Acid Base Hydrolysis by 177P72yh

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									Acid Base Hydrolysis

   Worked Examples
                         pH
                              
   pH   log 10 H   log 10 1 H   10  pH

• Determine the pH of a solution which contains 4
  x 10-5 M (moles/litre) H+
• The decimal part is always positive
• Log 4.0 = 0.602
• log (4 x 10-5) = -5 + 0.602 = -4.398
• pH = -log[H+] = -(-4.398) = 4.398
• Or log 0.00004 = -4.398 –ve log = 4.398
                        pH
                   log 1 H   10
  pH   log 10 H   
                           10
                                          pH



• Find the hydrogen ion concentration
  corresponding to pH 5.643
• pH = -log[H+] = 5.643
• log[H+] = -5.643 = (0.357) + (-6.000)
• Antilog of 0.357 = 2.28 x 10 -6
• [H+] = 2.28 x 10-6
                     Hydrolysis
• Salt of a weak acid and strong base
• Calculate:
• 1) the hydrolysis constant,
• 2) the degree of hydrolysis and
• 3) the hydrogen concentration for 0.01 M solution of
  sodium acetate
• Na+ + Ac - + H2O ↔ Na+ + HAc + OH-


                         14
         K w 1.0 10
    Kh                 5.5 1010
         K a 1.82 105
                         Hydrolysis
• The degree of dissociation is given by
                     x2
            Kh 
                 1  x V
• Substituting for Kh and V = 1/c we obtain

                       x 2  0.01
      5.5  10 10   
                         1  x 
• Solving this quadratic equation, x = 0.000235 or
  0.0235%
                    Hydrolysis
• If the solution was completely hydrolysed, the
  concentration of acetic acid produced would be 0.01 M.
  But the degree of hydrolysis is 0.0235%. Therefore the
  concentration of acetic acid is 2.35 x 10 -6 M
• This is also equal to the hydroxide ion concentration
• [OH-] = 2.35 x 10-6
• M pOH = 5.63
• pH = pKw – pOH i.e. pH = 14 – 5.63 = 8.37
• The pH may also be determined using
• pH = ½pKw + ½pKa + ½log c
•      = 7.0 + 2.37 + ½(-2)
• pH = 8.37
                              Hydrolysis
• Salt of a strong acid and weak base
• Determine the pH of a 0.2 M NH4Cl solution
• NH4- + Cl- + H2O ↔ NH4OH + Cl- + H+


        
          H NH OH   Acid Base
              
                                                     
                                                       Kw
                  NH 
                          4
   Kh
                      4
                          
                                 Unhydrolysed Salt  K b
             x2
        
          1 - x V
• Since [NH4OH] and [H+] are equal
                 Hydrolysis
     Kh   
            H NH 4OH   H   K w
                              2

                        
                 NH 4          c      Kb

    H   
               c
                   Kw
                   Kb
     pH  1 pK w  1 pK b  1 log c
          2        2        2


• Ammonia in water Kb = 1.85 x 10-5 and
• pKb = 4.74
• pH = ½pKw - ½pKb - ½log c
                 Hydrolysis
• pH = ½(14) - ½(4.74) - ½(-0.6989)
• pH = 7.0 – 2.37 + 0.3495
• pH = 4.98
                            Hydrolysis
• Salt of a weak acid and weak base
• NH4+ + Ac- + H2O ↔ NH4OH + HAc
•
   Kh 
        a     a
                                
                                  NH 4OH  HAc 
                                   f                                     f HAc
                                    NH  Ac 
           NH 4OH         HAc                                  NH 4OH
                                               
          a NH   a Ac                   4
                                                     
                                                               f NH   f Ac 
              4                                                    4



        
          NH 4OH  HAc
   Kh
           NH  Ac 
                  4
                               
                                                 ........(1)
                                        ..........

   Kh 
            Base Acid 
          Hydrolysed Salt         2
                     Hydrolysis
• If x is the degree of hydrolysis of 1g mol. of salt
  dissolved in V litres of solution, then
• [MOH] = [HA] = x/V; [M+] = [A-] = (1 – x)/V
• Substituting the values in the equation 1 given in the
  previous slide
               x V x V          x2
    Kh                       
                               
         1  x V  1  x V 1  x 2   
• The degree of hydrolysis and consequently pH is
  independent of the concentration of the solution.
                         Hydrolysis
• Remembering the following equations must hold
  simultaneously

             
    K w  H   OH 

    Ka   
           H  Ac  and K  NH  OH 
                    
                                 4
                                         


             HAc          b
                                 NH 4OH

• Also, it can be shown that

   K h  K w K a  Kb
   pK h  pK w  pK a  pKb
                           Hydrolysis
• The hydrogen ion concentration of the hydrolysed
  solution is calculated in the following manner.

