Complexometric Reactions and Titrations

1.        The Chelon Effect

2.        EDTA Titrations

3.        EDTA Equilibria

4.        The Formation Constant

5.        Titration Curves

6.        Indicators

7.        EDTA Titration Curves

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 :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.

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 = Ag(NH3)+                        kf1 = [Ag(NH3)+]/[Ag+][NH3]

Ag(NH3)+ + NH3 = Ag(NH3)2+             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 = Ag(NH3)2+                   kf = [Ag(NH3)2+]/[Ag+][NH3]2

 

It is clear fro 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 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+ = Ag+ + 2 NH3                   kinst = [Ag+][NH3]2/[Ag(NH3)2+]

Therefore, we have:

 

Kinst = 1/kf

 

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.

 

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.

 

[M2+] = 0.10 M

[L] = 0.10 M

M2+ + L = ML2+

 

Before Equilibrium

0

0

0.1

Equation

M2+

L

ML2+

At Equilibrium

x

x

0.10 - x

 

 

 

 

 

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

 

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. Kf = 5.0x107

 

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

 

 

 

 

 

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

 

The Chelon Effect

 

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.

 

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 metal ions have good stability.

 

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.

 

The Formation Constant

 

Reaction eith 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- = CaY2-                 kf = 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. The question now is how to calculate the amount of Ca2+ at equilibrium? Let us look at what happens after CaY2- is formed: Of course, partial dissociation of the chelate will take place

 

CaY2- = Ca2+ + Y4- 

 

However, [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 Y4- which 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)

 

The Conditional Formation Constant

 

We have seen that for the reaction

 

Ca2+ + Y4- = 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. 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.

 

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- = MgY2- + In3- (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 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.

 

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 Mg2+-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 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+.

 

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.

 

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.

 

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- = CaY2-

 

Mmol 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- = Ca2+ + Y4-

 

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:

 

Before Equilibrium

0.050

0.050

0

Equation

CaY2-

Ca2+

Y4-

At Equilibrium

0.050 – x

0.050 + x

a4 x

 

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

 

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- = CaY2-

 

Mmol 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 dissociation of the complex

 

CaY2- = Ca2+ + Y4-

 

CT = [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:

 

Before Equilibrium

0.05

0

0

Equation

CaY2-

Ca2+

Y4-

At Equilibrium

0.05 – x

x

a4 x

 

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%

The assumption is valid

[Ca2+] = 1.7x10-6 M

 

pCa = 5.77

 

 Example

 

Calculate the titer of a 0.100 M EDTA solution in terms of mg CaCO3 (FW = 100.0) per mL EDTA

 

Solution

 

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.

Mg CaCO3 = 0.100 mmol x 100.0 mg/mmol = 10.0

Therefore, the titer og EDTA in terms of CaCO3 is 10.0 mg CaCO3/mL EDTA

 

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

 

 

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- = CaY2-                kf = 5.0x1010

 

Before Equilibrium

0.033

0

0.050

Equation

CaY2-

Ca2+

Y4-

At Equilibrium

0.033 – x

x

a4 (0.050 + x)

 

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. In fact this should be expected since kf is very large that its dissociation s negligible especially in presence of excess EDTA (common ion).

[Ca2+] = 3.9x10-11 M

pCa = 10.41

 

 

Titration Curves

 

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 [Mn+] ) and volume of EDTA. Let us look at this example:

 

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- = CaY2-                kf = 5.0x1010

1. After addition of 0 mL EDTA

 

[Ca2+] = 0.10

pCa = 1.00

 

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 since kf is very high and in presence of the common ion (Ca2+) 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 CaY2- formed = 2.5

[CaY2-] = 2.5/125 = 0.02 M

 

Before Equilibrium

0.02

0.06

0

Equation

CaY2-

Ca2+

Y4-

At Equilibrium

0.02 – x

0.06 + x

a4 x

 

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

 

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

 

Before Equilibrium

0.033

0.033

0

Equation

CaY2-

Ca2+

Y4-

At Equilibrium

0.033 – x

0.033 + x

a4 x

 

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

 

4. After addition of 100 mL EDTA

 

mmol EDTA added = 0.10 x 100 = 10

mmol Ca2+ left = 10 – 10 = ??

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- = CaY2-                kf = 5.0x1010

 

Before Equilibrium

0.05

0

0

Equation

CaY2-

Ca2+

Y4-

At Equilibrium

0.05 – x

x

a4 x

 

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%

The assumption is valid

[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- = CaY2-                kf = 5.0x1010

 

Before Equilibrium

0.04

0

0.02

Equation

CaY2-

Ca2+

Y4-

At Equilibrium

0.04 – x

x

a4 (0.02 + x)

 

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

 

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- = CaY2-                kf = 5.0x1010

 

Before Equilibrium

0.033

0

0.033

Equation

CaY2-

Ca2+

Y4-

At Equilibrium

0.033 – x

x

a4 (0.033 + x)

 

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

 

 

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 = Ag(NH3)+                       kf1 = [Ag(NH3)+]/[Ag+][NH3]

Ag(NH3)+ + NH3 = Ag(NH3)2+            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+]

 

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 adds up to unity (b0 + b1 + b2 = 1)

 

Derivation of a term for each fraction is straightforward and requires substitution in the CAg relation above. 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)

CAg /[Ag+] = ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)

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:

 

b1 = kf1 [NH3] / ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)

 

b2 = Kf1 kf2 [NH3]2/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)

 

 

Example

 

Calculate the concentration of the different ion species of silver for 0.010 M Ag+ in a 0.10 M NH3 solution. Assume that ammonia concentration will not change. Kf1 = 2.5x103, kf2 = 1.0x104

 

Solution

 

The concentration of silver ion species can be obtained as a function of ammonia concentration and formation constants from the relations above:

 

b0 = 1/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)

 

Substitution in the above equation yields:

 

b0 = 1/ ( 1 + 2.5x103 * 0.1 + 2.5x103 * 1.0x104 *( 0.10)2)

b0 = 4.0x10-6

b0 = [Ag+]/ CAg

4.0x10-6 = [Ag+]/0.010

[Ag+] = 4.0x10-8 M

 

In the same manner calculations give:

b1 = 1.0x10-3

b1 = [Ag(NH3)+]/ CAg

1.0x10-3 = [Ag(NH3)+]/ 0.010

[Ag(NH3)+] = 1.0x10-5 M

 

b2 = 1.0

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