Chemistry 2410: Problem Sheet 5 Solutions
1.
The acid dissociation constant for chloroacetic acid is 1.36 x 10-3. Calculate [H3O+] for
a solution containing 0.0200 M chloroacetic acid, neglecting the small amount of H3O+
produced from the dissociation of H2O. and using an iterative (successive
approximations) approach.
ClCH2COOH + H2O <========> H3O+ + ClCH2COOKa = 1.36 x 10-3
+
Let [H3O ] = x
x2
Ka =
(0.200− x )
Iterative process:
Let 0.200 – x0 = 0.0200:
x1 = √(1.36 x 10-3)(0.0200) = 0.00522
Let 0.200 – x1 = 0.0200 – 0.00522 = 0.0148: x2 = √(1.36 x 10-3)(0.0148) = 0.00448
Let 0.200 – x2 = 0.0200 – 0.00448 = 0.0155: x3 = √(1.36 x 10-3)(0.0155) = 0.00459
Let 0.200 – x3 = 0.0200 – 0.00459 = 0.0154: x4 = √(1.36 x 10-3)(0.0154) = 0.00458
Let 0.200 – x4 = 0.0200 – 0.00458 = 0.0154: x5 = √(1.36 x 10-3)(0.0154) = 0.00458
€
So [H3O+] = 0.00458 M
pH = -log(0.00458) = 2.339
2.
Calculate the fraction of dissociation, α, for the sample of chloroacetic acid described in
question [1].
[ClCH2COO-] = [H3O+] = 0.00458 M
H 3O + ]
[
1.36 x10−3
0.00458
[K a ]
α
=
=
α0 =
=
1
[H 3O + ] + K a (0.00458 +1.36 x10−3 )
[H 3O + ] + K a (0.00458 +1.36 x10−3 )
(
(
)
α 0 = 0.771
€
α 1 = 0.229
€
)
3.
Derive expressions for all fractions of dissociation for a phosphoric acid system.
H3PO4 + H2O <========> H3O+ + H2PO4H2PO4- + H2O <========> H3O+ + HPO42HPO42- + H2O <========> H3O+ + PO43-
Ka1 = 7.11 x 10-3
Ka2 = 6.34 x 10-8
Ka3 = 4.22 x 10-13
CT = [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]
−
2−
3−
H 2PO 4
HPO 4
PO 4
H 3PO 4 ]
[
α0 =
; α1 =
; α2 =
; α3 =
CT
CT
CT
CT
[
]
[H O ][H PO ] ==> H PO
=
−
+
a) K a1
€
3
2
4
[H 3PO 4 ]
4
4
€
€
€
€
K a1 H 3PO 4
+
−
2−
4
a2
2
4
a2
K a1 [H 3PO 4 ]
+ 2
+
3
3
2−
a3
4
4
a3
a2
a1
+
3
3
+ 3
4
3
Substitute equations a), b), and c) into CT:
K [H PO ] K K [H PO ] K K K [H PO ]
C T = [H 3PO 4 ] + a1 3 + 4 + a2 a1 3 2 4 + a3 a2 a1 33 4
[H 3O ]
[H 3O + ]
[H 3O + ]
[H O ]
]
[H O ]
3
+
3
€
€
[
3−
2−
C T = [H 3PO 4
€
] = [H[ O ] ]
K [H PO ] K
=
] [H O ] =
]
3
4
+ 3
€
[
3−
3
4
€
−
4
]
[H PO ]
[H O ]
[H O ][PO ] ==> PO = K [HPO ] = K K K [H PO ]
=
[ ] [H O ]
[HPO ]
[H O ]
−
+
c) K a3
2
2−
3
2
€
[
[H O ][HPO ] ==> HPO
=
+
b) K a2
[
+ 2
K a1 [H 3PO 4 ] [H 3O ] K a2K a1 [H 3PO 4 ] [H 3O ] K a3K a2K a1 [H 3PO 4 ]
+
+
3 +
2
3
+
[H 3O + ] [H 3O + ]2
[H 3O + ] [H 3O ]
[H 3O + ]
This can be rewritten as:
 H O+ 3 + K H O+ 2 + K K H O+ + K K K 
[ 3 ] a1[ 3 ] a2 a1[ 3 ] a3 a2 a1 
C T = [H 3PO 4 ]
3


