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Acid-Base Equilibria

Acid-Base Equilibria. Chapter 16. Equilibrium, Part II. Acids and Bases: A Brief Review (Ch 4). Acid tastes feels strong acid with [acid] > 1 dissolves protein detect: Base tastes feels strong base with [base] > 1 dissolves protein detect: Arrhenius definition

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Acid-Base Equilibria

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  1. Acid-Base Equilibria Chapter 16 Equilibrium, Part II Chem 1422, Chapter 16

  2. Acids and Bases: A Brief Review (Ch 4) • Acid • tastes • feels • strong acid with [acid] > 1 dissolves protein • detect: • Base • tastes • feels • strong base with [base] > 1 dissolves protein • detect: • Arrhenius definition • acids increase [H+] in water • bases increase [OH-] in water • acid + base → salt + water. sour squeeky clean changes litmus to red bitter slippery changes litmus to purple Chem 1422, Chapter 16

  3. Acids and Bases: A Brief Review (Ch 4) “H+” literally means a bare proton; single protons cannot actually exist in water! In water, the proton always attaches to one or more water molecules to form proton clusters. The simplest cluster is H3O+(aq). Larger clusters such as H5O2+(aq), H7O3+(aq), etc. also exist. Chem 1422, Chapter 16

  4. Acids and Bases: A Brief Review (Ch 4) These clusters are in dynamic equilibrium with one another in water: H+ + H2O → H3O+ H3O+ + H2O ⇌ H5O2+ H5O2+ + H2O ⇌ H7O3+ ... H2n+1On+ + H2O ⇌ H2n+3On+1+ The symbols H+(aq) and H3O+(aq) are synonymous and represent all of the possible proton clusters in water, known collectively as "hydronium ion". Chem 1422, Chapter 16

  5. Acids and Bases • Arrhenius definition • acids increase [H+] in water • bases increase [OH-] in water • Thishydronium/hydroxyl definition is restricted to aqueous solutions • Brønsted-Lowry definition • An acid donates “H+” (to a base) • A base accepts “H+” (from an acid) • Thisproton transfer definition is not restricted to aqueous solutions; but in this course, only acid/base aqueous solutions will be studied. Chem 1422, Chapter 16

  6. Proton Transfer Reactions in Water HCl(aq) + H2O(l) → Cl-(aq) + H3O+(aq) HCl donatesa proton to H2O. Therefore, HCl is a BL acid. H2O acceptsa proton from HCl. Therefore, H2O is a BL base. acid ionization reaction NH3(aq) + H2O(l) →NH4+(aq) + OH-(aq) H2O donates a proton to NH3. Therefore,H2O is a BL acid. NH3accepts a proton from H2O. Therefore,NH3 is a BL base. base hydrolysis reaction Water is one of many amphoteric substances which can behave as either an acid or a base. Chem 1422, Chapter 16

  7. Proton Transfer Equilibrium In solution, the proton transfer reaction is in equilibrium. A BL acid must have an ionizable proton, and can be a cation: NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) neutral: H3PO4(aq) + H2O(l) ⇌ H2PO4-(aq) + H3O+(aq) anion: HCO3-(aq) + H2O(l) ⇌ CO32-(aq) + H3O+(aq) A BL base can be a neutral: NH3(aq) + H2O(aq) ⇌ NH4+(aq) + OH-(aq) anion: SO42-(aq) + H2O(aq) ⇌ HSO4-(aq) + OH-(aq) not a cation (why?) Chem 1422, Chapter 16

  8. HAz(aq) + H2O(l) ⇌Az-1(aq) + H3O+(aq) acid ionization reaction base hydrolysis reaction Bz(aq) + H2O(l) ⇌HBz+1 + OH-(aq) base acid base acid AcidBase HCl Cl- HC2H3O2 C2H3O2- H3PO4 H2PO4- H2PO4- HPO42- HPO42- PO43- AcidBase NH4+ NH3 NH3 NH2- NH2- NH2- NH2- N3- AcidBase H2SO4 HSO4- HSO4- SO42- H3O+ H2O H2O HO- HO- O2- Conjugate BL Acid-Base Pairs acid base base acid Every acid/base equilibrium has two conjugate pairs. Some examples of conjugate acid/base pairs: tetraprotic monoprotic diprotictriprotic Chem 1422, Chapter 16

  9. Keq = Kc = Ka = [H+]eq [Az-1]eq [HAz]eq Relative BL Acid Strength HAz(aq) + H2O(l) ⇌Az-1(aq) + H3O+(aq) Keq acid ionization reaction Equilibrium constant Keq = Kc is symbolized as Ka after the name of the reaction: Ka is sometimes called the “acid dissociation constant”. Values of Ka for some weak acids are listed in Appendix Table D-1. Chem 1422, Chapter 16

