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Ionic Equilibrium - Career Launcher

VIEWS: 21 PAGES: 59

									Chemistry
Ionic equilibrium-III
Session Objectives
Session Objectives




1. Hydrolysis of salts

2. Buffer and buffer capacity
Buffer solution

Types of buffer solution




 a.    Acidic buffer:            CH3COOH and CH3COONa


 b.    Basic buffer:             NH4OH and NH4Cl


 c.    Salt or neutral buffer:    CH3COONH4
Animation of buffer
Explanation
Now, to find the pH, Let us consider

CH3 COOH       CH3 COO  H

     H  CH COO 
Ka    3         
       CH3COOH
                

Since CH3COOH is feebly ionized, so most of CH3COO–
will come from the salt.
Hence, [CH3COO–] = [Salt]

             CH COOH
 H   K  3       
         a
             CH COO 
              3      
Explanation

Taking log of both sides

 log H    logK  log [Acid]
                  a
                           [Salt]
                [Salt]
 pH  pKa  log
                [Acid]

This is known as Henderson’s equations.
Similarly, for basic buffer

                    [Salt]
 pOH  pKb  log
                   [Base]

 We know, pH + pOH = 14
Questions
Illustrative example 1
 Calculate the pH of a buffer solution
containing 0.1 M each of acetic acid and
sodium acetate. (Ka = 1.8 × 10–5)
What will be the change in pH on adding
(a) 0.01 moles of HCl to 1.0 L of solution?
(b) 0.01 moles of NaOH to 1.0 L of solution?
(Assume that no change in volume occurs on
the addition of HCl or NaOH)
Solution:
 We know
pH  pKa  log
               [Salt]
               [Acid]
                                   
                      or pH   log 1.8  105
                                               
                                                log
                                                     0.1
                                                     0.1
 pH = 4.745 + 0 = 4.745
Solution
(a) 0.01 moles of HCl give 0.01 moles
   of H+ which reacts with 0.01 moles
   of CH3COO– to form CH3COOH.
   Therefore,
 CH COO   0.1  0.01  0.09 M
  3      

 CH3COOH  0.1  0.01  0.11 M
          
                  [0.09]
 pH  4.745  log         4.745  0.087  4.658
                  [0.11]

Change in pH = 4.745 – 4.658 = 0.087 units

  Hence, pH will decrease.
Solution
(b) On adding 0.01 moles of NaOH to a
    litre of solution, 0.01 moles of OH– will
    react with 0.01 moles of CH3COOH.
    Therefore,

 [CH3COOH] = 0.1 – 0.01 = 0.09 M
 [CH3COO–] = 0.1 + 0.11 = 0.11 M
                    [0.11]
 pH  4.745  log           4.745  0.086  4.831
                    [0.09]
 Change in pH = 4.831 – 4.745 = 0.086 units


  Hence, pH will increase.
Illustrative example 2
Calculate the amount of NH3 and NH4Cl
required to prepare a buffer solution of pH
9.0 when total concentration of buffering
reagents is 0.6 mol L-1. pKb for NH3 = 4.7,
log 2 = 0.30 ?

Solution:
 NH3 and NH4CL form a basic buffer

 Let NH3   xM
         
    NH4Cl  (0.6  x)M
         

                  NH4Cl
  pOH  pKb  log         
                    NH3 
                        
Solution

                    0.6  x
14  pH  4.7  log
                       x
                  0.6  x
                           antilog 0.3  2
                     x

0.6 – x = 2x         3x = 0.6


     0.6
 x       0.2
      3

 NH3   0.2M
      

 NH4Cl  0.4M
      
Buffer Capacity
Buffer capacity is defined
quantitatively as number of moles of
acid or base added in 1 L of solution
as to change the pH by unity, i.e.

Buffer capacity


    Number of moles of acid or base added to 1 litre
  =
                    Change in pH
Questions
Illustrative example 3
The pH of a buffer is 4.75. When 0.01
mole of NaOH is added to 1 L of it, the
pH become 4.83. Calculate its buffer
capacity.
Solution:
                  Moles of base added to 1 litre solution
Buffer capacity =
                              Change in pH
                     0.01
                              0.125
                  4.83  4.75
Illustrative example 4
10 ml. of 10-5 M solution of sodium
hydroxide is diluted to one litre. The
pH of the resulting solution will be
nearly equal to:
(a) 6                (b) 7

