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Ch16 and 17 - Acids and Bases

Ch16 and 17 - Acids and Bases. Brady & Senese, 5th Ed. Index. 15.1. Brønsted-Lowry acids and bases exchange protons 15.2. Strengths of Brønsted acids and bases follow periodic trends 15.3. Lewis acids and bases involve coordinate covalent bonds

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Ch16 and 17 - Acids and Bases

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  1. Ch16 and 17 - Acids and Bases Brady & Senese, 5th Ed

  2. Index 15.1. Brønsted-Lowry acids and bases exchange protons 15.2. Strengths of Brønsted acids and bases follow periodic trends 15.3. Lewis acids and bases involve coordinate covalent bonds 15.4. Elements and their oxides demonstrate acid-base properties 15.5. pH is a measure of the acidity of a solution 15.6. Strong acids and bases are fully dissociated in solution

  3. Brønsted Acid/Base Reactions Transfer H+ • Products differ by one H+ from the reactants to form conjugate • Conjugate acid-base pairs differ by one H+. • HCN(aq) + OH-(aq)↔H2O(l) + CN-(aq) • Note that in the conjugate pairs, the acid has one more H+ than its conjugate base Brønsted base conjugate acid Brønsted acid conjugate base 15.1. Brønsted-Lowry acids and bases exchange protons

  4. Identify the Conjugate Partner for Each Learning Check Cl- NH4+ C2H3O2- HCN F- 15.1. Brønsted-Lowry acids and bases exchange protons

  5. Your Turn! How many of the following pairs are conjugate pairs: • HCN/CN- ii.HCl/Cl- iii. H2S/S2- • 1 • 2 • 3 • None of them are conjugate 15.1. Brønsted-Lowry acids and bases exchange protons

  6. Amphoteric Substances • Amphoteric (aka amphiprotic) substances are able to act either as Brønsted acid or Brønsted base • The following materials are amphoteric: • Water • Amino acids • Anions of polyprotic acids 15.1. Brønsted-Lowry acids and bases exchange protons

  7. Acid/Base Strengths In Aqueous Solution • Hydronium ion (H3O+) is the strongest acid in solution: stronger acids react completely with water to give H3O+ • Hydroxide ion (OH-) is the strongest possible base in solution: stronger bases react completely with water to give OH- • This leveling effect causes very strong acids/bases to have nearly equal strength in aqueous solution 15.1. Brønsted-Lowry acids and bases exchange protons

  8. Conjugate Pairs Have Reciprocal Strengths • The stronger the acid, the weaker its conjugate base • The stronger the base, the weaker its conjugate acid • Strong acids are ionized 100%: their anions are extraordinarily poor bases- most are essentially neutral 15.1. Brønsted-Lowry acids and bases exchange protons

  9. Acid Strength Binary Acids: Acidity increases from left to right in a period, increasing electronegativitymakes the H-X bond more polar (weaker) Acidity increases from top to bottom in a group, the increasing length of the H-X bond leads to poor orbital overlap and a weaker bond Oxoacids: Acidity increases as number of oxygens increases (presence of high EN oxygen draws electrons away from other bonds) Oxoacids with at least 2 more oxygens than hydrogens tend to be strong. Others tend to be weak. 15.2. Strengths of Brønsted acids and bases follow periodic trends 9

  10. δ- δ- δ+ δ+ δ- δ+ δ+ δ+ δ+ δ- δ- Learning Check • Which is stronger? • H2S or H2O • CH4 or NH3 • HF or HI 15.2. Strengths of Brønsted acids and bases follow periodic trends

  11. Lewis Acids and Bases • Lewis acids (electron pair acceptors) • molecules & ions with incomplete valence shells • molecules or ions with central atoms that can accommodate additional electrons • Lewis bases (electron pair donors) • molecules & ions that have complete valence shells with unshared electrons • generally donate lone pairs or pi bond electrons 15.3. Lewis acids and bases involve coordinate covalent bonds

  12. Lewis Acid/Base Reactions • Lewis acids accept an electron pair to form coordinate covalent bonds • Lewis bases donate lone pairs of electron to form coordinate covalent bonds • Neutralization is the formation of a coordinate covalent bond between the donor and acceptor 15.3. Lewis acids and bases involve coordinate covalent bonds

