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  1. Homework Problems Chapter 15 Homework Problems: 4, 6, 8, 24, 34, 36, 38, 39, 42, 52, 56a, 62, 68, 70, 74, 80a, 89, 93, 100, 103a, 116, 117, 124

  2. CHAPTER 15 Acids and Bases

  3. Acids and Bases (Arrhenius) There are several definitions we can use for acids and bases: Arrhenius definition acid - A substance which, when added to water, forms H+ ion base - A substance which, when added to water, forms OH- ion Examples: HCl(aq)  H+(aq) + Cl-(aq) HF(aq)  H+(aq) + F-(aq) NaOH(s)  Na+(aq) + OH-(aq) NH3(aq) + H2O()  NH4+(aq) + OH-(aq) The advantages of the Arrhenius definition are that it is simple and easy to implement. The disadvantages are that it is tied in to a particular solvent (water) and is not a general definition.

  4. Acids and Bases (Bronsted-Lowry) Bronsted-Lowry definition acid - a proton (H+) donor; forms a conjugate base base - a proton (H+) acceptor; forms a conjugate acid H+ HF(aq) + H2O()  H3O+(aq) + F-(aq) acid base conjugate acid conjugate base In the Bronsted theory, in an acid-base reaction an acid donates a proton to form a conjugate base, while a base accepts a proton to form a conjugate acid. In addition, in Bronsted theory acids form hydronium ion (H3O+ ion) instead of hydrogen ion (H+ ion) when added to water.

  5. Examples of Acids HCl(aq) + H2O()  H3O+(aq) + Cl-(aq) acid: HCl conjugate base: Cl- base: H2O conjugate acid: H3O+ CH3COOH(aq) + H2O()  H3O+(aq) + CH3COO-(aq) acid: CH3COOH conjugate base: CH3COO- base: H2O conjugate acid: H3O+ The second reaction goes in both directions, so has an equili-brium constant, the acid ionization constant KC = Ka = [H3O+] [CH3COO-] [CH3COOH]

  6. Examples of Bases NaOH(s)  Na+(aq) + OH-(aq) Considered an ionization reaction in Bronsted theory, not an acid-base reaction. NH3(aq) + H2O()  NH4+(aq) + OH-(aq) base: NH3 conjugate acid: NH4+ acid: H2O conjugate base: OH- The reaction goes in both directions, so has an equilibrium con-stant, the base ionization constant KC = Kb = [NH4+] [OH-] [NH3]

  7. Amphoteric Substances Some substances can act as either Bronsted acids or Bronsted bases. Substances that can act as either acids or bases depending on what they are reacting with are called amphoteric. For example: H2O (water) (acid) NH3(aq) + H2O()  NH4+(aq) + OH-(aq) (base) CH3COOH(aq) + H2O()  H3O+(aq) + CH3COO-(aq) HCO3- (hydrogen carbonate ion) (acid) NH3(aq) + HCO3-(aq)  NH4+(aq) + CO32-(aq) (base) CH3COOH(aq) + HCO3-(aq)  H2CO3(aq) + CH3COO-(aq)

  8. We can picture the reaction of a Bronsted acid with a Bronsted base in several ways.

  9. Given any substance, the conjugate base for the substance is formed by removing a proton (H+), and the conjugate acid is formed by adding a proton (H+). Example: H2O The conjugate base of H2O is OH- The conjugate acid of H2O is H3O+ Example: HSO4- The conjugate base of HSO4- is SO42- The conjugate acid of HSO4- is H2SO4

  10. Example Chlorous acid (HClO2) is a weak acid. Pyridine (C5H5N) is a weak base. Indicate the behavior of these two substances when added to water, according to Bronsted theory.

