1 / 60

Chapter 16 Acids and Bases

Chapter 16 Acids and Bases. Contents in Chapter 16. 16-1 Arrhenius Theory: A Brief Review 16-2 Brønsted–Lowry Theory of Acids and Bases 16-3 Self-Ionization of Water and the pH Scale 16-4 Strong Acids and Strong Bases 16-5 Weak Acids and Weak Bases 16-6 Polyprotic Acids

leo-harris
Download Presentation

Chapter 16 Acids and Bases

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 16Acids and Bases

  2. Contents in Chapter 16 16-1 Arrhenius Theory: A Brief Review 16-2 Brønsted–Lowry Theory of Acids and Bases 16-3 Self-Ionization of Water and the pH Scale 16-4 Strong Acids and Strong Bases 16-5 Weak Acids and Weak Bases 16-6 Polyprotic Acids 16-7 Ions as Acids and Bases 16-8 Molecular Structure and Acid–Base Behavior 16-9 Lewis Acids and Bases

  3. H2O NaOH(s) → Na+(aq) + OH-(aq) 16-1 Arrhenius Theory: A Brief Review • Acid: A substance that provides H+ in aqueous solution, e.g., HCl • Base (alkalis): A substance that provides OH– in aqueous solution, e.g., NaOH H2O HCl(g) → H+(aq) + Cl-(aq) Na+(aq) + OH-(aq)+ H+(aq) + Cl-(aq) → H2O(l) + Na+(aq) + Cl-(aq) H+(aq)+ OH-(aq) → H2O(l) • Arrhenius theory did not handle non OH– bases such as ammonia (NH3). • Neutralization reaction: Combination of hydrogen ions (H+) and hydroxide ions (OH–) to form water.

  4. Conjugate base of NH4+ Conjugate Acid of NH3 Conjugate base of H2O Conjugate acid of OH− 16-2 Brønsted–Lowry Theory of Acids and Bases*** • Definition: • Acid: the substance act as H+ donor • Base: the substance act as H+ acceptor

  5. Ionization constants*** • Acid ionization constant CH3COOH + H2O CH3COO– + H3O+ Conjugate acid Conjugate base Conjugate base Conjugate acid • Base ionization constant Conjugate base Conjugate acid NH3 + H2O  NH4+ + OH– Conjugate acid Conjugate base *H2O is an amphiprotic (amphoteric) substance, act as either an acid or a base.

  6. Strengths of conjugate acid-base pairs • The stronger an acid, the weaker its conjugate base. (The stronger a base, the weaker its conjugate acid.) • An acid-base reaction is favored in the direction from the stronger member to the weaker member of each conjugate acid-base pair.

  7. Brønsted–Lowry acid–base reaction: weak acid • CH3COOH is only slightly ionized. • Reverse reaction proceeds to a greater extent than does the forward reaction. • H3O+ is a stronger acid than CH3COOH and CH3COO− is a stronger base than H2O.

  8. Brønsted–Lowry acid–base reaction: strong acid • HCl is essentially completely ionized. • The forward reaction proceeds almost to completion. • H3O+ is a weaker acid than HCl and Cl− is a much weaker base than H2O.

  9. More about strengths of acid and bases • Leveling (solvent) effect: The solvent's ability to level the effect of a strong acid (or strong base) dissolved in it. e.g., HI and HBr are leveled to the same acidic strength in H2O. • Differentiating (solvent) effect: The solvent's ability to differentiate the acidic (or basic) strength. Example: • Ka (Kb) values are used to compare the strengths of weak acids (bases).

  10. 16-3 Ionization of Water and the pH Scale • Self ionization of water Ion-product of water, KW: KW = [H3O+][OH–] At 25oC, KW = 1x10–14 KW applies to all aqueous solutions—acids, bases, salts, and nonelectrolytes—not just to pure water.

  11. p function: –log • pH = –log [H3O+] [H3O+] = 10–pH pOH = –log [OH–] [OH–] = 10–pOH • Acidic solution: [H3O+] > [OH–] pH < pOH Basic solution: [H3O+] < [OH–] pH > pOH • Aqueous solution at 25oC: pKW = pH + pOH = 14.00 • Aqueous solution at 25oC: acidic solution: [H3O+] > 1.0×10–7 pH < 7.00 basic solution: [H3O+] < 1.0×10–7 pH > 7.00 neutral solution: [H3O+] = 1.0×10–7 pH = 7.00

  12. The pH scale and pH values of some common materials • pH = −1 ([H3O+] ≈ 10 M) and pH = 15 ([OH−] ≈ 10 M) are possible. • The pH scale is useful only in the range 2 < pH < 12, because the molarities of H3O+ and OH− in concentrated acids and bases may differ from their true activities.

