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

2. 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

3. 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. 3

4. Example
A 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

5. 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 = x60 = 144.9
% urea = (144.9/200)x100 = 72.5% 5

6. 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+ + H2BO3- Borate is a strong conjugate base which can be titrated with a standard HCl solution. 6

7. Example A 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 = % protein = (218.8/300) x 100 = 72.9% 7

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)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.

13. Let us look at the complexation reaction of Ag+ with NH3:
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)+ kf1 = [Ag(NH3)+]/[Ag+][NH3]
Ag(NH3)+ + NH3 D 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 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 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+ D 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. Complexometric Reactions, Cont….
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 D ML2+ Kf = ( 0
[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

21. Silver ion forms a stable 1:1 complex with trien
Silver ion forms a stable 1:1 complex with trien. Calculate the silver ion concentration at equilibrium when 25 mL of M silver nitrate is added to 50 mL of 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

22. Kf = ( 0. 25/75 – x )/(x. 50/75 + x) Assume 0
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-, 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.

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 Complexes
Cation
KMY
Ag+
2.1 x 107
Cu2+
6.3 x 1018
Mg2+
4.9 x 108
Zn2+
3.2 x 1016
Ca2+
5.0 x1010
Cd2+
2.9 x 1016
Sr2+
4.3 x 108
Hg2+
6.3 x 1021
Ba2+
5.8 x 107
Pb2+
1.1 x 1018
Mn2+
6.2 x1013
Al3+
1.3 x 1016
Fe2+
2.1 x1014
Fe3+
1.3 x 1025
Co2+
2.0 x1016
V3+
7.9 x 1025
Ni2+
4.2 x1018
Th4+
1.6 x 1023

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

30. CT = [H4Y] + [H3Y-] + [H2Y2-] + [HY3-] + [Y4-]
The question now is how to calculate the amount of Ca2+ at equilibrium?
CaY2- D 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)

32. 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. 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.

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

36. Complexometric Titrations, Cont… Complexometric Indicators
Lecture 33
Complexometric Titrations, Cont…
Complexometric Indicators

37. MgIn- (Color 1) + Y4- D MgY2- + In3- (Color 2)
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)

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.

42. 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:

43. 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.

47. 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.

49. Example Find the concentrations of all species in solution at equilibrium resulting from mixing 50 mL of M Ca2+ with 50 mL of M EDTA adjusted to pH 10. a4 at pH 10 is kf = 5.0x1010 Solution Ca2+ + Y4- g CaY2- mmol Ca2+ added = x 50 = 10.0 mmol EDTA added = x 50 = 5.00 mmol Ca2+ excess = 10.0 – 5.00 = 5.00 [Ca2+]excess = 5.00/100 = M mmol CaY2- formed = 5.00 [CaY2-] = 5.00/100 = CaY2- D Ca2+ + Y4-

50. 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:

51. Kf = [CaY2-]/[Ca2+][Y4-] 5. 0x1010 = (0. 05 – x)/((0. 050 + x)
Kf = [CaY2-]/[Ca2+][Y4-] 5.0x1010 = (0.05 – x)/(( x)* a4 x) assume that 0.05>>x x = 5.6x10-11 Relative error will be very small value The assumption is valid [Ca2+] = x = M [CaY2-] = – x = 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. a4 at pH 10 is 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

53. Ca2+ + Y4- D CaY2- 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:

54. Kf = [CaY2-]/[Ca2+][Y4-] 5. 0x1010 = (0. 05 – x)/(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% [Ca2+] = 1.7x10-6 M pCa = 5.77

55. Complexometric Reactions, Cont…
Lecture 34
Complexometric Reactions, Cont…
EDTA Titrations

56. Calculate the titer of a 0
Calculate the titer of a M EDTA solution in terms of mg CaCO3 (FW = 100.0) per mL EDTA
The EDTA concentration is mmol/mL, therefore, the point here is to calculate the mg CaCO3 reacting with mmol EDTA. We know that EDTA reacts with metal ions in a 1:1 ratio. Therefore mmol EDTA will react with 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 g of CaCO3 in HCl and adjusting the pH to 10. The solution is then titrated with EDTA requiring 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 =

