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					                Chapter 16
     Acids and Bases




Johannes N. Bronsted         Thomas M. Lowry
1879-1947.                   1874-1936.
Both independently developed Bronsted-Lowry
                                               1
theory of acids and bases.
Acids and Bases: A Brief Review
   Classical Acids:
   Taste sour
   Donate H+ (called “H-plus” or “proton”)
   Turn litmus red
   Generally formed from H-Z, where Z = nonmetal

   Classical Bases:
   Taste bitter and feel soapy.
   Donate OH- (called “O-H-minus” or “hydroxide”)
   Turn litmus blue
   Generally formed from MOH, where M = metal

   Neutralization:                H+ in water is actually
   Acid + Base  Salt + water     in the form of H3O+,
   H-Z + MOH  MZ + HOH           “hydronium”           2
Brønsted-Lowry Acids and Bases
Proton Transfer Reactions
• Brønsted-Lowry acid/base definition:
     acid donates H+
     base accepts H+.
• Brønsted-Lowry base does not need to contain OH-.
• Consider HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq):
  – HCl donates a proton to H2O. Therefore, HCl is an acid.
  – H2O accepts a proton from HCl. Therefore, H2O is a base.
• Water can behave as either an acid or a base.
• Amphoteric substances can behave as acids and bases.

                                                          3
Brønsted-Lowry Acids and Bases
Conjugate Acid-Base Pairs
• Whatever is left of the acid after the proton is donated
  is called its conjugate base.
• Similarly, whatever remains of the base after it
  accepts a proton is called a conjugate acid.
• Consider HCl( aq)  H 2 O( l )  H 3O  ( aq)  Cl  ( aq)

   –After H2O (base) gains a proton it is converted into H3O+
   (acid). Therefore, H2O and H3O+ are conjugate acid-base
   pairs.
   –After HCl (acid) loses its proton it is converted into Cl-
   (base). Therefore HCl and Cl- are conjugate acid-base pairs.
•Conjugate acid-base pairs differ by only one proton.4
Brønsted-Lowry Acids and Bases
Conjugate Acid-Base Pairs

 In each of following Bronsted-Lowry Acid/Base reactions,
 Which is acid? base? conjugate acid? conjugate base?

         HCl + H2O  H3O+ + Cl-
          acid       base    conj. acid    conj. base

        NH3 + H2O  NH4+ + OH-
         base      acid     conj. acid    conj. base


   CH3NH2 + H2SO4  CH3NH3+ + HSO4-
       base          acid       conj. acid              conj. base
                                                                 5
Brønsted-Lowry Acids and Bases
                  Relative Strengths of Acids and
                    Bases
                  • The stronger the acid, the
                    weaker the conjugate base.
                  • The stronger the base, the
                    weaker the conjugate acid.
                  • H+ is the strongest acid that
                    can exist in equilibrium in
                    aqueous solution.
                  • OH- is the strongest base
                    that can exist in equilibrium
                    in aqueous solution.
                                             6
Strong and Weak Acids and Bases
    Strong acids: completely ionized in water:
    HCl, HNO3, H2SO4 (also HBr, HI)
    Strong bases: completely ionized in water:
    MOH, where M = alkali
    M(OH)2, where M = alkaline earth
    Weak acids: incompletely ionized in water:
    any acid that is not strong - acetic acid, etc.
    Ka is finite.
    Weak bases: incompletely ionized in water:
    any base that is not strong – NH3, etc.
    Kb is finite.                           7
The Autoionization of Water
The Ion Product of Water
• In pure water the following equilibrium is established
     H2O(l) + H2O(l)                  H3O+(aq) + OH-(aq)

                                 -
                [H3O ][OH ]
         K eq          2
                                         but [H2O]2 = constant
                  [H2O]
                                         
    K eq[H 2O]  K w  [H 3O ][OH ]
                  2                             -


                                             14
    K w  [H 3O ][OH ]  1.0 10
                              -                     (at 25oC)


     This is called the autoionization of water
                                                                8
 The Autoionization of Water
In pure water at 25oC, [H3O+][OH-] = 1 x 10-14
and also, [H3O+] = [OH-] = 1 x 10-7
(From now on, for simplification, let’s use the abbreviation:
 [H+] = [H3O+] which means [H]+ = [OH-] = 1 x 10-7

