 1.0х10 -7 M [OH - ] = 1x10 -7 neutral b. [H + ] = 6.7х10 -4 M[OH - ] = 1.5x10 -11 acidic c. [H + ] = 1.9х10 -11 M[OH - ] = 5.3x10 -4 basic d. [H + ] = 2.3 M[OH - ] = 4.3x10 -15 very acidic

39. Values of K W as a function of temperature are as follows: a. Is the autoionization of water exothermic or endothermic? b. Calculate [H + ] and [OH - ] in a neutral solution at 50˚C. Temp ( 0 C)KwKw 01.14x10 -14 251.00x10 -14 352.09x10 -14 402.92x10 -14 505.47x10 -14

39. Values of K W as a function of temperature are as follows: a. Is the autoionization of water exothermic or endothermic? b. Calculate [H + ] and [OH - ] in a neutral solution at 50˚C. As temp increases, autoionization increases, therefore endothermic. [H + ]. [OH - ] = 5.47x10 -14 [H + ] = [OH - ] = 2.34x10 -7 Temp ( 0 C)KwKw 01.14x10 -14 251.00x10 -14 352.09x10 -14 402.92x10 -14 505.47x10 -14

Ch14.3 The pH Scale pH = –log[H + ] Ex1) What is the pH of a neutral solution?

pH = –log[H + ] Ex1) What is the pH of a neutral solution? [H + ] = 1.0x10 -7 pH = –log[1x10 -7 ] pH = 7.00 pOH = –log[OH – ] pH + pOH = 14

Ex2) pH of human blood is 7.41 at 25 o C. Calc pOH, [H + ], and [OH – ]

Ch14.4 pH of Strong Acids To calc the pH, ID the major species involved. Ex3) a. Calc the pH of 0.10M HNO 3. Major species: b. Calc the pH of 1.0x10 -10 M HCl.

Ch14.5 pH of Weak Acids Ex4) Find the pH of a 1.0M HF solution. K a = 7.2x10 -4

Ch14.5 pH of Weak Acids Ex4) Find the pH of a 1.0M HF solution. K a = 7.2x10 -4 HF + HOH  H 3 O + + F - Initial: Equilibrium: 1. Decide major species:

Ex4) Find the pH of a 1.0M HF solution. K a = 7.2x10 -4 HF + HOH  H + + F - Initial: Equilibrium: 1. Decide major species: HF and H 2 O. 2. Decide if the major species can contribute H ions: HF  H + + F - K a = 7.2x10 -4 H 2 O  H + + OH - K w = 1x10 -14

Ex4) Find the pH of a 1.0M HF solution. K a = 7.2x10 -4 HF + HOH  H + + F - Initial: Equilibrium: 1. Decide major species: HF and H 2 O. 2. Decide if the major species can contribute H ions: HF  H + + F - K a = 7.2x10 -4 H 2 O  H + + OH - K w = 1x10 -14 (Usually one will dominate the H + addition. In this case its HF. 3. Decide initial concentrations:

Ex4) Find the pH of a 1.0M HF solution. K a = 7.2x10 -4 HF + HOH  H + + F - Initial:[1.00] ‘excess’ 0 0 Equilibrium: 1. Decide major species: HF and H 2 O. 2. Decide if the major species can contribute H ions: HF  H + + F - K a = 7.2x10 -4 H 2 O  H + + OH - K w = 1x10 -14 (Usually one will dominate the H + addition. In this case its HF. 3. Decide initial concentrations: HF = 1.00M, F - = 0, H + = 10 -7 ~ 0. 4. Decide concentration at equilibrium.

Ex4) Find the pH of a 1.0M HF solution. K a = 7.2x10 -4 HF + HOH  H + + F - Initial:[1.00] ‘excess’ 0 0 Equilibrium: [1.00 –x] [x] [x] 1. Decide major species: HF and H 2 O. 2. Decide if the major species can contribute H ions: HF  H + + F - K a = 7.2x10 -4 H 2 O  H + + OH - K w = 1x10 -14 (Usually one will dominate the H + addition. In this case its HF. 3. Decide initial concentrations: HF = 1.00M, F - = 0, H + = 10 -7 ~ 0. 4. Decide concentration at equilibrium. HF = (1.00 – x), F - = (0 + x), H + = (0 + x) 5. Put into K a and solve.

HW#57. Using the K a values given in Table14.2, calculate the conc of all species present and the pH of a 1.5M HOCI soln. K a = 3.5x10 -8 Ch14 HW#2 p713 47,49,57,59,60

57. Using the K a values given in Table14.2, calculate the concentrations of all species present and the pH of a 1.5M HOCI solution. Given: K a = 3.5x10 -8 HOCl + HOH  H + + OCl - Initial:[1.5] ‘excess’ 0 0 Equilibrium: [1.5 –x] [x] [x] 1. Decide major species: HOCl and H 2 O. 2. Decide if the major species can contribute H ions: HOCl  H + + OCl - K a = 3.5x10 -8 Dominates H 2 O  H + + OH - K w = 1x10 -14 3. Decide initial concentrations: HOCl = 1.5M, OCl - = 0, H + = 10 -7 ~ 0. 4. Decide concentration at equilibrium. HOCl = (1.5 – x), OCl - = (0 + x), H + = (0 + x) 5. Put into K a and solve.

Ch14 HW#2 p713 47,49,57,59,60 47. What are the major species present in 0.250 M solutions of each of the following acids? Calculate the pH of each of these solutions. a. HCI + HOH  H + + Cl - pH = –log[H + ] Init: Eq: b. HBr (a strong acid)

Ch14 HW#2 p713 47,49,57,59,60 47. What are the major species present in 0.250 M solutions of each of the following acids? Calculate the pH of each of these solutions. a. HCI + HOH  H + + Cl - pH = –log[H + ] Init: [0.250] [0] [0] pH = –log[0.250] Eq: [0] [0.250] [0.250] pH = 0.602 b. HBr (a strong acid) same

49. Calculate the pH of each of the following solutions of a strong acid in water. a. 0.10M HCI pH = –log[H + ] b. 5.0M HCI c. 1.0x10 -10 M HCI

49. Calculate the pH of each of the following solutions of a strong acid in water. a. 0.10M HCI pH = –log[H + ] pH = 1 b. 5.0M HCI pH = 0.699 c. 1.0x10 -10 M HCI pH = 10

57. Using the K a values given in Table14.2, calculate the concentrations of all species present and the pH of a 1.5M HOCI solution. Given: K a = 3.5x10 -8 HOCl + HOH  H + + OCl - Initial:[1.5] ‘excess’ Equilibrium: 1. Decide major species: 2. Decide if the major species can contribute H ions: 3. Decide initial concentrations: 4. Decide concentration at equilibrium. 5. Put into K a and solve.

