Download Complexometric Reactions and Titrations

Lecture 31
Acid-Base Titartions, Cont…
Kjeldahl Analysis
Complexometric Reactions
1
Kjeldahl Analysis
An application of acid-base titrations that finds
an important use in analytical chemistry is
what is called Kjeldahl nitrogen analysis.
This analysis is used for the determination of
nitrogen in proteins and other nitrogen
containing compounds. Usually, the quantity
of proteins can be estimated from the
amount of nitrogen they contain. The
Kjeldahl analysis involves the following
steps:
2
1. Digestion of the nitrogen containing compound and
converting the nitrogen to ammonium hydrogen
sulfate. This process is accomplished by
decomposing the nitrogen containing compound
with sulfuric acid.
2. The solution in step 1 is made alkaline by addition of
concentrated NaOH which coverts ammonium to
gaseous ammonia , and the solution is distilled to
drive the ammonia out.
3. The ammonia produced in step 2 is collected in a
specific volume of a standard acid solution (dilute)
where neutralization occurs.
4. The solution in step 3 is back-titrated against a
standard NaOH solution to determine excess acid.
5. mmoles of ammonia are then calculated and related
3 to mmol nitrogen.
Example
A 0.200 g of a urea (FW = 60, (NH2)2CO) sample is
analyzed by the Kjeldahl method. The ammonia is
collected in a 50 mL of 0.05 M H2SO4. The excess
acid required 3.4 mL of 0.05 M NaOH. Find the
percentage of the compound in the sample.
2 NH3 + H2SO4 = (NH4)2SO4
½ mmol ammonia = mmol H2SO4 reacted
(NH2)2CO = 2 NH3
mmol urea = ½ mmol ammonia
mmol H2SO4 titrated = ½ mmol NaOH
4
mmol urea = mmol H2SO4 taken – ½ mmol NaOH
mmol urea = 0.05 x 50 – 1/2 x 0.05 x 3.4 = 2.415
mg urea = 2.415x60 = 144.9
% urea = (144.9/200)x100 = 72.5%
5
Modified Kjeldahl Analysis
In conventional Kjeldahl method we need two
standard solutions, an acid for collecting
evolved ammonia and a base for back-titrating
the acid. In a modified procedure, only a
standard acid is required. In this procedure,
ammonia is collected in a solution of dilute
boric acid, the concentration of which need not
be known accurately. The result of the reaction
is the borate which is equivalent to ammonia.
NH3 + H3BO3 g NH4+ + H2BO3Borate is a strong conjugate base which can be
titrated with a standard HCl solution.
6
Example
A 0.300 g feed sample is analyzed for its protein
content by the modified Kjeldahl method. If 25
mL of 0.10 M HCl is required for the titration
what is the percent protein content of the
sample (mg protein = 6.25 * mg N).
Solution
mmol N = mmol HCl
mmol N = 0.10 x 25 = 2.5
mmol N = 2.5
mg N = 2.5 x 14 = 35
mg protein = 35 * 6.25 = 218.8
% protein = (218.8/300) x 100 = 72.9%
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Complexometric Reactions
and Titrations
8
Complexes are compounds formed from
combination of metal ions with ligands
(complexing agents). A metal ionis an
electron deficient species while a
ligand is an electron rich, and thus,
electron donating species. A metal ion
will thus accept electrons from a ligand
where coordination bonds are formed.
Electrons forming coordination bonds
come solely from ligands.
9
A ligand is called a monodentate if it donates a
single pair of electrons (like :NH3) while a
bidentate ligand (like ethylenediamine,
:NH2CH2CH2H2N:) donates two pairs of
electrons. Ethylenediaminetetraacetic acid
(EDTA) is a hexadentate ligand. The ligand
can be as simple as ammonia which forms a
complex with Cu2+, for example, giving the
complex Cu(NH3)42+. When the ligand is a
large organic molecule having two or more of
the complexing groups, like EDTA, the ligand
is called a chelating agent and the formed
complex, in this case, is called a chelate.