  11. Example Chlorous acid (HClO2) is a weak acid. Pyridine (C5H5N) is a weak base. Indicate the behavior of these two substances when added to water, according to Bronsted theory. HClO2(aq) + H2O()  H3O+(aq) + ClO2-(aq) C5H5N(aq) + H2O()  C5H5NH+(aq) + OH-(aq)

  12. Strong and Weak Acids There are seven common strong acids: Binary strong acidsTernary strong acids HCl HClO3 HNO3 HBr HClO4 H2SO4 (1st proton) HI Sulfuric acid is special in that it is a strong acid with respect to the first proton and a weak acid with respect to the second proton. H2SO4(aq) + H2O()  HSO4-(aq) + H3O+(aq) HSO4-(aq) + H2O()  SO42-(aq) + H3O+(aq) Ka2 = [SO42-] [H3O+] = 1.2 x 10-2 [HSO4-]

  13. All other acids are weak acids. HNO2(aq) + H2O()  H3O+(aq) + NO2-(aq) Ka = [H3O+] [NO2-] = 4.5 x 10-4 [HNO2] CH3COOH(aq) + H2O()  H3O+(aq) + CH3COO-(aq) Ka = [H3O+] [CH3COO-] = 1.8 x 10-5 [CH3COOH] The larger the value of Ka, the stronger the acid. Therefore, HNO2 is a stronger acid than CH3COOH.

  14. Strong and Weak Bases There are seven common strong soluble bases: Group 1A strong basesGroup 2A strong bases LiOH Sr(OH)2 NaOH Ba(OH)2 KOH RbOH CsOH All other metal hydroxides are insoluble bases. They do not dissolve in water to an appreciable extent, but react as bases in acid-base reactions. Examples: Cu(OH)2, Al(OH)3, Pb(OH)2. Cu(OH)2(s) + 2 HCl(aq)  Cu2+(aq) + 2 Cl-(aq) + 2 H2O()

  15. Weak bases establish an equilibrium in water. NH3(aq) + H2O()  NH4+(aq) + OH-(aq) Kb = [NH4+] [OH-] = 1.8 x 10-5 [NH3] N2H4(aq) + H2O()  N2H5+(aq) + OH-(aq) Kb = [N2H5+] [OH-] = 8.9 x 10-7 [N2H4] Since Kb is larger for NH3 than for N2H4, NH3 is a stronger base than N2H4.

  16. Autoionization of Water In pure water there will be a small number of H3O+ and OH- ions. This is due to the autoionization reaction: H2O() + H2O()  H3O+(aq) + OH-(aq) Kw = [H3O+] [OH-] = 1.0 x 10-14 (at T = 25 C) Because the above reaction is endothermic (Hrxn = + 55.8 kJ/mol) the value for Kw increases as T increases.

  17. Equilibrium in Pure Water Since H2O() + H2O()  H3O+(aq) + OH-(aq) Kw = [H3O+] [OH-] = 1.0 x 10-14 (at T = 25 C) Substance Initial Change Equilibrium H3O+ 0.0 x x OH- 0.0 x x So (x) (x) = x2 = 1.0 x 10-14 x = 1.0 x 10-7 So in pure water at equilibrium at T = 25 C, [H3O+] = [OH-] = 1.0 x 10-7 M.

  18. Acidic, Basic, and Neutral Solutions We can use the above results to define what we mean by an acidic, basic, and neutral solution, using [H3O+] [OH-] = 1.0 x 10-14. acidic solution [H3O+] > [OH-] [H3O+] > 1.0 x 10-7 M [OH-] < 1.0 x 10-7 M neutral solution [H3O+] = [OH-] [H3O+] = 1.0 x 10-7 M [OH-] = 1.0 x 10-7 M basic solution [H3O+] < [OH-] [H3O+] < 1.0 x 10-7 M [OH-] > 1.0 x 10-7 M

  19. pH pH represents a convenient way of representing the concentration of hydronium ion in an aqueous solution. pH is defined as follows: pH = - log10[H+] = - log10[H3O+] For a neutral solution at T = 25 C pH = - log10[H3O+] = - log10(1.0 x 10-7) = 7.00 Note that the number of digits to the right of the decimal point is equal to the number of significant figures in the H3O+ concentration. For acidic and basic solutions acidic solution [H3O+] > 1.0 x 10-7 M means pH < 7.00 basic solution [H3O+] < 1.0 x 10-7 M means pH > 7.00 The further away the pH is from 7.00 the more acidic (if less than 7.00) or basic (if greater than 7.00) the solution.