  13. 16-4 Strong Acids and Strong Bases • The contribution due to the self-ionization of water can generally be ignored (unless the solution is extremely dilute), i.e., for strong acids and bases, dissociated completely. Therefore, [H3O+]  CHCl CHCl: initial concentration of HCl [OH–]  CNaOH CNaOH: initial concentration of NaOH.

  14. For extremely dilute solution of a strong acid and strong base, 1.0 x 10–8 M HCl for example:

  15. 16-5 Weak Acids and Weak Bases • Identifying Weak Acids and Bases

  16. [H3O+] from HA Degree of ionization = [HA] originally [H3O+] from HA 100% Percent ionization = [HA] originally • Percent Ionization (A weak acid HA for example) HA + H2O H3O+ + A-

  17. Equilibrium of monoprotic acid, HA HA + H2O  H3O+ + A– CHA– x x x Therefore • 5% rule: Assume CHA – x  CHA * CHA – x  CHA, using:

  18. Equilibrium of monobasic base, B B + H2O  HB+ + OH– CB– x x x Therefore • 5% rule: Assume CB – x  CB * CB – x  CB, using:

  19. 16-6 Polyprotic Acids

  20. Diprotic acid H2A + H2O  HA– + H3O+ HA–+ H2O  A2– + H3O+ For conjugate base: A2–+ H2O  HA– + OH– HA–+ H2O  H2A + OH– Ka1 x Kb2 = Kw Ka2 x Kb1 = Kw

  21. Triprotic acid Phosphoric acid for example: H3PO4 + H2O  H2PO4– + H3O+ H2PO4–+ H2O  HPO42– + H3O+ HPO42–+ H2O  PO43– + H3O+ Ka1 x Kb3 = Kw Ka2 x Kb2 = Kw Ka3 x Kb1 = Kw • Ionization constants for polyprotic acid progressively decrease: Ka1 > Ka2 > Ka3 > ….. • Except in very dilute solutions, essentially all of the H3O+ ions come from the first ionization step alone.

  22. EXAMPLE 16-9 Calculating Ion Concentrations in a Polyprotic Acid Solution For a 3.0 M H3PO4, calculate (a) [H3O+]; (b) [H2PO4–]; (c) [HPO42–]; (d) [PO43–]. Solve (a) assume that all H3O+ the forms in the first ionization step H3PO4 + H2O  H2PO4– + H3O+ Initial conc. 3.0 - - Change –x ＋x ＋x Equilibrium (3.0 – x) x x x2 = 0.021 x = [H3O+] = 0.14 M Check: (x/CH3PO4) x 100% = 4.7% < 5%, OK!!

  23. (b) H3PO4 + H2O  H2PO4– + H3O+ Equilibrium (3.0 – x) x x [H2PO4–]  [H3O+] =0.14 M (c) H2PO4–+ H2O  HPO42– + H3O+ Since [H2PO4–]  [H3O+], therefore [HPO42–] = 6.3 x 10–8 M

  24. (d) HPO42–+ H2O  PO43– + H3O+ Since [H3O+] = 0.14 M and [HPO42–] = 6.3 x 10–8 M [PO43–] = 1.9x 10–19 M

  25. A Somewhat Different Case: H2SO4 • (1)H2SO4 + H2O  HSO4– + H3O+ Ka1 103 • CH2SO4 CH2SO4 CH2SO4 • (2) HSO4–+ H2O  SO42– + H3O+ • CH2SO4－x x x • Concentrated solutions(>0.5 M H2SO4): H3O+ is predominated by first ionization step. e.g., 1.00 M H2SO4, [H3O+]  1.00 M. • Very dilute solutions(< 0.001 M H2SO4): both ionization steps are nearly completely dissociated, e.g., 0.001 M H2SO4, [H3O+]  0.002 M, [SO42–] 0.001 M. • Intermediate concentrations (0.001 M< CH2SO4 <0.5 M), first ionization step is completely dissociated, thesecond ionization step is partially dissociated.