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11
12
The tendency of complex formation is
controlled by the formation constant of the
reaction between the metal ion (Lewis acid)
and the ligand (Lewis base). As the formation
constant increases, the stability of the
complex increases.
Let us look at the complexation reaction of Ag+
with NH3:
Ag+ + NH3 D Ag(NH3)+
Ag(NH3)+ + NH3 D Ag(NH3)2+
13
kf1 = [Ag(NH3)+]/[Ag+][NH3]
kf2 = [Ag(NH3)2+]/[Ag(NH3)+][NH3]
Kf1 x kf2 = [Ag(NH3)2+]/[Ag+][NH3]2
Now look at the overall reaction:
Ag+ + 2 NH3 D Ag(NH3)2+
kf = [Ag(NH3)2+]/[Ag+][NH3]2
It is clear from inspection of the values of the kf
that:
Kf = kf1 x kf2
For a multistep complexation reaction we will
always have the formation constant of the
overall reaction equals the product of all step
14 wise formation constants
The formation constant is also called the
stability constant and if the equilibrium is
written as a dissociation the equilibrium
constant in this case is called the
instability constant.
Ag(NH3)2+ D Ag+ + 2 NH3
kinst = [Ag+][NH3]2/[Ag(NH3)2+]
Therefore, we have:
Kinst = 1/kf
15
Stability of Metal-Ligand Complexes
The stability of complexes is influenced by a
number of factors related to the ligand and
metal ions.
1. Nature of the metal ion: Small ions with high
charges lead to stronger complexes.
2. Nature of the ligand: The ligands forming
chelates impart extra stability (chelon
effect). For example the complex of nickel
with a multidentate ligand is more stable
than the one formed with ammonia.
3. Basicity of the ligand: Greater basicity of the
ligand results in greater stability of the
complex.
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4. Size of chelate ring: The formation of
five- or six-membered rings provides
the maximum stability.
5. Number of metal chelate rings: The
stability of the complex is directly
related to the number of chelate rings
formed between the ligand and metal
ion. Greater the number of such rings,
greater is the stability.
7. Steric effects: These also play an
important role in the stability of the
complexes.
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Lecture 32
Complexometric Reactions, Cont….
Calculations
EDTA Equilibria
18
Example
A divalent metal ion reacts with a ligand to form a
1:1 complex. Find the concentration of the metal
ion in a solution prepared by mixing equal
volumes of 0.20 M M2+ and 0.20 M ligand (L). kf =
1.0x108.
The formation constant is very large and
essentially the metal ions will almost
quantitatively react with the ligand.
The concentration of metal ions and ligand will be
half that given as mixing of equal volumes of the
ligand and metal ion will make their
concentrations half the original concentrations
since the volume was doubled.
19
[M2+] = 0.10 M, [L] = 0.10 M
M2+ + L D ML2+
Kf = ( 0.10 –x )/x2
Assume 0.10>>x since kf is very large
1.0x108 = 0.10/x2,
x = 3.2x10-5
Relative error = (3.2x10-5/0.10) x 100 = 3.2x10-2 %
The assumption is valid.
[M2+] = 3.2x10-5 M
20
Silver ion forms a stable 1:1 complex with trien.
Calculate the silver ion concentration at equilibrium
when 25 mL of 0.010 M silver nitrate is added to 50 mL
of 0.015 M trien. Kf = 5.0x107
Ag+ + trien D Ag(trien)+
mmol Ag+ added = 25x0.01 = 0.25
mmol trien added = 50x0.015 = 0.75
mmol trien excess = 0.75 – 0.25 =
0.50
[Trien] = 0.5/75 M
[Ag(trien)+] = 0.25/75 M
21
Kf = ( 0.25/75 – x )/(x * 0.50/75 + x)
Assume 0.25/75>>x since kf is very large
5.0x107 = (0.25/75)/(x * 0.50/75)
x = 1.0x10-8
Relative error = (1.0x10-8/(0.25/75)) x 100 = 3.0x10-4 %
The assumption is valid.
