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Zumdahl’s Chapter 14

Zumdahl’s Chapter 14. Acids Bases. Acid-Base Models Acidity and K a pH, pOH, and pK a Calculating pH Dominant Sources Polyprotic Acids % dissociation Bases and K b. Conjugate Acids and Bases Structure and Strength Oxoacids Lewis Acids Acid-Base Solution Al Gore Rhythms.

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Zumdahl’s Chapter 14

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  1. Zumdahl’sChapter 14 Acids Bases

  2. Acid-Base Models Acidity and Ka pH, pOH, and pKa Calculating pH Dominant Sources Polyprotic Acids % dissociation Bases and Kb Conjugate Acids and Bases Structure and Strength Oxoacids Lewis Acids Acid-Base Solution Al Gore Rhythms Chapter Contents

  3. Acid-Base Models

  4. Caustic Characteristics • Greek: kaustikos kaiein“to burn” • Acids & bases corrode nearly everything • HA(aq) + H2O(l)  H3O+(aq) + A–(aq) • HA “acid” and A–“conjugate base” • H2O “base” and H3O+“conjugate acid” • H3O+“hydronium ion”no free protons! • Ka = [H+] [A–] / [HA] “acid dissociation contant”

  5. Acidity and Ka • Strong acid: Ka  means [HA]eq = 0 • Weak acid: Ka  0 or [HA]eq  [HA]0 • A–(aq) + H3O+(aq)  HA(aq) + H2O(l) • K=Ka–1 or weak acid is associated with a strong conjugate base and vice versa. • Acids HSO4–HC2H3O2HCN • Ka 1.210–2 1.810–5 6.210–10

  6. pH, pOH, and pKa • pX  – log10(X) • Reduces Kw = 1.01210–14 to pKw = 13.995 to a manageable entity for comparison • Actually pKw = 14 is usually memorized. • 2 H2O(l)  H3O+(aq) + OH–(aq) K = Kw • pH+pOH=pKw because Kw = [H+][OH–] • Acid (pH<7), Neutral (pH=7), Base (pH>7) • E.g., pH(0.1 M acetic acid) = 2.87

  7. Strong acids [H+]eq = [HA]0 pH = p[HA]0 Strong bases [OH–]eq = [BOH]0 pOH = p[BOH]0 pH = 14 – pOH Don’t be fooled by [HA]0< 10–7 Weak Acids Ka = [H+][A–] / [HA] Ka = x2 / ( [HA]0 – x) Ka x2 / [HA]0 x  ( Ka [HA]0 )½ pH = – log10(x) Weak Bases x  ( Kb [MOH]0 )½ pH = 14 - px Calculating pH

  8. Dominant Sources • If more than one H+ source (multiple acids or a polyprotic acid), pKa will usually differ by a few units, and the largest will determine [H+]eq. E.g., 0.01M H3PO4 • pKa: 2.12, 7.21, and 12.32, respectively • Use Ka for 1st H only: pH = 2.24 (quadratic formula) • Determines H3PO4, H2PO4–, and H+ • Use other two pKa to find HPO42– and PO43–

  9. % Dissociation • Weak electrolyte present in molecular form mostly with few ions & will have a small % dissociation unless diluted. • 100%  [ionic form] / [initial amount] • E.g., 0.1 M acetic acid = [initial amount] • [acetate ion]  (0.10 Ka)½ = 0.0013 M • % dissoc.  100%  (Ka/0.1)½ = 1.3% • 0.01 M gives 4.2% shows more ionization upon dilution (but pushes the “5% approx.”)

  10. Bases and Kb • Strong base: [OH–]eq = [MOH]0 • where M=metal and pH = 14 - pOH • Weak bases often amines, :NR3 • where R = H and/or organic groups • Amine steals H+ from H2O to leave OH– • Kb = [NR3H+] [OH–] / [:NR3] • For NH3(ammonia, not an amine), Kb = 1.810–5 • Same pH rules & approximations apply.

  11. Conjugate Acids and Bases • “Salt of a weak (acid/base) is itself a weak (base/acid).” • Example of a weak base, NH3 • NH4+ + OH– NH3 + H2O K = Kb–1 • H2O  H+ + OH– K = Kw(add eqns) • NH4+  NH3 + H+ Kconjugate = Kw / Kb • Kacid(NH4+) = 10–14/1.810–5 = 5.510–10 • Weaker parent yields stronger conjugate!