  13. + - - : : : : Learning Check Identify the Lewis acid and base in the following • NH3 + H+↔NH4+ • F- + BF3↔BF4- 15.3. Lewis acids and bases involve coordinate covalent bonds

  14. Hydrated Metal Ions Can Act as Weak Acids • Electron deficiency of metal cations causes them to inductively attract electrons from the hydrating water molecules • M(H2O)mn+ + H2O↔M(H2O)m-1OH(n-1)+ + H3O+ • The higher the charge density (greater charge concentrated in smaller radius) the more acidic the metal. • Acidity increases left to right in a period. • Acidity decreases top to bottom in a group. 15.4. Elements and their oxides demonstrate acid-base properties

  15. Electron density drawn toward Al3+ Nearby H2O acts as base H2O H3O+ Al(H2O)63+ Al(H2O)5OH2+ The acidic behavior of the hydrated Al3+ ion.

  16. Oxides Can Be Acidic or Basic • Most metal oxides are basic • Non-metal oxides are acidic anhydrides • SO2+H2O →H2SO3 15.4. Elements and their oxides demonstrate acid-base properties

  17. Auto-ionization of Water (Kw) • Water ionizes to a very small extent (Kw=1.0x10-14 at room temperature) according to the following reaction: • H2O(l) + H2O(l)↔ H3O+(aq) + OH-(aq) • Since water is present in all aqueous solutions, the Kw equilibrium exists in all aqueous solutions. • Kw=[H3O+ ][OH-] • Kw = 1.0 x 10-14 at 25°C • When [H3O+]=[OH-], the solution is neutral. 15.5. pH is a measure of the acidity of a solution

  18. pH and Kw • p -log • pHis defined for aqueous solutions only, and is temperature dependent • pH=-log[H3O+] • 10-pH = [H3O+] • It derives from the auto ionization of water. • Kw=[H3O][OH-] • pKw= pH + pOH • pH>7 is basic; pH=7 is neutral; pH<7 is acidic 15.5. pH is a measure of the acidity of a solution

  19. Figure 18.5 The pH values of some familiar aqueous solutions pH = -log [H3O+]

  20. Indicators Help Us Estimate pH 15.5. pH is a measure of the acidity of a solution

  21. Complete the following with the missing data at 25 deg C Learning Check 11.51 3.1×10-12 4.3×10-10 4.64 6.7×10-13 12 3.92×10-9 5.59 15.5. pH is a measure of the acidity of a solution

  22. Strong Acids Ionize 100% in Water • As the substances are placed into water, they form H3O+ . • Water’s contribution to [H3O+] is negligible • The pH can be calculated from the concentration of H3O+ produced by the strong acid • The reaction of strong acids occurs irreversibly, so we show the reaction with a → instead of using a double arrow 15.6. Strong acids and bases are fully dissociated in solution

  23. Learning Check What is the pH of 0.1M HCl • HCl(aq) + H2O(l) →H3O+(aq) + Cl-(aq) • 0.1 - 0 0 I • -0.1 - 0.1 0.1 C • 0 - 0.1 0.1 end • pH = -log(0.1) = 1 15.6. Strong acids and bases are fully dissociated in solution

  24. Strong Bases Dissociate 100% In Water • They are strong electrolytes that form OH- when dissolved • pOH can be calculated from the [OH-] from the solution • Water’s contribution is negligible if the base is sufficiently concentrated [OH-]>10-7M 15.6. Strong acids and bases are fully dissociated in solution

  25. Learning Check What is the pH of 0.5M Ca(OH)2? • Ca(OH)2(aq) → Ca2+(aq) + 2OH-(aq) • 0.5 0 0 I • -0.5 +0.5 + 0.5×2 C • 0 0.5 1.0 end • pOH = -log(1.0) =0 • pH = 14 15.6. Strong acids and bases are fully dissociated in solution

  26. Index 16.1. Ionization constants can be defined for weak acids and bases   16.2. Calculations can involve finding or using Ka and Kb   16.3. Salt solutions are not neutral if the ions are weak acids or bases   16.4. Simplifications fail for some equilibrium calculations 16.5. Buffers enable the control of pH   16.6. Polyprotic acids ionize in two or more steps   16.7. Acid-base titrations have sharp changes in pH at the equivalence point 26