57. Using the K a values given in Table14.2, calculate the concentrations of all species present and the pH of a 1.5M HOCI solution. Given: K a = 3.5x10 -8 HOCl + HOH  H + + OCl - Initial:[1.5] ‘excess’ 0 0 Equilibrium: [1.5 –x] [x] [x] 1. Decide major species: HOCl and H 2 O. 2. Decide if the major species can contribute H ions: HOCl  H + + OCl - K a = 3.5x10 -8 Dominates H 2 O  H + + OH - K w = 1x10 -14 3. Decide initial concentrations: HOCl = 1.5M, OCl - = 0, H + = 10 -7 ~ 0. 4. Decide concentration at equilibrium. HOCl = (1.5 – x), OCl - = (0 + x), H + = (0 + x) 5. Put into K a and solve.

59. Calculate the concentration of all species present and the pH of a 0.020 M HF solution.

59. Calculate the concentration of all species present and the pH of a 0.020 M HF solution. K a = 7.2x10 -4 HF + HOH  H + + F - Initial:[0.20] Equilibrium:

59. Calculate the concentration of all species present and the pH of a 0.020 M HF solution. HF + HOH  H + + F - Initial:[0.20] ‘excess’ 0 0 Equilibrium: [0.20 –x] [x] [x] 1. Decide major species: HF and H 2 O. 2. Decide if the major species can contribute H ions: HF  H + + F - K a = 7.2x10 -4  Dominates H 2 O  H + + OH - K w = 1x10 -14 3. Decide initial concentrations: HF = 0.20M, F - = 0, H + = 10 -7 ~ 0. 4. Decide concentration at equilibrium. HF = (0.20 – x), F - = (0 + x), H + = (0 + x) 5. Put into K a and solve.

60. Calculate the pH of a 0.20 M iodic acid solution ( HIO 3, K a = 0.17 ). HIO 3 + HOH  H + + IO 3 - Initial:[0.20] Equilibrium:

60. Calculate the pH of a 0.20M solution iodic acid ( HIO 3, K a = 0.17 ). HIO 3 + HOH  H + + IO 3 - Initial:[0.20] ‘excess’ 0 0 Equilibrium: [0.20 –x] [x] [x] 1. Decide major species: HIO 3 and H 2 O. 2. Decide if the major species can contribute H ions: HIO 3  H + + IO 3 - K a = 1.7x10 -1  Dominates H 2 O  H + + OH - K w = 1x10 -14 3. Decide initial concentrations: HIO 3 - = 0.20M, IO 3 - = 0, H + = 10 -7 ~ 0. 4. Decide concentration at equilibrium. HIO 3 - = (0.20 – x), IO 3 - = (0 + x), H + = (0 + x) 5. Put into K a and solve.

Ch14.5B More pH Calcs Ex1) The hypochlorite ion ( OCl - ) is a strong oxidizing agent used in household bleaches, and forms in swimming pools treated with chlorine. It has a high affinity for protons and forms the weakly acidic hypochlorous acid ( HOCl, K a = 3.5x10 -8 ). Calc the pH of a 0.100M soln.

Ex1) ( HOCl, K a = 3.5x10 -8 ). Calc the pH of a 0.100M soln. 1. Major species: HOCl and HOH 2. H + contribution:

Ex1) ( HOCl, K a = 3.5x10 -8 ). Calc the pH of a 0.100M soln. 1. Major species: HOCl and HOH 2. H + contribution: HOCl  H + + OCl - K a = 3.5x10 -8 H 2 O  H + + OH - K w = 1x10 -14 3. Decide initial concentrations:

Ex1) ( HOCl, K a = 3.5x10 -8 ). Calc the pH of a 0.100M soln. 1. Major species: HOCl and HOH 2. H + contribution: HOCl  H + + OCl - K a = 3.5x10 -8 H 2 O  H + + OH - K w = 1x10 -14 3. Decide initial concentrations: (HOCl will dominate) HOCl  H + + OCl - Init:[0.100] 0 0 4. Concentration at equilibrium:

Ex1) ( HOCl, K a = 3.5x10 -8 ). Calc the pH of a 0.100M soln. 1. Major species: HOCl and HOH 2. H + contribution: HOCl  H + + OCl - K a = 3.5x10 -8 H 2 O  H + + OH - K w = 1x10 -14 3. Decide initial concentrations: (HOCl will dominate) HOCl  H + + OCl - Init:[0.100] 0 0 Eq:[0.100-x] [0+x] [0+x] 4. Concentration at equilibrium: 5. K a and solve:

Ex2) Calc the pH of a soln that contains 1.00M HCN ( K a = 6.2x10 -10 ) and 5.00M HNO 2 ( K a = 4.0x10 -4 ). Also calc the conc of cyanide ion, CN -, in this soln at eq. Ch14 HW#3 p713 61,63,64

61. Calculate the pH of a 0.50M solution of chlorous acid ( HCIO 2, K a = 1.2x10 -2 ) HClO 2 + HOH  H + + ClO 2 - Initial:[0.50] Equilibrium:

61. Calculate the pH of a 0.50M solution of chlorous acid ( HCIO 2, K a = 1.2x10 -2 ) HClO 2 + HOH  H + + ClO 2 - Initial:[0.50] ‘excess’ 0 0 Equilibrium: [0.50 –x] [x] [x] 1. Decide major species: HIO 3 and H 2 O. 2. Decide if the major species can contribute H ions: HClO 2  H + + IO 3 - K a = 1.2x10 -2  Dominates H 2 O  H + + OH - K w = 1x10 -14 3. Decide initial concentrations: HClO 2 = 0.50M, ClO 2 - = 0, H + = 10 -7 ~ 0. 4. Decide concentration at equilibrium. HClO 2 = (0.50 – x), ClO 2 - = (0 + x), H + = (0 + x) 5. Put into K a and solve.

