6. Indicators
8. Fractions of Dissociating Species in Polyligand Complexes
Complexes are compounds formed from combination of metal ions with ligands (complexing agents). A metal is an electron deficient species while a ligand is an electron rich, and thus, electron donating species. A metal will thus accept electrons from a ligand where coordination bonds are formed. Electrons forming coordination bonds come solely from ligands.
A ligand is called a monodentate if it donates a single pair of electrons (like :NH_{3}) while a bidentate ligand (like ethylenediamine, :NH_{2}CH_{2}CH_{2}H_{2}N:) 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 Cu^{2+}, for example, giving the complex Cu(NH_{3})_{4}^{2+}. 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.
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 NH_{3}:
Ag^{+} + NH_{3} =
Ag(NH_{3})^{+} k_{f1}
= [Ag(NH_{3})^{+}]/[Ag^{+}][NH_{3}]
Ag(NH_{3})^{+} + NH_{3}
= Ag(NH_{3})_{2}^{+} k_{f2}
= [Ag(NH_{3})_{2}^{+}]/[Ag(NH_{3})^{+}][NH_{3}]
K_{f1}
x k_{f2} = [Ag(NH_{3})_{2}^{+}]/[Ag^{+}][NH_{3}]^{2}
Now look at the overall reaction:
Ag^{+} + 2 NH_{3} =
Ag(NH_{3})_{2}^{+} k_{f}
= [Ag(NH_{3})_{2}^{+}]/[Ag^{+}][NH_{3}]^{2}
It is clear fro inspection of the values of the k_{f} that:
K_{f}
= k_{f1} x k_{f2}
For a multistep complexation reaction we will always have the formation constant of the overall reaction equals the product of all step 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(NH_{3})_{2}^{+}
= Ag^{+} + 2 NH_{3} k_{inst
}= [Ag^{+}][NH_{3}]^{2}/[Ag(NH_{3})_{2}^{+}]
Therefore, we have:
K_{inst}
= 1/k_{f}
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 M^{2+} and 0.20 M Ligand (L). k_{f} = 1.0x10^{8}.
Solution
The formation constant is very high 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.
[M^{2+}] = 0.10 M
[L] = 0.10 M
M^{2+} + L = ML^{2+}
Before Equilibrium |
0 |
0 |
0.1 |
Equation |
M^{2+}^{} |
L |
ML^{2+}^{} |
At Equilibrium |
x |
x |
0.10 - x |
K_{f} = ( 0.10 x )/x^{2}
Assume 0.10>>x since k_{f} is very large
1.0x10^{8} = 0.10/x^{2}
x = 3.2x10^{-5}
Relative error = (3.2x10^{-5}/0.10) x 100 = 3.2x10^{-2} %
The assumption is valid.
[M^{2+}] = 3.2x10^{-5} M
Example
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 trine. K_{f} = 5.0x10^{7}
Solution
Ag^{+} + trien = Ag(trien)^{+}
Mmol Ag^{+} added = 25x0.01 = 0.25
Mmol trien added = 50x0.015 = 0.75
The reaction occurs in a 1:1 ratio
Mmol trien excess = 0.75 0.25 = 0.50
[Trien] = 0.5/75
[Ag(trien)^{+}] = 0.25/75
Before Equilibrium |
0 |
0.50/75 |
0.25/75 |
Equation |
Ag^{+}^{} |
trien |
Ag(trien)^{+}^{} |
At Equilibrium |
x |
0.50/75 + x |
0.25/75 - x |
K_{f} = ( 0.25/75 x )/(x * 0.50/75 + x)
Assume 0.25/75>>x since k_{f} is very large
5.0x10^{7} = (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
We have seen earlier that large multidentate ligands can form complexes with metal ions. These complexes are called chelates. The question is which is more stable a chelate formed from a chelating agent with four chelating groups or a complex formed from the same metal with four moles of ligand having the sale donating group? This can be simply answered by looking at the thermodynamics of the process. We know from simple thermodynamics that spontaneous processes are favored if an increase in entropy results. Now look at the dissociation of the chelate and the complex mentioned above, dissociation of the chelate will give two molecules (the metal ion and the chelating agent) while dissociation of the complex will give five molecules (the metal ion and four ligand molecules). Therefore, dissociation of the complex results in more disorder and thus more entropy. The dissociation of the complex is thus more favored and therefore the chelate is more stable as its dissociation is not favored.