58. Example Find the concentration of Ca2+ in a 20 mL of 0
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 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 = M mmol CaY2- = 4.0 [CaY2-] = 4.0/120 = M Ca2+ + Y4- D CaY2- kf = 5.0x1010

59. Kf = [CaY2-]/[Ca2+][Y4-] 5x1010 = (0. 033 – x)/(x. a4(0
Kf = [CaY2-]/[Ca2+][Y4-] 5x1010 = (0.033 – x)/(x* a4( 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
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 kf = 5x1010 Solution Again, we should remember that EDTA reactions with metal ions are 1:1 reactions. Therefore, we have: Ca2+ + Y4- D CaY2- kf = 5.0x After addition of 0 mL EDTA [Ca2+] = 0.10 M pCa = 1.00

61. 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

62. Kf = [CaY2-]/[Ca2+][Y4-] 5x1010 = (0. 02 – x)/((0. 06 + x)
Kf = [CaY2-]/[Ca2+][Y4-] 5x1010 = (0.02 – x)/(( x) * a4 x) assume that 0.02>>x x = 1.9x10-11 The assumption is valid even without verification. [Ca2+] = x10-11 = 0.06 M pCa = 1.22

63. 3. After addition of 50 mL EDTA mmol EDTA added = 0. 10 x 50 = 5
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 = 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 = M

64. Kf = [CaY2-]/[Ca2+][Y4-] 5x1010 = (0. 033 – x)/((0. 033 + x)
Kf = [CaY2-]/[Ca2+][Y4-] 5x1010 = (0.033 – x)/(( x)* a4 x) assume that 0.033>>x x = 5.7x10-11 The assumption is valid even without verification. [Ca2+] = x10-11 = M pCa = 1.48

65. 4. After addition of 100 mL EDTA mmol EDTA added = 0
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

66. Kf = [CaY2-]/[Ca2+][Y4-] 5x105 = (0. 05 – x)/(x. a4 x) assume that 0
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 = 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

67. Kf = [CaY2-]/[Ca2+][Y4-] 5x1010 = (0. 04 – x)/(x. a4(0
Kf = [CaY2-]/[Ca2+][Y4-] 5x1010 = (0.04 – x)/(x* a4( x) ) assume that 0.02>>x x = 1.1x10-10 The assumption is valid [Ca2+] = 1.1x10-10 M pCa = 9.95

68. 6. After addition of 200 mL EDTA mmol EDTA added = 0
6. After addition of 200 mL EDTA mmol EDTA added = 0.10 x 200 = 20 mmol EDTA excess = 20 – 10 = 10 CT = 10/300 = M mmol CaY2- = 10 [CaY2-] = 10/300 = M Ca2+ + Y4- D CaY2- kf = 5.0x1010

69. Kf = [CaY2-]/[Ca2+][Y4-] 5x1010 = (0. 033 – x)/(x. a4(0
Kf = [CaY2-]/[Ca2+][Y4-] 5x1010 = (0.033 – x)/(x* a4( x) ) assume that 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 D Ag(NH3)+ kf1 = [Ag(NH3)+]/[Ag+][NH3]
Ag(NH3)+ + NH3 D 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: 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)

72. 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)

73. b1 = kf1 [NH3] / ( 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)

74. Example Calculate the concentration of the different ion species of silver for 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

75. b0 = 1/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)
Substitution in the above equation yields:
b0 = 1/ ( x103 * x103 * 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

76. b2 = 0. 999 or 1. 0 if we consider significant figures
b2 = or 1.0 if we consider significant figures. b2 = [Ag(NH3)2+]/ CAg 1.0 = [Ag(NH3)2+]/ [Ag(NH3)2+] = 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