(c) 8                (d) 9

Solution:
Moles of NaOH in 10ml

 105  10  103
 107
Solution
Moles of NaOH in 103 ml diluted solution

       7     103
 10              105
              10
NaOH  Na  OH
       
 105                      105

[OH ]  105
 pOH  5
 pH  14  5  9




                                    Hence, answer is (d).
Illustrative example 5
Calculate the pH value of the mixture
containing 50 c.c M - HCl and 30 c.c.
M-NaOH solution assuming both to be
completely ionised

 Solution:

   HCl                  
                NaOH  NaCl  H2O
          3
50110         301103
moles           moles


pH of the solution depends on the conc. of the
 remaining subs tan ce
Solution
                       (50  103  30  103 )
 [HCl] re m ainin g 
                                 80
                                103

                        20  103
                                  0.25  [H ]
                          0.08


  pH   log[H ]
        log0.25  0.602
Solubility and solubility product
Any sparingly soluble salt will dissolve in
very small amount and whatever
dissolves will ionize into respective ions.
At certain temperature solubility of a salt
is fixed.

 Thus
 AgCl (s)       Ag+ + Cl
     Ag  Cl 
 K           
     [AgCl (s)]

[AgCl] may be taken as constant,
thus K·[AgCl] = [Ag+]·[Cl–]
Solubility and solubility product
or       Ksp = [Ag+]·[Cl–]

Where, Ksp = Solubility product
in saturated solution.

 In saturated solution,
 the ionic product, Ki = Ksp


 If Ki < Ksp   the solution is unsaturated.


If Ki > Ksp the solution is super saturated and
precipitation occurs.
Different cases of solubility
product
a. Electrolyte of the type AB2/A2B

AB2  s       A 2  2B
               s       2s
Where s =solubility of the salt in mole/L
                       2
Ksp   A 2  B 
             

    s  (2s)2  4s3

PbCl2       Pb2  2Cl

                           2
Ksp    Pb2  Cl  = s × (2s)2 = 4s3
                  
b. Electrolyte of the type AB3
 AB3  s        A 3  3B
                 s           3s
                         3
  Ksp   A 3  B 
               

   s(3s)3  27s4

               Ksp
  s  4
               27
  AlI3  s      Al3  3I 
                         3
   Ksp     Al3  I          = s × (3s)3 = 27s4
                  
c. Mixture of electrolytes AB and A`B

A 'B  s          A'  B
                    s     (s  s ')

 Ksp   A   B 
    1      Total

        s(s  s ')    (I)
  A 'B  s        A '  B
                     s'    s ' s

  Ksp   A '  B 
     2           Total
           s '(s  s ')      (II)
 Comparing (i) and (ii) Now solving these three equations
 Ksp    s               we can find s and if K sp1 and Ksp2
    1
             (III)
 Ksp    s'
      2
Question
Illustrative example 6
The concentration of Ag+ ion in a
saturated solution of Ag2CrO4 at 20° C
is 1.5 × 10–4 M. Determine the
solubility product of Ag2CrO4 at 20°C.
Solution:
 Ag2CrO4          2Ag  CrO42
    Ag   1.5  104 M
        

   CrO    2   1.5  104
       4                    0.75  104 M
                    2
                   2
    Ksp   Ag  CrO42 
                        

   (1.5  104 )2  0.75  104

   1.69  1012
Solubility of a salt in presence of
common ion
Let us find out the solubility of Ag2CrO4
(Ksp =1.9 x 10–12) in 0.1 M Ag NO3 solution.

 In water

Ag2 CrO4               
             2Ag  CrO42
    S          2S      S

Ksp  (2S)2 S  4S3  1.9  1012

       1.9  1012
S  3              0.78  104 mol3 / L3
            4
In AgNO3

                  
AgNO3  Ag  NO3
        
                0.1
Solubility of a salt in presence of
common ion

Ag2CrO4                 
              2Ag  CrO42
               2S`       S`
                   
 Ksp  [Ag ]2[CrO4 ]

1.9  1012  (2s` 0.1)2S
                        `
  S` 0.1
    3      2
 S` and S` can be neglected

  1.9  1012  102 x S`
  S  1.9  1010
    `

 Solubility decreases in presence of common ion
Question
Illustrative example 7
The solubility of Mg(OH)2 in pure
water is 9.57 × 103 gL–1. Calculate its
solubility in gL–1 in 0.02 M Mg(NO3)2
solution.