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 metal ions have good stability.
EDTA can be regarded as H_{4}Y where in solution we will have, in addition to H_{4}Y, the following species: H_{3}Y^{-}, H_{2}Y^{2-}, HY^{3-}, and Y^{4-}. The amount of each species depends on the pH of the solution where:
a_{4} = [Y^{4-}]/C_{T} where C_{T} = [H_{4}Y] +
[H_{3}Y^{-}] + [H_{2}Y^{2-}] + [HY^{3-}]
+ [Y^{4-}]
a_{4} = k_{a1}k_{a2}k_{a3}k_{a4}/([H^{+}]^{4}
+ k_{a1} [H^{+}]^{3} + k_{a1}k_{a2}[H^{+}]^{2}
+ k_{a1}k_{a2}k_{a3}[H^{+}] + k_{a1}k_{a2}k_{a3}k_{a4})
The species Y^{4-} is the ligand species in EDTA titrations and thus should be looked at carefully.
Reaction eith EDTA with a metal ion to form a chelate is a simple reaction. For example, EDTA reacts with Ca^{2+} ions to form a Ca-EDTA chelate forming the basis for estimation of water hardness. The reaction can be represented by the following equation:
Ca^{2+} + Y^{4-} = CaY^{2-} k_{f} = 5.0x10^{10}
K_{f} = [CaY^{2-}]/[Ca^{2+}][Y^{4-}]
The formation constant is very high and the reaction between Ca^{2+} and Y^{4-} can be considered quantitative. Therefore, if equivalent amounts of Ca^{2+} and Y^{4- }were mixed together, an equivalent amount of CaY^{2-} will be formed. The question now is how to calculate the amount of Ca^{2+} at equilibrium? Let us look at what happens after CaY^{2-} is formed: Of course, partial dissociation of the chelate will take place
CaY^{2-} = Ca^{2+} + Y^{4-}
However, [Ca^{2+}] # [Y^{4-}] at this point since the amount of Y^{4-} is pH dependent and Y^{4-} will disproportionate to form all the following species, depending on the pH
C_{T} = [H_{4}Y]
+ [H_{3}Y^{-}] + [H_{2}Y^{2-}] + [HY^{3-}]
+ [Y^{4-}]
Where, C_{T} is the sum of all species derived from Y^{4-} which is equal to [Ca^{2+}].
Therefore, the [Y^{4-}] at equilibrium will be less than the [Ca^{2+}] and in fact it will only be a fraction of C_{T} where:
a_{4} = [Y^{4-}]/C_{T}
a_{4} = k_{a1}k_{a2}k_{a3}k_{a4}/([H^{+}]^{4} + k_{a1} [H^{+}]^{3} + k_{a1}k_{a2}[H^{+}]^{2} + k_{a1}k_{a2}k_{a3}[H^{+}] + k_{a1}k_{a2}k_{a3}k_{a4})
The Conditional Formation Constant
We have seen that for the reaction
Ca^{2+} + Y^{4-} = CaY^{2-} k_{f} = 5.0x10^{10}
We can write the formation constant expression
K_{f} = [CaY^{2-}]/[Ca^{2+}][Y^{4-}]
However, we do not know the amount of Y^{4-} at equilibrium but we can say that since a_{4} = [Y^{4-}]/C_{T}, then we have:
[Y^{4-}] = a_{4}C_{T}
Substitution in the formation constant expression we get:
K_{f} = [CaY^{2-}]/[Ca^{2+}]a_{4}C_{T} or at a given pH we can write
K_{f}^{'} = [CaY^{2-}]/[Ca^{2+}]C_{T}
Where K_{f}^{' } is called the conditional formation constant. It is conditional since it is now dependent on pH.
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 k_{f} 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.
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 (H_{3}In) in a Mg^{2+} reaction with EDTA:
MgIn^{-} (Color 1) + Y^{4-} = MgY^{2-} + In^{3-} (Color 2)
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 Cr^{3+} 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.
2. Lack of
a suitable indicator
This is the most sever 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 Mg^{2+}-EDTA titration has excellent indicators that show very good change in color at the end point. Look at the following situations:
a. A little of a known standard Mg^{2+} 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 Mg^{2+}.
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 Mg^{2+} in the Mg-EDTA complex thus releasing Mg^{2+} 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 Mg^{2+} in the complex.
c. If it is not easy to get a Mg-EDTA complex, just add a little Mg^{2+} to the EDTA titrant. Standardize the EDTA and start titration. At the very first point of EDTA added, some Mg^{2+} 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.
