Acid-Base by langkunxg

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									 Acid-Base Equilibrium

Equilibrium chemistry of acids and
              bases
Simple Acid Base Chemistry

• An acid reacts with a base to give a salt plus
  water.
• HCl + NaOH  NaCl + H2O
• Net Ionic Equation
• H+ + OH-  H2O
• Arrhenius Definition of Acid and Base
  – Acid produces H+ ions in solution
  – Base produces OH+ ions in solution
Another Definition of Acid and Base

•   NH3(g) + HCl(g)  NH4Cl(s)
•   NH3(aq) + HCl(aq)  NH4Cl(aq)
•   Net Ionic Equation
•   NH3(aq) + H+(aq)  NH4+(aq)
•   Bronsted-Lowry Definition
    – Acid is a proton donor
    – Base is a proton acceptor
An Acid and its Conjugate Base

•   HCl  Cl-
•   NH4+  NH3
•   HC2H3O2  C2H3O2-
•   H2O  OH-
•   H3O+  H2O
•   H3PO4  H2PO4-
•   H2PO4-  HPO42-
An Base and its Conjugate Acid

•   OH-  H2O
•   NH3  NH4+
•   C2H3O2-  HC2H3O2
•   Cl-  HCl
•   H2O  H3O+
•   H2PO4-  H3PO4
•   HPO42-  H2PO4-
The Hydronium Ion, H3O+

                                      +

H+   +   O   H            H   O   H

         H                    H
Other Related Species
Strong Acids and Bases

• Strong Acid
  – HCl + H2O  H3O+ + Cl-
• Other Strong Acids
  – HBr, HI, HNO3, HClO3, HClO4, H2SO4
• Strong Base
  – NaOH  Na+ + OH-
• Other Strong Bases
  – KOH, Ca(OH)2,
Weak Acids and Bases

• Weak Acids
  – HF + H2O  H3O+ + F-
  – NH4+ + H2O  H3O+ + NH3
• Weak Bases
  – NH3 + H2O  NH4+ + OH-
  – F- + H2O  HF + OH-
Relative Strengths of Acids and Bases



The stronger                The stronger
the acid the                the base the
weaker the                  weaker the
conjugate base.             conjugate acid.


                           H- + H2O  H2 + OH-
Autoionization of Water
                                                 +
                                                                 -
 O   H   +   O       H           H   O       H       +   O   H
 H               H                   H


                                                         14
Kw  [H3 O ][OH ]  [H ][OH ] 1.00 10

Neutral Solutions

[H3O ]  [OH  ]  1.00 10 7
The pH Scale

•   pH = -log[H3O+] = -log[H+]
•   pOH = -log[OH-]
•   Neutral Solution
•   pH = -log(1.0 x 10-7) = 7.00
•   pH < 7 is acidic, pOH > 7
•   pH > 7 is basic, pOH < 7
•   pKw = pH + pOH
Measurement of pH

• Colored indicators that change color as the
  pH changes
• These indicators are either weak acid or
  weak bases themselves
• pH meter and a glass electrode
• The glass electrode develops a potential that
  is proportional to the log of the hydrogen
  ion concentration
Weak Acids

• HX + H2O  H3O+ + X-
• HX  H+ + X-
• Acid Dissociation Constant

       [H3O ][X  ]        [H  ][X  ]
  Ka                  Ka 
          [HX]                [HX]
Acid Dissociation Constants
Calculation of Ka from pH Measurements
  HC2H3O2 + H2O  H3O+ + C2H3O2-
• A solution of acetic acid is prepared by
  dissolving 0.01 mole of acetic acid in water
  in a 100mL volumetric flask. The solution
  pH was determined to be 2.87.
• The initial concentration is 0.10M
• The equilibrium hydronium ion
  concentration is:
               pH      2.87           3
   [H3O ]  10       10        1.3  10 M
Finish the Calculation of Ka
             HC2H3O2       H 3O +   C2H3O2-

