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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 & ions with multiple bonds that can be shifted to accept electrons • 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. Auto-ionization of Water (Kw) • Water ionizes to a very small extent (Kw=10-14 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 = 10-14 at 25°C • When [H3O+]=[OH-], the solution is neutral. 15.5. pH is a measure of the acidity of a solution

  15. 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

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

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

  18. 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 6.6×10-13 12.18 3.9×10-9 5.593 15.5. pH is a measure of the acidity of a solution

  19. 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

  20. 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

  21. 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

  22. 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) =0.00 • pH = 14.00 15.6. Strong acids and bases are fully dissociated in solution

  23. 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 23

  24. 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 24

  25. 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 25

  26. 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 26

  27. Learning Check Determine the pH of 0.10M 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 27

  28. 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 28

  29. Learning Check: Determine the pH of 0.10M 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 29

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