63. Calculate the PH of a solution containing a mixture of 0.10 M HCl and 0.10 M HOCl. ( HCl, K a = > 1, HCIO 2, K a = 1.2x10 -2 ) HCl + HClO 2 + HOH  H + + Cl - + ClO 2 - Initial: [0.10] [0.10] ‘excess’ 0 0 0 Equilibrium: 1. Decide major species: HCl, HClO 2 and H 2 O. 2. Decide if the major species can contribute H ions: HCl  H + + CI - K a = HClO 2  H + + IO 3 - K a = H 2 O  H + + OH - K w = 3. Decide initial concentrations: HCl = 0.50M, Cl - = 0, H + = 10 -7 ~ 0. 4. Decide concentration at equilibrium. pH = –log[H + ]

63. Calculate the PH of a solution containing a mixture of 0.10 M HCl and 0.10 M HOCl. ( HCl, K a = > 1, HCIO 2, K a = 1.2x10 -2 ) HCl + HClO 2 + HOH  H + + Cl - + ClO 2 - Initial: [0.10] [0.10] ‘excess’ 0 0 0 Equilibrium: [0] [0.10] [0.10] 1. Decide major species: HCl, HClO 2 and H 2 O. 2. Decide if the major species can contribute H ions: HCl  H + + CI - K a > 1 Dominates HClO 2  H + + ClO 2 - K a = 1.2x10 -2 H 2 O  H + + OH - K w = 1x10 -14 3. Decide initial concentrations: HCl = 0.50M, Cl - = 0, H + = 10 -7 ~ 0. 4. Decide concentration at equilibrium. HCl = (0), Cl - = (0.10), H + = (0.10) pH = –log[H + ] pH = –log[0.10] = 1.0

64. (Mod) Calculate the PH of a solution containing 1.0M HF and 1.0 M HOC 6 H 5. ( HF, K a = 7.2x10 -4, HOC 6 H 5, K a = 1.6x10 -10 ) HF + HOC 6 H 5 + HOH  H + + F - + OC 6 H 5 - Initial: [1.0] [1.0] ‘excess’ 0 0 0 Eq: 1. Decide major species: HF, HOC 6 H 5 and H 2 O. 2. Decide if the major species can contribute H ions: HF  H + + F - K a = HOC 6 H 5  H + + OC 6 H 5 - K a = H 2 O  H + + OH - K w = 3. Decide initial concentrations: HF = 1.0M, F - = 0, H + = 10 -7 ~ 0. 4. Decide concentration at equilibrium. 5. K a and solve:

64. Calculate the PH of a solution containing 1.0M HF and 1.0 M HOC 6 H 5. ( HF, K a = 7.2x10 -4, HOC 6 H 5, K a = 1.6x10 -10 ) HF + HOC 6 H 5 + HOH  H + + F - + OC 6 H 5 - Initial: [1.0] [1.0] ‘excess’ 0 0 0 Eq: [1.0 – x] [x] [x] 1. Decide major species: HF, HOC 6 H 5 and H 2 O. 2. Decide if the major species can contribute H ions: HF  H + + F - K a = 7.2x10 -4 Dominates HOC 6 H 5  H + + OC 6 H 5 - K a = 1.6x10 -10 H 2 O  H + + OH - K w = 1x10 -14 3. Decide initial concentrations: HF = 1.0M, F - = 0, H + = 10 -7 ~ 0. 4. Decide concentration at equilibrium. HF = [1.0 – x], F - = (x), H + = (x) 5. K a and solve:

64. HF + HOC 6 H 5 + HOH  H + + F - + OC 6 H 5 - Eq: [1.0 – x] [x] [x] 2. Decide if the major species can contribute H ions: HF  H + + F - K a = 7.2x10 -4 Dominates HOC 6 H 5  H + + OC 6 H 5 - K a = 1.6x10 -10 H 2 O  H + + OH - K w = 1x10 -14 4. Decide concentration at equilibrium. HF = [1.0 – x], F - = (x), H + = (x) 5. K a and solve:

Ch14.5C Percent Dissociation Ex3) Calc the % dissoc of acetic acid ( K a = 1.8x10 -5 ) in each: a. 1.00M HC 2 H 3 O 2 b. 0.100M HC 2 H 3 O 2

Ex4) In a 0.100M soln of lactic acid ( HC 3 H 5 O 3 ), 3.7% dissociates. Calc K a. Ch14 HW#4 p713 65,69

65. Calculate the percent dissociation of the acid in each of the following solutions. a. 0.50 M acidic acid b. 0.050 M acidic acid c. 0.0050 M acidic acid d. Use the Le Chatelier’s principal to explain why percent increases as the concentration of a weak acid decreases. ( HC 2 H 3 O 2 K a = 1.8x10 -5 ) HC 2 H 3 O 2  H + + C 2 H 3 O 2 - K a = 1.8x10 -5 Initial: [0.50] 0 0 Eq: [0.50 – x] [x] [x]

Ch14 HW#4 p713 65,69 65. Calculate the percent dissociation a. 0.50 M acidic acid b. 0.050 M acidic acid c. 0.0050 M acidic acid d. Use the Le Chatelier’s principal to explain why percent increases HC 2 H 3 O 2  H + + C 2 H 3 O 2 - K a = 1.8x10 -5 Initial: [0.50] 0 0 [0.050] 0 0 [0.0050] 0 0

69. A 0.15 M solution of a weak acid is 3.0% dissociation calculate K a. HA  H + + A - Initial: [0.15] 0 0 Eq: [0.15 – x] [x] [x]

Ch14.6 Bases All group 1 hydroxides are strong bases: These group 2 hydroxides are strong bases: Ex1) Calc the pH of a 5.0x10 -2 M NaOH soln.

B (aq) + H 2 O (l) BH + (aq) + OH - (aq) K b =

Ex2) (AP ?) Give a detailed explanation of the basic properties of ammonia.