  20. pOH and pK We can use the “p” notation as a general symbol to indicate that we are taking - log10 of something. In particular pH = - log10[H3O+] pOH = - log10[OH-] pK = - log10 K By reversing the above definitions we get the following relationships [H3O+] = 10-pH [OH-] = 10-pOH K = 10-pK

  21. Relationship Between pH and pOH There is a simple relationship between pH and pOH. [H3O+] [OH-] = 1.0 x 10-14 -log10[H3O+] + ( - log10[OH-]) = - log10(1.0 x 10-14) pH + pOH = 14.00 (at 25 C) The more general relationship, true at all temperatures, is pH + pOH = pKw

  22. Relationship Between pH and Concentration If we know the concentration of hydronium ions in solution we can find the pH of the solution (and vice versa). We can also find pOH and the concentration of hydroxide ions in solution. Example: A particular solution has pH = 4.62 at T = 25. C. What are the concentrations of H3O+ and OH- in the solution?

  23. Relationship Between pH and Concentration Example: A particular solution has pH = 4.62 at T = 25. C. What are the concentrations of H3O+ and OH- in the solution? [H3O+] = 10-pH = 10-4.62 = 2.4 x 10-5 M [H3O+] [OH-] = 1.0 x 10-14, so [OH-] = 1.0 x 10-14 = 1.0 x 10-14 = 4.2 x 10-10 M [H3O+] 2.4 x 10-5 We could also find the OH- concentration as follows: pH + pOH = 14.00 pOH = 14.00 - pH = 14.00 - 4.62 = 9.38 [OH-] = 10-pOH = 10-9.38 = 4.2 x 10-10 M

  24. pH for Solutions of Strong Acids or Strong Bases Because strong acids and strong soluble bases are strong electrolytes, and so completely dissociate in solution, finding the value for pH is relatively simple. We may use the following procedure: 1) Use the information in the problem to find the concentration of H3O+ (strong acid) or OH- (strong base). 2) Find the pH a) For strong acids, find the pH directly b) For strong bases, find the pOH, then use (at T = 25 C) pH + pOH = 14.00 ; pH = 14.00 - pOH to find the pH

  25. Example: Find the pH for the following solutions, at T = 25 C a) A 3.5 x 10-3 M solution of HBr, a strong acid b) A solution formed by dissolving 0.200 moles of Sr(OH)2, a strong soluble base, in water, to form a solution with a final volume of 500.0 mL.

  26. a) A 3.5 x 10-3 M solution of HBr, a strong acid Reaction is HBr(aq) + H2O()  H3O+(aq) + Br-(aq) [H3O+] = 3.5 x 10-3 mol HBr 1 mol H3O+ = 3.5 x 10-3 M L soln 1 mol HBr pH = - log10(3.5 x 10-3) = 2.46

  27. b) A solution formed by dissolving 0.200 moles of Sr(OH)2, a strong soluble base, in water, to form a solution with a final volume of 500.0 mL. Reaction is Sr(OH)2(s)  Sr2+(aq) + 2 OH-(aq) [OH-] = 0.200 mol Sr(OH)22 mol OH- = 0.800 M 0.5000 L soln 1 mol Sr(OH)2 pOH = - log10(0.800) = 0.097 pH = 14.00 - pOH = 14.00 - 0.097 = 13.90