[Ag+] = 1.0x10-8 M
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EDTA Titrations
Ethylenediaminetetraacetic acid disodium salt
(EDTA) is the most frequently used chelate in
complexometric titrations. Usually, the
disodium salt is used due to its good
solubility. EDTA is used for titrations of
divalent and polyvalent metal ions. The
stoichiometry of EDTA reactions with metal
ions is usually 1:1. Therefore, calculations
involved are simple and straightforward.
Since EDTA is a polydentate ligand, it is a
good chelating agent and its chelates with
23 metal ions have good stability.
24
EDTA Equilibria
EDTA can be regarded as H4Y where in solution
we will have, in addition to H4Y, the following
species: H3Y-, H2Y2-, HY3-, and Y4-. The
amount of each species depends on the pH
of the solution where:
a4 = [Y4-]/CT where:
CT = [H4Y] + [H3Y-] + [H2Y2-] + [HY3-] + [Y4-]
a4 = ka1ka2ka3ka4/([H+]4 + ka1 [H+]3 + ka1ka2[H+]2 +
ka1ka2ka3[H+] + ka1ka2ka3ka4)
The species Y4- is the ligand species in EDTA
titrations and thus should be looked at
carefully.
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The Formation Constant
Reaction of EDTA with a metal ion to form a
chelate is a simple reaction. For example, EDTA
reacts with Ca2+ ions to form a Ca-EDTA chelate
forming the basis for estimation of water
hardness. The reaction can be represented by
the following equation:
Ca2+ + Y4- = CaY2kf = 5.0x1010
Kf = [CaY2-]/[Ca2+][Y4-]
The formation constant is very high and the
reaction between Ca2+ and Y4- can be considered
quantitative. Therefore, if equivalent amounts of
Ca2+ and Y4- were mixed together, an equivalent
amount of CaY2- will be formed.
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Formation Constants for EDTA Complexes
Cation
KMY
Cation
KMY
Ag+
Mg2+
Ca2+
Sr2+
Ba2+
Mn2+
Fe2+
Co2+
Ni2+
2.1 x 107
4.9 x 108
5.0 x1010
4.3 x 108
5.8 x 107
6.2 x1013
2.1 x1014
2.0 x1016
4.2 x1018
Cu2+
Zn2+
Cd2+
Hg2+
Pb2+
Al3+
Fe3+
V3+
Th4+
6.3 x 1018
3.2 x 1016
2.9 x 1016
6.3 x 1021
1.1 x 1018
1.3 x 1016
1.3 x 1025
7.9 x 1025
1.6 x 1023
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Minimum pH for effective titrations
of various metal ions with EDTA.
The question now is how to calculate the amount of
Ca2+ at equilibrium?
CaY2- D Ca2+ + Y4However, [Ca2+] # [Y4-] at this point since the amount of
Y4- is pH dependent and Y4- will disproportionate to
form all the following species, depending on the pH
CT = [H4Y] + [H3Y-] + [H2Y2-] + [HY3-] + [Y4-]
Where, CT is the sum of all species derived from Y4which is equal to [Ca2+].
Therefore, the [Y4-] at equilibrium will be less than the
[Ca2+] and in fact it will only be a fraction of CT
where:
a4 = [Y4-]/CT
a4 = ka1ka2ka3ka4/([H+]4 + ka1 [H+]3 + ka1ka2[H+]2 + ka1ka2ka3[H+] + ka1ka2ka3ka4)
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The Conditional Formation Constant
We have seen that for the reaction
Ca2+ + Y4- D CaY2- kf = 5.0x1010
We can write the formation constant expression
Kf = [CaY2-]/[Ca2+][Y4-]
However, we do not know the amount of Y4- at
equilibrium but we can say that since a4 = [Y4-]/CT,
then we have:
[Y4-] = a4CT
Substitution in the formation constant expression we
get:
Kf = [CaY2-]/[Ca2+]a4CT or at a given pH we can write
Kf' = [CaY2-]/[Ca2+]CT
Where Kf' is called the conditional formation constant.