  27. Weak Acids & (Ka) Weak acids only partially ionize The equilibrium constant Ka is used to indicate the degree of ionization Ka are tabulated in table 16.1, as are pKa Ka=10-pKa pKa=-log(Ka) Acid strength increases as Ka increases and pKa decreases For Conjugate Pairs Kw=Ka x Kb 16.1. Ionization constants can be defined for weak acids and bases 27

  28. Bases & Kb Weak bases only partially ionize The equilibrium constant Kb is used to indicate the degree of ionization Kb and pKb are tabulated in table 16.2 Kb=10-pKb pKb=-log(Kb) Base strength increases as Kb increases and pKb decreases For Conjugate Pairs Kw=Ka x Kb 16.1. Ionization constants can be defined for weak acids and bases 28

  29. Determining the pH Of Aqueous Weak Acid Solutions Dominant equilibrium is Ka reaction write the net ionic equation look up the Ka value for the acid set up ICE table solve for x Calculate pH from the hydronium concentration at equilibrium 16.2. Calculations can involve finding or using Ka and Kb 29

  30. Simplifications: Dropping x Term Solving ICE tables sometimes requires the quadratic equation Other times simplifications may be made If the K for the reaction is small, the degree of ionization will also be small The concentration of a reactant may sometimes be simplified to the initial concentration You must perform a proof to show that the dropped x was justified (less than 5% error) 16.2. Calculations can involve finding or using Ka and Kb 30

  31. Learning Check Determine the pH of 0.1M solutions of: HC2H3O2 Ka=1.8×10-5 HCN Ka=6.2×10-10 0.1M N/A 0 0 -x -x +x +x X=1.34(10-3)M (0.1-x)≈0.1 N/A x x pH=2.87 0.1M N/A 0 0 -x -x +x +x X=7.87(10-6)M (0.1-x) ≈ 0.1 N/A x x pH=5.10 16.2. Calculations can involve finding or using Ka and Kb 31

  32. Determining The pH Of Base Solutions: Dominant equilibrium is Kb reaction write the net ionic equation set up ICE table for starting quantities solve for x Calculate pOH from the [OH-] at equilibrium, and convert to pH 16.2. Calculations can involve finding or using Ka and Kb 32

  33. Learning Check: Determine the pH of 0.1M solutions of: N2H4 Kb=1.7×10-6 NH3 Kb=1.8×10-5 0.1M N/A 0 0 -x -x +x +x (0.1-x) ≈ 0.1 N/A x x X=4.12(10-4)M pOH=3.38 pH=10.62 0.1M N/A 0 0 -x -x +x +x (0.1-x) ≈ 0.1 N/A x x X=1.34(10-3)M pOH=2.87 pH=11.13 16.2. Calculations can involve finding or using Ka and Kb 33

  34. Percent Ionization Weak acids ionize some amount less than 100%. Generally calculated from Ka and ICE table Percentage ionized varies with starting concentration, even for the same acid 16.2. Calculations can involve finding or using Ka and Kb 34

  35. Learning Check Determine the % ionization of 0.2M solution of HC2H3O2 Ka=1.8×10-5 0.2M N/A 0 0 -x -x +x +x (0.2-x) ≈ 0.2 N/A x x x=1.90×10-3M 0.95 % ionized 16.2. Calculations can involve finding or using Ka and Kb 35

  36. Learning Check Determine the % ionization of 0.1M solution of HC2H3O2 Ka=1.8×10-5 0.1M N/A 0 0 -x -x +x +x (0.1-x) ≈ 0.1 N/A x x x=1.34 x 10-3M 1.3 % ionized 16.2. Calculations can involve finding or using Ka and Kb 36

  37. Buffers enable the control of pH Are a combination of a weak acid and its conjugate weak base or a weak base and its conjugate acid Are a special case of the common ion effect (the presence of the conjugate base inhibits the ionization of the Brønsted acid) Are resistant to changes in the pH Any acid added to the mixture is reacted by the base of the buffer pair added bases are reacted by the conjugate acid 16.5. Buffers enable the control of pH 37

