2A. Chemical Equilibria Aqueous Solutions and Chemical by gmj10717

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									II Computer Titrations
2A. Chemical Equilibria
 Aqueous Solutions and Chemical Equilibria.
1. Equilibrium--the rate of a forward and reverse processes (reactions) are equal

2. Electrolytes form ions when dissolve in solvents. Partially, completely---weak, strong




 Slide 1
2A1 Acid-Base Equilibria
       Brnsted-Lowry theory--An acid donates protons while a base
accepts protons

       A conjugate base is formed when an acid loses a proton and
a conjugated acid is formed when a base accepts a proton

                     acid1=base1+proton




Slide 2                                                      Fig 9-2, p.232
2A2 Equilibrium-constant expression
                        wW  xX  yY  zZ   Pure liquid, solid
                            [Y ] y [ Z ]z   will not appear
                        K
                           [W ]w [ X ] x




Slide 3                                                   Table 9-2, p.235
2B. Ion-product constant for water Kw




Slide 4                                 Table 9-3, p.237
 2C. Acid-Base Dissociation constants, Ka ,Kb
                                                                  
                                                                                     
                                                                                [ NH 4 ][OH  ]
                  NH 3  H 2O  NH  OH                                    Kb 
                                                         4
                                                                                    NH 3 
                                               [ NH 3 ][ H 3O  ]     
                  NH  H 2O  NH 3  H 3O K a 
                          4
                                                     NH 4  
                                                                                                       
      a. dissociation-constants for conjugated acid/base pairs
                                                                                                                 
                                          
                     [ NH 3 ][ H 3O ] [ NH 4 ][OH ]
              KaKb                                 H 3O  OH   K w
                          NH 4  
                                         NH 3     
                                           10        K w 1.00 10 14
              K a  5.70 10                     Kb                   1.75 10 5
                                                      K a 5.70 10 10
     b. Hydronium ion concentration for weak acid solutions
          HA  H 2O  H 3O  A                          
                                                             Ka   
                                                                    H O A  water dissociation is suppressed
                                                                            3
                                                                                       


                                                             HA
          A   H O 
              -
                          3
                              
                                           cHA  A  HA HA  cHA  H 3O  

          then K 
                      H O  H O   K H O  K c
                                            2
                                                                   2                           
                                                                                                               0
                     c  H O 
                                   3
                      a                                     3                      a   3               a HA
                              HA            3

                     K a  K a  4 K a cHA
                     
                                2
                  
          H 3O 
                               2
Slide 5                                                     
          On the other hand , if H 3O   cHA H 3O   K a cHA                            
  2C1. Excel Solver and Goal Seek functions
  1. Excel Solver to solve quadratic equations
  --set equation as 0 by adjusting X ([H3O+]) in the condition of x>=0
     
a. A -  H 3O 
H O   K H O  K c
  3
         2
                    a       3
                                 
                                                 a HA            0                   K a  1.75 x10 5    cHA  0.01

                   
      b. H 3O   [ A ]  [OH  ] Charge balance

      H O   K O   HKO 
        3
              

               H
                  HA    a
                                
                                     K  1.75 x10    c  0.01
                                                         w
                                                                                 a
                                                                                                5
                                                                                                      HA
                            3                        3

                        
      H O   K (cHOH O )  HKO 
                                                         
              
        3
                        a       HA
                                             
                                                 3
                                                  Practice on solver0.xls w
                                                                              
                                     3                                3

      H O   K H O  K c
        3
               2
                        a        3
                                         
                                                         a HA      Kw  0




  Slide 6
 2C1. Excel Solver and Goal Seek functions

2. Excel Solver to solve complex Eq. problems


                        
                           2
K sp  Mg 2 OH   7.1x10-14                                 
                                               K w  H 3O  OH   1x10 14
                      