  10. Strong Acids HCl, HBr, HI, HNO3, HClO3, HClO4, H2SO4 Relative BL Acid Strength HAz(aq) + H2O(l) ⇌Az-1(aq) + H3O+(aq) Ka acid ionization reaction The value of Ka is a measure of the strength of HAz (relative to baseH2O(l)): Weak acid: equilibrium lies far to the left (Ka << 1) Moderate acid: equilibrium is midway (Ka≈ 1) Strong acid: equilibrium lies far to the right (Ka >> 1) There are only 7 common neutral acids which donate almost all (>99.99%; Ka > 104) of their protons to H2O: Chem 1422, Chapter 16

  11. [HBz+1]eq [OH-]eq Kb = [Bz]eq Strong Bases: O2-, NH2-, NH2-, N3-, S2-, H- OH- (from soluble hydroxides), is moderately strong. Relative BL Base Strength H2O(l) + Bz(aq)⇌OH-(aq) + HBz+1(aq) Keq = Kb base hydrolysis reaction The strongest BL bases have Kb >> 1: Kb is sometimes called the “base hydrolysis constant”. Values of Kb for a few weak bases are listed in Appendix Table D-2. Chem 1422, Chapter 16

  12. Kc = Kw = [H+][OH-] H2O(l) + H2O(l) H3O+(aq) + OH-(aq) “water ionization constant” The Autoionization of Water Water is amphoteric, so in all aqueous solutions, including pure water, the following equilibrium is always established: At 25 oC (“room temperature”), Kw = 1.0 × 10-14. However, the autoionization of water is endothermic, so for example, at 37 oC, Kw = 2.36 × 10-14 (in accordance with LeChatelier's principle). Chem 1422, Chapter 16

  13. Kc = Kw = [H+][OH-] H2O(l) + H2O(l) H3O+(aq) + OH-(aq) “water ionization constant” The Autoionization of Water Water is amphoteric, so in all aqueous solutions, including pure water, the following equilibrium is always established: = 1×10-14 In aqueous chemistry, many numbers (e.g., equilibrium constants, concentrations) are very small, so a short-hand has been invented for small numbers. Chem 1422, Chapter 16

  14. The p Scale Define: pH = – log10[H+] so [H+] = 10-pH (“anti-log”) The symbol "p" in pH means "negative power of 10" You can calculate the negative log (and anti-log) of any number: pOH = – log[OH-] [OH-] = 10-pOH pKw = – log Kw Kw = 10-pKw pKa = – log Ka Ka = 10-pKa pKb = – log Kb Kb = 10-pKb On most calculators, the log x and 10x functions are together. Chem 1422, Chapter 16

  15. The p Scale H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq) Kw = [H+][OH-] pKw = -log{[H+][OH-]} = {-log[H+]} + {-log[OH-]} In all aqueous solutions pKw = pH + pOH In pure water and all other neutral aqueous solutions: [H+] = [OH-] and pH = pOH Kw = [H+][OH-] = x2, x = Kw , pH = pOH = ½pKw At 25 oC, pKw = 14.00 so in a neutral solution pH = pOH = 7.00 At 37 oC, pKw = 13.63 so in a neutral solution pH = pOH = 6.81 Chem 1422, Chapter 16

  16. pH @ 25 oC pH Scale Acids and bases are common in our everyday life, and [H+] is a very important factor in bio-metabolism. Thus, the pH of household items is vital information. Chem 1422, Chapter 16

  17. pH @ 25 oC pH Scale The pH and pOH scales do not end at 0 and 14. For example, a 10 M HCl solution has a pH of -1 and a pOH of 15. Chem 1422, Chapter 16

  18. KaKb = Kw For any conjugate acid/base pair: pKa + pKb =pKw For any conjugate acid/base pair: Conjugate Acid/Base Pairs Consider the conjugate acid base pair HAz and Az-1. acid ionization reaction HAz(aq) + H2O(l) ⇌ H3O+(aq) + Az-1(aq) Ka base hydrolysis reaction Az-1(aq) + H2O(l) ⇌HAz(aq) + OH-(aq) Kb Kw = KaKb H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq) Example: for HS-, Ka = 1×10-19 (appendix D-1)so for S2-, Kb = Kw/Ka = 1×105 (a strong base) Chem 1422, Chapter 16