  38. Simplification Of Buffer pH Calculations – Henderson-Hasslebach Equation 16.5. Buffers enable the control of pH 38

  39. Learning Check What is the pH of a buffer made from 0.5M HF (Ka=6.8×10-4) and 0.2M NaF? pKa=-log(6.8×10-4) pH=2.8 16.5. Buffers enable the control of pH 39

  40. Buffer Characteristics Buffer capacity is the ability to compensate for added acid or base, defined for each component acid buffer capacity=moles of conjugate base present base buffer capacity= moles of acid present. If acid is added, remove from base and add to the acid of the pair If base is added, remove from acid of the pair and add to the acid of the pair. 16.5. Buffers enable the control of pH 40

  41. Your Turn! Which of the following could be used as a buffering pair? NH4Cl/NH3 HCl/NaCl NaOH/H2O All of these 16.5. Buffers enable the control of pH 41

  42. Buffer Characteristics Consider 500. mL of a sodium acetate/acetic acid buffer in which the concentration of sodium acetate is 0.500M and that of acetic acid is 0.100M. What is the pH after 10.0 mL of 0.100M NaOH is added? (Ka acetic acid= 1.8×10-5) 16.5. Buffers enable the control of pH 42

  43. Selecting A Buffer To choose buffer system, try to find a system whose pKa is close to desired system pH (pH = pKa is ideal) Then, calculate necessary ratio of components to obtain desired pH. Consider toxicity of the materials! 16.5. Buffers enable the control of pH 43

  44. Learning Check Choose a buffer system to maintain a pH of 5.5. Describe its composition completely. select acid with pKa≈5.5. thus, look for Ka ≈3.2×10-6 H2C8H4O4, phthalic acid has Ka2=3.9×10-6. We could use NaHC8H4O4 and Na2C8H4O4. 16.5. Buffers enable the control of pH 44

  45. Polyprotic Acids Have more than one ionizable hydrogen Each successive ionization has a specific ionization constant (Ka) Each successive proton is significantly less acidic than the last For H3PO4: H3PO4(aq) + H2O(l)↔H3O+(aq) + H2PO4-(aq) Ka1 H2PO4-(aq) + H2O(l)↔H3O+(aq) + HPO42-(aq) Ka2 HPO42-(aq)+ H2O(l)↔H3O+(aq) + PO43-(aq) Ka3 Ka1 >> Ka2 >> Ka3 16.6. Polyprotic acids ionize in two or more steps 45

  46. Learning Check: What is the pH of a 0.1M solution of H3PO4? Ka1=7.1×10-3; Ka2=6.3×10-8; Ka3=4.5×10-13 16.6. Polyprotic acids ionize in two or more steps 46

  47. Titration The controlled addition of one substance (titrant) to a known quantity of another substance (analyte) until the stoichiometric requirements are met Equivalence point - the volume of titrant needed to achieve an equimolar concentration of titrant and of titrate Endpoint - the volume of titrant necessary to achieve the stoichiometric ratio of reactants The endpoint may be the last of several equivalence points in the case of polyprotic acids 16.7. Acid-base titrations have sharp changes in pH at the equivalence point 47

  48. Curve for a strong acid-strong base titration

  49. pH = 8.80 at equivalence point pKa of HPr = 4.89 methyl red [HPr] = [Pr-] Titration of 40.00mL of 0.1000M HPr with 0.1000M NaOH Curve for a weak acid-strong base titration

  50. PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000M NaOH: PLAN: The amounts of HPr and Pr- will be changing during the titration. Remember to adjust the total volume of solution after each addition. SOLUTION: (a) Find the starting pH using the methods of Chapter 18. Amount (mol) HPr(aq) + OH-(aq) Pr-(aq) + H2O(l) Calculating the pH During a Weak Acid- Strong Base Titration (a) 0.00mL (b) 30.00mL (c) 40.00mL (d) 50.00mL Ka = [Pr-][H3O+]/[HPr] [Pr-] = x = [H3O+] [Pr-] = x = [H3O+] x = 1.1x10-3 ; pH = 2.96 (b) Before addition 0.004000 - 0 - Addition - 0.003000 - - After addition 0.001000 0 0.003000 -

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