Transcript Document

Chemistry
Ionic equilibrium-II
Session Objectives
Session Objectives
1. pH of weak acids
2. pH of mixture of two strong acids
3. pH of mixture of strong and weak acids
4. Dissociation of polybasic acids
5. pH of mixture of two weak acids
For mixture of two strong acids
Let us consider a mixture containing
200 ml 0.1 M HCl and 500 ml 0.2 M H2SO4.
Since both are strong electrolytes,
H
HCl 
200 × 0.1×10–3

Cl
0.02 mole
= 0.02 mole
H2SO4 

500 × 0.2 × 10–3
= 0.1 mole
2H
2 × 0.1 mole

SO42
For mixture of two strong acids
 Total
ion concentration,
[H ]Total 
0.02  0.2
 0.314
0.7
pH = -log H+ 
= -log 0.314 = 0.503
  Total
Note: For mixture of any two acids or bases, degree of
ionization of water is taken negligible because of common
ion effect.
Mixture of strong and weak acids
Let us consider a mixture of strong acid
(HA) and a weak acid (HX) of concentration
C1 and C2 respectively.
Now, for strong acid
HA 
 H  A 
C1
C1
C1
For weak acid,
HX
C2 1   
H
C2 

X
C2 
Dissociation constant of weak acid,
Ka 
[H ]Total [X ]
[HX]

C1  C2   C2 
C2 1   
Mixture of strong acid and weak acid
Ka 
[H ]Total [X ]
[HX]

C1  C2   C2 
C2 1   
where,   degree of dissociation of weak acid.
 pH = -log(C1 + C2)
Question
Illustrative example 1
A solution contains 0.10 M H2S and
0.3 M HCl. Calculate the concentration
of [HS–] and [S–2] ions in the solution.
For H2S,
Ka1 = 1.0 x 10-7
Ka2 = 1.3 x 10-13
Solution:
HCl 
 H  Cl
0.3
H 2S
0.3
H+ + HS–
[H ][HS ]
Ka 
 1  107
1
[H2S]
Solution
HS–
Ka 
H+ + S–2
[H ][S2 ]
2
[HS ]
 1.3  1013
[H ] co min g from H2S is negligibly small due to
common ion effect
[H ]  0.3M.
 10
7
0.3  [HS ]

0.1
 [HS ]  3.33  108
Solution
Considering [HS–] dissociates to
a very small extent
1.3  1013 
 [S
2
0.3  [S2 ]
3.33  108
4.33  1021
]
 14.43  1021
0.3
 1.443  1020
Dissociation of polybasic acids
Acids giving more than one hydrogen
ion per molecule are ‘polybasic’ or
‘polyprotic’ acids.
Examples are H2C2O4, H2CO3, H2S,
H3PO4, H3AsO4, etc.
These dissociate in stages . For example,
H2S
HS
H  HS
H  S
H  HS 

K1    
H2S
H  S 

K2    
HS 


Dissociation of polybasic acids
Normally K2 << K1.
To calculate hydrogen ion concentration,
only the first step should be considered
as the H+ obtained from successive dissociation can
be neglected, but to calculate the concentration of
[HS ],[S2 ] etc.
then both the equilibria have to be considered.
pH of mixture of two weak acids
Let HA and HB are two weak acids.
HA
C1(1  1 )
H
C11
HB
C2(1  2 )
H
C22
A
C11


B
C22
[H ]Total [A  ]
(C   C22 )C11
Ka1 
 1 1
   (1)
[HA]
C1(1  1)
[H ]Total [B ]
Ka2 
[HB]
(C11  C22 )C22

   (2)
C2(1  2 )
pH of mixture of two weak acids
Dividing (1) by (2)
Ka1
Ka2
1

2
1 , 2  1
Solving this equation we can calculate [H ]Total
and hence pH
Question
Illustrative example 2
A solution is prepared by mixing 0.2M
HCOOH with 0.5 M CH3 COOH.
Given KaCH3COOH=1.8 x 10–5 ,
KaHCOOH =2.1x10-4
Calculate [HCOO–] , [CH3COO–] and
pH of the solution.
Solution:
H
HCOOH
0.2(11 )
(0.21  0.52 )

