Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 9

The hydrosphere

PROBLEMS/SOLUTIONS
1.

Use Fig. 1.4 to estimate the residence time of water in the oceans and on the continents. In both cases,
indicate limitations in interpreting these results. Keeping in mind the limitations, comment on the
significance of residence time values in the three compartments of the environment.

Solution
Refer to Figure 1.4 (the water cycle) and Example 1.1 in Chapter 1.
Oceans
Residence time,  = steady state amount of water in oceans  flux in (or flux out)
Calculation with flux in:
flux in

= 3.5x1016 kg y-1 + 1.2x1016 kg y-1 + 3.86x1017 kg y-1
(lakes and rivers)
(pore water)
(precipitation)
= 4.33x1017 kg y-1

The steady state amount of water in the oceans is given as 1.37x1021 kg, so


= 1.37x1021 kg  4.33x1017 kg y-1 = 3164 y

Calculation with flux out:

Flux out

= 4.23x1017 kg y-1

( + biosphere uptake - variable, unknown)

= 4.23x1017 kg y-1


= 1.37x1021 kg  4.23x1017 kg y-1 = 3238 y

The average residence time for water in the oceans is approximately 3200 y.
Continents
Residence time,  = steady state amount of water associated with continents  flux in (or flux out)
Calculation with flux in:
flux in

= 1.1x1017 kg y-1
(precipitation)

The steady state amount of water associated with the continents is
1.27x1017 kg + 4.2x1018 kg + 5.3x1018 kg + 2.9x1019 kg = 3.9x1019
(lakes & rivers) (pore and groundwater)
(ice)


= 3.9x1019 kg  1.1x1017 kg y-1 = 355 y

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 9

The hydrosphere

Calculation with flux out:
Flux out = 3.5x1016 kg y-1 + 1.2x1016 kg y-1 + 7.3x1016 kg y-1 (+ biosphere uptake, variable)
= 1.2x1017 kg y-1


= 3.9x1019 kg  1.2x1017 kg y-1 = 325 y

The average residence time for water in association with the continents is approximately 330 y.
Limitations of the calculated results arise due to the incomplete information with respect to the biological
considerations and our inability to use exact data (for inputs, outputs and steady state amounts) for such
large and complex heterogeneous systems. As well, with the recent warming trend for the average Earth
temperature and subsequent melting of the Antarctic ice sheet, a considerable amount of water (flux in) has
also been omitted in the above calculation. The values calculated are averages and in specific geographical
areas, the cycling of the water would occur much faster or slower depending on the specific environmental
factors.
There is a significant degree of difference of water retention in the three compartment of the environment.
The atmosphere cycles water very quickly, on the order of days, while the continents on average are in the
low hundreds of days and the oceans for thousands of days.
2.

Determine the water availability (in m3 y-1 pc) in your country using websites such as,
/>accessed
November, 2016, and compare it with values for other countries as are shown on Figure 9.3.


Solution
There will be many different responses depending on the country chosen, and a number of countries data
are provided for comparison.
Country
water availability
Country
water availability
cubic meters
cubic meters
Canada
94 353
Brazil
48 314
United Kingdom
2465
Ireland
13 673
Guyana
316 689
Iraq
3287
New Zealand
86 554
Nigeria
2514
India
1880
Poland
1596
The data from this website generally appear to be in good agreement with the data on Figure 9.3. The

variation of the results of water availability is striking. Very populated countries like India have relatively
low amounts of water available compared to less populated countries like Canada, which also has an
abundant of fresh water resources.

3.

The concentration of titanium in the South Pacific ocean, near the island nation of Fiji, has been
found to be approximately 3.0 × 10 –9 mol L–1. Calculate the concentration in ppm, ppb, or ppt as
appropriate.

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 9

The hydrosphere

Solution
The density of ocean water is 1.025 kg L-1 (at 15C). Assume a 1.00 L volume at 15C.
There are 3.0x10-9 moles of Ti (M.W. = 47.88 g mol-1) in 1.00 L, or in 1.025 kg of sea water.
3.0x10-9 mol x 47.88 g mol-1 = 1.4364x10-7 g
or ( x 1000 mg g-1)
or ( x 1000 g mg-1)
or ( x 1000 ng g-1)

=
1.4364x10-4 mg
=
1.4364x10-1 g

= 143.64 ng

of Ti per 1.025 kg sea water
of Ti per 1.025 kg sea water
“
“

The most appropriate unit appears to be ng kg-1.
143.64 ng  1.025 kg = 140.1 ng kg-1
The concentration of titanium in the South Pacific is 140 ppt (ng kg -1) which is a part per trillion by weight.
4.