Example
Find the concentrations of all species in solution at equilibrium resulting from mixing 50 mL of 0.200 M Ca^{2+} with 50 mL of 0.100 M EDTA adjusted to pH 10. a_{4} at pH 10 is 0.35. k_{f} = 5.0x10^{10}
Solution
Ca^{2+} + Y^{4-} = CaY^{2-}
Mmol Ca^{2+} added = 0.200 x 50 = 10.0
Mmol EDTA added = 0.100 x 50 = 5.00
Mmol Ca^{2+} excess = 10.0 5.00 = 5.00
[Ca^{2+}]_{excess }= 5.00/100 = 0.050 M
mmol CaY^{2-} formed = 5.00
[CaY^{2-}]
= 5.00/100 = 0.050
CaY^{2-} = Ca^{2+} + Y^{4-}
C_{T} = [H_{4}Y]
+ [H_{3}Y^{-}] + [H_{2}Y^{2-}] + [HY^{3-}]
+ [Y^{4-}]
K_{f} = [CaY^{2-}]/[Ca^{2+}]a_{4}C_{T}
[Ca^{2+}] = C_{T}
Using the same type of calculation we are used to perform, one can write the following:
Before
Equilibrium |
0.050 |
0.050 |
0 |
Equation |
CaY^{2-}^{} |
Ca^{2+}^{} |
Y^{4-}^{} |
At
Equilibrium |
0.050
x |
0.050 +
x |
a_{4} x |
K_{f} = [CaY^{2-}]/[Ca^{2+}][Y^{4-}]
5.0x10^{10} = (0.05 x)/((0.050 + x)* a_{4} x)
assume that 0.05>>x
x = 5.6x10^{-11}
Relative error will be very small value
The assumption is valid
[Ca^{2+}] = 0.050 + x = 0.050 M
[CaY^{2-}] = 0.050 x = 0.050 M
[Y^{4-}] = 0.35 * 5.6x10^{-11} = 1.9x10^{-11} M
Example
Calculate the pCa of a solution at pH 10 after addition of 100 mL of 0.10 M Ca^{2+} to 100 mL of 0.10 M EDTA. a_{4} at pH 10 is 0.35. k_{f} = 5.0x10^{10}
Solution
Ca^{2+} + Y^{4-} = CaY^{2-}
Mmol Ca^{2+} = 0.10 x 100 = 10
Mmol EDTA = 0.10 x 100 = 10
Mmol CaY^{2- }= 10
[CaY^{2-}] = 10/200 = 0.05 M
Therefore, Ca^{2+ }will be produced from partial dissociation of the complex
CaY^{2-} = Ca^{2+} + Y^{4-}
C_{T} = [H_{4}Y]
+ [H_{3}Y^{-}] + [H_{2}Y^{2-}] + [HY^{3-}]
+ [Y^{4-}]
K_{f} = [CaY^{2-}]/[Ca^{2+}]a_{4}C_{T}
[Ca^{2+}] = C_{T}
5.0x10^{10} = 0.05/([Ca^{2+}]^{2} x 0.35)
[Ca^{2+}] = 1.7x10^{-6} M
pCa = 5.77
Using the same type of calculation we are used to perform, one can write the following:
Before
Equilibrium |
0.05 |
0 |
0 |
Equation |
CaY^{2-}^{} |
Ca^{2+}^{} |
Y^{4-}^{} |
At
Equilibrium |
0.05 x |
x |
a_{4} x |
K_{f} = [CaY^{2-}]/[Ca^{2+}][Y^{4-}]
5.0x10^{10} = (0.05 x)/(x* a_{4} x)
assume that 0.05>>x
x = 1.7x10^{-6}
Relative error = (1.7x10^{-6}/0.05) x 100 = 3.4x10^{-3}%
The assumption is valid
[Ca^{2+}] = 1.7x10^{-6} M
pCa = 5.77
Example
Calculate the titer of a 0.100 M EDTA solution in terms of mg CaCO_{3} (FW = 100.0) per mL EDTA
Solution
The EDTA concentration is 0.100 mmol/mL, therefore, the point here is to calculate the mg CaCO_{3} 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 CaCO_{3}.
