Zumdahl's Chapter 14

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					Zumdahl’s
Chapter 14
       Acids
       Bases
Chapter Contents
  Acid-Base Models                 Conjugate Acids and
  Acidity and Ka                    Bases
  pH, pOH, and pKa                 Structure and
  Calculating pH
                                     Strength
        Dominant Sources           Oxoacids
             Polyprotic Acids      Lewis Acids
        % dissociation             Acid-Base Solution
    Bases and Kb                    Al Gore Rhythms
Acid-Base Models
              Definition   Example    Definition   Example
               of Acid      of Acid    of Base     of Base

     Svante   Donates                 Donates
                             HCl                    NaOH
  Arrhenius     H+                     OH–

   Brøsted-   Donates                 Accepts      NH3 or
                             HCl
     Lowry      H+                      H+          H2O

      G.N.    Accepts                 Donates
                             BF
                             •3                     : NH3
     Lewis    electron                electron
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”
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–    HC2H3O2 HCN
  Ka          1.210–2 1.810–5 6.210–10
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
Calculating pH
    Strong acids              Weak Acids
        [H+]eq = [HA]0            Ka = [H+][A–] / [HA]
        pH = p[HA]0               Ka = x2 / ( [HA]0 – x)
    Strong bases                  Ka  x2 / [HA]0
        [OH–]eq = [BOH]0          x  ( Ka [HA]0 )½
        pOH = p[BOH]0             pH = – log10(x)
        pH = 14 – pOH         Weak Bases
    Don’t be fooled by            x  ( Kb [MOH]0 )½
     [HA]0 < 10–7                  pH = 14 - px
                    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–
% 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.”)
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.
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!
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
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
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!
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.
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
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
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!
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
     H3 + 610–3 H2 – 5.810–3 H – 3.610–8 = 0

Hout = (3.6 10–8 + 5.810–3 Hin – 610–3 Hin2)1/3

     Hin (the guess)          Hout (the refinement)
         5.310–3                    5.610–3
         5.610–3                   5.5810–3
        5.5810–3                   5.5810–3

 [H+] = 5.5810– 3 M     and by B = H2 / (H + 2K2),
   [B2+] = 5.0410– 3 M     and by C = K2 B / H,
   [C+] = 2.7 10– 3 M which makes [A3+] = 4.710– 3 M
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.
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
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)
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.
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.
Acid-Base Algorithm
  Know the significant      Solve controlling K
   species for the               Write K expression.
   conditions.                   Set initial conditions
  Ignore the rest.               for smallest change.
                                 Define all species in
  For quantitative
                                  terms of x (change).
   reactions, keep only
                                 Find x solving K exp.
   the products.
                                 Check any approx.
  Find acids & bases.           Find pH & [species]

				
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