32 It is conditional since it is now dependent on pH.
Titration Curves
In most cases, a titration is performed by addition of
the titrant (EDTA) to the metal ion solution adjusted
to appropriate pH and in presence of a suitable
indicator. The break in the titration curve is
dependent on:
1. The value of the formation constant.
2. The concentrations of EDTA and metal ion.
3. The pH of the solution
As for acid-base titrations, the break in the titration
curve increases as kf increases and as the
concentration of reactants is increased. The pH
effect on the break of the titration curve is such that
sharper breaks are obtained at higher pH values.
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34
35
Minimum pH for effective titrations
of various metal ions with EDTA.
Lecture 33
Complexometric Titrations, Cont…
Complexometric Indicators
36
Indicators
The indicator is usually a weaker chelate
forming ligand. The indicator has a
color when free in solution and has a
clearly different color in the chelate.
The following equilibrium describes the
function of an indicator (H3In) in a Mg2+
reaction with EDTA:
MgIn- (Color 1) + Y4- D MgY2- + In3- (Color 2)
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Problems Associated with Complexometric Indicators
There could be some complications which may render
some complexometric titrations useless or have
great uncertainties. Some of these problems are
discussed below:
1. Slow reaction rates
In some EDTA titrations, the reaction is not fast enough
to allow acceptable and successful determination of
a metal ion. An example is the titration of Cr3+ where
direct titration is not possible. The best way to
overcome this problem is to perform a back titration.
However, we are faced with the problem of finding a
suitable indicator that is weaker than the chelate but
is not extremely weak to be displaced at the first
drop of the titrant.
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2. Lack of a suitable indicator
This is the most severe problem in EDTA
titrations and one should be critical
about this issue and pay attention to
the best method which may be used to
overcome this problem. First let us take
a note of the fact that Mg2+-EDTA
titration has excellent indicators that
show very good change in color at the
end point. Look at the following
situations:
42
a. A little of a known standard Mg2+ is
added to the metal ion of interest. Now
the indicator will form a clear cut color
with magnesium ions. Titration of the
metal ion follows and after it is over,
added EDTA will react with Mg-In
chelate to release the free indicator,
thus changing color. This procedure
requires performing the same titration
on a blank containing the same amount
of Mg2+.
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44
b. A blank experiment will not be
necessary if we add a little of Mg-EDTA
complex to the metal ion of interest.
The metal ion will replace the Mg2+ in
the Mg-EDTA complex thus releasing
Mg2+ which immediately forms a good
color with the indicator in solution. No
need to do any corrections since the
amount of EDTA in the added complex
is exactly equal to the Mg2+ in the
complex.
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c. If it is not easy to get a Mg-EDTA complex,
just add a little Mg2+ to the EDTA titrant.
Standardize the EDTA and start titration. At
the very first point of EDTA added, some
Mg2+ is released forming a chelate with the
indicator and thus giving a clear color.
It is wise to consult the literature for suitable
indicators of a specific titration. There are a
lot of data and information on titrations of all
metals you may think of. Therefore, use this
wealth of information to conduct successful
EDTA titrations.
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48
Example
Find the concentrations of all species in solution at
equilibrium resulting from mixing 50 mL of 0.200 M
Ca2+ with 50 mL of 0.100 M EDTA adjusted to pH 10.