[OH  ]  H 3O   2[ Mg 2 ] Charge balance
constraints

[ Mg 2 ] 
                K sp
                         H O   OH  and c  0
                                   K  w

              OH      
                      2       3           

                                                    Practice on solver1.xls




  Slide 7
2C1. Excel Solver and Goal Seek functions




 3. Finding the Solubility of Mg(OH)2 with Goal Seek




               Practice on goal seek1.xls
Slide 8
          2C2. Buffer solutions
          A buffer is a mixture of a weak acid and its conjugate base or
           a weak base and its conjugate acid that resists changes of pH
                                                      [ H 3O  ][ A ]
            HA  H 2O  H 3O   A K a 
                                                          HA
                                    [OH  ][ HA]
                                       
            A  H 2O  OH  HA K b 
                                         A                
            HA  cHA  H 3O   OH   A   cNaA  H 3O   OH  
            HA  cHA A   cNaA
            Then from acid dissociation constant

            H O   K
                  3
                          
                              a
                                  cHA
                                  c NaA
                                        when K a  10 3 and c HA and c NaA are large enough

             The Henderson - Hasselbalch equation

                             
            - log H 3O    log K a  log
                                                  c NaA
                                                  cHA
                                                                            c
                                                        i.e. pH  pK a  log NaA
                                                                             cHA
           Effects of dilution, added acids and bases, till conditions break
Slide 9
 2C3. Composition of buffer solutions as a function of pH--Alpha values

cT  cHA  c NaA Ka 
                      H 3O  A        
                                      A 
                                                  
                                            K a HA
                                                      HA  H 3O  A                                     
                         HA                H 3O 
                                                               Ka      
0   
       HA          1      
                               A 


            cT                    cT
              K HA
(1)cT  HA  a                          HA        cT
                                                                   0       
                                                                              HA             H O       


                                                                                             K  H O 
                                                                                                       3
                                                                                                                
               H 3O                                  Ka                           cT
                                                           1
                                                            
                                                                                                   a        3
                                                         
                                                    H 3O

        
          H O A   A  A   c
                             
                                                                 Ka
                                                                  1   
                                                                         A  
                                                                                       
                                  H O   1                                                          
                 3                                           T
(2)cT                                                         
             K  a                      3                cT     H 3O   K a
                                       Ka      Practice alpha weak acid.xls
     Buffer Capacity--the number of moles of strong acid or base that 1 L
     of the buffer solution can absorb without changing pH by more than 1
                                           dcb    dca
                                             
 Slide 10                                 dpH     dpH
pKa=4.74




 Slide 11
            Practice alpha weak acid.xls   Fig 9-5, p.258
2C5 An inverse master equation approach
In this approach, the titration curve is developed as a function of the
extent of titration, or fraction titrated, , where
      Vb cb
                                   Vb cb and Va ca are the number of
      Va ca
                             moles of the base added and the acid
                             to be titrated
  The equivalence point in the titration occurs when =1.
 a. for NaOH+HCl=NaCl+H2O           OH   H 3O   2 H 2O
 There is a charge balance--
                                                        Vc                     Vc                      Kw
 Na     H 3O    Cl    OH   and [Na + ]  b b
                                                                   
                                                                     Cl    a a
                                                                                         OH -  
                                                                                               
                                                       V V
                                                          a    b              V V
                                                                                 a    b              H 3O  
                                                                                                            
 Vb cb
         H 3O    a a 
Va  Vb 
                      Vc      Kw
                   V  V  H O 
                                                 Vb cb   Vc
                                                Va  Vb Va  Vb 
                                                                      
                                                        a a   H 3 O    K w /  H 3O  
                                                                                               
                      a  b   3 
Vb cb
       1
           
            H 3O    K w /  H 3O   Va  Vb 
                                                 1
                                                                           
                                                          H 3O    K w /  H 3O  
                                                                                    
                                                                                       
                                                                                          H 3O    K w /  H 3O   Vb cb
                                                                                                                  
Va ca                       Va ca                                     ca                             (Va ca )cb
 Slide 12
                                                                              H O
                                                                               