  28. Weak Acids or Weak Bases For problems involving solutions containing a single weak acid or weak base we proceed as we do other equilibrium problems. 1) Start with the following information Balanced chemical reaction Initial conditions Value for Ka (weak acid) or Kb (weak base) 2) Set up the problem using the “ICE” method 3) Find [H3O+] (weak acid) or [OH-] (weak base) 4) Find pH (for weak base, first find pOH, then use pH + pOH = 14.00 to find pH)

  29. Example: Find the pH of a 0.100 M solution of HNO2, a weak acid (Ka = 4.5 x 10-4), at T = 25 C

  30. Find the pH of a 0.100 M solution of HNO2, a weak acid (Ka = 4.5 x 10-4), at T = 25 C HNO2(aq) + H2O()  H3O+(aq) + NO2-(aq) Ka = [H3O+] [NO2-] = 4.5 x 10-4 [HNO2] Substance Initial Change Equilibrium H3O+ 0.00 x x NO2- 0.00 x x HNO2 0.100 - x 0.100 - x (x) (x) = 4.5 x 10-4 (0.100 - x) There are two ways to proceed...

  31. 1) Assume x is small (in this case, x << 0.100) (x) (x)  x2 = 4.5 x 10-4 (0.100 - x) 0.100 so x2 = (0.100)(4.5 x 10-4) = 4.5 x 10-5 x = (4.5 x 10-5)1/2 = 6.7 x 10-3 [H3O+] = 6.7 x 10-3 M ; pH = - log10(6.7 x 10-3) = 2.17 Is 6.7 x 10-3 << 0.100? Yes (at least 10 times smaller). 2) Solve the quadratic x2 = (0.100 - x)(4.5 x 10-4) = (4.5 x 10-5) - (4.5 x 10-4) x x2 + (4.5 x 10-4) x - (4.5 x 10-5) = 0 x = 6.5 x 10-3 ; - 6.9 x 10-3 [H3O+] = 6.5 x 10-3 M ; pH = - log10(6.5 x 10-3) = 2.19

  32. Percent Dissociation The percent dissociation of a weak acid is defined as % dissociation = amount dissociated x 100% initial amount For strong acids the percent dissociation is ~ 100%. For weak acids we can use the equilibrium concentrations to find the percent dissociation. For the above example, initial HNO2 = 0.100 M amount dissociated = 6.7 x 10-3 M % dissociation = 6.7 x 10-3 x 100% = 6.7 % 0.100

  33. pH Calculations involving Weak Bases Just as we can find concentrations and pH values for solutions of weak acids, we can do the same for solutions of weak bases. Example: Find the pH of a 0.100 M solution of NH3, a weak base (Kb = 1.8 x 10-5), at T = 25 C

  34. Find the pH of a 0.100 M solution of NH3, a weak base (Kb = 1.8 x 10-5), at T = 25 C NH3(aq) + H2O()  NH4+(aq) + OH-(aq) Kb = [NH4+] [OH-] = 1.8 x 10-5 [NH3] Substance Initial Change Equilibrium NH4+ 0.00 x x OH- 0.00 x x NH3 0.100 - x 0.100 - x (x) (x) = 1.8 x 10-5 (0.100 - x)

  35. Assume x is small (in this case, x << 0.100) (x) (x)  x2 = 1.8 x 10-5 (0.100 - x) 0.100 so x2 = (0.100)(1.8 x 10-5) = 1.8 x 10-6 x = (1.8 x 10-6)1/2 = 1.3 x 10-3 [OH-] = 1.3 x 10-3 M ; pOH = - log10(1.3 x 10-3) = 2.87 pH = 14.00 - 2.87 = 11.13 Is 1.3 x 10-3 << 0.100? Yes (at least ten times smaller) If we solve using the quadratic formula, we get pH = 11.12

  36. Polyprotic Acid A polyprotic acid has two or more ionizable protons that can be donated in an acid-base reaction. Monoprotic (one ionizable proton) HCl, HNO2, CH3COOH Diprotic (two ionizable protons) H2CO3, H2SO4 Triprotic (three ionizable protons) H3PO3 For polyprotic acids one can talk about the acid dissociation constant for each proton.

