Since pKa2 is less than pKa1 the ZnOH+ species will have a very limited stability domain, resulting in a low concentration relative to all other species for all pH values.. The reactions
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PROBLEMS/SOLUTIONS
1 Plot the distribution (versus pH) of the zinc aquo complex and the first four deprotonated species of
this complex The pKa values are pKa1 = 9.2, pKa2 = 7.9, pKa3 = 11.3, pKa4 = 12.3 Note the unusual
feature that pKa2 is smaller than pKa1 Comment on how this affects the distribution
Solution
The reactions are shown without the waters of hydration and uses H+ as a short form for the hydronium ion 1) Zn2+ + H2O ZnOH+ + H+
[ZnOH+][H+]
Ka1 = - = 2.51 x 10-10
[Zn2+]
2) ZnOH+ + H2O Zn(OH)2 + H+
[Zn(OH)2][H+]
Ka2 = - = 1.26 x 10-8
[ZnOH+]
3) Zn(OH)2 + H2O Zn(OH)3- + H+
[Zn(OH)3-][H+]
Ka3 = - = 5.01 x 10-12
[Zn(OH)2]
4) Zn(OH)3- + H2O Zn(OH)42- + H+
[Zn(OH)42-][H+]
Ka4 = - = 5.01 x 10-13
[Zn(OH)3-]
5) Ksp = [Zn2+][OH-]2 = 3.0 x 10-16
From the pKa values given, at pH less than 7 expect that the Zn2+ will be fully hydrated and protonated
CZn = [Zn2+] + [ZnOH+] + [Zn(OH)2] + [Zn(OH)3-] + [Zn(OH)42-]
αZn2+ = [Zn2+] ÷ CZn, αZnOH+ = [ZnOH+] ÷ CZn, αZn(OH)2 = [Zn(OH)2] ÷ CZn,
αZn(OH)3- = [Zn(OH)3-] ÷ CZn, and αZn(OH)42- = [Zn(OH)42-] ÷ CZn
Rewrite the above equations for each species in terms of [H+]
a) From 5 (and using Kw) [Zn2+] = 3 x 1012[H+]2
b) From 1, with substitution [ZnOH+] = 3 x 1012Ka1[H+]
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c) From 2, with substitution [Zn(OH)2] = 3 x 1012Ka1Ka2
3 x 1012Ka1Ka2Ka3
d) From 3, with substitution [Zn(OH)3-] = -
[H+]
3 x 1012Ka1Ka2Ka3Ka4
e) From 4, with substitution [Zn(OH)42-] = -
[H+]2
Table of Results: (solved for the [H+] given)
[Zn2+] [ZnOH+] [Zn(OH)2] [Zn(OH)3-] [Zn(OH)42-] CZn
pH =1 3x1010 7.5x101 9.5x10-6 4.8x10-16 2.4x10-27 3.0x1010
pH =2 3x108 7.5x100 9.5x10-6 4.8x10-15 2.4x10-25 3.0x108
pH =3 3x106 7.5x10-1 9.5x10-6 4.8x10-14 2.4x10-23 3.0x106
pH =4 3x104 7.5x10-2 9.5x10-6 4.8x10-13 2.4x10-21 3.0x104
pH =5 3x102 7.5x10-3 9.5x10-6 4.8x10-12 2.4x10-19 3.0x102
pH =6 3x100 7.5x10-4 9.5x10-6 4.8x10-11 2.4x10-17 3.0
pH =7 3x10-2 7.5x10-5 9.5x10-6 4.8x10-10 2.4x10-15 3.0x10-2
pH =8 3x10-4 7.5x10-6 9.5x10-6 4.8x10-9 2.4x10-13 3.2x10-4
pH =9 3x10-6 7.5x10-7 9.5x10-6 4.8x10-8 2.4x10-11 1.3x10-5
pH =10 3x10-8 7.5x10-8 9.5x10-6 4.8x10-7 2.4x10-9 1.0x10-5
pH =11 3x10-10 7.5x10-9 9.5x10-6 4.8x10-6 2.4x10-7 1.4x10-5
pH =12 3x10-12 7.5x10-10 9.5x10-6 4.8x10-5 2.4x10-5 8.1x10-3
pH =13 3x10-14 7.5x10-11 9.5x10-6 4.8x10-4 2.4x10-3 2.9x10-3
pH =14 3x10-16 7.5x10-12 9.5x10-6 4.8x10-3 2.4x10-1 2.4x10-1
Graph of vs pH for the zinc aquo complexes
0.00
0.20
0.40
0.60
0.80
1.00
1.20
Zn2+
Zn(OH)2
pH
Zn(OH)3
-ZnOH+
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Individual α results for graph at specified pH value
[Zn2+] [ZnOH+] [Zn(OH)2] [Zn(OH)3-] [Zn(OH)42-]
pH 1 1.00 0.00 0.00 0.00 0.00
pH 2 1.00 0.00 0.00 0.00 0.00
pH 3 1.00 0.00 0.00 0.00 0.00
pH 4 1.00 0.00 0.00 0.00 0.00
pH 5 1.00 0.00 0.00 0.00 0.00
pH 6 1.00 0.00 0.00 0.00 0.00
pH 7 1.00 0.00 0.00 0.00 0.00
pH 8 0.95 0.02 0.03 0.00 0.00
pH 9 0.23 0.06 0.71 0.00 0.00
pH 10 0.00 0.01 0.94 0.05 0.00 pH 11 0.00 0.00 0.65 0.33 0.02
pH 12 0.00 0.00 0.12 0.59 0.29 pH 13 0.00 0.00 0.00 0.17 0.83 pH 14 0.00 0.00 0.00 0.02 0.98 Plot the data versus pH using an appropriate program Since pKa2 is less than pKa1 the ZnOH+ species will have a very limited stability domain, resulting in a low concentration relative to all other species for all pH values This is evident on the graph above 2 The stepwise formation constants for the complexes Pb(OH)+ (aq) and Pb(OH)2(aq) from Pb2+ (aq) are 2.0 × 106 and 4.0 × 104, respectively The reactions can be written in simple form as and Calculate the pKa1 and pKa2 values for deprotonation of the aquo complex of lead(II), and determine the fractional concentration of the two most important species at pH 7.0 Solution As in the previous solution we again use H+ to indicate the hydronium ion [PbOH+][H+] [Pb(OH)2][H+] Ka1 = - Ka2 = -