Mg CaCO_{3} = 0.100 mmol x 100.0 mg/mmol = 10.0
Therefore, the titer og EDTA in terms of CaCO_{3} is 10.0 mg CaCO_{3}/mL EDTA
Example
An EDTA solution is standardized against high purity CaCO_{3} by dissolving 0.3982 g of CaCO_{3} 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 CaCO_{3} = mmol EDTA
mg/FW = Molarity x V_{mL}
398.2/100.0 = M x 38.26
M_{EDTA} = 0.1041
Example
Find the concentration of Ca^{2+} in a 20 mL of 0.20 M solution at pH 10 after addition of 100 mL of 0.10 M EDTA. a_{4} at pH 10 is 0.35. k_{f} = 5x10^{10}
Solution
Initial mmol Ca^{2+ } = 0.20 x 20 = 4.0
mmol EDTA added = 0.10 x 100 = 10
mmol EDTA
excess = 10 4.0 = 6.0
C_{T} = 6.0/120 = 0.050 M
Mmol CaY^{2-
} = 4.0
[CaY^{2-}] = 4.0/120 = 0.033 M
Ca^{2+} + Y^{4-} =
CaY^{2-} k_{f}
= 5.0x10^{10}
Before
Equilibrium |
0.033 |
0 |
0.050 |
Equation |
CaY^{2-}^{} |
Ca^{2+}^{} |
Y^{4-}^{} |
At
Equilibrium |
0.033
x |
x |
a_{4} (0.050 + x)_{} |
K_{f} = [CaY^{2-}]/[Ca^{2+}][Y^{4-}]
5x10^{10} = (0.033 x)/(x* a_{4}(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. In fact this should be expected since k_{f} is very large that its dissociation s negligible especially in presence of excess EDTA (common ion).
[Ca^{2+}] = 3.9x10^{-11} M
pCa = 10.41
We can derive a titration curve for the titration of a metal ion with EDTA n the same manner as done for acid-base titrations discussed in previous sections. The titration curve in this case is a relation between pM ( - log [M^{n+}] ) and volume of EDTA. Let us look at this example:
Example
Find pCa in a 100 mL solution of 0.10 M Ca^{2+} at pH 10 after addition of 0, 25, 50, 100, 150, and 200 mL of 0.10 M EDTA. a_{4} at pH 10 is 0.35. k_{f} = 5x10^{10}
Solution
Again, we should remember that EDTA reactions with metal ions are 1:1 reactions. Therefore, we have:
Ca^{2+}
+ Y^{4-} = CaY^{2-} k_{f}
= 5.0x10^{10}
1. After addition of 0 mL EDTA
[Ca^{2+}] = 0.10
pCa = 1.00
2. After addition of 25 mL EDTA
Initial mmol Ca^{2+} = 0.10 x 100 = 10
Mmol EDTA added = 0.10 x 25 = 2.5
Mmol Ca^{2+} left = 10 2.5 = 7.5
[Ca^{2+}]_{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 since k_{f} is very high and in presence of the common ion (Ca^{2+}) the amount of f calcium from dissociation of the chelate will even be much smaller. However, let us calculate the amount of calcium released from the chelate:
Mmol CaY^{2-}
formed = 2.5
[CaY^{2-}] = 2.5/125 = 0.02 M
Before
Equilibrium |
0.02 |
0.06 |
0 |
Equation |
CaY^{2-}^{} |
Ca^{2+}^{} |
Y^{4-}^{} |
At
Equilibrium |
0.02 x |
0.06 + x |
a_{4} x |
K_{f} = [CaY^{2-}]/[Ca^{2+}][Y^{4-}]
5x10^{10} = (0.02 x)/((0.06 + x)* a_{4} x)
assume that 0.02>>x
x = 1.9x10^{-11}
The assumption is valid even without verification.
[Ca^{2+}] = 0.06 + 1.9x10^{-11} = 0.06 M
pCa = 1.22
3. After addition of 50 mL EDTA
mmol EDTA added = 0.10 x 50 = 5.0
mmol Ca^{2+}
left = 10 5.0 = 5.0
[Ca^{2+}]_{left} = 5.0/150 = 0.033 M
We will see by similar calculation as in step above that the amount of Ca^{2+} 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 CaY^{2-}
formed = 5.0
[CaY^{2-}] = 5.0/150 = 0.033 M
Before
Equilibrium |
0.033 |
0.033 |
0 |
Equation |
CaY^{2-}^{} |
Ca^{2+}^{} |
Y^{4-}^{} |
At
Equilibrium |
0.033
x |
0.033 +
x |
a_{4} x |
K_{f} = [CaY^{2-}]/[Ca^{2+}][Y^{4-}]
5x10^{10} = (0.033 x)/((0.033 + x)* a_{4} x)
assume that 0.033>>x
x = 5.7x10^{-11}
The assumption is valid even without verification.