  37. Example: H2CO3 1st proton H2CO3(aq) + H2O()  H3O+(aq) + HCO3-(aq) Ka1 = [H3O+][HCO3-] = 4.3 x 10-7 [H2CO3] 2nd proton HCO3-(aq) + H2O()  H3O+(aq) + CO32-(aq) Ka2 = [H3O+][CO32-] = 5.6 x 10-11 [HCO3-] For a polyprotic acid Ka1 > Ka2 > Ka3...

  38. Calculations Involving Polyprotic Acids It would seem like calculations with polyprotic acids should be complicated, since there are several sources of H3O+ ions. However, there is usually a big enough difference in the values of the Kas (acid dissociation constants) that only the first dissociation needs to be considered for polyprotic acid solutions. Procedure (diprotic acid): 1) Calculate equilibrium concentrations using the first ionization constant. 2) Calculate equilibrium concentrations using the second ionization constant, using the results from the first calculation for the initial conditions. 3) If there are any significant changes in concentrations involving the first ionization, go back and recalculate concentrations using the previous results for the initial conditions.

  39. Relationship Between Ka and Kb We may find a general relationship between the values for Ka and Kb for a Bronsted acid/conjugate base pair of substances. We proceed as follows: Let HA be a weak monoprotic acid. A- is the conjugate base. HA(aq) + H2O() H3O+(aq) + A-(aq) Ka = [H3O+] [A-] [HA] A-(aq) + H2O() HA(aq) + OH-(aq) Kb = [HA] [OH-] [A-] Ka Kb = [H3O+] [A-] [HA] [OH-] = [H3O+] [OH-] = Kw [HA] [A-] Ka Kb = Kw pKa + pKb = pKw Ka Kb = 1.0 x 10-14 pKa + pKb = 14.00, at T = 25. C

  40. Example: The value for the acid ionization constant for acetic acid (CH3COOH) is Ka = 1.8 x 10-5 at T = 25. C. What is the value for Kb for the acetate ion (CH3COO-)?

  41. Example: The value for the acid ionization constant for acetic acid (CH3COOH) is Ka = 1.8 x 10-5 at T = 25. C. What is the value for Kb for the acetate ion (CH3COO-)? Ka Kb = Kw = 1.0 x 10-14 for an acid/conjugate base pair Kb = Kw = (1.0 x 10-14) = 5.6 x 10-10 Ka (1.8 x 10-5) Note that we could prepare a solution that initially only contained the conjugate base if we added a salt such as NaCH3COO, KCH3COO, etc. NaCH3COO(aq)  Na+(aq) + CH3COO-(aq) Calculations would be done in the same way as any other weak acid or weak base solution.

  42. General Statements About Acid and Base Strength We may use the above results to make the following general statement concerning the relative strengths of acids and their conjugate bases. This is based on the relationship Ka Kb = Kw = 1.0 x 10-14. The stronger the acid the weaker the conjugate base. Example: Which is a stronger base, F- or CN-? Ka(HF) = 3.5 x 10-4 Ka(HCN) = 4.9 x 10-10 Since HF is a stronger acid than HCN, F- is a weaker base than CN-. Kb(F-) = (1.0 x 10-14)/(3.5 x 10-4) = 2.9 x 10-11 Kb(CN-) = (1.0 x 10-14)/(4.9 x 10-10) = 2.0 x 10-5

  43. Acid-Base Properties of Salts Recall that a salt is an ionic compound formed from the reaction of an acid with a base. We have the following four possibilities: 1) Salt of a strong acid and a strong base. Example: HCl + NaOH  NaCl + H2O (Na+ and Cl-) No acid-base properties for the salt. Solutions will be neutral. 2) Salt of a strong acid and a weak base. Example: HCl + NH3  NH4Cl (NH4+ and Cl-) NH4+ will act as a weak acid. Solutions will be acidic. 3) Salt of a weak acid and a strong base. Example: HF + KOH  KF + H2O (K+ and F-) F- will act as a weak base. Solutions will be basic.

