Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 13

Metals and Semi-metals in the hydrosphere

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

Ka2 =

Zn(OH)2 + H+
[Zn(OH)2][H+]
---------------------- = 1.26 x 10-8
[ZnOH+]
Zn(OH)3- + H+

3) Zn(OH)2 + H2O

[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)3- = [Zn(OH)3-] ÷ CZn,

and

αZn(OH)2 = [Zn(OH)2] ÷ CZn,

αZn(OH)42- = [Zn(OH)42-] ÷ CZn


Rewrite the above equations for each species in terms of [H+]
a) From 5 (and using Kw)
b) From 1, with substitution

[Zn2+] = 3 x 1012[H+]2
[ZnOH +] = 3 x 1012Ka1[H+]

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 13

Metal and semi-metals in the hydrosphere

c) From 2, with substitution

[Zn(OH) 2] = 3 x 1012Ka1Ka2

d) From 3, with substitution

3 x 1012Ka1Ka2Ka3
[Zn(OH) 3 ] = -----------------------[H+]

e) From 4, with substitution

3 x 1012Ka1Ka2Ka3Ka4
[Zn(OH) 4 ] = -----------------------------[H+]2

-


2-

Table of Results: (solved for the [H+] given)

pH =1
pH =2
pH =3
pH =4
pH =5
pH =6
pH =7
pH =8
pH =9
pH =10
pH =11
pH =12
pH =13
pH =14

[Zn2+]

[ZnOH+]

3x1010
3x108
3x106
3x104
3x102
3x100

3x10-2
3x10-4
3x10-6
3x10-8
3x10-10
3x10-12
3x10-14
3x10-16

7.5x101
7.5x100
7.5x10-1
7.5x10-2
7.5x10-3
7.5x10-4
7.5x10-5
7.5x10-6
7.5x10-7
7.5x10-8
7.5x10-9
7.5x10-10
7.5x10-11
7.5x10-12

[Zn(OH)2] [Zn(OH)3-] [Zn(OH)42-]
9.5x10-6
9.5x10-6
9.5x10-6
9.5x10-6
9.5x10-6

9.5x10-6
9.5x10-6
9.5x10-6
9.5x10-6
9.5x10-6
9.5x10-6
9.5x10-6
9.5x10-6
9.5x10-6

4.8x10-16
4.8x10-15
4.8x10-14
4.8x10-13
4.8x10-12
4.8x10-11
4.8x10-10
4.8x10-9
4.8x10-8
4.8x10-7
4.8x10-6
4.8x10-5
4.8x10-4
4.8x10-3

2.4x10-27
2.4x10-25
2.4x10-23
2.4x10-21
2.4x10-19

2.4x10-17
2.4x10-15
2.4x10-13
2.4x10-11
2.4x10-9
2.4x10-7
2.4x10-5
2.4x10-3
2.4x10-1

CZn
3.0x1010
3.0x108
3.0x106
3.0x104
3.0x102
3.0
3.0x10-2
3.2x10-4
1.3x10-5
1.0x10-5
1.4x10-5
8.1x10-3
2.9x10-3
2.4x10-1

Graph of  vs. pH for the zinc aquo complexes

Zn(OH)2


1.20


1.00

Zn2+

Zn(OH)42-

0.80
0.60

Zn(OH)3-

ZnOH+

0.40
0.20

pH
117

0.00
-0.20 0

5

10

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 13

Metals and Semi-metals in the hydrosphere

Individual α results for graph at specified pH value

pH
pH
pH
pH
pH
pH
pH
pH
pH
pH
pH
pH
pH
pH

1
2
3
4
5
6

7
8
9
10
11
12
13
14

[Zn2+] [ZnOH+] [Zn(OH)2] [Zn(OH)3-]
1.00 0.00
0.00
0.00
1.00 0.00
0.00
0.00
1.00 0.00
0.00
0.00
1.00 0.00
0.00
0.00
1.00 0.00
0.00
0.00
1.00 0.00
0.00
0.00
1.00 0.00
0.00

0.00
0.95 0.02
0.03
0.00
0.23 0.06
0.71
0.00
0.00 0.01
0.94
0.05
0.00 0.00
0.65
0.33
0.00 0.00
0.12
0.59
0.00 0.00
0.00
0.17
0.00 0.00
0.00
0.02

[Zn(OH)42-]
0.00
0.00
0.00
0.00
0.00
0.00

0.00
0.00
0.00
0.00
0.02
0.29
0.83
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.

Ka1

[PbOH+][H+]
= -----------------[Pb2+]


Kf1

[PbOH+]
= ----------------[Pb2+][OH-]

[PbOH+]
2.0 x 10 = -------------------[Pb2+] x Kw/[H+]
6

Ka2

[Pb(OH) 2][H+]
= -------------------[PbOH +]

Kw
[OH-] = --------[H +]
(Kw = 1 x 10-14)

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 13

Metal and semi-metals in the hydrosphere

6

2.0 x 10 x 1 x 10


-14

[PbOH+] [H+]
= -------------------- = 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 Pb 2+ ion and the first
deprotonated (PbOH+) species.

using

Ka1

[PbOH+] [H+]
= -------------------- = 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

 0.5



NTA3





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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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 13

Metals and Semi-metals in the hydrosphere

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

K f´ = 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 Mn 2+ (a Borderline metal) is
not that much different than Mg2+ (a Type A metal). Following this trend, the stability constant for Ni 2+
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 Cu 2+ but not as far as Fe3+ along the
diagonal arrow. The log Kf for Hg2+ is predicted to be around 13.5.

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 13
6.

Metal and semi-metals in the hydrosphere

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 µg L-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 = ---------------[Ca2+][FA]


for

Ca2+ + FA

CaFA

Use a value of 5.0 mmol g-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 mg L-1 x 5.0 mmol g-1 = 0.06 µmol L-1 = 6.0 x 10-8 M
Ca2+ = 9 x 10-4 M

and
at equilibrium:

[CaFA]
1.2 x 103 = -----------------------------------------------------(9 x 10-4 M - [CaFA] ) (6 x 10-8 M - [CaFA])
assume [Ca2+] >>> [CaFA]
[CaFA] = 3.1 x 10 -8 M

solve:

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]4d 105s2

and that of Cd2+ is:

[Kr] 4d 10

Cd in the Periodic table is in the same group (12 or 2B) as Zn and Hg. Zn 2+ 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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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 13

Metals and Semi-metals in the hydrosphere

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
oxidizing
Cd2+
CdOH+
CdCl+
CdO
CdHCO3+

8.

pH = 7 – 9

reducing
CdS
Cd

oxidizing
Cd(OH)2
Cd2+
CdCl+
CdCl2
CdCl3CdCl42Cd(OH)Cl
CdHCO3+

reducing
CdS
Cd

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 Zn 2+ < 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 (HCO 3-). 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.

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