CH3COOH
H
0.21
 CH3COO
(0.21  0.52 )
0.5(12 )
K1  2.1  10
 HCOO
4
0.52
(0.21  0.52 )1

......(1)
(1  1)
Solution
K2  1.8  10
5
(0.21  0.52 )2

(1  2 )
......(2)
1 , 2  1
1

 11.67
2
From (2)
2.1  104
1

5
2
1.8  10
1  11.67 2
1.8  105  (0.2  11.67 2  0.52 ) 2
 2.834  2
2
Solution
1.8  105
 2 
 2.52  103
2.834
 1  11.67  2.5  103
 2.94  102
 H 
 
total 
(0.21  0.52 )
 (5.88  103  1.26  103 )
 7.14  103
Solution
 pHmix  3  log7.14  2.15
HCOO   0.2  5.88  103
1


CH COO   0.5  1.25  103
2
 3

Hydrolysis of Salts
NH4OH  H  Cl 
NH4Cl  H2O
HCOONa  H2O
HCOOH  OH  Na
A. Hydrolysis of a salt of weak acid and strong base
The hydrolysis reaction is
At eqm.
CH3COO  H2O
CCh
CH3COOH  OH
Ch
where C = concentration of salt
h = degree of hydrolysis.
Ch
Hydrolysis of Salts
Hydrolysis constant
Kh 
OH 
Ch2



(1  h)
CH COO 
 3

CH3COOH
Kh  ch2

h  1
Kw
Kh 
 Ch2
Ka
OH   Ch


Kw . C

Ka
Kw
Kw
h 
or h 
K a .C
K a .C
2
Hydrolysis of Salts
1
1
1



log OH   log Kw  log Ka  log C
2
2
2
pOH 
1
pKw  log Ka  logC
2
1
1
1
pH  pKw pKa  logC
2
2
2
Hydrolysis of Salts
B. Salt hydrolysis of strong acid
and weak base
For NH4Cl  NH4  Cl
NH4  H2O
NH4OH  H
Similarly,
Kw
Kh 
 Ch2
Kb
Kw
h
Kb  C
where Kb =
H   Ch  K w  C
 
Kb
1
1
1
pH  pKw  pKb  logC
2
2
2
Dissociation constant of
weak base.
Hydrolysis of Salts
C. Hydrolysis for a salt of
weak acid and weak base
CH3COONH4  CH3COO  NH4
At eqm.
At eqm.
CH3COO
C 1  h

NH4  H2O
C 1  h
H2O
CH3COOH  OH
Ch
Ch
NH4 OH  H
Ch
Ch
Hydrolysis of Salts
Kh
CH3 COOH NH4 OH


CH COO  NH 
 3
 4
Kw
c 2h2


K a  K b c 2 1  h 2
Kw
h

=
1-h
Ka ×Kb
Hydrolysis of Salts
Now, to calculate the pH
CH3COOH
C 1  h
H  CH3COO
Ch
Ch
For pH,
H  CH COO 
H  Ch
3



   
Ka 
C 1  h
CH3COOH
1  h



 H  Ka 

 
h
Kw  Ka
Kb
1
1
1
pH  pKw  pKa  pKb
2
2
2
Questions
Illustrative example 3
Calculate the pH at the equivalence
point of the titration between 0.1 M
CH3COOH (50 ml) with 0.05 M
NaOH . Ka(CH3COOH) = 1.8 × 10–5.
Solution:
At the equivalence point,
CH3COOH

NaOH 
 CH3COONa  H2O
50  0.1  103
V  0.05  103
Let V ml NaOH is required to reach the equivalence point.
At the equivalence point,
N1V1
50  0.1  103


N2 V2
V  0.05  103
Solution
V = 100 ml
0.1  50
 CH3COONa 
 0.033
150
CH3COO  H2O
C(1  h)
Kw
Kh 
 Ch2
Ka
Kw
h
Ka  C
CH3COOH  OH
Ch
Ch
[OH ]  C . h
Solution
K
[H ] 
 w 
[OH ] Ch