[Ca^{2+}] = 0.033+ 5.7x10^{-11} = 0.033 M
pCa = 1.48
4. After addition of 100 mL EDTA
mmol EDTA added = 0.10 x 100 = 10
mmol Ca^{2+} left = 10 10 = ??
This is the equivalence point. The only source for Ca^{2+} is the dissociation of the Chelate
Mmol CaY^{2-}
formed = 10
[CaY^{2-}] = 10/200 = 0.05 M
Ca^{2+} + Y^{4-}
= CaY^{2-} k_{f}
= 5.0x10^{10}
Before
Equilibrium |
0.05 |
0 |
0 |
Equation |
CaY^{2-}^{} |
Ca^{2+}^{} |
Y^{4-}^{} |
At
Equilibrium |
0.05 x |
x |
a_{4} x |
K_{f} = [CaY^{2-}]/[Ca^{2+}][Y^{4-}]
5x10^{5} = (0.05 x)/(x* a_{4} x)
assume that 0.05>>x
x = 1.7x10^{-6}
Relative error = (1.7x10^{-6}/0.05) x 100 = 3.4x10^{-3}%
The assumption is valid
[Ca^{2+}] = 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
C_{T} = 5.0/250 = 0.02 M
Mmol CaY^{2-
} = 10
[CaY^{2-}] = 10/250 = 0.04 M
Ca^{2+} + Y^{4-} =
CaY^{2-} k_{f}
= 5.0x10^{10}
Before
Equilibrium |
0.04 |
0 |
0.02 |
Equation |
CaY^{2-}^{} |
Ca^{2+}^{} |
Y^{4-}^{} |
At
Equilibrium |
0.04 x |
x |
a_{4} (0.02 + x)_{} |
K_{f} = [CaY^{2-}]/[Ca^{2+}][Y^{4-}]
5x10^{10} = (0.04 x)/(x* a_{4}(0.02 + x) )
assume that 0.02>>x
x = 1.1x10^{-10}
The assumption is valid
[Ca^{2+}] = 1.1x10^{-10} M
pCa = 9.95
6. After addition of 200 mL EDTA
mmol EDTA
added = 0.10 x 200 = 20
mmol EDTA
excess = 20 10 = 10
C_{T} = 10/300 = 0.033 M
Mmol CaY^{2-
} = 10
[CaY^{2-}] = 10/300 = 0.033 M
Ca^{2+} + Y^{4-} =
CaY^{2-} k_{f}
= 5.0x10^{10}
Before
Equilibrium |
0.033 |
0 |
0.033 |
Equation |
CaY^{2-}^{} |
Ca^{2+}^{} |
Y^{4-}^{} |
At
Equilibrium |
0.033
x |
x |
a_{4} (0.033 + x)_{} |
K_{f} = [CaY^{2-}]/[Ca^{2+}][Y^{4-}]
5x10^{10} = (0.033 x)/(x* a_{4}(0.033 + x) )
assume that 0.033>>x
x = 5.7x10^{-11}
The assumption is undoubtedly valid
[Ca^{2+}] = 5.7 x10^{-11} M
pCa = 10.24
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^{+} + NH_{3} =
Ag(NH_{3})^{+} k_{f1}
= [Ag(NH_{3})^{+}]/[Ag^{+}][NH_{3}]
Ag(NH_{3})^{+} + NH_{3} = Ag(NH_{3})_{2}^{+} k_{f2} = [Ag(NH_{3})_{2}^{+}]/[Ag(NH_{3})^{+}][NH_{3}]
We have Ag^{+}, NH_{3}, Ag(NH_{3})^{+}, and Ag(NH_{3})_{2}^{+} all present in solution at equilibrium where
C_{Ag}
= [Ag^{+}] + [Ag(NH_{3})^{+}] + [Ag(NH_{3})_{2}^{+}]
The fraction of each Ag^{+} species can be defined as:
b_{0} =
[Ag^{+}]/ C_{Ag}
b_{1} =
[Ag(NH_{3})^{+}]/ C_{Ag}
b_{2} =
[Ag(NH_{3})_{2}^{+}]/ C_{Ag}
As seen for fractions of a polyprotic acid dissociating species, one can look at the b values as b_{0} for the fraction with zero ligand (free metal ion, Ag^{+}), b_{1} as the fraction of the species having one ligand (Ag(NH_{3})^{+}) while b_{2} as the fraction containing two ligands (Ag(NH_{3})_{2}^{+}).