Ex3) Calc the pH of a 15.0M soln of NH 3. ( K b =1.8x10 -5 )

Ex4) Calc the pH of a 1.0M soln of methylamine, CH 3 NH 2. ( K b =4.38x10 -4 ) Ch14 HW#5 p714 75,77,79(a,c),81,85,89a

75. Write the reaction of the corresponding K b equilibrium expression for each of the following substances acting as bases in water. a. NH 3 b. C 5 H 5 N 77. Use table 14.3 to help order the following bases from strongest to weakest. NO 3 -,H 2 O,NH 3,C 5 H 5 N

Ch14 HW#5 p714 75,77,79(a,c),81,85,89a 75. Write the reaction and the corresponding K b equilibrium expression for each of the following substances acting as bases in water. a. NH 3(aq) + HOH (l) NH 4 + (aq) + OH - (aq) b. C 5 H 5 N 77. Use table 14.3 to help order the following bases from strongest to weakest. NO 3 -,H 2 O,NH 3,C 5 H 5 N

Ch14 HW#5 p714 75,77,79(a,c),81,85,89a 75. Write the reaction and the corresponding K b equilibrium expression for each of the following substances acting as bases in water. a. NH 3(aq) + HOH (l) NH 4 + (aq) + OH - (aq) b. C 5 H 5 N (aq) + HOH (l) C 5 H 5 NH + (aq) + OH - (aq) 77. Use table 14.3 to help order the following bases from strongest to weakest. NO 3 -,H 2 O,NH 3,C 5 H 5 N (Might consider writing each out and predicting.)

77. Use table 14.3 to help order the following bases from strongest to weakest. NO 3 - (aq) + H 2 O (l)  HNO 3 (aq) + OH - (aq) K b = negligible H 2 O (l)  H + (aq) + OH - (aq) K b = 1x10 -14 NH 3(aq) + HOH (l)  NH 4 + (aq) + OH - (aq) K b = 1.8x10 -5 C 5 H 5 N (aq) + HOH (l)  C 5 H 5 NH + (aq) + OH - (aq) K b = 1.7x10 -9

79. Use table 14.3 to help answer the following questions. a. Which is a stronger base, NO 3 - or NH 3 ? c. Which is a stronger base, OH - or NH 3 ? 81. Calculate the pH of the following solutions. a. 0.10 M NaOH b. 1.0x10 -10 M NaOH c. 2.0 M NaOH

79. Use table 14.3 to help answer the following questions. a. Which is a stronger base, NO 3 - or NH 3 ? c. Which is a stronger base, OH - or NH 3 ? 81. Calculate the pH of the following solutions. a. 0.10M NaOH pH is dominated by the OH - donated by _______ b. 1.0x10 -10 M NaOH pH is dominated by the OH - donated by ______ c. 2.0M NaOH pH is dominated by the OH - donated by _______ a. K w = [H + ][OH - ] b. K w = [H + ][OH - ] c. K w = [H + ][OH - ]

79. Use table 14.3 to help answer the following questions. a. Which is a stronger base, NO 3 - or NH 3 ? c. Which is a stronger base, OH - or NH 3 ? 81. Calculate the pH of the following solutions. a. 0.10M NaOH pH is dominated by the OH - donated by the NaOH. b. 1.0x10 -10 M NaOH pH is dominated by the OH - donated by ______ c. 2.0M NaOH pH is dominated by the OH - donated by _______ a. K w = [H + ][OH - ] 1.0x10 -14 = [H + ][0.10] [H + ] = 1.0x10 -14 pH = 13.00 b. K w = [H + ][OH - ] c. K w = [H + ][OH - ]

79. Use table 14.3 to help answer the following questions. a. Which is a stronger base, NO 3 - or NH 3 ? c. Which is a stronger base, OH - or NH 3 ? 81. Calculate the pH of the following solutions. a. 0.10M NaOH pH is dominated by the OH - donated by the NaOH. b. 1.0x10 -10 M NaOH pH is dominated by the OH - donated by the HOH. c. 2.0M NaOH pH is dominated by the OH - donated by ______ a. K w = [H + ][OH - ] 1.0x10 -14 = [H + ][0.10] [H + ] = 1.0x10 -14 pH = 13.00 b. K w = [H + ][OH - ] 1.0x10 -14 = [H + ][1.0x10 -7 ] [H + ] = 1.0x10 -7 pH = 7.00 c. K w = [H + ][OH - ]

79. Use table 14.3 to help answer the following questions. a. Which is a stronger base, NO 3 - or NH 3 ? c. Which is a stronger base, OH - or NH 3 ? 81. Calculate the pH of the following solutions. a. 0.10M NaOH pH is dominated by the OH - donated by the NaOH. b. 1.0x10 -10 M NaOH pH is dominated by the OH - donated by the HOH. c. 2.0M NaOH pH is dominated by the OH - donated by the NaOH. a. K w = [H + ][OH - ] 1.0x10 -14 = [H + ][0.10] [H + ] = 1.0x10 -14 pH = 13.00 b. K w = [H + ][OH - ] 1.0x10 -14 = [H + ][1.0x10 -7 ] [H + ] = 1.0x10 -7 pH = 7.00 c. K w = [H + ][OH - ] 1.0x10 -14 = [H + ][2.0] [H + ] = 5.0x10 -15 pH = 14.30

85. Calculate the concentration of an aqueous KOH solution that has pH = 10.50.

85. Calculate the concentration of an aqueous KOH solution that has pH = 10.50. 10.50 = -log[H + ] [H + ] = 3.2x10 -11 1x10 -14 = [3.2x10 -11 ][OH - ] [OH - ] = 3.2x10 -4 M

89. Calculate [OH - ], [H + ], and the pH of 0.20M solution of: a. Triethylamine [(C 2 H 5 ) 3 N, K b = 4.0 x 10 -4 ] pH is dominated by the OH - donated by ______

89. Calculate [OH - ], [H + ], and the pH of 0.20M solution of: a. Triethylamine [(C 2 H 5 ) 3 N, K b = 4.0 x 10 -4 ] pH is dominated by the OH - donated by (C 2 H 5 ) 3 N