   Initial      0.10       0.00       0.00

   Change     0.0013 0.0013        0.0013

   Equil.      0.099      0.0013    0.0013

Ka 
      H3O  C2 H3O2  0.00130.0013
                 

                                        1.8 10
                                                  5

       [HC2 H3O2 ]          0.099
Calculate the pH from Ka
  HC2H3O2 + H2O  H3O+ + C2H3O2-
• A solution of acetic acid is prepared by
  dissolving 0.005 mole of acetic acid in
  water in a 100mL volumetric flask. What is
  the pH of the solution.
• The initial concentration is 0.05M

  Ka 
          H3O  C2 H3 O2 
                      

                               1.8 10
                                          5

           [HC2 H3O2 ]
Finish the Calculation of pH
            HC2H3O2      H 3O +     C2H3O2-

  Initial     0.050      0.00        0.00

  Change       x          x          x

  Equil.     0.05 x       x          x

          x2     x2            5
  Ka                1.8  10
       0.05  x 0.05
Solution of the Equation
        x2     x2
Ka          
     0.05  x 0.05
 x   3O  0.05Ka  0.051.8 10
      H                                    5
                                                   0.00095
 pH   log[H3 O ]   log(9.5  10 4 )  3.02

                0.00095
 % Ionization          100%  1.9%
                 0.050
  If the percent ionization is greater 5% you
  should not make the assumption and use the
  quadratic equation.
Polyprotic Acids

• H3PO4 + H2O  H3O+ + H2PO4- Ka1 
                                                H3O  H2 PO4 
                                                          


                                                      H3PO4 

• H2PO4 + H2O  H3
       -           O+   + HPO4    2-
                                       Ka2 
                                                H3O  HPO4  
                                                        2


                                                  H2PO4   




• HPO4 + H2O  H3
      2-           O+   + PO43-
                                       Ka3   
                                               H O PO 
                                                  3
                                                               3
                                                                4

                                                 HPO     2
                                                           4
Various Polyprotic Acids
Weak Bases

• Weak Base + H2O  Conjugate acid + OH-
• NH3 + H2O  NH4+ + OH-


 Kb 
         NH4 OH  
         


           NH3 
Various Weak Bases
Relationship Between Ka and Kb

• NH3 + H2O  NH4+ + OH-
• NH4+ + H2O  H3O+ + NH3

Ka K b   
           H O NH   NH OH   H O OH  K
             3
                 
                         3
                             
                             4
                                      
                                                      

              NH         NH 
                                              3           w
                     4           3



 pK a  pKb  pKw  14.00
      K                                   Kw
  Ka  w                             Kb 
      Kb                                  Ka
Comparison of Ka’s and Kb’s
Acid-Base Properties of Salts

• Hydrolysis Reactions
• X- + H2O  HX + OH-
• R-NH3+ + H2O  H3O+ + R-NH2
Different Types of Salts

• SA/SB Salt                • WA/SB Salt
  – Neither the cation or     – The anion hydrolyzes
    the anion hydrolyze       – pH > 7
  – pH = 7                  • WA/WB Salt
• SA/WB Salt                  – Both cation and anion
  – The cation hydrolyzes       hydrolyze
  – pH < 7                    – pH depends on Ka and
                                Kb of the acid and base
Factors Affecting Acid Strength

•   H---X bond
•   Polarity of the bond
•   Strength of the bond
•   Stability of conjugate base
•   Why is HF a weak acid?
Binary Acids     4A        5A      6A     7A

                 CH4       NH3 H2O HF
      Period 2
                 No acid




                                                 Acid Strength
                 or base    Weak          Weak
                 properties base          acid

                 SiH4      PH3 H2S HCl
      Period 3
                 No acid
                 or base    Weak Weak Strong
                 properties base acid acid
                          Base Strength
                       Acid Strength
Oxyacids

•   Y-O-H, electronegativity of Y
•   I-O-H < Br-O-H < Cl-O-H
•   HYOn, number of oxygens, oxidation state
•   HClO < HClO2 < HClO3 < HClO4
Carboxylic Acids