   We define    pH = -log [H+]    and   pOH = -log [OH-]

 In pure water at 25oC, pH = pOH = 7.00
                           pH + pOH = 14
                           pKw = 14
               Acidic solutions have pH < 7.00
               Basic solutions have pH > 7.00          9
The pH Scale
         Conc. Drano




         Battery acid
                        10
The pH Scale
  (a) For [H+] = 3.4 x 10-5M, calculate pH, pOH and [OH-]
      pH = 4.47, pOH = 9.53, [OH-] = 2.95 x 10-10

  (b) For [OH-] = 4.4 x 10-3M, calculate pH, pOH, and [H+]
      pOH = 2.36, pH = 11.64, [H+] = 2.27 x 10-12

  (c) For pH= 8.9, calculate, pOH, [H+], [OH-]
      pOH = 5.1, [H+] = 1.26 x 10-9, [OH-] = 7.94 x 10-6

  (d) For pOH= 3.2, calculate pH, [H+], [OH-]
     pH = 10.8, [H+] = 1.58 x 10-11, [OH-] = 6.3 x 10-4

  The pH meter is the most accurate way to measure pH
     values of solutions.
                                                            11
Use this “decision tree” to calculate pH
values of solutions of specific solutions.
 1. Is it pure water? If yes, pH = 7.00.
 2. Is it a strong acid? If yes, pH = -log[HZ]
 3. Is it a strong base? If yes, pOH = -log[MOH]
     or pOH = -log (2 x [M(OH)2])
 4. Is it a weak acid? If yes, use the relationship
       Ka = x2/(HZ – x), where x = [H+]
 5. Is it a weak base? If yes, use the relationship
       Kb = x2/(base – x), where x = [OH-]
 6. Is it a salt (MZ)? If yes, then decide if it is neutral, acid,
    or base; calculate its K value by the relationship
    KaKb = Kw, where Ka and Kb are for a conjugate system;
    then treat it as a weak acid or base.
 7. Is it a mixture of a weak acid and its weak conjugate base?
    It is a buffer; see next chapter.                            12
2. Strong Acids
   Calculate the pH of 0.2 M HCl .
   pH = -log[H+] = -log[0.2] = 0.70

3. Strong Bases
    Calculate the pH of 0.2 M NaOH .
    pOH = -log[OH-] = -log[0.2] = 0.70
    pH = 14 – 0.70 = 13.30

    Calculate the pH of 0.2 M Ba(OH)2.
    pOH = -log[OH-] = -log[0.4] = 0.40
    pH = 13.60


                                         13
4. Weak Acids
Weak acids are only partially ionized in solution, and
 are in equilibrium:
      HA(aq) + H2O(l)              H3O+(aq) + A-(aq)
(shorthand):   HA(aq)        H+(aq) + A-(aq)

          [H 3O  ][A - ]                 [H  ][A - ]
     Ka                     OR:     Ka 
              [HA]                          [HA]
                                         (shorthand)


   •Ka is the acid dissociation constant.
                                                         14
4. Weak Acids




                15
4. Weak Acids
 Calculate pH of 0.2 M solution of acetic acid, HC2H3O2
 From previous slide, Ka= 1.8 x 10-5
 From now on, we’ll use the shorthand notation for HC2H3O2              = HAc
 Denote the acetate ion, C2H3O2- as “Ac-”

 Equilibrium is then:          HAc         H+ + Ac-
       Initial conc (M):        0.2            0          0
              change:           -x            +x         +x
        at equilibrium:        0.2-x           x         x
                                  2
        [H  ][Ac  ]     x
    Ka=                                        But, this is a quadratic equation!!
          [HAc]         0.2  x
                 We can assume x is small if Ka < 10-4.
  Then,   x2/0.2= 1.8 x 10-5           x = 1.90 x 10-3 = [H+] = [Ac-]
                                       pH = -log (1.90 x 10-3) = 2.72
                                                                              16
4. Weak Acids
Polyprotic Acids
• Polyprotic acids have more than one ionizable proton.
• The protons are removed in steps not all at once:
H2SO3(aq)          H+(aq) + HSO3-(aq) Ka1 = 1.7 x 10-2