                                                                                       
                                                                                             K w /  H 3O  
                                                                                                               
       H O                                                    
                                                                                 3
                                                                           1
                                     H 3O   K w /  H 3O  
                                                             
                                                           
           3         K w /  H 3O                                                           ca
  1                                                               
                       ca                           cb
                                                                           1
                                                                               H O
                                                                                3
                                                                                       
                                                                                             K w /  H 3O  
                                                                                                               
                                                                                               cb

                                                                      The effect of titrant
                                                                      concentration can be
                                                                      easily seen by changing
                                                                      the concentration of the
                                                                      base




Slide 13
b. for HCl +NaOH =NaCl+H2O
                                                                                H O
                                                                                 
                                                                                        
                                                                                              K w /  H 3O  
                                                                                                                
                                                                 
                                                                                   3
               H 3O   K w /  H 3O 
                                   
                                           H 3O   K w /  H 3O  
                                                                           1
   Va ca                                                                                cb
        1                                                          
   Vb cb                 cb                           ca
                                                                             1
                                                                                H O
                                                                                 3
                                                                                        
                                                                                              K w /  H 3O  
                                                                                                                
                                                                                                ca




                                                                        Titration curve for
                                                                        NaOH with HCl


Slide 14
 2C6 Titration of weak acids and bases
 a. Distribution diagram
In 4A2, we discussed  values
of a weak monoprotic acid:
    HA  H O      
                         A   K
                                                

 
                               H O  K
                                                                   a

          K  H O 
                     3
 0                                      1                     
     c   T   a
                          c  3                   T         3               a


b. Titration of a weak acid with
a strong base
NaOH  HA  NaA  H 2O
or OH   HA  A  H 2O
Na  H O   A  OH 
     
                 3
                                                     



Na   V  V , A    c   V  V
        Vc      b b            Vc   
                                                 1 T           1
                                                                           a a

             a               b                                         a         b


           
               H O  K /H O      3
                                             
                                                       w           3
                                                                           

                         1
   Vb cb                          ca
       
   Va ca
                     1
                                 
                        H 3O   K w / H 3O                             
Slide 15
                                 cb
c. Titration of a weak base with a strong acid
NH 4OH  HCl  NH 4 Cl  H 2O
                     
NH 4OH  H 3O   NH 4  2 H 2O
dissociation constants :

NH 3  H 2O  NH  OH                           
                                                              
                                                                NH OH    NH   H O 
                                                                                                                 


                                                                                      H O  K
                                                                      4                         4               3
                                                         Kb
                                 4
                                                                  NH    3      c
                                                                                    0
                                                                                            T               3
                                                                                                                
                                                                                                                            a

    
NH  H 2O  NH 3  H 3O                          
                                                              
                                                                NH H O    NH   K
                                                                                


                                                                   NH                 H O  K
                                                                      3     3                           3               a
    4                                                    Ka                            1                           
                                                                          4        c                T           3               a

NH  H O   Cl  OH 
       
       4        3
                                                       



Cl   VV cV , NH    c  
              a a                   
                                     4           0 T          0
                                                                 Vb cb
                                                                Va  Vb
           a         b


OH   HKO 
                    w
                         
                3


                0       
                           H O  K /H O 
                                 3
                                         
                                                     w        3
                                                                  


  Vc                             cb
 a a 
  Vb cb
                    1
                             
                       H 3O   K w / H 3O                     
                                ca

Slide 16
2C7 polyfunctional acids and bases
           H 2CO3  H 2O  H 3O  HCO                   
                                                                 K a1   
                                                                          H O HCO 
                                                                                    3
                                                                                                            
                                                                                                             3
                                                                                                                         Stochiometry in
                                                         3
                                                                                    H 2CO3                             the text p.407-414