  26. Check: (x/CH2SO4) x 100% = 2.2% < 5%, OK!!

  27. 16-7 Ions as Acids and Bases • Hydrolysis: The reaction between an ion and water. (1) NaCl(aq) Na+(aq) + Cl–(aq) Neutral Na+(aq) + H2O  Cl–(aq) + H2O  (2) NH4Cl(aq) NH4+(aq) + Cl–(aq) Acidic NH4+(aq) + H2O  NH3(aq) + H3O+(aq) Cl–(aq) + H2O  (3) CH3COONa(aq) Na+(aq) + CH3COO–(aq) Basic Na+(aq) + H2O  CH3COO–(aq) + H2O  CH3COOH(aq) + OH–(aq) (4) CH3COONH4(aq) NH4+(aq) + CH3COO– (aq) ????? NH4+(aq) + H2O  NH3(aq) + H3O+(aq) CH3COO–(aq) + H2O  CH3COOH(aq)+ OH–(aq) X X X X

  28. The pH of Salt Solutions

  29. (a) NaOCl(aq) Na+(aq) + OCl–(aq)Basic Na+(aq) + H2O  OCl–(aq) + H2O  HOCl(aq) + OH–(aq) X (b) KCl(aq) Na+(aq) + Cl–(aq)Neutral K+(aq) + H2O  Cl–(aq) + H2O  X X (c) NH4NO3(aq) NH4+(aq) + NO3–(aq)Acidic NH4+(aq) + H2O  NH3(aq) + H3O+(aq) NO3–(aq) + H2O  X

  30. 16-8 Molecular Structure and Acid–Base Behavior • Strengths of Binary Acids • Homolytic dissociation vs. Heterolytic dissociation • Homolytic dissociation • HX  H + X D(H–X) • Heterolytic dissociation • HX  H+ + X– D(H+X– ) • Bond dissociation energy for the gas phase ionization reaction D(H+X– ) = D(H–X) + IE(H) + ΔHea

  31. Bond dissociation energies (kJ/mol ) and Ka values for some binary acids

  32. Comparing binary acids of X in the same row* • The higher polarity of the bond (the larger ΔEN (electronegativity difference)), the stronger acid. • Small ΔEN has more covalent character • Large ΔEN has more ionic character Molecule CH4 NH3 H2O HF ΔEN 0.4 0.9 1.4 1.9 Acidity: CH4 < NH3 < H2O < HF

  33. Comparing binary acids of X the same group The larger bond length (the larger X radius, the weaker H—X bond) the stronger acid. Molecule HF HCl HBr HI BE (kJ/mol) 565 431 364 297 Anion radius (pm) 136 181 195 216 Ka6.6x10–4 ~106 ~108 ~109 Acidity: HF < HCl < HBr < HI *** HF(aq) is a weak acid Other example: Acidity: H2O < H2S < H2Se < H2Te

  34. Strengths of Oxoacids (H–O–EOn) • Comparing the EN of E • The larger EN (electronegativity) of E, the weaker H–O bond, the stronger acid. Molecule H–OI H–OBr H–OCl EN 2.5 2.8 3.0 Ka 2.3x10–11 2.5x10–9 2.9x10–8 Acid strength HOI < HOBr < HOCl More examples: Acid strength: H2SeO3 < H2SO3 HBrO4< HClO4

  35. Comparing the n of On • The more number (n) of terminal O, the weaker protonated H–O bond, the stronger acid. *O is the element has second higher electronegativity. More examples: Ka1(H2SO3)< Ka1(H2SO4)

  36. Strengths of R–COOH vs. R–OH • Ethoxide ion is a much stronger base than is acetate ion. • The stronger the conjugate base, the weaker the corresponding acid.

  37. Strengths of carboxylic acids (R–COOH) • Comparing electron-donating ability of R • Electron-donating ability: –C2H5 > –CH3 > –H • The higher electron-donating ability, the weaker acid strength. Example Acidity: CH3CH2–COOH < CH3–COOH < H–COOH Ka: CH3CH2–COOH < CH3–COOH < H–COOH pKa: CH3CH2–COOH > CH3–COOH > H–COOH

  38. Comparing electron-withdrawing ability of R • Electron withdrawing ability: Cl > Br > I • The higher electron withdrawing ability, the stronger acid strength Example

More Related