  19. Strengths of Conjugate Pairs in Water The stronger the acid, the weaker its conjugate base. KaKb=Kw For the 7 strong acids, their conjugate bases are so weak that, for 6 of them, their reactions with water are negligible (Kb≈ 0) and they are called spectator ions: Cl-, Br-, I-, NO3-, ClO3-, ClO4- Exception: HSO4- is also a very weak base, but it is not a spectator because it is also a weak acid (Ka = 1.2×10-2). Chem 1422, Chapter 16

  20. Strengths of Conjugate Pairs in Water The stronger the base, the weaker its conjugate acid. KaKb=Kw The soluble IA and IIA metal hydroxides are strong bases. Their conjugate acids, the IA and IIA metal ions, are so weak (Ka ≈ 0) they are spectator ions: Li+, Na+, K+, Rb+, Cs+, Ca2+, Sr2+, Ba2+ By contrast, the rest of the metal hydroxides are weak bases, so their conjugate acids (the metal ions themselves) are relatively strong acids (Al3+ is quite a strong acid!). More about this later. Chem 1422, Chapter 16

  21. Strength of Hydroxyl OH-(aq) is a strong base, but how strong is it? Consider the following equations involving the conjugate pair HA/A- and their constants Ka and Kb: 1 H2O(l) + HA(aq) ⇌ H3O+(aq) + A-(aq) Ka 2 OH-(aq) + HA(aq) ⇌ H2O(l) + A-(aq) Kb-1 = KaKw-1 Equation 1 defines the strength Ka of acid HA with respect to base H2O, but it also defines the strength of base H2O with respect to acid HA: “Kb(H2O)”. Equation 2 defines the strength “Kb(OH-)” of base OH- with respect to acid HA. Chem 1422, Chapter 16

  22. “Kb(OH-)” = Kw-1 “Kb(H2O)” Strength of Hydroxyl OH-(aq) is a strong base, but how strong is it? Consider the following equations involving the conjugate pair HA/A- and their constants Ka and Kb: 1 H2O(l) + HA(aq) ⇌ H3O+(aq) + A-(aq) Ka 2 OH-(aq) + HA(aq) ⇌ H2O(l) + A-(aq) Kb-1 = KaKw-1 “Kb(H2O)” = Ka “Kb(OH-)” = KaKw-1 Since the two “Kb” values are in reference to the same acid HA, they can be compared directly: Hyroxyl is a stronger base than water by a factor of Kw-1 Chem 1422, Chapter 16

  23. pH Measurement pH is measured most accurately with a pH meter. Chem 1422, Chapter 16

  24. pH Measurement pH is measured most accurately with a pH meter. Certain dyes (called indicators) change color as pH changes. Indicators are less precise than pH meters, because they do not show a sharp color change as pH changes. Chem 1422, Chapter 16

  25. pH Calculations The pH of any aqueous solution can be calculated, but the method of calculation depends on what is dissolved in the solution. Each of the following requires a slightly different method: • Strong acids • Strong bases • Weak acids • Weak bases • Salts Chem 1422, Chapter 16

  26. pH Calculations - Strong Acids The seven common strong acids are HCl, HBr, HI, HNO3, HClO3, HClO4, and H2SO4. All of these acids are monoprotic (even H2SO4 is treated as monoprotic in calculations). Strong acids are strong electrolytes because they ionize “completely” in solution. That is, equilibrium lies so far to the right that the acid ionization reaction is treated as a completion reaction: HNO3(aq) + H2O(l) → H3O+(aq) + NO3-(aq) or HNO3(aq) → H+(aq) + NO3-(aq) Chem 1422, Chapter 16

  27. pH Calculations -Strong Acids For a strong monoprotic acid, [H+] is equal to the molarity of the acid. Example: HCl(aq) → H+(aq) + Cl-(aq) For a 0.0346 M hydrochloric acid solution: [Cl-] = [H+] = 0.0346 M and pH = -log(0.0346) = 1.46 Example: H2SO4(aq) → H+(aq) + HSO4-(aq) What is the concentration of a sulfuric acid solution with a pH of 3.26? The concentration of each ion is equal to the initial concentration of the acid: [H+] = [HSO4-] = 10-3.26 = 5.49 × 10-4 M (Note: weak acid HSO4- contributes an insignificant amount of H+ to the solution). Chem 1422, Chapter 16

  28. pH Calculations -Strong Bases • Soluble Hydroxide salts: • IA metal hydroxides of Li, Na, K, Rb, Cs • IIA metal hydroxides of Ca, Sr, Ba (not Be, Mg) • All other metal hydroxides are only slightly soluble and pH depends on their solubility (chapter 17). • Soluble hydroxides are strong electrolytes which dissolve completely in solution NaOH(aq) → Na+(aq) + OH-(aq) Ca(OH)2(aq) → Ca2+(aq) + 2 OH-(aq) Chem 1422, Chapter 16