           HCO3  H 2O  H 3O   CO32 K a 2                      
                                                                      H O CO                       2


                                                                        HCO 
                                                                                3                       3
                                                                                                
                                                                                                3




     0     
              H n A  H 3O                n
                                                   1   
                                                           H           n 1    A   
                                                                                          K H O     a1        3
                                                                                                                         n 1


                  cT                D                                   cT                                       D

     2     
              H   n2 A
                             2
                                   K               
                                            a1 K a 2 H 3O           
                                                                   n2
                                                                                ... n          
                                                                                                  A   K   n
                                                                                                                            a1   K a 2 ...K an
                    cT                               D                                                   cT                        D
     D  H 3O            K H O   K K H O 
                        n
                                   a1           3
                                                     n 1
                                                                        a1      a2              3
                                                                                                          n2
                                                                                                                       ...  K a1 K a 2 ...K an

                          2 
                                 H O  K /H O    3
                                                             
                                                                            w               3
                                                                                                    

                             1          2
       Vc                                         ca
      b b 
       Va ca
                                  1
                                                    
                                     H 3O   K w / H 3O                                  
                                              cb
Slide 17
Titration curves for diprotic acids




Slide 18
Titration curves for difunctional bases
                1  2 0 
                              H O  K /H O 
                                   3
                                       
                                               w   3
                                                       


      Va ca                    cb
          
      Vb cb
                   1 3
                              
                                   
                     H O  K w / H 3O            
                           ca




Slide 19
 2D. Solubility-product constant, Ksp
     Ba ( IO3 ) 2 ( s )  Ba 2 ( aq )  2 IO3 (aq )

     K
               Ba IO 
                    2          2
                               3
                                         K Ba( IO3 ) 2 ( s )  K sp  Ba    2
                                                                                    IO 
                                                                                        2

           Ba( IO3 ) 2 ( s)                                                          3


Example 9-3 P.239
How many grams of Ba(IO3)2 can be dissolved in 500 mL of water at 25C
Ksp=1.57x10-9
           IO   2Ba 
                
                3
                               2


           K  IO  Ba   2Ba  Ba   4Ba   1.57 10
                          2        2            2   2   2       2 3        9
               sp        3
                                         1/ 3
                    1.57 10 9 
           Ba  
               2
                               
                                                7.32  10  4 M
                         4      
           in 500 mL
           no.mol  7.32 10  4  0.5
           mass Ba ( IO3 ) 2  7.32  10  4  0.5  487
Slide 20    0.178 g
AgNO3  KSCN  AgSCN ( s )  KNO3
or Ag   SCN   AgSCN ( s )                            According to the charge balance
Ag  K   NO  SCN 
                          
                            3
                                            



 K   V  V NO   V                                                  SCN   Ag 
    
         V cSCN  SCN 
                            V                       Ag 
                                                            c Ag              
                                                                                   K      sp
                                                                                           
                                                             VAg 
                                       3
           SCN       Ag                       SCN 


Ag  V                                                                                                                               
     
          VSCN  cSCN                VAg  c Ag              K sp       VSCN  cSCN           VAg  c Ag         K sp
                                                                                                                            Ag 
           SCN 
                    VAg           VSCN   VAg            Ag    
                                                                          VSCN   VAg        VSCN   V Ag        Ag   
                                                                                                                               
VSCN  cSCN 
                       K sp
                      
                          
                                Ag 
                                       
                                                              K sp
                                                                  
                                                                        Ag 
                                                                                              
                                                                                                  
                  1  Ag                  VSCN  cSCN    Ag                                
 c Ag  V Ag              cSCN          c V                   c Ag                        
                                              Ag    Ag
                                                                                                 
