13-4-2011
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Complex ion equilibria and
solubility
A Lewis base is an electron pair donor which
can react with a Lewis acid (electron pair
acceptor) to form a compound called “ a
complex). Example:
Ag+ (aq) + 2NH3 (aq) D Ag(NH3)2+
The compound Ag(NH3)2+ is called a complex.
The formation constant of a complex is
usually large, and the assumption that most
reactants are converted to products is
reasonable.
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In many cases, the formation of the complex is
observed for many reactions. Example:
Co(H2O)6+2 is pink while when adding HCl to its
solution it turns blue as a result of formation
of CoCl42- complex
Also copper salts in water are usually faint
blue due to hydrated ion Cu(H2O)42+,
however, upon addition of ammonia, a
precipitate of Cu(OH)2 is formed which then
dissolves in ammonia to produce a dark blue
color due to formation of Cu(NH3)42+.
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Formation Constant
For the reaction:
Ag+ (aq) + 2NH3 (aq) D Ag(NH3)2+
The formation constant ( also called stability
constant) can be written as:
Kf = [Ag(NH3)2+]/{[Ag+][NH3]2}
For this reaction, Kf = 1.5*107
Therefore, it is really safe to assume that a
small amount (x) is left out of reactants.
4
A 0.2 mole quantity of CuSO4 is added to a liter
of 1.2 M NH3 solution. What is the
concentration of Cu2+ ions at equilibrium?
Kf (Cu(NH3)42+) = 5*1013
[Cu2+] = 0.2 M, since we have 1L soln
Perform the calculation in two steps
1. Since Kf is very large, we can practically
assume all reactants are converted to
products (i.e. a quantitative reaction)
2. Use equilibrium constant (Kf) to calculate
the concentrations as usual
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1. Quantitative step:
Cu2+
+ 4NH3 g Cu(NH3)42+
initial
0.2
final
0
2. Equilibrium step:
Initial
change
Conc
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1.2
0.4
0
0.2
Cu2+ +
4NH3 D
Cu(NH3)42+
0
+x
x
0.4
+4x
0.4+4x
0.2
-x
0.2-x
Kf = (0.2 – x)/{x*(0.4 + 4x)4}
Since Kf is very large, we can safely
assume that 0.2>>x
5*1013 = 0.2/{x*(0.4)4}
X= 1.6*10-13, which is really very small as
compared to 0.2
X = [Cu2+] = 1.6*10-13M
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If 1.54 g of AgNO3 are dissolved in 900
mL of 0.3 M NH3, what are the
concentrations of Ag+, Ag(NH3)2+, and
NH3 at equilibrium? FW (AgNO3) = 170
g/mol, and Kf (Ag(NH3)2+) = 1.5*107
First, calculate the concentration of Ag+
[Ag+]i = {1.54/170)}/0.900 L = 0.01 M
Now, we can perform the two steps
procedure, the quantitative and the
equilibrium steps:
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1. Quantitative step:
Ag+ (aq) + 2NH3 (aq) g Ag(NH3)2+
initial
0.01
0.30
0
final
0
0.28
0.01
2. Equilibrium step:
Ag+ (aq) + 2NH3 (aq) D Ag(NH3)2+
Initial
0
0.28
0.01
change
+x
+2x
-x
Conc
x
0.28+2x
0.01-x
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Kf = [Ag(NH3)2+]/{[Ag+][NH3]2}
Kf = (0.01 – x)/{x*(0.28 +2x)2}
Kf is very large, therefore we can safely
assume that 0.01>>x
1.5*107 = 0.01/{x*(0.28)2}
X = 8.5*10-9 , which is very small as
compared to 0.01
X = [Ag+] = 8.5*10-9 M, [NH3] = 0.28 M, and
[Ag(NH3)2+] = 0.01M
10
Calculate the concentrations of Cd2+, CN-, and
Cd(CN)4- at equilibrium when 0.475 g of
Cd(NO3)2 (FW = 236.5 g/mol) is dissolved in
500 mL of 0.50 M KCN solution. Kf
(Cd(CN)42-) = 7.1*1016
Solution:
Perform the calculation in two steps
1. Since Kf is very large, we can practically
assume all reactants are converted to
products (i.e. a quantitative reaction)
2. Use equilibrium constant (Kf) to calculate
the concentrations as usual
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First, find the concentration of Cd2+ in
solution:
[Cd2+] = mol/L
No. mol Cd2+ = g/FW = 0.475/236.5 = 2*10-3
[Cd2+] = (2*10-3 mol/0.500 L) = 4*10-3 M
Now, we can proceed by the two steps
approach, as usual.
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1. Quantitative step:
Cd2+
+
4CN- g Cd(CN)42-
initial
4*10-3
0.50
final
0
0.484
2. Equilibrium step:
Cd2+
Initial
change
Conc
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0
+x
x
+ 4CN-
0
4*10-3
g Cd(CN)42-
0.484
4*10-3
+4x
-x
0.484+4x 4*10-3 -x
7.1*1016 = (4*10-3 – x)/{x*(0.484 + 4x)4}
Since Kf is very large, we can safely
assume that 4*10-3 >>x
7.1*1016 = 4*10-3/{x*(0.484)4}
X= 1.0*10-18, which is really very small as
compared to 4*10-3
X = [Cd2+] = 1.0*10-18 M
[CN-] = 0.484 + 4X = 0.484 M
[Cd(CN)42-] = 4*10-3 – x = 4*10-3 M
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Amphoteric hydroxides
These are hydroxides that can act as bases
by giving off hydroxides and as acids by
accepting and reacting with hydroxides:
Examples: Al(OH)3, Pb(OH)2, Cr(OH)3,
Zn(OH)2, Cd(OH)2
Al(OH)3(s) D Al3+ + 3OHAl(OH)3(s) + OH- D Al(OH)4-(aq)
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Selected Problems
1, 3, 5, 6, 8-16, 17, 19, 21-23, 27, 29, 31,
33, 34, 35, 38, 39, 40, 41, 42, 44-46, 53,
54.
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