 HSO3-(aq)          H+(aq) + SO32-(aq)    Ka2 = 6.4 x 10-8

•It is always easier to remove the first proton in a
polyprotic acid than the second.
•Therefore, Ka1 > Ka2 > Ka3 etc.
•Most H+(aq) at equilibrium usually comes from the
first ionization (i.e. the Ka1 equilibrium).         17
4. Weak Acids
Polyprotic Acids




                   18
Carbonic acid, H2CO3, Ka1=4.3 x 10-7
    H2CO3  H+ + HCO3-             Ka1 = 4.3 x 10-7

     HCO3-       H+ + CO32-       Ka2 = 5.6 x 10-11

Since Ka1>>Ka2, nearly all H+ ions come from 1st equilibrium.
Therefore, the 1st equilibrium determines the pH.
Calculate pH of 0.4 M H2CO3 solution.
Ka = x2/HZ-x
4.3 x 10-7 = x2/0.4
x = [H+] = 4.15 x 10-4
pH = -log (4.1 x 10-4) = 3.38


                                                          19
5. Weak Bases
• Weak bases remove protons from substances.
• There is an equilibrium between the base and the
  resulting ions:
      Weak base + H2O         conjugate acid + OH-
   Example:
       NH3(aq) + H2O(l)       NH4+(aq) + OH-(aq)


  The base dissociation constant, Kb is defined as:
                              -
             [ NH ][OH ]
        Kb           4
                [ NH3 ]
                                                      20
5. Weak Bases
• The larger Kb the stronger the base.




                                         21
5. Weak Bases
What is pH of 0.15 M solution of methylamine, NH2CH3, a weak base?

For convenience, denote it as “B”
 Then:       B + H 2O              BH+ +        OH- Kb =4.4 x 10-4
 Initially: 0.15                     0              0
 change:     -x                     +x              +x
 At equil: 0.15-x                    x               x

                 x2     x2                     4
Then:                      4.4x10
               .15  x .15
                            3           
          x  8.12 x 10  [OH ]
        pOH  log[OH  ]  log(8.12x1 0 3 )  2.09
        pH  14  pOH  14  2.09  11.9                             22
Relationship Between Ka and Kb
• For a conjugate acid-base pair
                  Ka  Kb = Kw (constant)
• Therefore, the larger the Ka, the smaller the Kb. That
  is, the stronger the acid, the weaker the conjugate
  base.
• Taking negative logarithms:
                  -log Ka- log Kb= -log Kw
                      pKa + pKb = pKw

 For HAc (acetic acid), the pKa = -log (1.8 x 10-5) = 4.74.
 Thus, the pKb for Ac- (acetate 0 is 14 – 4.74 = 9.26
                                                      23
6. Salts
 Salts may be acidic, basic or neutral.
 Salts made by a strong acid and a strong base are neutral,
 e.g., NaCl, KNO3.
 Salts made by a weak acid and a strong base are weakly basic,
 e.g., sodium acetate, NaAc, NaHCO3.
 Salts made by a strong acid and a weak base are weakly acidic,
 e.g., NH4Cl.

 Calculate the pH of a 0.35 M solution of sodium acetate.
 Since Ka Kb = Kw, where Ka = 1.8 x 10-5 for acetic acid,
 Then Kb for NaAc (sodium acetate) is
         Kb = Kw/Ka = 1.00 x 10-14/1.8 x 10-5 = 5.56 x 10-10.
 Now treat this as a weak base problem,
 Kb = x2/base = 5.56 x 10-10 = x2/0.35
 x = [OH-] = 1.39 x 10-5
                                                                24
 pOH = 4.85 and pH = 9.14

				
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