                                                                                               
                            K sp
                           Ag     Ag 
                                                   
                       1
     VSCN  cSCN                c Ag 
                   
        c Ag  VAg         K sp
                           Ag     Ag 
                                                   
                       1
 Slide 21                       cSCN 
           Indicator
           Ag   Cl   AgCl ( s )
           Ag  
               
                        K sp  1.82 10 10  1.35 10 5 M
           2 Ag   CrO4   Ag 2CrO4 ( s )(red )
                       2



           CrO  2
                          K sp
                                  
                                      1.2 10 12
                                                      6.6 10 3 M
                   4
                        Ag  1.35 10 
                             2               5 2




Slide 22
Calculate the molar solubility of thallium(I) sulfide in a H2S saturated solution (0.1 M
H2S) at (a)pH 1 and (b)pH 4. Given that Ksp =6.0x10-22, and Ka1, Ka2 are 9.6x10-8,
1.3x10-14 respectively for H2S.




Camium sulfide (Ksp=1x10-27)is less soluble than thallium (I) sulfide (Ksp=6x10-22).
Find the conditions under which Cd2+ and Tl+ can be separately quantitatively with H2S
from a solution that is 0.1 M in each cation (1:1000).




Slide 23
 2E. Stepwise and overall formation constants for complex ions
         A complex ion consists of a central metal ion that can accept a
 pair of electrons and ligands that can donate a pair of electrons
         Stepwise formation constants are symbolized by K1, K2 and so
 forth.                       2          Ni(CN ) 
                                    CN  Ni (CN )    
                                                                  K1 
                                                                                                     
                         Ni
                                                                               Ni CN 
                                                                                       2            



                         Ni (CN )   CN   Ni (CN ) 2                    K2 
                                                                                                 Ni (CN ) 2 
                                                                               Ni(CN ) CN                         


                                             
                         Ni (CN ) 2  CN  Ni (CN )               
                                                                           K 
                                                                                  Ni(CN )                       


                                                                               Ni(CN ) CN 
                                                                                                                  3
                                                                  3            3                                       
                                                                                                          2

                                            
                         Ni (CN )  CN  Ni (CN )                 2
                                                                            K 
                                                                                   Ni(CN )                      2


                                                                                Ni(CN ) CN 
                                                                                                                  4
                                     3                            4                4                                      
                                                                                                              3


   Overall formation constants are designated by the symbol n
            Ni 2  2CN   Ni (CN ) 2                 2  K1 K 2 
                                                                                                Ni(CN ) 2 
                                                              Ni CN                           2               2


                 2           
                       3CN  Ni (CN )           
                                                       KK K 
                                                                 Ni(CN )                                             
                                                                                                                       3

                                                                Ni CN 
            Ni                                   3        3            1       2       3                 2              3


                 2           
                       4CN  Ni (CN )           2
                                                        KK K K 
                                                                     Ni(CN )                                                 2
                                                                                                                               4

                                                                    Ni CN 
            Ni                                   4            4            1       2        3    4                2            4
Slide 24
   2E1 An alpha value is the fraction of the total metal concentration
   existing in that form:         1                                                            1  L 
M                                                              ML 
      1  1  L    2  L    3  L   ...   n  L            1  1  L    2  L    3  L   ...   n  L 
                              2           3                 n                                  2           3                 n



                                 2  L                                                          n  L
                                        2                                                                n

 ML                                                             ML 
       1  1  L    2  L    3  L   ...   n  L             1  1  L    2  L    3  L   ...   n  L 
    2                           2           3                 n       n                          2           3                 n