[H 3O + ]


so:
3


H 3O + ]
H 3PO 4 ] 
[
[

α0 =
=
3
2
+
+
+


CT
 [H 3O ] + K a1 [H 3O ] + K a2K a1 [H 3O ] + K a3K a2K a1 
−


+ 2
H 2PO 4
K
H
O
[
]
a1
3


α1 =
=
 [H O + ] 3 + K [H O + ] 2 + K K [H O + ] + K K K 
CT
a1
3
a2 a1
3
a3 a2 a1 
 3
2−


HPO 4
K a2K a1 [H 3O + ]


α2 =
=
3
2
+
+
+


CT
 [H 3O ] + K a1 [H 3O ] + K a2K a1 [H 3O ] + K a3K a2K a1 
3−


PO 4
K a3K a2K a1


α3 =
=
 [H O + ] 3 + K [H O + ] 2 + K K [H O + ] + K K K 
CT
a1
3
a2 a1
3
a3 a2 a1 
 3
[
]
[
]
[
]
+
4.
€
€
€
€
€
€
Calculate the concentrations of all phosphate species in a 0.0300 M H3PO4 solution at pH
8.0.
from question 3:
3


H 3O + ]
H 3PO 4 ] 
[
[

α0 =
=
3
2
+
+
+


CT
 [H 3O ] + K a1 [H 3O ] + K a2K a1 [H 3O ] + K a3K a2K a1 
−


+ 2
H 2PO 4
K
H
O
[
]
a1
3


α1 =
=
 [H O + ] 3 + K [H O + ] 2 + K K [H O + ] + K K K 
CT
a1
3
a2 a1
3
a3 a2 a1 
 3
2−


+
HPO 4
K a2K a1 [H 3O ]


α2 =
=
3
2
+
+
+


CT
 [H 3O ] + K a1 [H 3O ] + K a2K a1 [H 3O ] + K a3K a2K a1 
3−


PO 4
K a3K a2K a1


α3 =
=
 [H O + ] 3 + K [H O + ] 2 + K K [H O + ] + K K K 
CT
a1
3
a2 a1
3
a3 a2 a1 
 3
€
]
[
]
[
]
CT = 0.0300 M
pH = 8.0, so [H3O+] = 10-8.0 = 1 x 10-8 M
3
2
Let D = [H 3O + ] + K a1 [H 3O + ] + K a2K a1 [H 3O + ] + K a3K a2K a1
D = (1 x 10-8)3 + (7.11 x 10-3)(1 x 10-8)2 + (6.34 x 10-8)(7.11 x 10-3)(1 x 10-8) +
(4.22 x 10-13)(6.34 x 10-8)(7.11 x 10-3)
D = 5.22 x 10-18
€
 C H O + 3  0.0300 1.0 x10−8 3
] =
(
) = 5.7 x 10-9 M
T[ 3
[H 3PO 4 ] =
−18

D
5.22 x10


 C K H O + 2  0.0300 7.11x10−3 1.0 x10−8 2
] =
(
)(
) = 4.1 x 10-3 M
T a1 [ 3
−
H 2PO 4 = 
−18


D
5.22 x10


 C K K [H O + ]  0.0300(6.34 x10−8 )( 7.11x10−3 )(1.0 x10−8 )
T a 2 a1
3
2−
=
= 2.6 x 10-2 M
HPO 4 = 
−18

D
5.22 x10


−13
−8
−3
 C T K a 3K a 2K a1  0.0300( 4.22 x10 )(6.34 x10 )( 7.11x10 )
3−
PO 4 = 
= 1.1 x 10-6 M
=
−18

D

5.22 x10
[
]
[
]
[
€
[
]