Solution:
        Solubility of Mg(OH)2 in pure water
        = 9.57 × 10–3 gL–1

           9.57  103
                      mol L1  1.65  104 mol L1
               58
         Mg(OH)2      Mg2  2OH
                                2
         Ksp  Mg2  OH 
                                 1.65  104 (2  1.65  104 )2
Solution
 17.97  1012
Let the solubility ‘s’ of Mg(OH)2 in
presence of Mg(NO3)2

  Mg(NO3 )2 is completely ionized.
Mg(NO3 )2  Mg2  2NO3
            
  0.02 M                 0.02 M

Mg2    s ' 0.02
     

OH    2s '
     
                           2
 Ksp    Mg2  OH 
                    
Solution
                     2
  s ' 0.02 2s '
  s ' 0.02   4s '2    s '  0.02
                                       
 0.02  4  s '2

  0.08  s '2  17.97  1012

         17.97  1012
 s' 
             0.08
     14.99  106 M

 Solubility in gL–1 = 14.99 × 10–6 × 58

  = 8.69 × 10–4 gL–1
Solubility of a salt forming
complex
 Let the solubility of AgCl is s mol L–1

AgCl(s)           Ag (aq.)  Cl (aq.)
                  Sx           S

 Ag  2NH3           [Ag(NH3 )2 ] (aq.)
 S x     C 2x             x

 Ksp  (S  x)S
                  x
Kf 
        (S  x)(C  2x)2

  Kf=Eqm. Constant for the formation of complex

 Solubility of salt increases due to complex formation.
Question
Illustrative example 8
A solution of Ag+ ion at a concentration
of 4 x 10–3 M just fails to yield a precipitate
of AgCl with a concentration of 1 x 10–3 M
of Cl– ion when the concentration of NH3 in
the solution is 2 x 10–2 M at equilibrium.
Calculate the magnitude of the equilibrium
constant for the reaction
         
Ag(NH3 )2   Ag  2NH3
Ksp AgCl at 250 C  1  1010
Solution:
 To make AgCl soluble in solution, maximum conc. of Ag+
 in the solution,
          Ksp     1010
     
[Ag ]                 107 [Ag ]max  107M
         [Cl ]       3
                  10
Solution
So, rest of Ag+ forms complex with NH3
  Ag         2NH3                
                           Ag(NH3 )2
4103  x    2102 2x        x

Now, 4  103  x  107
                      x  4  103  107
                       4  103
                   4  103
  Kf 
        107  (2  102  8  103 )2
             4  103
                             2.78  108
        107  1.44  104
 Eqm. cons tan t for the given reaction
       1
 K f`     0.36  108  3.6  109
       Kf
Selective precipitation
Let us consider a solution containing
0.01M Cl– and 0.02 M SO4–2
AgNO3 is being added to this solution.

Let us find out Ki=ionic product for AgCl and Ag2SO4
As   Ki > Ksp precipitate forms

  Ksp AgCl = 10–10

Ksp Ag2SO4 = 10–13

 AgCl(s)     Ag  Cl

 Now, [Ag ][Cl ]  Ksp  1010

         1010
 [Ag ]         108M
          102
Selective precipitation
Ag2SO4               
            2Ag  SO42
           
 [Ag ]2[SO42 ]  1013

           1013
 [Ag ]                2.24  106
           2  102


For precipitation of AgCl, lesser conc. of Ag+ is required.
So it will precipitate first and Ag2SO4 gets precipitated
only when [Ag+] becomes greater than 2.24 x 10–6 M
Selective precipitation
We can find out % of Cl– precipipated
when Ag2SO4 starts precipitating

 Then

        
                       Ksp AgCl
   [Cl ]remaining 
                        [Ag ]
                         1010
                                     4.46  105
                       2.24  106

        [Cl ] initial 102 M
                    4.46  105
 % Cl remaining           2
                                  100  0.446%
                        10
 Cl precipitated  99.55%
Question
Illustrative example 9
A solution has Zn+2 and Cu+2 each At 0.2M.
The Ksp of ZnS and H2S are 1 x 10–22 and
of CuS is 8 x 10–37. If the solution is made
1M in H3O+ and H2S gas is passed until the
solution is saturated, should a precipitate
form? Also report the ionic concentration
product for ZnS and CuS in solution when
first ion starts precipitating.
Solution:
ZnS     Zn2  S2 Ksp  1 x1022
CuS     Cu2  S2 Ksp  8 x1037
H2S    2H  S2   Ksp  1 x1022
Solution

As 1M solution of H3O is saturated with H2S,
        Ksp    1022
[S2 ]   2         1022
        [H ]     1

  Now, Ki ZnS  [Zn2 ][S2 ]
                    0.2  1022
                    2  1023
   Ki  Ksp
   No precipitates of ZnS will form
   Ki CuS  [Cu2 ][S2 ]
            0.2  1022
            2  1023
Solution