  Slide 25
     Mass balance
                                                        cT , L  [ ML]  [ L]
                                                         cLVL                 cV
cT , L  [ ML]  [ L]                                            [ ML]   L L L
                                                        VL  VM              VL  VM
 cLVL          c V
          ML M M  [ L]                                     cLVL           c V
VL  VM       VL  VM                                            , [ ML]  M M  [ M ]
                                                             cM VM          VL  VM
      cLVL
                                                        cLVL      V  VM  cM VM                       cV        VL  VM
     cM VM                                                         x L         V V       [M ]   L L L      x
                                                                                                                   c V
                                                        VL  VM       cM VM       L                    VL  VM   
            VL  VM                           VL  VM                                  M                            M M
  cLVL                           cM VM
          x              ML
                                       [ L]  x
                                               c V   1  VL  VM [ M ]   
VL  VM       cM VM           VL  VM           M M                               L
                                                                  cM VM
               V            V
   ML  L [ L]  M [ L]                                        c V [ M ] VM
             cM VM         cM VM                          1 L L                    [ M ]   L
                                                                cM VM cL        cM VM
   [ L]             [ L]
 1 
             ML                                                [M ] 
       cL 
                                                                                  [M ]
                    cM                                 1   L 
                                                                            1
                                                                     cL         cM
             [ L]
      ML                                                          [M ]
             cM                                                 1
                                                                   cM
          [ L]                                          
       1                                                              [M ]
           cL                                                1L 
                                                                        cL



 Slide 26
2E2 Inverse master equations for complexation titrations
                                    ML 
                                            L                                            ML  2 ML 
                                                                                                             L
                          VL c L        cM                                      VL cL                    2
                                                                                                             cM
For formation of ML                            For formation of ML 2             
                          VM cM    1
                                      L                                       VM cM          1
                                                                                                    L
                                      cL                                                            cL

                                       ML  2 ML  3 ML 
                                                                 L                                               ML  2 ML  3 ML  ...  n ML 
                                                                                                                                                         L
                            VL cL                  2         3
                                                                 cM                                VL c L                    2         3           n
                                                                                                                                                         cM
For formation of ML 3                                               For formation of ML n            
                            VM cM                 1
                                                       L                                         VM cM                          1
                                                                                                                                       L
                                                       cL                                                                              cL

               VL cL
            
               VM cM

                    1
                       M 
                              cM
            
                1L          
                                M 
                                   cL
                 [ L]
            L 
                 cT , L
                            1
            
 Slide 27       1  K ML M 
2E3 EDTA




Slide 28   p.458
 2E4 pH dependence of the
       EDTA titrations




Conditional formation
      constants

In fact
              [CaY 2 ]       [CaY 2 ]
K CaY                     
            [Ca 2 ][Y 4 ] [Ca 2 ] 4 cT , L
                           [CaY 2 ]
K   '
            4 K CaY   
                          [Ca 2 ]cT , L
    CaY




Slide 29
   Calculate the equilibrium concentration of Ni2+ in a solution with an analytical NiY2-
   concentration of 0.0150 M at pH (a)3.0 and (b) 8.0
              Ni 2   Y 4  NiY 2
                          [ NiY 2 ]
              K NiY                       4.2 x1018
                        [ Ni 2  ][Y 4 ]
              [ NiY 2 ]  0.0150  [ Ni 2 ]  0.0150
              [ Ni 2 ]  [Y 4 ]  [ HY 3 ]  [ H 2Y 2 ]  [ H 3Y  ]  [ H 4Y ]  cT
                          [ NiY 2 ]    [ NiY 2 ]   [ NiY 2 ]
              K NiY                              
                        [ Ni ][Y ] [ Ni ] 4 cT [ Ni 2 ]2  4
                            2     4      2


                                     0.015
              (a ) K NiY                            4.2 x1018
                             [ Ni 2 ]2 2.5 x10 11
              [ Ni 2 ]  1.2 x10 15 M
                                    0.015
              (b) K NiY                            4.2 x1018
                             [ Ni 2 ]2 5.4 x10 3
Slide 30      [ Ni 2 ]  8.1x10 10 M
   Calculate the equilibrium concentration of Ni2+ in a solution that was
   prepared by mixing 50.00 mL of 0.03000 M Ni2+ with 50.00 mL of 0.05000 M
   EDTA. The mixture was buffered to a pH of 3.0.




Slide 31

								
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