a4 at pH 10 is 0.35. kf = 5.0x1010
Solution
Ca2+ + Y4- g CaY2mmol Ca2+ added = 0.200 x 50 = 10.0
mmol EDTA added = 0.100 x 50 = 5.00
mmol Ca2+ excess = 10.0 – 5.00 = 5.00
[Ca2+]excess = 5.00/100 = 0.050 M
mmol CaY2- formed = 5.00
[CaY2-] = 5.00/100 = 0.050
CaY2- D Ca2+ + Y449
CT = [H4Y] + [H3Y-] + [H2Y2-] + [HY3-] + [Y4-]
Kf = [CaY2-]/[Ca2+]a4CT
[Ca2+] = CT
Using the same type of calculation we are used to
perform, one can write the following:
50
Kf = [CaY2-]/[Ca2+][Y4-]
5.0x1010 = (0.05 – x)/((0.050 + x)* a4 x)
assume that 0.05>>x
x = 5.6x10-11
Relative error will be very small value
The assumption is valid
[Ca2+] = 0.050 + x = 0.050 M
[CaY2-] = 0.050 – x = 0.050 M
[Y4-] = 0.35 * 5.6x10-11 = 1.9x10-11 M
51
Example
Calculate the pCa of a solution at pH 10 after
addition of 100 mL of 0.10 M Ca2+ to 100 mL
of 0.10 M EDTA. a4 at pH 10 is 0.35. kf =
5.0x1010
Solution
Ca2+ + Y4- = CaY2mmol Ca2+ = 0.10 x 100 = 10
mmol EDTA = 0.10 x 100 = 10
mmol CaY2- = 10
[CaY2-] = 10/200 = 0.05 M
Therefore, Ca2+ will be produced from partial
52 dissociation of the complex
Ca2+ + Y4- D CaY2CT = [H4Y] + [H3Y-] + [H2Y2-] + [HY3-] + [Y4-]
Kf = [CaY2-]/[Ca2+]a4CT
[Ca2+] = CT
5.0x1010 = 0.05/([Ca2+]2 x 0.35)
[Ca2+] = 1.7x10-6 M
pCa = 5.77
Using the same type of calculation we
are used to perform, one can write the
following:
53
Kf = [CaY2-]/[Ca2+][Y4-]
5.0x1010 = (0.05 – x)/(x* a4 x)
assume that 0.05>>x
x = 1.7x10-6
Relative error = (1.7x10-6/0.05) x 100 = 3.4x10-3%
[Ca2+] = 1.7x10-6 M
pCa = 5.77
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Lecture 34
Complexometric Reactions, Cont…
EDTA Titrations
55
Calculate the titer of a 0.100 M EDTA solution in
terms of mg CaCO3 (FW = 100.0) per mL
EDTA
The EDTA concentration is 0.100 mmol/mL, therefore,
the point here is to calculate the mg CaCO3 reacting
with 0.100 mmol EDTA. We know that EDTA reacts
with metal ions in a 1:1 ratio. Therefore 0.100 mmol
EDTA will react with 0.100 mmol CaCO3.
mmol CaCO3 = mmol EDTA
mg CaCO3/100 = 0.1 * 1
mg CaCO3 = 10.0
56
Example
An EDTA solution is standardized against high
purity CaCO3 by dissolving 0.3982 g of
CaCO3 in HCl and adjusting the pH to 10. The
solution is then titrated with EDTA requiring
38.26 mL. Find the molarity of EDTA.
Solution
EDTA reacts with metal ions in a 1:1 ratio.
Therefore,
mmol CaCO3 = mmol EDTA
mg/FW = Molarity x VmL
398.2/100.0 = M x 38.26,
MEDTA = 0.1041
57
Example
Find the concentration of Ca2+ in a 20 mL of 0.20 M
solution at pH 10 after addition of 100 mL of 0.10
M EDTA. a4 at pH 10 is 0.35. kf = 5x1010
Solution
Initial mmol Ca2+ = 0.20 x 20 = 4.0
mmol EDTA added = 0.10 x 100 = 10
mmol EDTA excess = 10 – 4.0 = 6.0
CT = 6.0/120 = 0.050 M
mmol CaY2- = 4.0
[CaY2-] = 4.0/120 = 0.033 M
Ca2+ + Y4- D CaY2kf = 5.0x1010
58
Kf = [CaY2-]/[Ca2+][Y4-]
5x1010 = (0.033 – x)/(x* a4(0.050 + x) )
assume that 0.033>>x
x = 3.9x10-11
The assumption is valid by inspection of the values
and no need to calculate the relative error. [Ca2+] =
3.9x10-11 M
pCa = 10.41
59
Example
Find pCa in a 100 mL solution of 0.10 M Ca2+ at pH
10 after addition of 0, 25, 50, 100, 150, and 200
mL of 0.10 M EDTA. a4 at pH 10 is 0.35. kf =
5x1010
Solution
Again, we should remember that EDTA reactions
with metal ions are 1:1 reactions. Therefore, we
have:
Ca2+ + Y4- D CaY2kf = 5.0x1010
1. After addition of 0 mL EDTA
[Ca2+] = 0.10 M
pCa = 1.00
60
2. After addition of 25 mL EDTA
Initial mmol Ca2+ = 0.10 x 100 = 10
mmol EDTA added = 0.10 x 25 = 2.5
mmol Ca2+ left = 10 – 2.5 = 7.5
[Ca2+]left = 7.5/125 = 0.06 M
In fact, this calcium concentration is the major source
of calcium in solution since the amount of calcium
coming from dissociation of the chelate is very
small, especially in presence of Ca2+ left in solution.