Kw
KwKa
C
1014  1.8  105
3.3  102
 5.45  1018
 2.33  1019
pH   log[H ]  9  log 2.33  8.63
Illustrative example 4
Calculate the percentage hydrolysis of
decinormal solution of ammonium acetate,
given Ka = 1.75 × 10–5, Kb = 1.80 × 10–5
and Kw = 1 × 10–14. What will be the
change in degree of hydrolysis when 2 L of
water is added to 1 L of the above
solution?
Solution:
Since CH3COONH4 is a salt of weak acid
and weak base
Kw
h2
Kh 

Ka  Kb (1  h)2
Solution
h


1h

Kw
Ka  Kb
1  1014
1.75  105  1.8  105
 5.63  103
 h  5.63  103
[h  1]
Since degree of hydrolysis is not affected by the
concentration in this case. So on dilution there will be no
change in degree of hydrolysis.
Illustrative example 5
Hydrolysis constant of Zn+2 is 1 × 10–9
(a) Calculate pH of a 0.001 M
solution of ZnCl2.
(b) What is the basic dissociation constant
of Zn(OH)+?
Solution:
Zn2
 H2O
103(1  h)
Zn(OH)  H
103h
108h
Kh  1  109  103h2  h  1
Solution
h 
109
103
 103
H   103h  106
 
pH  6
Zn(OH)
103h
Zn2  OH
103
103
Zn2  OH 
11

 

10
5
Kb 


10
Zn(OH) 
6
10


Illustrative example 6
When 0.20 M acetic acid is neutralised
with 0.20 M NaOH in 0.50 litre of water,
the resulting solution is slightly alkaline,
calculate the pH of the resulting solution
(Ka for acetic acid = 1.8 x 10-5)
Solution:
CH3 COOH  NaOH 
 CH3COONa  H2O
moles
0.2  0.5
CH3COO  H2O
0.1(1  h)
0.1h
0.2  0.5
0.2  0.5
CH3COOH  OH
0.1h
Solution
Kw 0.1h2
Kh 

 0.1h2
Ka
1h
 h  1
h
1014
1.8  105  0.1
OH   0.1h


 0.745  104
pOH  5.128
pH  8.872
Illustrative example 7
Calcium lactate is a salt of a weak
organic acid and represented as
Ca(Lac)2. A saturated solution of Ca
(Lac)2 contains 0.13 mol of this salt in
0.50 litre solution. The pOH of this
solution is 5.60. Assuming a complete
dissociation of the salt, calculate Ka of
lactic acid.
Solution:
Ca(Lac)2 
 Ca2  2Lac1
0.13mol
2  0.13 mol.
Lac   0.26  0.52 M