89. Calculate [OH - ], [H + ], and the pH of 0.20M solution of: a. Triethylamine [(C 2 H 5 ) 3 N, K b = 4.0 x 10 -4 ] pH is dominated by the OH - donated by (C 2 H 5 ) 3 N 1x10 -14 = [H + ][0.0089] [H + ] = 1.1x10 -12 M pH = -log[1.1x10 -12 ] pH = 11.95

Ch14.7 Polyprotic Acids H 2 CO 3(aq) H + (aq) + HCO 3 - (aq) K a1 = HCO 3 - (aq) H + (aq) + CO 3 -2 (aq) K a2 =

H 3 PO 4(aq) H + (aq) + H 2 PO 4 - (aq) K a1 = H 2 PO 4 - (aq) H + (aq) + HPO 4 -2 (aq) K a2 = HPO 4 -2 (aq) H + (aq) + PO 4 -3 (aq) K a3 = Write a general statement comparing the values of the K a ’s: (For a weak polyprotic acid)

Ex1) Calc the pH of a 5.0M H 3 PO 4 soln and the eq concs of H 3 PO 4, H 2 PO 4 -, HPO 4 -2, and PO 4 -3.

Ex2) Calc the pH of a 1.0M H 2 SO 4 soln. Ch14 HW#6 p714 91,93,95,99,101

91. Calculate the pH of a 0.20M C 2 H 5 NH 2 solution ( K b = 5.6x10 -4 ) C 2 H 5 NH 2(aq) + HOH (l) C 2 H 5 NH 3 + (aq) + OH - (aq) 1x10 -14 = [H + ][OH - ]

Ch14 HW#6 p714 91,93,95,99,101 91. Calculate the pH of a 0.20M C 2 H 5 NH 2 solution ( K b = 5.6x10 -4 ) C 2 H 5 NH 2(aq) + HOH (l) C 2 H 5 NH 3 + (aq) + OH - (aq) 1x10 -14 = [H + ][0.011]

Ch14 HW#6 p714 91,93,95,99,101 91. Calculate the pH of a 0.20M C 2 H 5 NH 2 solution ( K b = 5.6x10 -4 ) C 2 H 5 NH 2(aq) + HOH (l) C 2 H 5 NH 3 + (aq) + OH - (aq) 1x10 -14 = [H + ][0.011] [H + ] = 9.4x10 -13 M pH = -log[9.4x10 -13 ] pH = 12.02

93. Calculate the percent ionization in each of the following solutions. ( K b = 1.8x10 -5 ) a. 0.10M NH 3 NH 3(aq) + HOH (l) NH 4 + (aq) + OH - (aq) b. 0.010M NH 3

95. Codeine ( C 18 H 21 NO 3 ) is a derivative of morphine that is used as an analgesic, narcotic, or antitussive. It was once commonly used in cough syrups but now is available only by prescription because of addictive properties. If the pH of a 1.7x10 -3 M solution of codeine is 9.59, calculate K b. C 18 H 21 NO 3(aq) + HOH (l) C 18 H 22 NO 3 + (aq) + OH - (aq) pOH = -log[OH - ]

95. Codeine ( C 18 H 21 NO 3 ) is a derivative of morphine that is used as an analgesic, narcotic, or antitussive. It was once commonly used in cough syrups but now is available only by prescription because of addictive properties. If the pH of a 1.7x10 -3 M solution of codeine is 9.59, calculate K b. C 18 H 21 NO 3(aq) + HOH (l) C 18 H 22 NO 3 + (aq) + OH - (aq) 4.41 = -log[OH - ] [OH - ] = 3.9x10 -5

99. Using the K a values in table 14.4, and only the first dissociation step, calculate the pH of 0.10M solutions of each: a. H 3 PO 4 H 3 PO 4(aq) H + (aq) + H 2 PO 4 - (aq) K a1 = 7.5x10 -3 b. H 2 CO 3 H 2 CO 3(aq) H + (aq) + HCO 3 - (aq) K a1 = 4.3x10 -7

101. Calculate the pH of a 2.0 M H 2 SO 4 solution. H 2 SO 4(aq) H + (aq) + HSO 4 - (aq) K a1 = large pH = -log[2.0]pH = -0.3 This is the pH of the solution. HSO 4 - (aq) H + (aq) + SO 4 -2 ( aq) K a2 = 1.2x10 -2 But this is true if HSO 4 - (aq) is the only ion present.

Ch14.8 Acid-Base Properties of Salts Ex1) NaCl  Major species: Ex2) NaC 2 H 3 O 2  Major species: Eqn: K b =

Ex2) NaC 2 H 3 O 2  Na + (aq) + C 2 H 3 O 2 - (aq) Major species: Na +, C 2 H 3 O 2 -, H 2 O Eqn: C 2 H 3 O 2 - (aq) + H 2 O (l) HC 2 H 3 O 2(aq) + OH - (aq) Acetic acid, K a :

Ex3) Calc the pH of a 0.30M NaF soln. K a for HF is 7.2x10 -4. NaF (aq)  Na + (aq) + F - (aq) Major species: Na +, F -, H 2 O Fluoride forms a weak acid: F - (aq) + H 2 O (l)  HF (aq) + OH - (aq) leaving soln slightly basic

Ex4) Calc the pH of a 0.10M NH 4 Cl soln. K b for NH 3 is 1.8x10 -5.

Ex5) Calc the pH of a 0.10M AlCl 3 soln. K a for Al(H 2 O) 6 +3 is 1.4x10 -5.