       H   K A 
          
                      
                        HA  K         xV                x
                                   a             Ka 
                  a         
                                      1  x  V        1  x 
       but x 1  x   K h

       H   K
              
                       a    K h  K w  K a Kb
       or pH  1 pK w  1 pK a  1 pK b
               2        2        2
                     Hydrolysis
• If the ionisation constant of the acid and base are equal,
  i.e. Ka = Kb, pH = ½pKw = 7 and the solution is neutral,
  although the hydrolysis may be considerable.
• Ka > Kb, pH < 7
• Kb > Kc, pH > 7
• The pH of a solution of ammonium acetate is given by:
• pH = 7.0 + 2.37 – 2.37 = 7.0 and the solution is
  approximately neutral
• On the other hand, for a solution of ammonium formate
• pH = 7.0 +1.88 – 2.37 = 6.51
• Formic acid; Ka = 1.77 x 10-4; pKa = 3.75 and the solution
  reacts slightly acid.
                      pH Buffers
• A solution of 0.0001 M HCl should have a pH of 4, but
  this solution is extremly sensitive to traces of alkali from
  the glass container and ammonia in the air.
• Likewise a solution of 0.0001 M NaOH should have a pH
  of 10 but this solution is sensitive to carbon dioxide in the
  air.
• An aqueous solution of KCl pH = 7
• Likewise, an aqueous solution of NH4Ac pH = 7
• The addition of 1 mL of 1M HCl to 1 litre of KCl changes
  the pH to ≈ 3
• The addition of 1 mL of 1M HCl to 1 litre of NH4Ac does
  not change the pH much at all.
                    pH Buffers
• This is because the H+ ions added are mopped up by the
  acetate ions: H+ + Ac- → HAc (the equilibrium lies very
  much to the right hand side of this equation).
• The ammonium acetate it is behaving as a buffer and
  resisting changes in pH. A buffer possesses reserved
  acidity and reserved alkalinity.
                          pH Buffers
• For a buffer formed from a weak acid (HA) and the salt
  of a weak acid (MA). HA ↔ H+ + A-
• Making an approximation of activities ≈ concentrations

                 H   
                           HA   K
                             A 
                                        a


• If the concentration of the acid = c a
• The concentration of the salt = cs
• Then the concentration of the un-dissociated portion of
  the acid = ca – [H+]
• The solution must be electrically neutral [A-] = cs + [H+]
                       pH Buffers

                  H  
                           
                                  
                             ca  H 
                                       Ka
                                  
                             cs  H 


• The quadratic equation can be solved for [H+]
• It can also be simplified by considering that in a mixture
  of weak acid with its salt, the dissociation constant of the
  acid is repressed by the common ion effect. The
  hydrogen ion concentration is negligibly small in
  comparison with ca and cs
• Equation reduces to:

                H 
                       ca
                         Ka       H          Acid   K
                                               Salt  a
                        cs
                     pH Buffers
• Or
        pH  pK a  log
                         Salt 
                        Acid 
       Similarly
            -  Base  K
       OH   Salt  b     or pOH  pKb  log
                                                 Salt 
                                                Base
• If the concentrations of an acid and salt are equal (i.e.
  half neutralised), then pH = pKa
• For acetic acid, Ka = 1.82 x 10-5, pKa = 4.74
• At half titre point a 0.1 M HAc solution would have a pH
  of 4.74. This would also be true for higher and lower
  concentrations of HAc, e.g 1.0 M and 0.05 M HAc
                     pH Buffers
• If we add a small concentration of H+ ions they would
  combine with acetate ions to form acetic acid.
• H+ + CH3COO- ↔ CH3COOH
• Similarly a small concentration of OH- added will
  combine with H+ ions from dissociation of HAc and form
  water. More acetic will dissociate to replace the H+ ions
  depleted in this manner.
• Example: Calculate the pH of a solution produced by
  adding 10 mL of 1M HCl to 1 liter of a solution which is
  0.1M in acetic acid and 0.1M in sodium acetate.
• The pH of the acetic acid – sodium acetate buffer is
  given by:
                   pH Buffers

   pH  pK a  log
                    Salt   4.74  0.00  4.74
                   Acid 
• Neglecting the volume change from 1000 to 1010 mL.
• The HCl reacts with acetate ion forming pratically
  undissociated acetic acid.
• H+ + CH3COO- ↔ CH3COOH
• Therefore [CH3COO-] = 0.1- 0.01 = 0.09
• And [CH3COOH] = 0.1 + 0.01 = 0.11
• pH = 4.74 + log 0.09/0.11 = 4.74 – 0.09 = 4.65
                      pH Buffers
• The addition of the strong acid to the buffer change the
  pH by 4.74 - 4.65 = 0.09 pH
• Adding 10 mL of 1M HCl to 1 litre of pure water (pH 7),
  the pH would have changed from 7 to –log(0.01) = 2, by
  5 pH unites.
• A solution that contains equal concentrations of acid and
  salt, or a half neutralized solution, has the maximum
  buffer capacity. Other mixtures will also have
  considerable buffer capacity, but the pH will differ slightly
  from the half-neutralized acid.
• E.g. a quarter neutralized solution of acid, [Acid] = 3[Salt]
• pH = pKa + log ⅓
•     = pKa – 0.477
                     pH Buffers
• E.g. a three quarter neutralized acid solution, 3[Acid] =
  [Salt]
• pH = pKa + log 3
•     = pKa + 0.477
• Generally speaking the buffering capacity is maintained
  for mixtures within the range:
• 1 acid : 10 salt and 10 acid : 1 salt
• pH = pKa  1
• The concentration of the acid is usually of the order 0.05-
  0.20 M.
• Similar remarks apply to buffers of weak bases
  (NH3/NH4Cl)
                   pH Buffers
pH of 0.2 M HAc, 0.2 M NaAc Buffer Mixtures 10 mL Vol.

      HAc (x mL)   NaAc (y mL)    pH
      9.5          0.5            3.42
      9.0          1.0            3.72
      8.0          2.0            4.05
      7.0          3.0            4.27
      6.0          4.0            4.45
      5.0          5.0            4.63
      4.0          6.0            4.80
      3.0          7.0            4.99
      2.0          8.0            5.23
      1.0          9.0            5.57

								
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