[Pb2+] [PbOH+] [PbOH+] Kw Kf1 = - [OH-] = -
[Pb2+][OH-] [H+]
[PbOH+]
2.0 x 106 = - (Kw = 1 x 10-14)
[Pb2+] x Kw/[H+]
Trang 4[PbOH+] [H+]
2.0 x 106 x 1 x 10-14 = - = Ka1 = 2.0 x 10-8
[Pb2+]
pKa1 = 7.7
pKa2 = 9.4 (calculated in the same way)
At pH = 7 the two most important species will be the fully protonated hydrated Pb2+ ion and the first deprotonated (PbOH+) species
[PbOH+] [H+]
using Ka1 = - = 2.0 x 10-8
[Pb2+]
at [H+] = 1 x 10-7 [PbOH+]/[Pb2+] = 0.20
Assuming these are the only two species at pH 7, the fractional amount of each is:
PbOH+ = 0.17 and Pb2+ = 0.83
3 Without calculation, sketch a distribution diagram, versus pH, for uncomplexed NTA species in water,
covering the entire pH range from 0 to 14 What will be the most important species in water with near neutral pH?
Solution
1.0
NTA
NTA
NTA2 NTA3
0.5
0.0
0 2 4 6 8 10 12 14
pH
Above is an approximate species distribution diagram based on when pH = pKa and the acid and base forms
of a particular species (e.g NTA in this case) are equal Assuming only the two species at any one time, α must = 0.5 for both (marked with •) There may be a small error in this assumption when the difference in
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pKa values is small, as in the case of the first two values here The lines are roughly sketched in to show the approximate diagram It is clear from this diagram that NTA2- is the most important species at near neutral
pH values
4 A freshwater sample contains 160 ppb Cu2+ and 4.3 ppm fulvic acid which has 4.0 mmol g-1 sites available for complexation Calculate the equilibrium concentrations of free and complexed copper and
of free fulvic acid Use the conditional stability constant in Table 13.3
Solution (refer also to Examples 13.2 and 13.3)
Cu2+ + FA CuFA Kf´ = 1 x 104 (M.W Cu = 63.546 g mol-1, Appendix B.1)
160 µg L-1 = 2.52x10-6 mol L-1 Cu2+ (Total Cu2+ concentration)
Effective concentration of fulvic acid, i.e., sites available for complexing Cu
4.3 mg L-1 x 4.0 mmol g-1 = 17.2 µmol L-1 (concentration of available sites, [FAu])
1 x 104 = [CuFA] / [Cu2+] [FAu]
[CuFA] = Total Cu2+ - uncomplexed (free) [Cu2+] (u)
1 x 104 = (2.52x10-6 - u) / u x 1.72x10-5
u = 2.15x10-6 mol L-1 (uncomplexed (free) [Cu2+])
[CuFA] = 2.52x10-6 - 2.15x10-6 = 3.7x10-7 mol L-1
[FAu] = 1.72x10-5 - 3.7x10-7 = 1.68x10-5 mol L-1 ( or 98% free)
The equilibrium concentrations of free Cu2+, complexed copper, and of free fulvic acid are: 2.15x10-6 mol
L-1, 3.7x10-7 mol L-1, and 1.68x10-5 mol L-1, respectively
5 Examine Table 13.4, which reports data related to the stability of complexes between various metals
and NTA Comment on the relative values in the table in terms of the environmental classification of
metals Predict the approximate pKf values for nickel(II) and mercury(II) based on that classification
Solution
The ions, such as Ca2+ and Mg2+ (Type A metals) that are found in relatively high concentrations in most natural waters, have relatively small stability constants for complex formation with NTA On the other hand, metal ions such as Cu2+, Zn2+ and Pb2+ (Borderline metals) are of concern because they may be toxic
at elevated levels and they have larger stability constants for complex formation with NTA The trend of increasing stability follows the diagonal arrow on Figure 13.2, which is why Mn2+ (a Borderline metal) is not that much different than Mg2+ (a Type A metal) Following this trend, the stability constant for Ni2+ will be greater than that for Zn2+ and less than, but close to that for Cu2+ The log Kf for Ni2+ is predicted to