Ex6) Predict acidic, basic, or neutral: a. NH 4 C 2 H 3 O 2 b. NH 4 CN c. Al 2 (SO 4 ) 3 Ch14 HW#7 p715 103,105,107,109b

103. Arrange the following 0.10M solutions in order of most acidic to most basic. KOH,KCl,KCN,NH 4 Cl, HCl

Ch14 HW#7 p715 103,105,107,109b 103. Arrange the following 0.10M solutions in order of most acidic to most basic. KOH,KCl,KCN,NH 4 Cl, HCl 1 HCl  H + + Cl - (Releases a flood of H ions) 2 NH 4 Cl  NH 4 + + Cl - (Ammonium ions grab some OH ions, leaving soln slightly acidic) 3 KCl  K + + Cl - (Neither grab from water ions) 4 KCN  K + + CN - (Cyanate ions will grab some H ions, leaving soln slightly basic) 5 KOH  K + + OH - (Releases a flood of OH ions)

105. Given that the K a value for acetic acid is 1.8x10 -5 and the K a value for hypochlorous acid is 3.5x10 -8, (weaker acid) which is a stronger base, OCl - or C 2 H 3 O 2 ? (K a ). ( K b ) = K w OCl - : C 2 H 3 O 2 :

105. Given that the K a value for acetic acid is 1.8x10 -5 and the K a value for hypochlorous acid is 3.5x10 -8, (weaker acid) which is a stronger base, OCl - or C 2 H 3 O 2 ? (K a ). ( K b ) = K w OCl - : (stronger base) C 2 H 3 O 2 :

107. Sodium azide ( NaN 3 ) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a 0.010 M solution of NaN 3. The K a value for the hydrazoic acid ( HN 3 ) is 1.9x10 -5. NaN 3(aq)  Na + (aq) + N 3 - (aq) N 3 - (aq) forms weak acid [Na] = [HN 3 ] = [OH - ] =

107. Sodium azide ( NaN 3 ) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a 0.010 M solution of NaN 3. The K a value for the hydrazoic acid ( HN 3 ) is 1.9x10 -5. NaN 3(aq)  Na + (aq) + N 3 - (aq) N 3 - (aq) forms weak acid N 3 - (aq) + H 2 O (l)  HN 3(aq) + OH - (aq) Eq: 0.010M – x ‘excess’ x x [Na] = [HN 3 ] = [OH - ] =

107. Sodium azide ( NaN 3 ) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a 0.010 M solution of NaN 3. The K a value for the hydrazoic acid ( HN 3 ) is 1.9x10 -5. NaN 3(aq)  Na + (aq) + N 3 - (aq) N 3 - (aq) forms weak acid N 3 - (aq) + H 2 O (l)  HN 3(aq) + OH - (aq) Eq: 0.010M – x ‘excess’ x x [Na] = 0.010M [HN 3 ] = 2.3x10 -6 M [OH - ] = 2.3x10 -6 M

109. Calculate the pH of each of the following solutions. b. 0.050 M NaCN NaCN (aq)  Na + (aq) + CN - (aq) CN - (aq) forms weak acid CN - (aq) + H 2 O (l)  HCN (aq) + OH - (aq) Eq: 0.050M – x ‘excess’ x x

109. Calculate the pH of each of the following solutions. b. 0.050 M NaCN NaCN (aq)  Na + (aq) + CN - (aq) CN - (aq) forms weak acid CN - (aq) + H 2 O (l)  HCN (aq) + OH - (aq) Eq: 0.050M – x ‘excess’ x x [OH - ] = 8.9x10 -4 MpOH = - log[8.9x10 -4 M] = 3.04 pH = 10.95

Ch14.9 – 14.11 Structure of Acids and Bases Binary acids: X – H Oxyacids: H – O – X

Oxides – Nonmetal oxides react w water to form acids. Ex1) SO 3(g) + HOH  SO 2(g) + HOH  CO 2(g) + HOH 

Oxides – Metal oxides react w water to form bases. Ex2) CaO (s) + HOH  K 2 O (s) + HOH 

Lewis Theory Lewis acid – electron pair acceptor Lewis base – electron pair donator Ch14 HW#8 p715 115(a-e),117,121,125,127

115. Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced chemical equations for the reactions causing the solution to be acidic or basic. The relevant K a and K b values are found in tables 14.2 and 14.3. a. NaNO 3 b. NaNO 2 c. C 5 H 5 NHClO 4 d. NH 4 NO 2 e. KOClf. NH 4 OCl

a. NaNO 3 Na + + HOH (l)  NaOH + H + NaOH is a strong base so Na + is a weak Lewis acid NO 3 - + HOH (l)  HNO 3 (aq) + OH - (aq) HNO 3 is a strong acid so NO 3 - is a weak Lewis base Neither of these equations happen! neutral b. NaNO 2

Na + + HOH (l)  NaOH + H + NaOH is a strong base so Na + is a weak Lewis acid NO 2 - (aq) + H 2 O (l)  HNO 2 (aq) + OH - (aq) HNO 2 (K a = 4.0x10 -4 ) weak LA so NO 2 - is a stronger Lewis base 2 nd equation happens. NO 2 - is a stronger base than Na + is an acid basic c. C 5 H 5 NHClO 4

c. C 5 H 5 NHClO 4 C 5 H 5 NH + + H 2 O (l)  C 5 H 5 N + H + C 5 H 5 N (K b = 1.7x10 -9 ) somewhat weak LB C 5 H 5 NH + (K a = 5.9x10 -6 ) is a somewhat weak LA This equation happens ClO 4 - + H 2 O (l)  HClO 4 (aq) + OH - (aq) HClO 4 is a strong acid, so ClO 4 - is a weak Lewis base This equation doesn’t happen. C 5 H 5 NH + is a stronger acid than ClO 4 - is a base. acidic d. NH 4 NO 2

NH 4 + + H 2 O (l)  NH 3 + H + NH 4 + : (K a = 5.6x10 -10 ) weak Lewis Acid NO 2 - (aq) + H 2 O (l)  HNO 2(aq) + OH - (aq) HNO 2 (K a = 4.0x10 -4 ) moderate acid?, K a. K b = K w so then NO 2 - (K b = 2.5x10 -11 ) is a weaker Lewis base NH 4 + is a stronger acid than NO 2 - is a base. acidic e. KOCl

K + + HOH (l)  KOH + H + KOH is a strong base, so K + is a weak Lewis acid This reaction doesn’t happen OCl - + HOH (l)  HOCl (aq) + OH - (aq) Hypochlorous acid (K a = 3.5x10 -8 ) K a. K b = K w  K b = 2.9x10 -7 is a middle, weak LB This reaction happens. OCl - is a stronger base than K + is an acid Basic

117. Place the species in each of the following groups in order of increasing acid strength. Explain the order you chose for each group. a. HIO 3, HBrO 3 b. HNO 2, HNO 3 c. HOCl, HOI d. H 3 PO 4, H 3 PO 3