  12. Conjugate pH Calculations • Salts of strong parents are neutral; those of very weak bases, very acidic: • Fe(H2O)63+ Fe(H2O)5OH2+ + H+ • pKa1 = 2.2 or Ka1 = 6.310–3 (pKb = 11.8) • pH of 3 M Fe(NO3)3 ? (HNO3 strong; ignore!) • Ka1 = [FeOH2+] [H+] / [Fe3+] = x2 / (3-x) • x  [3Ka1]½ = 0.14 (5% OK)  pH = 0.86 • pKa2 = 3.5, pKa3 = 6, pKa4 = 10

  13. NON-Dominant Situations • In that Fe complex, pKa1 and pKa2 are very close (2.2 and 3.5), so the second conjugate may be important: • Fe(H2O)5OH2+ Fe(H2O)4(OH)2+ + H+ • This will certainly be the case at lower formal iron concentrations, F, since % dissociation is greater there. Try F=0.01

  14. Simplifing Approximations • Even at F=0.01, we can safely ignore pKa3, pKa4, and pKw, since they are dominated by pKa1 and pKa2. • Abbreviate those equilibria as • A3+ B2+ + H+ K1 = B H / A • B2+  C+ + H+ K2 = C H / A • But those are 2 eqns in 4 unknowns!

  15. 2 Additional Conditions! • Conservation of iron requires that • F = A + B + C • Conservation of charge must include a spectator anion for A3+ such as 3 Cl– • 3 A3+ + 2 B2+ + C+ + H+ = 3 F (conc. Cl–) • Notice that higher charges get scaled by their charge for proper balance.

  16. Solve 4 Nonlinear Equations • Original equations: • F = A + B + C • 3F = 3A + 2B + C + H • B H = K1 A • C H = K2 B • Remove A by substituting from 1st eqn • A = F – B – C

  17. Solve 3 Nonlinear Equations • Remaining after substitution • 3F = 3F – 3B – 3C + 2B + C + H • B H = K1 ( F – B – C ) • C H = K2 B • Simplify • H = B + 2C • B H = K1 F – K1 B – K1 C • C H = K2 B  substitute C = K2 B / H

  18. Solve 2 Nonlinear Equations • H = B + 2 K2 B / H • B H = K1 F – K1 B – K1 K2 B / H • Simplify • H2 = B H + 2 K2 B  B = H2 / (H + 2K2) • B H2 = K1 F H – K1 B H – K1 K2 B • Substitute for B • H3 + K1 H2 – K1 (F – K2)H – 2 K1 K2 F = 0 • Only H remains as an unknown!

  19. Solve 1 Cubic Equation • Use the dominant solution as first guess in an iteration to the non-dominant one. • A3+ B2+ + H+ K1 = B H / A • K1 = x2 / (F – x) • x2 + K1 x – F K1 = 0 • x = ( – K1 + [K12 + 4F K1]½) / 2 • x = 5.310–3  H • F = 0.01, K1 = 610–3, and K2 = 310– 4

  20. H3 + 610–3 H2 – 5.810–3 H – 3.610–8 = 0Hout = (3.6 10–8 + 5.810–3 Hin – 610–3 Hin2)1/3  [H+] = 5.5810– 3M and by B = H2 / (H + 2K2), [B2+] = 5.0410– 3M and by C = K2 B / H, [C+] = 2.7 10– 3M which makes [A3+] = 4.710– 3M

  21. More Conjugate Calculations • HCO3– H+ + CO32– pKa2 = 10.25 • pH of 0.10 M Na2CO3 ? (NaOH strong; ignore) • CO32– + H2O  HCO3– + OH– K = Kw/Ka2 • K = [HCO3–] [OH–] / [CO32–] = x2 / (0.10-x) • x  [0.10  1.810–4 ]½ = 4.210–3(5% OK) • pOH = 2.37  pH = 14.00 – 2.37 = 11.63 • Sodium carbonate is in laundry soap.

  22. Acid Structure and Strength • Higher polarity liberates protons. • While all hydrogen halides are strong acids (save HF), acidity lessens with halide size. • Even non-adjacent electronegativity drains e– density, liberating protons. • HClO is weak but HClO4 is strong. • (the H is attached to only 1 of the oxygens) • Cl3CCO2H is much stronger than CH3CO2H

  23. Oxoacids • German: O2 = sauerstoff or the stuff that makes sour oxoacids, –X–O–H • True when X is a fairly electronegative non-metal (halogen, N, P, S) or a metal in a high oxidation state (e.g., H2CrO4) • True as nonadjacent electronegativity aids polarity (e.g., CH3CO2H)

  24. Oxygen-containing Bases • M – O – H • Metals in lower oxidation states tend to give up the OH– and not the H+ • Sr(OH)2 Sr2+ + 2 OH– • Al(OH)3 is basic; Al3+ is acidic. • Water is amphoteric too, acting as a base by becoming H3O+, for example. • Brønsted definition: water accepts the proton.

  25. Lewis Acids and Bases • Water is both acid & base by Lewis too. • Base via H2O: + H+ H3O+ • donates lone pair e– to proton • Acid via H2O + S2– OH– + HS– • accepts the one of sulfide’s electrons • Lewis’ is the most general definition • E.g., Cu2+ + 6 :NH3  Cu(NH3)62+ • Cu’s the acid and :NH3’s the base.

  26. Know the significant species for the conditions. Ignore the rest. For quantitative reactions, keep only the products. Find acids & bases. Solve controlling K Write K expression. Set initial conditions for smallest change. Define all species in terms of x (change). Find x solving K exp. Check any approx. Find pH & [species] Acid-Base Algorithm

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