 Here, Ki  Ksp
  CuS starts precipitating

  Ki ZnS  2  1023  Ki CuS
Class exercise
Class exercise 1
The solubility of A2X3 is s mol dm–3. Its
solubility product is
(a) 6s4             (b) 64s4

(c) 36s5            (d) 108 s5

Solution:

A2X3        2A+3 + 3X-2
            2s     3s

           2   3
Ksp = 2s 3s =108 s5




 Hence, the answer is (d).
Class exercise 2
A compound M(OH)y has a Ksp of
 4 × 10–6 and the solubility is 10–2 M.
The value of y should be
(a) 1                      (b) 2

(c) 3                      (d) 4

Solution:

MOHy      M+y + Y OH-
             s         ys
                       y
Ksp = M y+  OH- 
                
                
Solution



 4 × 10– 6 = 10–2 (y × 10– 4)y
 (y × 10–2)y = 4 × 10– 4
 If we put y = 2, then only Ksp can be 4 × 10–6
 Hence y = 2.




 Hence, the answer is (b).
Class exercise 3
pH of a mixture containing 0.1 M and
0.2 M HX will be
Given pKb(X–) = 4
(a) 4 + log 2       (b) 10 + log 2
(c) 4 – log 2       (d) 10 – log 2

Solution:
                [Salt]
 pOH=pKb +log
               [Base]
             0.1
     = 4+log
             0.2
            1
     = 4+log = 4-log 2         pH = 14 – pOH = 10 + log 2
            2


 Hence, the answer is (d).
Class exercise 4
For a basic indicator Kb is 1 × 10–10.
It changes colour when the indicator
is 10–5 M. The pH of solution at this
point is
(a) 10              (b) 4
(c) 2               (d) 8
 Solution:
Since colour change happen at the end point of titration
and it is assumed that all indicator has been consumed
during the process
 HpH  H+ + pH-
       

It is assumed that at the end point [Hph] = [Ph–]
Solution

                     [Ph- ]
Hence, pOH= pKb +log
                     [4Ph]

 pOH = 10
  pH = 14 – 10 = 4




 Hence, the answer is (b).
Class exercise 5
For CH3COOH, pKa = 4.74 . A buffer
solution of acetic acid 0.1 M and lead
acetate (xM) has
pH = 5.04. The value of x is
(a) 0.2              (b) 0.1

(c) 0.02             (d) 0.05
Solution:
For an acidic buffer,
            CH COO- 
             3      
 pH=pKa+log         
            CH3COOH
                CH COO- 
                 3      
5.04 = 4.74+log         
                   0.1
Solution

 CH COO- 
  3      
         = Antilog 0.3
    0.1

  CH COO-  =0.199  0.2
   3
          
           

 For  CH3COO2 Pb  Pb+2 +2CH3COO-
                     

                    CH COO- 
                     3      
  CH3COO2 Pb = 
               
                              = 0.1
                        2




 Hence, the answer is (b).
Class exercise 6
Saccharin (Ka = 210–12) is a weak
acid represented by formula HSac.
A 4 × 10–4 mole of saccharin is
dissolved in 200 ml water of pH 3.
Assuming no change in volume,
calculate the concentration of Sac–
ions in the resulting solution at
equilibrium.

 Solution:

                 HSac             H+   +Sac–
Intial conc.     0.002          0.001     0
conc. at eqm.   (0.002–x)     (0.001+x)   x
Solution

Given
[H+] = 0.001 M                         
                                       
              -4                       
          4×10 ×1000
[HSac] =            = 2×10-3M = 0.002 M
              200                      

            [H+] [Sac-]
       Ka =
              [HSac]
               (0.001+ x) (x)
           =
                 (0.002 - x)
               (0.001+ x) x
 2×10–12 =
                (0.002 - x)
        x = 410-12 M
[Sac- ]eqm = 4×10-12M
Class exercise 7
Calculate the amount of NH3 and NH4Cl
required to prepare a buffer solution of
pH 9.0 when total concentration of
buffering reagents is 0.6 mol L–1. pKb for
NH3 = 4.7, log 2 = 0.30.
 Solution:
NH3 and NH4Cl forms a basic buffer
NH+  =x
 4
           NH+ 
              4
pOH=pKb +log     
            NH4OH
                x
5= 4.74+log
              0.6 - x
  x
        = Antilog0.3=1.999  2
0.6 - x
Solution

 x = 1.2 – 2x
      1.2
 x=       = 0.4
       3
 NH+ =0.4 M (approx.)
  4
    

 NH OH =0.2 M (approx.)
  4 
      
Thank you

								
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