  29. pH Calculations -Strong Bases • The pH of a strong base is derived from the initial [OH-] and pOH: pH = pKw – pOH. • Example: 0.0346 M KOH ([OH-] = 0.0346 M) • pOH = -log(0.0346) = 1.46 • at 25 oC, 14 - 1.46 = 12.54 = pH • Example: 0.0346 M Ca(OH)2 ([OH-] = 2×0.0346 M) • pOH = -log(0.0692) = 1.16 • at 25 oC, pH = 12.84 Chem 1422, Chapter 16

  30. pH Calculations -Strong Bases • Other strong bases include oxide ion, sulfide ion, etc. • Example: 0.0346 M Na2O(aq) • Na2O(aq) → 2Na+(aq) + O2-(aq) • O2-(aq) + H2O(l) → 2OH-(aq) • pOH = -log(2×0.0346) = 1.16 • at 25 oC, pH = 12.84 Chem 1422, Chapter 16

  31. HAz(aq) H+(aq) + Az-1(aq) [H+] [Az-1] Ka = [HAz] pH Calculations -Weak Acids Weak acids are only partially ionized in aqueous solution; un-ionized acid molecules are in equilibrium with hydronium ions and the conjugate base: Values of Ka are found in Appendix table D-1. Chem 1422, Chapter 16

  32. pH Calculations -Weak Acids Chem 1422, Chapter 16

  33. pH Calculations -Weak Acids Use the ICE table to calculate the pH of a monoprotic weak acid solution, initial concentration Ca. HA(aq) H+(aq) + A-(aq) Ka << 1 Qi = 0 0 < x < Ca pH = -log(x) Chem 1422, Chapter 16

  34. [H+] = pH Calculations -Weak Acids Approximation: if Ka < 10-2 then x will be small, and if Ca is much (> 102 times) larger than Ka, then x << CasoCa-x  Ca Ca >> Ka HA(aq) H+(aq) + A-(aq) Ka << 1 pH = -log(x) For example: calculate 0.15 - 0.00005 BIG± small ≈ BIG = 0.14995 = 0.15 to 2 sig figs Chem 1422, Chapter 16

  35. Sample Problems Ca = 0.158, Ka = 1.5 ×10-5 (from appendix D) Since Ca >> Ka, we can use the approximation 1. Calculate the pH of a 0.158 M solution of butanoic acid. pH = -log(1.54×10-3) = 2.81 compare this to the precise values:[H+] = 1.532×10-3, pH = 2.8147 Chem 1422, Chapter 16

  36. Sample Problems 2. A solution of HF has pH 2.15. What is Ca? Appendix D: Ka = 6.8×10-4 x = [H+] = 10-2.15 = 7.08×10-3 Ka= x2/(Ca - x) Ca = (x2 /Ka)+ x Ca = {(7.08×10-3)2 / 6.8×10-4} + 7.08×10-3 = 0.0808 M No approximation necessary Chem 1422, Chapter 16

  37. Sample Problems 3. The pH of a 0.172 M weak acid solution is 4.81. What is Ka for this acid? Ca = 0.172 x = [H+] = 10-4.81 = 1.55×10-5 Ka= x2/(Ca - x) = (1.55×10-5)2/(0.l72 - 1.55×10-5) = 1.40×10-9 No approximation necessary Chem 1422, Chapter 16

  38. HA(aq) H+(aq) + A-(aq) Weak Acids Percent ionization of a monoprotic acid is defined as the ratio of [H+]eq to Ca For example: 0.05 M acetic acid, %I = 1.90% 0.15 M acetic acid, %I = 1.10% Chem 1422, Chapter 16

  39. Weak Acids %I of Acetic Acid Chem 1422, Chapter 16

  40. Polyprotic Acids • Polyprotic acids have more than one ionizable proton. • The protons are removed in successive equilibria: H2SO3(aq) ⇌ H+(aq) + HSO3-(aq) Ka1 = 1.7 × 10-2 HSO3-(aq) ⇌ H+(aq) + SO32-(aq) Ka2 = 6.4 × 10-8 H3PO4(aq) ⇌ H+(aq) + H2PO4-(aq) Ka1 = 7.5 × 10-3 H2PO4-(aq) ⇌ H+(aq) + HPO42-(aq) Ka2 = 6.2 × 10-8 HPO42-(aq) ⇌ H+(aq) + PO43-(aq) Ka3 = 4.2 × 10-13 Chem 1422, Chapter 16

