Complex Ion Equilibrium
Before we get to complex ion equilibrium (yes, another K!) a few things
about Lewis acids and bases.

Lewis acid – a substance that can accept a pair of electrons from another
atom to forma a new bond

Lewis base – a substance that can donate a pair of electrons to another
atom to forma new bond
A +
B: 
Acid Base
B→A
Adduct
So, in a Lewis acid-base reaction a molecule or ion donates a pair of
electrons to another molecule or ion. The product is referred to as an acidbase adduct. The bond formed is called a coordinate covalent bond
(you’ve heard the name before and yes, you are right we did not cover it
during bonding – I saved it for now!)
Here are some examples
Lewis acid
H+
H+
BF3
Lewis base
H2O
H3N:
H3N:
Adduct
H3 O +
NH4+
H3N:BF3
Metal cations interact with water to form hydrated cations (the metal is
surrounded by water molecules – this is in your notes. You may recall our
discussion regarding aluminum)
Fe2+ forms six coordinate covalent bonds to water:
Fe2+ (aq) + 6 H2O (l)  [Fe(H2O)6]+ (aq)
Transition metals are well known for forming these types of ‘colorful’ ions (I
know you guys mumbling bad things about me now – just stop – color is
your friend!) which are usually called complex ions or coordination
complexes.
Here is copper (II) ion and ammonia. This is a good one to remember.
Notice it is ‘dark blue’ as opposed to the ‘light blue’ hydrated Cu2+ ion.
Here is a simple video clip of this reaction:
http://www.youtube.com/watch?v=VLP6ILYeJO4
Okay, so, let’s use our copper example to further discuss complex ions and
coordination compounds.
Complex ion
[Cu(NH3)4]2+
Coordination compound
[Cu(NH3)4]SO4
Cooper (II) ion reacts with aqueous ammonia and water to produce copper
(II0 hydroxide and ammonium ion:
Cu2+ (aq) + 2 NH3 (aq) + 2 H2O (l)  Cu(OH)2 (s) + 2 NH4+ (aq)
Adding more ammonia converts the copper (II) hydroxide the very stable
complex ion [Cu(NH3)4]2+ which has a very distinctive deep blue color:
Cu(OH)2 (s) + 4 NH3 ⇌ [Cu(NH3)4]2+ (aq) + 2 OH- (aq)
Nitrogen contributes a pair of electrons to form the covalent bond (Lewis
base!) Copper is the central ion (or atom). Bonded to the metal are
molecules or ions, in this case ammonia, called ligands. The number of
ligands attached to the metal defines the coordination number of the
metal.
In our example, the [Cu(NH3)4]2+ ion can not exist by itself in a solid state,
therefore the 2+ charge must be balanced by an anion with a total charge of
2- such as [Cu(NH3)4]SO4 or [Cu(NH3)4]Cl2, otherwise known as
coordination compounds.
OK, as far as coordination compounds go that is all you need to know.
There is much more information related to coordination compounds that is
not covered on the AP examination. However, I would take a look in my
textbook at some of the simpler geometries, particularly for [M(H2O)6]n+ ions
(it’s octahedral).
Solubility and Complex Ions
Consider the following reaction:
AgCl (s) + 2 NH3 (aq) ⇌ [Ag(NH3)2]+ (aq) + Cl- (aq)
Let’s look at dissolving AgCl (s) as a two step process. First, we know AgCl
dissolves minimally in water, giving Ag+ (aq) and Cl- (aq) ions. Second, the
Ag+ ion combines with NH3 to give the ammonia complex. Lowering [Ag+]
through complexation with NH3 shifts the solubility equilibrium to the right,
so more solid AgCl dissolves:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Ksp = 1.8 x 10-10
Ag+ (aq) + 2 NH3 (aq) ⇌ [Ag(NH3)2]+ (aq)
Kformation = 1.6 x 107
The equilibrium constant for the formation of a complex ion such as
[Ag(NH3)2]+ is called the formation constant (yet another K!). The large
value of this equilibrium constant means that the equilibrium lies well to the
right. Combining the two K values yields the net equilibrium constant for
dissolving AgCl in aqueous ammonia.
Knet = Ksp x Kformation = (1.8 x 10-10)(1.6 x 107) = 2.9 x 10-3
AgNH   Cl 


2.9 x 10
3
3 2
NH 3 2

Sample problem
What is the value of the equilibrium constant, Knet, for dissolving AgBr (Ksp =
5.4 x 10-13) in a solution containing the thiosulfate ion, S2O32-? (Ag+ (aq) +
2 S2O32- (aq) ⇌ [Ag(S2O3)2]3- (aq) Kformation = 2.0 x 1013) Explain why
AgBr dissolves readily on adding aqueous sodium thiosulfate to the solid.
Solution
AgBr (s) ⇌ Ag+ (aq) + Br- (aq)
Ksp = 5.4 x 10-13
Ag+ (aq) + 2 S2O32- (aq) ⇌ [Ag(S2O3)2]3- (aq)
Kformation = 2.0 x 1013
Net chemical equation
AgBr (aq) + 2 S2O32- (aq) ⇌ [Ag(S2O3)2]3- (aq) + Br- (aq)
Knet = Ksp x Kformation = 1.0 x 101
The value of Knet is greater than one, indicating the reaction favors the
products. AgBr is predicted to dissolve in aqueous Na2S2O3.
Let’s look at a related problem….
Calculate the mass of AgBr that can dissolve in 1.00 L of 0.500 M Na2S2O3.
Ksp for AgBr = 5.4 x 10-13
Just when you thought you got away without having to do an ICE table!
AgBr (aq) + 2 S2O32- (aq) ⇌ [Ag(S2O3)2]3- (aq) + Br- (aq)
I
C
E
0.500
-2x
0.500 -2x
0
+x
x
Knet = 1.0 x 101 = [[Ag(S2O3)2]3-] [Br-]/[ S2O32-]2
1.0 x 101 = x2 / (0.500 – 2x)2
3.16 = x / (0.500 – 2x)
X = 1.58 – 6.32x
X = 0.216 M Br- = 0.216 M AgBr
1.00 L 
0.216 mol AgBr 187.8 g

 40.6 g AgBr
1L
1L mol
0
+x
x