0.5
Solution
Lac  H2O
LacH  OH
0.52(1  h)
0.52h
0.52h
OH   0.52h  2.51  106


h
Ka 
 4.83  106
Kh 
1014

0.52  4.83  10
6

2
Kw
Ka
 0.52h2  h  1
 8.24  104
Class exercise
Class exercise 1
The hydrolysis constant for FeCl2 will be
Kw
(a) Kh =
Kb
(c) Kh =
Kw2
Kb2
Kw2
(b) Kw =
Kb
(d) Kh =
Kb
Kw2
Solution:
FeCl2 is the salt of strong acid and weak base.
Since FeCl2  Fe+2 + 2Cl-
Fe+2 + 2H2O
Fe(OH)2 + 2H+
Solution
Kh =
[Fe(OH)2] [H+]2
Fe(OH)2
[Fe+2]
Fe+2 +2OH-
[Fe+2][OH-]2 [Fe2+][OH] [H+]2
Kb =
=
×
[Fe(OH)2]
[Fe(OH)2] [H+]2
Kw2
Kh =
Kb
Hence, the answer is (b)
Class exercise 2
pH of a solution produced when an
aqueous solution of pH 6 is mixed with an
equal volume of an aqueous solution of
pH 3, will be
(a) 4.5
(b) 4.3
(c) 4.0
(d) 3.3
Solution:
pH = 6
[H+]=10-6 M
pH = 3
10-6 +10-3
[H ]Total =
2
+
Solution
1.001×10-3
=
=5.005×10-4
2
 pH = –log [H+]Total
= 4 – log 5.005
= 3.3
Hence, the answer is (d).
Class exercise 3
Which one of the following is true for
any diprotic acid, H2X?
(a) Ka2 > Ka1
(b) Ka1 > Ka2
(c) Ka1 = Ka2
(d) Ka2 =
1
Ka1
Solution:
H2X being a diprotic acid,
H2X
H+ + HX- Ka1
HXH+ + X-2 Ka2
Due to the ‘common ion effect’ dissociation of HX–
will be less.
 Ka2 > Ka1
Class exercise 4
Ka (CH3COOH) = 1.7 × 10–5 and
[H+] = 3.4 × 10–4. Then initial
concentrations of CH3COOH is
(a) 3.4 × 10–4
(b) 6.8 × 10–3
(c) 3.4 × 10–3
(d) 6.8 × 10–2
Solution:
CH3COOH
C(1–)
H+ + CH3COO-
C
C2
Ka=
 C2
1- 
C
Solution
[H+]=3.4×10-4 = C
 1.7  10–5 = 3.4 × 10–4  
 
C 
1.7×10-5
3.4×10-4
3.4×10-4
5×10-2
=5×10-2
=0.68×10-2 =6.8×10-3
Hence, the answer is (b).
Class exercise 5
0.001 M HCl is mixed with 0.01 M HCOOH
at 25° C. If Ka HCOOH = 1.7 × 10–4,
find the pH of the resulting solution.
Solution:
HCl  H+ + Cl0.001
0.001 0.001
H+ + HCOO-
HCOOH
Initial mole
10–2
At eqm.
10–2(1– )

 H+ 
= 102   103
  Total
10–2

+10-2
Solution
H+ 
HCOO- 
  Total 

Ka=
HCOOH


  102   103 


-2
10 1-  
10-2  +10-3
Ka
1.7 × 10–4 = 10–3 
[ 10–2 2 << 10–3 ]
H+ = C =10-2 ×.17 =1.7×10-3
 = 1.7 × 10–1


 H+ 
= 1.7×10-3 +10-3 =2.7×10-3
  Total
 pH = 3 – log 2.7 = 2.57
Class exercise 6
What is the percentage hydrolysis
of N NaCN solution when the
80
Ka HCN = 1.3 × 10–9,
K2 = 1 × 10–14?
(a) 2.48
(b) 5.26
(c) 9.6
(d) 8.2
Solution:
NaCN  Na+ + CN-
CN-
+
1
1- h
80
H2O
HCN
h
80
+
OHh
80
Solution
2
 h 


2
Kw
h
80
 
 Kh =
= 
1
Ka
80
1h
 
80
80×10-5
 h=
= 6.15×10-4
1.3
10-14
h2


-19 80
1.3×10
= 2.48 × 10–2
 Percentage hydrolysis = 2.48 × 10–2 × 102 = 2.48
Class exercise 7
Calculate the pH of a solution obtained
by mixing 100 ml of 0.1 M HCl with 9.9
ml of 1 M NaOH.
Solution:
HCl
+
100×0.1×10-3
100×10-4
NaOH  NaCl + H2O
9.9×1×10-3
99×10-4
Acid remaining = 1 × 10–4 moles
-4
1×10
+
H  =
×103 =9.09×10-4


109.9
pH = –log[H+] = 4 – log 9.09
= 3.04
Class exercise 8
Calculate the percentage hydrolysis of
3 × 10–3 M aqueous solution of NaOCN
(Ka HCON = 3.33 × 10–4 M).
Solution:
NaOCN  Na+ + OCNOCN-
+
3×10-3 1-h
H2O
HOCN
+
OH-
3×10-3 h 3×10-3 h
Kw 3×10-3 h2
Kh =
=
 3×10-3 h2
Ka
1-h
Solution
h=
10-14
3.33×10-4 ×3×10-3
= 1×10-8 =10-4
Percentage hydrolysis = 10-4 ×102 = 0.01
Thank you