1 Lecture 31 Acid-Base Titartions, Cont… Kjeldahl Analysis Complexometric Reactions.

1

Lecture 31

Acid-Base Titartions, Cont…

Kjeldahl Analysis

Complexometric Reactions

22

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:

33

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 to mmol nitrogen.

44

ExampleA 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

55

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%

66

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 NH4+ + H2BO3

-

Borate is a strong conjugate base which can be titrated with a standard HCl solution.

77

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%

8

Complexometric Reactions and Titrations

9

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.

10

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)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.

13

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]

14

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 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 wise formation constants

15

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

16

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.

17

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.

18

Lecture 32

Complexometric Reactions, Cont….

Calculations

EDTA Equilibria

19

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.

20

[M2+] = 0.10 M, [L] = 0.10 M

M2+ + L 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

21

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 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

22

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

23

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.

25

EDTA Equilibria

EDTA can be regarded as H4Y where in solution we

will have, in addition to H4Y, the following

species: H3Y-, H2Y

2-, HY3-, and Y4-. The amount of

each species depends on the pH of the solution where:

4 = [Y4-]/CT where:

CT = [H4Y] + [H3Y-] + [H2Y

2-] + [HY3-] + [Y4-]

4 = 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.

27

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- = 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.

28

Formation Constants for EDTA ComplexesCationKMYCationKMY

Ag+2.1 x 107Cu2+6.3 x 1018

Mg2+4.9 x 108Zn2+3.2 x 1016

Ca2+5.0 x1010Cd2+2.9 x 1016

Sr2+4.3 x 108Hg2+6.3 x 1021

Ba2+5.8 x 107Pb2+1.1 x 1018

Mn2+6.2 x1013Al3+1.3 x 1016

Fe2+2.1 x1014Fe3+1.3 x 1025

Co2+2.0 x1016V3+7.9 x 1025

Ni2+4.2 x1018Th4+1.6 x 1023

29Minimum pH for effective titrations of various metal ions with EDTA.

30

The question now is how to calculate the amount of Ca2+ at equilibrium?

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-] + [H2Y

2-] + [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:

4 = [Y4-]/CT

4 = ka1ka2ka3ka4/([H+]4 + ka1 [H

+]3 + ka1ka2[H+]2 + ka1ka2ka3[H

+] + ka1ka2ka3ka4)

32

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 4 = [Y4-]/CTthen we have:

[Y4-] = 4CT

Substitution in the formation constant expression we get:

Kf = [CaY2-]/[Ca2+]4CT 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.

33

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.

35Minimum pH for effective titrations of various metal ions with EDTA.

36

Lecture 33

Complexometric Titrations, Cont…

Complexometric Indicators

37

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)

40

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:

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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+.

45

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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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. 4 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-

50

CT = [H4Y] + [H3Y-] + [H2Y

2-] + [HY3-] + [Y4-]

Kf = [CaY2-]/[Ca2+]4CT

[Ca2+] = CT

Using the same type of calculation we are used to perform, one can write the following:

51

Kf = [CaY2-]/[Ca2+][Y4-]

5.0x1010 = (0.05 – x)/((0.050 + x)*4 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

52

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. 4 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

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Ca2+ + Y4- CaY2-

CT = [H4Y] + [H3Y-] + [H2Y

2-] + [HY3-] + [Y4-]

Kf = [CaY2-]/[Ca2+]4CT

[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:

54

Kf = [CaY2-]/[Ca2+][Y4-]

5.0x1010 = (0.05 – x)/(x*4 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

55

Lecture 34

Complexometric Reactions, Cont…

EDTA Titrations

56

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

57

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

58

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. 4 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

59

Kf = [CaY2-]/[Ca2+][Y4-]

5x1010 = (0.033 – x)/(x*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. [Ca2+] = 3.9x10-11

M

pCa = 10.41

60

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. 4 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 M pCa = 1.00

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2. After addition of 25 mL EDTAInitial mmol Ca2+ = 0.10 x 100 = 10


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:


[CaY2-] = 2.5/125 = 0.02 M

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Kf = [CaY2-]/[Ca2+][Y4-]

5x1010 = (0.02 – x)/((0.06 + x) *4 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

63



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:


[CaY2-] = 5.0/150 = 0.033 M

64

Kf = [CaY2-]/[Ca2+][Y4-]

5x1010 = (0.033 – x)/((0.033 + x)*4 x)


x = 5.7x10-11

The assumption is valid even without verification.

[Ca2+] = 0.033+ 5.7x10-11 = 0.033 M

pCa = 1.48

65



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- CaY2- kf = 5.0x1010

66

Kf = [CaY2-]/[Ca2+][Y4-]

5x105 = (0.05 – x)/(x*4 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



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

67

Kf = [CaY2-]/[Ca2+][Y4-]

5x1010 = (0.04 – x)/(x*4(0.02 + x) )

assume that 0.02>>x

x = 1.1x10-10

The assumption is valid

[Ca2+] = 1.1x10-10 M

pCa = 9.95

68



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

69

Kf = [CaY2-]/[Ca2+][Y4-]

5x1010 = (0.033 – x)/(x*4(0.033 + x) )


x = 5.7x10-11

The assumption is undoubtedly valid

[Ca2+] = 5.7 x10-11 M

pCa = 10.24

70

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

+]

71

The fraction of each Ag+ species can be defined as:

0 = [Ag+]/ CAg

1 = [Ag(NH3)+]/ CAg

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

As seen for fractions of a polyprotic acid dissociating species, one can look at the values

as 0 for the fraction with zero ligand (free metal

ion, Ag+), 1 as the fraction of the species having

one ligand (Ag(NH3)+) while 2 as the fraction

containing two ligands (Ag(NH3)2+).

The sum of all fractions will necessarily add up to unity (0 + 1 2 = 1)

72

For the case of 0, we make all terms as a function of

Ag+ since 0 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)

73

The inverse of this equation gives:

0 = 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:

0 = 1/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)

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

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

2)

74

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 Ag(NH3)2+ kf = kf1*kf2 = 2.5*107

[NH3]left = 0.08 M

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0 = 1/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)

Substitution in the above equation yields:

0 = 1/ ( 1 + 2.5x103 * 0.08 + 2.5x103 * 1.0x104 *( 0.08)2)

0 = 6.2x10-6

0 = [Ag+]/ CAg

6.2x10-6 = [Ag+]/0.010

[Ag+] = 6.2x10-8 M

In the same manner calculations give:

1 = 1.2x10-3

1 = [Ag(NH3)+]/ CAg

1.2x10-3 = [Ag(NH3)+]/ 0.010

[Ag(NH3)+] = 1.2x10-5 M

•

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2 = 0.999 or 1.0 if we consider significant

figures.

2 = [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

1 Lecture 31 Acid-Base Titartions, Cont… Kjeldahl Analysis Complexometric Reactions.

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Transcript of 1 Lecture 31 Acid-Base Titartions, Cont… Kjeldahl Analysis Complexometric Reactions.