117. Place the species in each of the following groups in order of increasing acid strength. Explain the order you chose for each group. a. HIO 3, HBrO 3 HIO 3 < HBrO 3 I less electronegative As the EN of the central atom increases, the acid strength increases. b. HNO 2, HNO 3 HNO 2 < HNO 3 more oxygens As the #O’s of the oxyanion increases, the acid strength increases. c. HOCl, HOI d. H 3 PO 4, H 3 PO 3

117. Place the species in each of the following groups in order of increasing acid strength. Explain the order you chose for each group. a. HIO 3, HBrO 3 HIO 3 < HBrO 3 I less electronegative As the EN of the central atom increases, the acid strength increases. b. HNO 2, HNO 3 HNO 2 < HNO 3 more oxygens As the #O’s of the oxyanion increases, the acid strength increases. c. HOCl, HOI HOCl > HOI I less electronegative d. H 3 PO 4, H 3 PO 3 H 3 PO 4 > H 3 PO 3 more oxygens

121. Will the following oxides give acidic, basic, or neutral solutions when dissolved in water? Write reactions to justify your answers. a. CaO + HOH  b. SO 2 + HOH  c. Cl 2 O + HOH 

121. Will the following oxides give acidic, basic, or neutral solutions when dissolved in water? Write reactions to justify your answers. a. CaO + HOH  Ca(OH) 2 b. SO 2 + HOH  H 2 SO 3 c. Cl 2 O + HOH  H 2 Cl 2 O 2 (HOCl?) Cl 2 O + HOH  2HOCl

125. Aluminum hydroxide is an amphoteric substance. It can act as either a BrØnsted-Lowry base or a Lewis acid. Write a reaction showing Al(OH) 3 acting as a base toward H + and as an acid toward OH -. Al(OH) 3  127. Would you expect Fe 3+ or Fe 2+ to be the stronger Lewis acid? Explain.

125. Aluminum hydroxide is an amphoteric substance. It can act as either a BrØnsted-Lowry base or a Lewis acid. Write a reaction showing Al(OH) 3 acting as a base toward H + and as an acid toward OH -. Al(OH) 3 + 3H +  Al + (aq) + 3HOH (l) Al(OH) 3 + OH -  Al(OH) 4 127. Would you expect Fe 3+ or Fe 2+ to be the stronger Lewis acid? Explain. Fe 3+ Fe 2+

125. Aluminum hydroxide is an amphoteric substance. It can act as either a BrØnsted-Lowry base or a Lewis acid. Write a reaction showing Al(OH) 3 acting as a base toward H + and as an acid toward OH -. Al(OH) 3 + 3H +  Al + (aq) + 3HOH (l) Al(OH) 3 + OH -  Al(OH) 4 127. Would you expect Fe 3+ or Fe 2+ to be the stronger Lewis acid? Explain. Fe 3+ is stronger. More positive charge attracts more hydroxide. Fe 3+ + 3HOH  Fe(OH) 3 + 3H + Fe 2+ : Fe 2+ + 2HOH  Fe(OH) 2 + 2H +

Ch14 Rev#1 p712 25,30b,43,58,62,74,92,114,119d,122,128 25. Consider the reaction of acetic acid in water CH 3 CO 2 H(aq) + H 2 O ↔ CH 3 CO 2 - (aq) + H 3 O + (aq) where K a = 1.8 х 10 -5. a. Which two bases are competing for the proton? b. Which is the stronger base? c. In light of your answer to b, why do we classify the acetate ion (CH 3 CO 2 - ) as a weak base? Use an appropriate reaction to justify your answer.

Ch14 Rev#1 p712 25,30b,43,58,62,74,91,115f,119d,122,128 25. Consider the reaction of acetic acid in water CH 3 CO 2 H(aq) + H 2 O ↔ CH 3 CO 2 - (aq) + H 3 O + (aq) where K a = 1.8 х 10 -5. a. Which two bases are competing for the proton? CH 3 CO 2 H  CH 3 CO 2 - + H + H 2 O + H +  H 3 O + b. Which is the stronger base? With K a < 1 reaction favors reactants, so top reaction is favored CH 3 CO 2 - is the stronger base. c. In light of your answer to b, why do we classify the acetate ion (CH 3 CO 2 - ) as a weak base? Use an appropriate reaction to justify your answer.

30. For each of the following aqueous reactions, identify the acid, the base, the conjugate base, and the conjugate acid. b. H 2 O + HONH 3 + ↔ HONH 2 + H 3 O + 43. Calculate [H + ] and[OH - ] for each solution at 25˚C. pH = -log[H + ] Identify each solution as neutral, acidic, or basic. [H + ]. [OH - ]=1x10 -14 a. pH = 7.40 (the normal pH of blood) b. pH = 15.3 c. pH = -1.0 d. pH = 3.20 e. pOH = 5.0 pOH = -log[OH - ] f. pOH = 9.60

30. For each of the following aqueous reactions, identify the acid, the base, the conjugate base, and the conjugate acid. b. H 2 O + HONH 3 + ↔ HONH 2 + H 3 O + Base Acid CB CA 43. Calculate [H + ] and[OH - ] for each solution at 25˚C. pH = -log[H + ] Identify each solution as neutral, acidic, or basic. [H + ]. [OH - ]=1x10 -14 a. pH = 7.40 (the normal pH of blood) b. pH = 15.3 c. pH = -1.0 d. pH = 3.20 e. pOH = 5.0 pOH = -log[OH - ] f. pOH = 9.60

30. For each of the following aqueous reactions, identify the acid, the base, the conjugate base, and the conjugate acid. b. H 2 O + HONH 3 + ↔ HONH 2 + H 3 O + Base Acid CB CA 43. Calculate [H + ] and[OH - ] for each solution at 25˚C. pH = -log[H + ] Identify each solution as neutral, acidic, or basic. [H + ]. [OH - ]=1x10 -14 a. pH = 7.40 [H + ] = 3.98x10 -8 M [OH - ]= 2.5x10 -7 M Slightly Basic b. pH = 15.3 [H + ] = 5.01x10 -16 M [OH - ]= 20.0 Highly Basic c. pH = -1.0 [H + ] = 10.0M [OH - ]= 1x10 -15 M Highly Acidic d. pH = 3.20 [H + ] = 6.31x10 -4 M [OH - ]= 1.6x10 -11 M Acidic e. pOH = 5.0 [OH - ]= 1.0x10 -5 M [H + ] = 1.0x10 -9 M Basic f. pOH = 9.60 [OH - ]= 2.51x10 -10 M [H + ] = 3.98x10 -5 M Acid

