# 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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