  41. Polyprotic Acids • Polyprotic acids have more than one ionizable proton. • The protons are removed in successive equilibria: H2SO3(aq) ⇌ H+(aq) + HSO3-(aq) Ka1 = 1.7 × 10-2 HSO3-(aq) ⇌ H+(aq) + SO32-(aq) Ka2 = 6.4 × 10-8 • In polyprotic acids it is increasingly difficult to remove successive protons: Ka1 > Ka2 > Ka3etc. • Usually Ka1 >> Ka2, soapproximatelyall of the [H+]eq comes from the first ionization. • In problems, treat any polyprotic acid (including H2SO4) as a monoprotic acid. Chem 1422, Chapter 16

  42. Polyprotic Acids Note that the first two dissociation contants of citric acid are similar in magnitude, so citric acid cannot be treated as monoprotic; pH calculation of a citric acid solution involves coupled equilibria. Chem 1422, Chapter 16

  43. Weak Bases Weak bases accept a few protons from water: NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Kb << 1 Chem 1422, Chapter 16

  44. Types of Weak Bases • Neutralbases have lone pairs to attract and attach protons. A huge number of these contain nitrogen; nitrogenous bases are called amines. • Amines are related to ammonia (NH3) with one or more H atoms replaced with C groups: CH3NH2 (methylamine), (CH3)2NH (dimethylamine), (CH3)3N (trimethylamine). • Plants produce amines (alkaloids) such as caffeine, nicotine, opium, heroin, cocaine, etc. • Amines in the brain are neurotransmitters and CNS drugs • Amines in the blood act as hormones • About 20 amino acids (with -NH2 groups and –COOH groups) make up all proteins. • Amines called nucleic acids make up the genetic coding molecules DNA, RNA, tRNA, etc. Chem 1422, Chapter 16

  45. Types of Weak Bases • Neutralbases have lone pairs to attract and attach protons. A huge number of these contain nitrogen; nitrogenous bases are called amines. Pyridine Aniline Cytosine Chem 1422, Chapter 16

  46. Types of Weak Bases • Neutralbases have lone pairs to attract and attach protons. <name>amine <name>ammonium cation is a conjugate acid Chem 1422, Chapter 16

  47. HOCl(aq) + H2O(l) OCl-(aq) + H+(aq) Ka = 3.0×10-8 OCl-(aq) + H2O(l) HOCl(aq) + OH-(aq) Kb = ? Types of Weak Bases • Anionic bases have lone pairs andnegative charges to attract and attach protons. • Anionic bases are the conjugates of weak acids. • Example: OCl- (hypochlorite) is the conjugate base of weak acid HOCl (hypochlorous acid): Kb = Kw/Ka = 1.0×10-14/ 3.0×10-8 = 3.3×10-7 For a conjugate acid/base pair, KaKb = Kw Chem 1422, Chapter 16

  48. H2CO3(aq) + H2O(l) HCO3-(aq) + H+(aq) Ka1 = 4.3×10-7 HCO3-(aq) + H2O(l) CO32-(aq) + H+(aq) Ka2 = 5.6×10-11 CO32-(aq) + H2O(l) HCO3-(aq) + OH-(aq) Kb1 HCO3-(aq) + H2O(l) H2CO3(aq) + OH-(aq) Kb2 = Kw/Ka1 Types of Weak Bases • Anionic bases have lone pairs andnegative charges to attract and attach protons. • Anionic bases are the conjugates of weak acids. • Example: CO32- is the conjugate base of HCO3- which is the conjugate base of H2CO3 = Kw/Ka2 Identify the correct conjugate pair! Chem 1422, Chapter 16

  49. B(aq) + H2O(l) HB+(aq) + OH-(aq) A-(aq) + H2O(l) HA(aq) + OH-(aq) Weak Base Problems Quadratic equation? Graphing calculator? pOH = -log(x) pH = 14 - pOH If Kb < 10-2 and Cb > ~100Kb (very common) then x << CbsoCb - x is  Cb Chem 1422, Chapter 16

  50. B(aq) + H2O(l) HB+(aq) + OH-(aq) A-(aq) + H2O(l) HA(aq) + OH-(aq) Weak Bases Percent ionization is defined as: 100 times the ratio of [OH-] to Cb where x = [OH-] If the approximation is valid, then 0.1 M NH3(aq) (“NH4OH”, Kb = 1.8×10-5), %I = 1.3% Chem 1422, Chapter 16

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