58. A solution with a total volume of 250.0mL is prepared by diluting 20.0 mL of glacial acetic acid with water. Calculate [H + ] and the pH of this solution. Assume that glacial acetic acid is pure liquid acetic acid with a density of 1.05 g/cm 3. (K a = 1.8х10 -5 ) CH 3 CO 2 H (aq) + H 2 O ↔ CH 3 CO 2 - (aq) + H 3 O + (aq) Initial: Eq:

58. A solution with a total volume of 250.0mL is prepared by diluting 20.0 mL of glacial acetic acid with water. Calculate [H + ] and the pH of this solution. Assume that glacial acetic acid is pure liquid acetic acid with a density of 1.05 g/cm 3. (K a = 1.8х10 -5 ) CH 3 CO 2 H (aq) + H 2 O ↔ CH 3 CO 2 - (aq) + H 3 O + (aq) Initial: [0.0014] ‘excess’ 0 0 Eq: [0.0014 –x] [x] [x]

62. Monochloroacetic acid, HC 2 H 2 ClO 2, is a skin irritant that is used in “chemical peels” intended to remove the top layer of dead skin from the face and ultimately improve the complexion. K a = 1.35 х 10 -3. Calculate the pH of a 0.10M solution of monochloroacetic acid. HC 2 H 2 ClO 2(aq) + H 2 O ↔ C 2 H 2 ClO 2 - (aq) + H 3 O + (aq) Initial: [0.10] ‘excess’ 0 0 Eq: [0.10 –x] [x] [x]

74. One mole of a weak acid HA was dissolved in a 2.0 L of water. After the system had come to equilibrium, the concentration of HA was found to be 0.45 M. Calculate the K a for HA. HA (aq) + H 2 O ↔ H + (aq) + A - (aq) Initial: [0.5] ‘excess’ 0 0 Eq: [0.45] [0.05] [0.05]

92. Calculate the pH of a 0.050M ( C 2 H 5 ) 2 NH 2 soln ( K b = 1.3х10 -3 ). ( C 2 H 5 ) 2 NH 2 (aq) + HOH (l) ( C 2 H 5 ) 2 NH 3 + (aq) + OH - (aq) What led me to this reaction?

92. Calculate the pH of a 0.050M ( C 2 H 5 ) 2 NH 2 soln ( K b = 1.3х10 -3 ). ( C 2 H 5 ) 2 NH 2 (aq) + HOH (l) ( C 2 H 5 ) 2 NH 3 + (aq) + OH - (aq) 1x10 -14 = [H + ][OH - ] [H + ] = pH =

92. Calculate the pH of a 0.050M ( C 2 H 5 ) 2 NH 2 soln ( K b = 1.3х10 -3 ). ( C 2 H 5 ) 2 NH 2 (aq) + HOH (l) ( C 2 H 5 ) 2 NH 3 + (aq) + OH - (aq) 1x10 -14 = [H + ][0.0081] [H + ] = 1.24x10 -12 M pH = -log[1.24x10 -12 ] pH = 11.91

115. Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced chemical equations for the reactions causing the solution to be acidic or basic. The relevant K a and K b values are found in tables 14.2 and 14.3. f. NH 4 OCl NH 4 + + H 2 O (l)  NH 3 + H + NH 4 + : (K a = 5.6x10 -10 ) OCl - + HOH (l)  HOCl (aq) + OH - (aq) Hypochlorous acid (K a = 3.5x10 -8 ) K a. K b = K w  OCl - (K b = 2.9x10 -7 )

115. Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced chemical equations for the reactions causing the solution to be acidic or basic. The relevant K a and K b values are found in tables 14.2 and 14.3. f. NH 4 OCl NH 4 + + H 2 O (l)  NH 3 + H + NH 4 + : (K a = 5.6x10 -10 ) weaker LA This reaction doesn’t happen OCl - + HOH (l)  HOCl (aq) + OH - (aq) Hypochlorous acid (K a = 3.5x10 -8 ) K a. K b = K w  OCl - (K b = 2.9x10 -7 )is a middle, weak LB This reaction happens. OCl - is a stronger base than NH 4 + is an acid Basic

119. Place the species in each of the following groups in order of increasing acid strength. d. NH 4 +, PH 4 + ( bond energies: N─H, 391 kJ/mol; P─H, 322 kJ/mol)

122. Will the following oxides give the acidic, basic, or neutral solutions when dissolved in water? a. Li 2 O b. CO 2 c. SrO Write reactions to justify your answer. a. Li 2 O + HOH  b. CO 2 + HOH  c. SrO + HOH  128. Use the Lewis acid-base model to explain the following reaction. CO 2(g) + H 2 O (l) H 2 CO 3(aq)

122. Will the following oxides give the acidic, basic, or neutral solutions when dissolved in water? a. Li 2 O b. CO 2 c. SrO Write reactions to justify your answer. a. Li 2 O + HOH  2LiOH b. CO 2 + HOH  H 2 CO 3 c. SrO + HOH  Sr(OH) 2 128. Use the Lewis acid-base model to explain the following reaction. CO 2(g) + H 2 O (l) H 2 CO 3(aq) K a for carbonic acid = 4.3x10 -7

Ch14.1 Acids and Bases Acids – taste sour Arrhenius Theory of Acids – they produce hydrogen ions in aqueous solutions. Bases – Taste bitter, feel slippery.

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Ch14.1 Acids and Bases Acids – taste sour Arrhenius Theory of Acids – they produce hydrogen ions in aqueous solutions. Bases – Taste bitter, feel slippery.

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Presentation on theme: "Ch14.1 Acids and Bases Acids – taste sour Arrhenius Theory of Acids – they produce hydrogen ions in aqueous solutions. Bases – Taste bitter, feel slippery."— Presentation transcript:

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