be near 12 Hg2+ is a Type B metal and is found just slightly beyond Cu2+ but not as far as Fe3+ along the diagonal arrow The log Kf for Hg2+ is predicted to be around 13.5
Trang 66 In lake water containing 0.9 mmol L–1 calcium and 12 µg L–1 fulvic acid, determine the fraction of the fulvic acid that is bound to calcium, assuming that this is the only metal present in a significant concentration The pH of the water is 5.0
Solution
Lake water: 0.9 mmol L-1 Ca2+ and 12 µgL-1 fulvic acid (FA), pH = 5
The conditional stability constant for Ca2+ and FA reported in Table 13.3 at pH 5 is:
[CaFA]
Kf = 1.2 x 103 = - for Ca2+ + FA CaFA
[Ca2+][FA]
Use a value of 5.0 mmolg-1 for the determination of binding sites on the FA (see Example 13.2) This leads
to a concentration of potential binding sites [FA] being:
0.012 mgL-1 x 5.0 mmol g-1 = 0.06 µmolL-1 = 6.0 x 10-8 M
and Ca2+ = 9 x 10-4 M
at equilibrium:
[CaFA]
1.2 x 103 = -
(9 x 10-4 M - [CaFA] ) (6 x 10-8 M - [CaFA])
assume [Ca2+] >>> [CaFA]
solve: [CaFA] = 3.1 x 10-8 M
The fraction of fulvic acid bound to calcium is then given by:
[CaFA] 3.1 x 10-8 M
- = - = 0.52
[FA] 6 x 10-8 M
7 Note the electronic structure of cadmium(II) and indicate the group (type A, borderline, or type B) in
which it should be placed Predict the principal inorganic forms of the element in fresh and sea water under both oxidizing and reducing conditions
Solution
The electronic structure of cadmium is: [Kr]4d105s2
and that of Cd2+ is: [Kr] 4d10
Cd in the Periodic table is in the same group (12 or 2B) as Zn and Hg Zn2+ is a borderline metal ion, while
Hg2+ is a type B metal ion Cd would have a similar ionic index as these two species, but would have a covalent index that falls between Zn2+ and Hg2+ The ion would then likely fall in a category close to the
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line separating borderline and type B metal ions One would also expect the toxicity to be intermediate between that of Zn and Hg
Inorganic forms (note: the only common oxidation states for Cd are 0 and +2)
Fresh water Sea water
pH = 4 – 6.5 pH = 7 – 9
oxidizing reducing oxidizing reducing
Cd2+ CdS Cd(OH)2 CdS
CdCl4 2-Cd(OH)Cl CdHCO3
8 Artificial and natural wetlands are frequently used as catchment basins for urban run-off and storm
water The sediments in these basins have some ability to extract and retain soluble metals from the water as it flows through the pond For lead, cadmium, and zinc, use information in this chapter and elsewhere to predict the affinity of each of these metals for the carbonate, the iron oxide, and the organic matter fractions of sedimentary material suspended and settled in the pond
Solution
Each of these metals will usually be found in the +2 oxidation state in an aqueous solution They all fall within the borderline category, but the relative type B character increases in the order Zn2+ < Cd2+ < Pb2+ Because borderline metal ions form complexes with a variety of ligand types, it would be expected that they would react with carbonate species (through an oxygen atom) Under near neutral pH conditions, the common form of carbonate would be the hydrogen carbonate ion (HCO3-) More stable complexes could form with ligands having sulfur or nitrogen donor atoms Such ligands are present on the insoluble organic matter fraction of pond sediments Binding to this sedimentary material is a means by which the metals can
be removed from the water column in a natural or constructed wetland Sedimentary iron(III) oxide can also form covalent bonds with metals thus remove them from solution