• R-CO2H
• R is hydrogen or a organic (contains carbon,
  hydrogen, and may be other elecments)
  group.
• Polar O-H bond
• Stability of the conjugate base
• Resonance
Lewis Acids and Bases

•   Acid is an electron pair receptor
•   Base is an electron pair donor
•   H3N: + BCl3  H3N- BCl3
•   Acid properties of metal cations
•   Fe3+ + 6CN:-  [Fe(CN)6]3-
•   Metal ion hydrolysis
•   Fe(H2O)63+ [Fe(H2O)5(OH)]2+ + H+
•   Smaller Higher charge ion are more acidic
16.28: Complete the following table by
calculating the missing entries. In each case
indicate whether the solution is acidic or basic.
                                             Acid
                                             or
    pH      pOH [H+]            [OH-]        Base
    5.15    8.85   7.1 x 10-6   1.4 x 10-9   Acid

     6.15   7.85   7.1 x 10-7 1.4 x 10-8     Acid

     5.74   8.26   1.8 x10-6 5.6 x 10-9      Acid

    12.96   1.04   1.1 x 10-13 9.2 x10-2 Base
16.35: Calculate the Concentration of an aqueous
solution of NaOH that has a pH of 11.50.
  pOH  14.00  pH  2.5
  OH   10
              2.5
                       3.2  10
                                   3




  NaOH   OH   3.2  10
                                       3
16.53: Calculate the percent ionization of hydrazoic
acid, HN3, in solutions of each of the following
concentrations (Ka is given in Appendix D); (a)
0.400 M; (c) 0.0400M (Ka = 1.9 x 10-5 )
                                [H  ][N3 ]
HN3  H+ + N3-             Ka 
                                 HN 3 
            
[H  ]  [N 3 ]  x   [HN3 ]  0.400  x

         x2       x2
 Ka                      x  0.4001.9 10 5   2.76 103
      0.400  x 0.400


 %I 
      N   100%  2.76  10
           3
                                 3
                                        100%  0.69%
        0.400             0.400
16.53: (c) 0.0400M (Ka = 1.9 x 10-5 )
                                  [H  ][N3 ]
HN3  H+ + N3-               Ka 
                                   HN 3 
[H  ]  [N 3 ]  x [HN3 ]  0.0400  x
            



    x2                5
            1.9  10             x 2 1.9 105 0.0400 1.9 105 x
0.0400  x
  x2 1.9 105 x  7.6 10 7  0
      1.9  10   5
                          1.9  10    
                                       5 2
                                               4(1)7.6  10 7 
 x
                                 21
  x  8.6 10 4

  %I 
       N   100%  8.6  10
            3
                                             4
                                                    100%  2.2%
         0.400                    0.040
16.76: An unknown salt is either KBr, NH4Cl,
KCN, or K2CO3. If a 0.100 M solution of the salt
has a pH of 11.15, what is the identity of the salt?

 The salt cannot be KBr the pH would be near 7 and it cannot
 be NH4Cl the pH would be less than 7.
 Calculate the pH of a 0.100M solution of one of the salts,
 let’s try KCN.
                                                HCN OH  Kw      1.0  1014
 CN-   + H2O  HCN +                 OH- Kb                                    2.0  10 5
                                                  CN 
                                                                K a 4.9  10
                                                                             10



                               x2                x2
 x  HCN   OH                                         x  0.102.0 10 5  1.4 10 3
                                              5
                    
                                      2.0 10 
                             0.1  x             0.1


 pOH   log   10                2.85   pH 14.00  pOH  14.00  2.85  11.15
                            3
             1.4
Based on their compositions and structures and on
conjugate acid-base relationships, select the stronger
base in each of the following pairs:
• (a) BrO- or ClO-
                            BrO-
• (b) BrO- or BrO2-
                            BrO-

• (c) HPO42- or H2PO4-
                            HPO42-

								
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