Equilibrium  Calculations with Complexes
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Q: 1 2 3
4 5

1

Give the formula for and arrange the following
complex ions in order of increasing stability:
tetramminecopper(II) ion
tetracyanomercurate(II) ion
tetramminezinc ion
tetracyanocadmate ion


2 
A solution is prepared by mixing 0.0200 M silver
nitrate (10.0 cm^{3}) with 1.04 M sodium cyanide (10.0 cm^{3}).
(1.)
Calculate the resulting concentration of silver ion.
(2.)
Predict what effect, if any, will occur if 1.00 M ammonia (1.00
cm^{3}) is added to the solution.


3 
What is the concentration
of copper(II) ion in a solution made by dissolving copper(II) sulfate
(0.100 mole) and ammonia (2 .00 mole) in water and making up to 500
mL? 

4 
(Advanced question) Unexposed silver
halides are removed from photographic film when they react with
a solution of Na_{2}S_{2}O_{3} and form
the complex Ag(S_{2}O_{3})_{2}^{3}.
What mass of sodium thiosulfate is required to prepare 1.00 L of
a solution that will dissolve 1.00 g of silver bromide by the formation
of this complex?


5 
(Advanced question) Slow addition of
sodium iodide to a mercury(II) nitrate solution results initially
in the precipitation of red mercury(II) iodide and then the dissolution
of this salt as a result of the reaction
HgI_{2}(s) + 2I^{}
[HgI_{4}]^{2}^{}
(1.)
Using K_{stab} for [HgI_{4}]^{2} (= 10^{30.28})
and K_{so} for HgI_{2} (=10^{10.37}), calculate
the equilibrium constant for this reaction.
(2.)
To 1.00 L of sodium iodide solution (0.200 M) is added 10.0 g of
solid Hg(NO_{3})_{2}. ½H_{2}O. Calculate
the final concentration of free (uncomplexed) mercury(II) ion?
(3.)
Will solid mercury(II) iodide be present at equilibrium in the above
reaction?


Equilibrium
 Calculations with Complexes (Answers) 

1

Stability is quantified
by the compound's K_{STAB} (the stability constant)
which is an expression incorporating the ratio of the concentration
of the compex and the concentrations of the dissociated metal and
ligand species. It always has the form
K_{STAB}
= [complex] /
([metal ion][ligand]^{n) }where n is the number of ligand
species in the complex. The larger its K_{STAB} value,
the more associated the complex  i.e. the more stable it is.
Since the complexes in
the question all have the same number of ligands (all have n = 4),
their K_{STAB} values have the same units, and can
be compared directly. Using K_{STAB} values from
the data sheet the complexes are listed in the following order of
increasing stability:
Zn(NH_{3})_{4}^{2+}
(K_{STAB} = 1 x 10^{9} M^{4})
Cu(NH_{3})_{4}^{2+}
(K_{STAB} = 1 x 10^{13}
M^{4})
Cd(CN)_{4}^{2}
(K_{STAB} = 1 x 10^{18} M^{4})
Hg(CN)_{4}^{2}
(K_{STAB} = 1 x 10^{41} M^{4})


2 
(1.)
The silver and cyanide ions will react to form the [Ag(CN)_{2}]^{}
complex:
Ag^{+} + 2CN^{}
[Ag(CN)_{2}]^{} K_{STAB} = 1 x 10^{20}
M^{2}
Note the use of square brackets in a reaction equation to
indicate a complex ion  square brackets are also used in
concentration calculations to denote concentration of a species
(this can be confusing!)
The amount of free Ag^{+} left in solution is found by
using the K_{STAB} of this complex.
The final volume is 20 mL, which is twice the volume of each of
the added solutions, so their concentration in the mixed solution
needs to be corrected by a factor of 0.50:
\
[Ag^{+}]_{init} = 0.500 x 0.0200 M = 0.0100 M
\
[CN^{}]_{init} = 0.500 x 1.04 M = 0.520 M
Next the equlibrium concentrations are calculated:
From the reaction equation we see that each mole of Ag^{+}
reacts with two moles of CN^{}, so the equilibrium concentration
of CN^{} is found:
[CN^{}]_{eqb} = 0.520 mol L^{1}  (2
x 0.0100 M) = 0.500 M
We assume all Ag^{+} initially present reacts to form the
complex [Ag(CN)_{2}^{}],
so [Ag(CN)_{2}^{}]_{eqb} = [Ag^{+}]_{init}
= 0.0100 M
Now K_{STAB} = [Ag(CN)_{2}^{}]
/ ([Ag^{+}] [CN^{}]^{2}) = 1 x 10^{20}
at equilibrium.
Rearranging the above expression for K_{STAB}:
[Ag^{+}] = [[Ag(CN)_{2}^{}]] / K_{STAB}
[CN^{}]^{2}
= 0.0100 / (1 x 10^{20} ) x (0.50 )^{2} = 4 x 10^{22}
M
(2.)
The addition of NH_{3} solution will produce no effect since
the K_{STAB} of [Ag(CN)_{2}]^{}
(1 x 10^{20} M^{2}) is much larger
than that of [Ag(NH_{3})_{2}]^{+} (1 x 10^{7}
M^{2})