However, let us calculate the amount of calcium
released from the chelate:
mmol CaY2- formed = 2.5
[CaY2-] = 2.5/125 = 0.02 M
61
Kf = [CaY2-]/[Ca2+][Y4-]
5x1010 = (0.02 – x)/((0.06 + x) * a4 x)
assume that 0.02>>x
x = 1.9x10-11
The assumption is valid even without verification.
[Ca2+] = 0.06 + 1.9x10-11 = 0.06 M
pCa = 1.22
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3. After addition of 50 mL EDTA
mmol EDTA added = 0.10 x 50 = 5.0
mmol Ca2+ left = 10 – 5.0 = 5.0
[Ca2+]left = 5.0/150 = 0.033 M
We will see by similar calculation as in step
above that the amount of Ca2+ coming from
dissociation of the chelate is exceedingly
small as compared to amount left. However,
for the sake of practice let us perform the
calculation:
mmol CaY2- formed = 5.0
[CaY2-] = 5.0/150 = 0.033 M
63
Kf = [CaY2-]/[Ca2+][Y4-]
5x1010 = (0.033 – x)/((0.033 + x)* a4 x)
assume that 0.033>>x
x = 5.7x10-11
The assumption is valid even without verification.
[Ca2+] = 0.033+ 5.7x10-11 = 0.033 M
pCa = 1.48
64
4. After addition of 100 mL EDTA
mmol EDTA added = 0.10 x 100 = 10
mmol Ca2+ left = 10 – 10 = 0
This is the equivalence point. The only source for Ca2+
is the dissociation of the Chelate
mmol CaY2- formed = 10
[CaY2-] = 10/200 = 0.05 M
Ca2+ + Y4- D CaY2- kf = 5.0x1010
65
Kf = [CaY2-]/[Ca2+][Y4-]
5x105 = (0.05 – x)/(x* a4 x)
assume that 0.05>>x,
x = 1.7x10-6
Relative error = (1.7x10-6/0.05) x 100 = 3.4x10-3%
[Ca2+] = 1.7x10-6 M, pCa = 5.77
5. After addition of 150 mL EDTA
mmol EDTA added = 0.10 x 150 = 15
mmol EDTA excess = 15 – 10 = 5.0
CT = 5.0/250 = 0.02 M
mmol CaY2- = 10
[CaY2-] = 10/250 = 0.04 M
Ca2+ + Y4- D CaY2- kf = 5.0x1010
66
Kf = [CaY2-]/[Ca2+][Y4-]
5x1010 = (0.04 – x)/(x* a4(0.02 + x) )
assume that 0.02>>x
x = 1.1x10-10
The assumption is valid
[Ca2+] = 1.1x10-10 M
pCa = 9.95
67
6. After addition of 200 mL EDTA
mmol EDTA added = 0.10 x 200 = 20
mmol EDTA excess = 20 – 10 = 10
CT = 10/300 = 0.033 M
mmol CaY2- = 10
[CaY2-] = 10/300 = 0.033 M
Ca2+ + Y4- D CaY2- kf = 5.0x1010
68
Kf = [CaY2-]/[Ca2+][Y4-]
5x1010 = (0.033 – x)/(x* a4(0.033 + x) )
assume that 0.033>>x
x = 5.7x10-11
The assumption is undoubtedly valid
[Ca2+] = 5.7 x10-11 M
pCa = 10.24
69
Fractions of Dissociating Species in Polyligand Complexes
When polyligand complexes are dissociated in
solution, metal ions, ligand, and intermediates are
obtained in equilibrium with the complex. For
example, look at the following equilibria
Ag+ + NH3 D Ag(NH3)+
Ag(NH3)+ + NH3 D Ag(NH3)2+
kf1 = [Ag(NH3)+]/[Ag+][NH3]
kf2 = [Ag(NH3)2+]/[Ag(NH3)+][NH3]
We have Ag+, NH3, Ag(NH3)+, and Ag(NH3)2+ all present
in solution at equilibrium where
CAg = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+]