The sum of all fractions will necessarily adds up to unity (b_{0 }+ b_{1 }+ b_{2} = 1)
Derivation of a term for each fraction is straightforward and requires substitution in the C_{Ag} relation above. For the case of b_{0}, we make all terms as a function of Ag^{+} since b_{0 }is a function of Ag^{+}. We use the equilibrium constants of each step:
C_{Ag}
= [Ag^{+}] + [Ag(NH_{3})^{+}] + [Ag(NH_{3})_{2}^{+}]
k_{f1} = [Ag(NH_{3})^{+}]/[Ag^{+}][NH_{3}]
[Ag(NH_{3})^{+}]
= k_{f1} [Ag^{+}][NH_{3}]
K_{f1} x k_{f2} = [Ag(NH_{3})_{2}^{+}]/[Ag^{+}][NH_{3}]^{2}
[Ag(NH_{3})_{2}^{+}]
= K_{f1} x k_{f2} [Ag^{+}][NH_{3}]^{2}
Substitution in the C_{Ag} relation gives:
C_{Ag} = [Ag^{+}] + k_{f1} [Ag^{+}][NH_{3}] + K_{f1} x k_{f2} [Ag^{+}][NH_{3}]^{2}
C_{Ag} = [Ag^{+}]( 1 + k_{f1} [NH_{3}] + K_{f1} k_{f2} [NH_{3}]^{2})
C_{Ag} /[Ag^{+}] = ( 1 + k_{f1} [NH_{3}] + K_{f1} k_{f2} [NH_{3}]^{2})
The inverse of this equation gives:
b_{0}
= 1/ ( 1 + k_{f1} [NH_{3}] + K_{f1} k_{f2} [NH_{3}]^{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:
b_{1} = k_{f1}
[NH_{3}] / ( 1
+ k_{f1} [NH_{3}]
+ K_{f1} k_{f2}
[NH_{3}]^{2})
b_{2}
= K_{f1} k_{f2}
[NH_{3}]^{2}/ ( 1 + k_{f1}
[NH_{3}] + K_{f1}
k_{f2} [NH_{3}]^{2})
Example
Calculate the concentration of the different ion species of silver for 0.010 M Ag^{+} in a 0.10 M NH_{3} solution. Assume that ammonia concentration will not change. K_{f1} = 2.5x10^{3}, k_{f2} = 1.0x10^{4}
Solution
The concentration of silver ion species can be obtained as a function of ammonia concentration and formation constants from the relations above:
b_{0}
= 1/ ( 1 + k_{f1} [NH_{3}] + K_{f1} k_{f2} [NH_{3}]^{2})
Substitution in the above equation yields:
b_{0} = 1/
( 1 + 2.5x10^{3} * 0.1 + 2.5x10^{3} * 1.0x10^{4} *(
0.10)^{2})
b_{0} = 4.0x10^{-6}
b_{0} = [Ag^{+}]/ C_{Ag}
4.0x10^{-6} = [Ag^{+}]/0.010
[Ag^{+}] = 4.0x10^{-8 }M
In the same manner calculations give:
b_{1} = 1.0x10^{-3}
b_{1} = [Ag(NH_{3})^{+}]/ C_{Ag}
1.0x10^{-3} = [Ag(NH_{3})^{+}]/ 0.010
[Ag(NH_{3})^{+}] = 1.0x10^{-5} M
b_{2} = 1.0
b_{2} = [Ag(NH_{3})_{2}^{+}]/ C_{Ag}
1.0 = [Ag(NH_{3})_{2}^{+}]/ 0.010
[Ag(NH_{3})_{2}^{+}] = 0.010 M
Therefore, it is clear that most Ag^{+} will be in the complex form Ag(NH_{3})_{2}^{+} since the formation constant is large for the overall reaction:
K_{f} = k_{f1}*k_{f2}
K_{f} = 2.5x10^{3} * 1.0x10^{4} = 2.5x10^{7}