Transcript Document

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 
3




Ka 
CH3COOH
Since CH3COOH is feebly ionized, so most of CH3COO–
will come from the salt.
Hence, [CH3COO–] = [Salt]
CH3COOH


 H  Ka
 
CH COO 
 3

Explanation
Taking log of both sides
[Acid]
 log H    logKa  log
 
[Salt]
[Salt]
pH  pKa  log
[Acid]
This is known as Henderson’s equations.
Similarly, for basic buffer
pOH  pKb  log
[Salt]
[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


0.1
[Salt]
5
pH  pKa  log
or pH   log 1.8  10
 log
[Acid]
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
pH  4.745  log
[0.09]
 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
pH  4.745  log
[0.11]
 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
0.6
x 
 0.2
3
 NH3   0.2M
NH4Cl  0.4M
3x = 0.6
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
 10
7
103

 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
50110
moles
3
 NaOH 
 NaCl  H2O
301103
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
A 3  3B
AB3  s 
s
Ksp  A

3s
3
3    
B
 
 s(3s)3  27s4
s 
4
Ksp
27
AlI3  s 
Al3  3I 
3    
Ksp   Al

I
 
3
= 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
we can find s and if Ksp1 and Ksp2
Ksp
s
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:
2Ag  CrO42
Ag2CrO4
 Ag   1.5  104 M


CrO
4

1.5  104

 0.75  104 M

2
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
S
2Ag  CrO42
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  CrO42
2S`
S`
 Ksp  [Ag ]2[CrO4 ]
1.9  1012  (2s`0.1)2S`
S` 0.1
 S`3 and S`2 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 '


 Ksp  Mg2  OH 



2
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
 s' 
17.97  1012
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.)
Ag  2NH3
S x
S
Sx
[Ag(NH3 )2 ] (aq.)
C 2x
x
Ksp  (S  x)S
Kf 
x
(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

7

7
[Ag
]

10
M
[Ag ] 

 10
max


3
[Cl ]
10
Solution
So, rest of Ag+ forms complex with NH3
Ag
4103  x
 2NH3
2102 2x

Ag(NH3 )2
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
8


2.78

10
107  1.44  104
 Eqm. cons tan t for the given reaction
K f`
1
 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
8
 [Ag ] 

10
M
2
10

Selective precipitation
Ag2SO4
2Ag  SO42
[Ag ]2[SO42 ]  1013

[Ag ] 
1013
2  102
 2.24  106
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

[Cl ]remaining 

Ksp AgCl
[Ag ]
1010
2.24  106
 4.46  105
[Cl ] initial 102 M
4.46  105
% Cl remaining 
 100  0.446%
2
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
22
10
[S2 ]   2 
 1022
1
[H ]
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 OHs
Ksp = M y+  OH- 



ys
y
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
(c) 4 – log 2
(b) 10 + log 2
(d) 10 – log 2
Solution:
[Salt]
[Base]
0.1
= 4+log
0.2
pOH=pKb +log
1
= 4+log = 4 - log 2
2
Hence, the answer is (d).
pH = 14 – pOH = 10 + log 2
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
Intial conc.
conc. at eqm.
0.002
(0.002–x)
H+
+Sac–
0.001
(0.001+x)
0
x
Solution
Given
[H+] = 0.001 M



-4


4×10 ×1000
-3
[HSac]
=
=
2×10
M
=
0.002
M


200
[H+] [Sac-]
Ka =
[HSac]
=
–12
2×10
(0.001+ x) (x)
(0.002 - x)
(0.001+ x) x
=
(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
5= 4.74+log
x
0.6 - x
x
= Antilog0.3=1.999  2
0.6 - x
Solution
x = 1.2 – 2x
x=
1.2
= 0.4
3
NH+ =0.4 M (approx.)
 4 
NH OH =0.2 M (approx.)
 4 
Thank you