70
The fraction of each Ag+ species can be defined
as:
b0 = [Ag+]/ CAg
b1 = [Ag(NH3)+]/ CAg
b2 = [Ag(NH3)2+]/ CAg
As seen for fractions of a polyprotic acid
dissociating species, one can look at the b
values as b0 for the fraction with zero ligand
(free metal ion, Ag+), b1 as the fraction of the
species having one ligand (Ag(NH3)+) while b2 as
the fraction containing two ligands (Ag(NH3)2+).
The sum of all fractions will necessarily add up to
unity (b0 + b1 + b2 = 1)
71
For the case of b0, we make all terms as a function of
Ag+ since b0 is a function of Ag+. We use the
equilibrium constants of each step:
CAg = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+]
kf1 = [Ag(NH3)+]/[Ag+][NH3]
[Ag(NH3)+] = kf1 [Ag+][NH3]
Kf1 x kf2 = [Ag(NH3)2+]/[Ag+][NH3]2
[Ag(NH3)2+] = Kf1 x kf2 [Ag+][NH3]2
Substitution in the CAg relation gives:
CAg = [Ag+] + kf1 [Ag+][NH3] + Kf1 x kf2 [Ag+][NH3]2
CAg = [Ag+]( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
[Ag+]/ CAg = 1/( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
72
The inverse of this equation gives:
b0 = 1/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
If we use the same procedure for the derivation of
relations for other fractions we will get the same
denominator but the nominator will change
according to the species of interest:
b0 = 1/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
b1 = kf1 [NH3] / ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
b2 = Kf1 kf2 [NH3]2/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
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Example
Calculate the concentration of the different ion
species of silver for 0.010 M Ag+ in a 0.10 M
NH3 solution. Kf1 = 2.5x103, kf2 = 1.0x104
Solution
Ag+ + 2NH3 D Ag(NH3)2+ kf = kf1*kf2 = 2.5*107
[NH3]left = 0.08 M
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b0 = 1/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
Substitution in the above equation yields:
b0 = 1/ ( 1 + 2.5x103 * 0.08 + 2.5x103 * 1.0x104 *( 0.08)2)
b0 = 6.2x10-6
b0 = [Ag+]/ CAg
6.2x10-6 = [Ag+]/0.010
[Ag+] = 6.2x10-8 M
In the same manner calculations give:
b1 = 1.2x10-3
b1 = [Ag(NH3)+]/ CAg
1.2x10-3 = [Ag(NH3)+]/ 0.010
[Ag(NH3)+] = 1.2x10-5 M
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b2 = 0.999 or 1.0 if we consider significant
figures.
b2 = [Ag(NH3)2+]/ CAg
1.0 = [Ag(NH3)2+]/ 0.010
[Ag(NH3)2+] = 0.010 M
Therefore, it is clear that most Ag+ will be in the
complex form Ag(NH3)2+ since the formation
constant is large for the overall reaction:
Kf = kf1*kf2
Kf = 2.5x103 * 1.0x104 = 2.5x107
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