  44. 4) Salt of a weak acid and a weak base. Example: HF + NH3  NH4F (NH4+ and F-) NH4+ will act as a weak acid. F- will act as a weak base. Solutions will be approximately neutral. Ka(HF) = 3.5 x 10-4 so Kb(F-) = 2.9 x 10-11 Kb(NH3) = 1.8 x 10-5 so Ka(NH4+) = 5.6 x 10-10 Since NH4+ is a stronger acid than F- is a base, the solution will be slightly acidic.

  45. Acid-Base Calculations for Salts We do acid-base problems for salts the same way as we do other weak acid or weak base problems. Example: What is the pH of a 0.100 M solution of ammonium chloride (NH4Cl), the salt of a strong acid and a weak base. Kb(NH3) = 1.8 x 10-5. Assume T = 25 C

  46. Example: What is the pH of a 0.100 M solution of ammonium chloride (NH4Cl), the salt of a strong acid and a weak base. Kb(NH3) = 1.8 x 10-5. Assume T = 25 C NH4Cl(s)  NH4+(aq) + Cl-(aq) NH4+(aq) + H2O()  H3O+(aq) + NH3(aq) Ka Kb = 1.0 x 10-14, so Ka = 1.0 x 10-14 = (1.0 x 10-14) = 5.6 x 10-10 Kb (1.8 x 10-5) Ka = [H3O+] [NH3] = 5.6 x 10-10 [NH4+] Substance Initial Change Equilibrium H3O+ 0.00 x x NH3 0.00 x x NH4+ 0.100 - x 0.100 - x

  47. Ka = [H3O+] [NH3] = (x) (x) = 5.6 x 10-10 [NH4+] (0.100 - x) Assume x << 0.100. Then x2 = 5.6 x 10-10 (0.100) x2 = (0.100) (5.6 x 10-10) = 5.6 x 10-11 x = (5.6 x 10-11)1/2 = 7.5 x 10-6 pH = - log10(7.5 x 10-6) = 5.13 Note that 7.5 x 10-6 << 0.100, so x is in fact small.

  48. Acid-Base Properties of Cations Metal cations can act as weak acids in solution. There are several ways in which these reactions can be written Example: Al3+ ion Al3+(aq) + H2O()  H+(aq) + Al(OH)2+(aq) Al(H2O)63+(aq) + H2O()  H3O+(aq) + Al(H2O)5(OH)2+ Generally speaking, metal cations act as weak acids when they are small and have multiple positive charges. Common examples of metal cations acting as weak acids are Co2+, Ni2+, Zn2+, Fe2+, Fe3+, and Al3+ ions.

  49. Factors Affecting Acid Strength We can often predict the relative strengths of weak acids by focussing on the factors that control acid strength. Binary acids 1) In the same column (group). Acid strength increases from top to bottom. Reason: The H - X bond strength decreases from top to bottom, so it is easier to break the bond and form H+ (or H3O+). Example: Group 7 binary acids. HF 567 kJ/mol Ka = 3.5 x 10-4 HCl 431 kJ/mol “strong acid” (Ka ~ 107) HBr 366 kJ/mol “strong acid” (Ka ~ 109) HI 299 kJ/mol “strong acid” (Ka ~ 1011)

  50. 2) In the same row. Acid strength increases from left to right. Reason: The electronegativity of the nonmetal increases from left to right, making the conjugate base more stable. Example: Second row. acid conj. base nonmetal EN CH4 CH3- EN(C) = 2.5 insoluble in water (Ka ~ 10-49) NH3 NH2- EN(N) = 3.0 “weak base” (Ka ~ 10-33) H2O OH- EN(O) = 3.5 “amphoteric” (Ka = 1.0 x 10-14) HF F- EN(F) = 4.0 “weak acid” (Ka = 3.5 x 10-4)