3 
First the initial concentrations of Cu^{2+}
and NH_{3} are obtained:
[Cu^{2+}]_{init} = 0.200 M
[NH_{3}]_{init} = 4.00 M
They react according to the equation:
Cu^{2+} + 4NH_{3} [Cu(NH_{3})_{4}]^{2+}
K_{STAB} = 1 x 10^{13} M^{4}
From this equation we see that each mole of
Cu^{2+} reacts with 4 moles of NH_{3}, so the equlibrium
concentration of NH_{3} is calculated:
[NH_{3}]_{eqb} = 4.00  (4 x
0.200) = 3.20 M
We assume all Cu^{2+} initially present
reacts to form the complex [Cu(NH_{3})_{4}]^{2+},
so [Cu(NH_{3})_{4}]^{2+}_{eqb} =
[Cu^{2+}]_{init} = 0.200 M. (This is typically justified
by the stability constant of complexes being very large).
Now K_{STAB} = [Cu(NH_{3})_{4}^{2+}]
/ [Cu^{2+}] [NH_{3}]^{4} = 1 x 10^{13}
M at equilibrium.
Rearranging for [Cu^{2+}]:
[Cu^{2+}] = [Cu(NH_{3})_{4}^{2+}]
/ K_{STAB} [NH_{3}]^{4}
= 0.2 00/ (1 x 10^{13} x (3.20)^{4})
= 2 x 10^{16} M


4

K_{SP}(AgBr)
= [Ag^{+}] [Br^{}] = 5.0 x 10^{13}
K_{STAB}(Ag(S_{2}O_{3})_{2}^{3})
= [Ag(S_{2}O_{3})_{2}^{3}] / ([Ag^{+}]
[S_{2}O_{3}^{2}]^{2}) = 10^{13.5}
K_{SP}
x K_{STAB} = 5.0 x 10^{13} x 10^{13.5}
= 5.0 x 10^{0.5} = 15.81
= ([Br^{}] [Ag(S_{2}O_{3})_{2}^{3}])
/ [S_{2}O_{3}^{2}]^{2}
Let x = [Ag^{+}]
at equilibrium
Let y = initial
[S_{2}O_{3}^{2}]
At equilibrium:
[Br^{}] = 5.32
x 10^{3} M
[Ag(S_{2}O_{3})_{2}^{3}]
= (5.32 x 10^{3}  x) M
[S_{2}O_{3}^{2}]
= y  2(5.32 x 10^{3}  x) M
\
15.81 = (5.32
x 10^{3}) (5.32 x 10^{3}  x) / ((y
 2(5.32 x 10^{3}  x)^{2})
Now we can assume that
x << 5.32 x 10^{3}, since practically all
silver is complexed at equilibrium (ie K_{STAB}(Ag(S_{2}O_{3})_{2}^{3})
is very large):
\
y  10.64
x 10^{3} = ((5.32 x 10^{3}) (5.32 x 10^{3})
/ 15.81)^{½ }=
5.32 x 10^{3} / 3.976 = 1.34 x 10^{3}
y = 1.34 x 10^{3}
+ 10.64 x 10^{3} = 1.198 x 10^{2} M
\
Mass of Na_{2}S_{2}O_{3}
required for 1.00 L of solution = 1.00 x 1.198 x 10^{2}
x 158.1 = 2 g (to 1 significant figure as required by the precision
of the K_{STAB}(Ag(S_{2}O_{3})_{2}^{3})).


5 
(1.)
HgI_{2}(s) + 2I^{}
HgI_{4}^{2}
Hg^{2+} + 2I^{} HgI_{2}(s)
K for HgI_{2}(s) + 2I^{}
HgI_{4}^{2} needs to be found, ie. [HgI_{4}^{2}]
/ [I^{}]^{2}
K_{STAB}(HgI_{4}^{2})
= [HgI_{4}^{2}] / ([Hg^{2+}] [I^{}]^{4})
= 10^{30.28}
K_{SP}(HgI_{2}) = [Hg^{2+}]
[I^{}]^{2}
K_{STAB} x K_{SP}
= [HgI_{4}^{2}] / [I^{}]^{2} =
10^{30.28} x 10^{10.37} = 10^{19.91}
\
K = 10^{19.91}
(2.)
The reaction is Hg^{2+} + 4I^{}
HgI_{4}^{2}
Initial [Hg^{2+}] = moles Hg(NO_{3})_{2}.
½H_{2}O / volume (L)
Volume = 1.00 L, so Initial
[Hg^{2+}] = mass / molar mass Hg(NO_{3})_{2}.
½H_{2}O = 10.0 / 333.6 M
Inital [I^{}] = 0.200 M
Inital [HgI_{4}^{2}] = 0
Equilibrium [Hg^{2+}] = x
Equilibrium [I^{}] = 0.20  4(10/333.6
 x) = 8.0 x 10^{2} M (assuming x <<
10/333.6, and ignoring it)
Equilibrium [HgI_{4}^{2}] =
10/333.6  x = 3.0 x 10^{2} M (assuming x
<< 10/333.6, and ignoring it)
K_{STAB} = (3.0 x 10^{2})
/ (x(8 x 10^{2})^{4}) = 10^{30.28}
x = (3.0 x 10^{2}) / (10^{30.28}
x 4.1 x 10^{5}) = 7.32 x 10^{2} / 10^{30.28}
= 3.8 x 10^{28} M
(3.)
Q = [Hg^{2+}] [I^{}]^{2} = (3.8 x 10^{28})
(8.0 x 10^{2})^{2} = 2.5 x 10^{30}.
This is less than K_{SP}(HgI_{2}),
so no precipitate will form.