The concentration of gold in the oceans averages approximately 2 × 10 –11 mol L–1. Calculate the total
mass in tonnes.

Solution
[Au] = 2 x 10-11 mol L-1 (in the oceans), assume average T of ocean water = 15ºC.
From Appendix A.1, the volume of the oceans is given as 1.35 x 1018 m3 = 1.35 x 1021 L
The total amount of gold in the oceans is then:
1.35 x 1021 L x 2 x 10-11 mol L-1 = 2.7 x 1010 mol
The atomic mass of gold is 196.97 g mol-1
The mass of gold is:
or in tonnes is:

2.7 x 1010 mol x 196.97 g mol-1 = 5.3 x 1012 g

5.3 x 1012 g ÷ 106 g t-1 = 5.3 x 106 t

The total mass of gold in the oceans is estimated to be 5.3 million tonnes.
5.


The principal anion and cation in Lake Huron, one of the Great Lakes in North America, are,
respectively, hydrogen carbonate and calcium. The concentration of the former is approximately
1.05 mmol L–1. Calculate the mass of solid calcium carbonate that would remain if 250 mL of Lake
Huron water is evaporated to dryness.

Solution
CaCO3 (F.W. 100.087 g mol-1)
Using the net ionic charge balance of the lake water, and considering only the principal anions and cations,
we can assume that the [Ca2+] is one half the [HCO3-] (see Section 11.2).

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 9

The hydrosphere

[HCO3-] = 2 x [Ca2+] = 1.05 mmol L-1

i.e.

[Ca2+] = 1.05 mmol L-1  2 = 0.525 mmol L-1

then

Processes that will occur during evaporation include:
HCO3-  H+ + CO32-


CO32- + Ca2+  CaCO3 (s)

and

The limiting reagent for these processes is the available amount of Ca 2+.
0.525 mmol L-1 x 0.250 L = 0.13125 mmol of Ca2+
Therefore there is a potential for 0.13125 mmol of CaCO3 to form.
0.13125 mmol x 100.087 g mol-1 = 13.1 mg
The amount of solid CaCO3 that will remain after 250 mL of lake water is evaporated is 13.1 mg.
6.

The value of Kw is 0.67 × 10–14 at 20ºC, 1.01 × 10–14 at 25ºC, and 1.45 × 10–14 at 30ºC. Calculate the
value of Kw at 10ºC, and determine the pH of pure water at that temperature.

Solution
Kw = 0.67 x 10-14 at 20ºC, Kw = 1.01 x 10-14 at 25ºC, Kw = 1.45 x 10-14 at 30ºC
The reaction being considered is:

2 H2O ↔ H3O+ + OH-

ΔHºrxn = [ΔHf°(H3O+) + ΔHºf(OH-)] – [2 x ΔHºf(H2O)]
ΔHºrxn = (–285.83 + –229.99) – (2 x –285.83)
ΔHºrxn = 55.84 kJ mol-1 (This value will be compared to the slope, obtained later in solution)
Sºrxn = [S°(H3O+) + S°(OH-)] – [2 x S°(H2O)]
Sºrxn = (69.91 + 10.75) – (2 x 69.91)
Sºrxn = –59.16 J mol-1 K -1 (This value will be compared to the intercept)
ΔGº = –RTlnKw
Which rearranges to:
Further rearrange to give:


&

ΔGº = ΔHºrxn – TΔS°rxn

–RTlnKw = ΔHºrxn – TΔS°rxn
RlnKw = ΔHº / T + ΔSº
y
= mx + b
(Note: m = –ΔHº, b = ΔSº , x = 1/T))
plot

RlnK vs. 1/T

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 9

The hydrosphere

-264
-265
-266

RlnK

-267
-268
-269

-270
-271
-272
0.0033
0.00332
0.00334
0.00336
0.00338
0.0034
0.00342
1/T

20°C 293 K
25°C 298 K
30°C 303 K

1/T = 0.0034129 K-1
1/T = 0.0033557 K-1
1/T = 0.0033003 K-1

Kw = 0.67 x 10-14
Kw = 1.01 x 10-14
Kw = 1.45 x 10-14

m = –57 030 J mol-1

Rln Kw = -271.4 J K-1 mol-1
Rln Kw = -268.0 J K-1 mol-1
Rln Kw = -264.9 J K-1 mol-1


b = –76.68 J K-1 mol-1

Now calculate Kw for T = 283 K.
Note that the values for the slope and intercept agree reasonably well with the values calculated from the
thermodynamic data from Appendix B.2.
The regression line for this plot is
RlnKw = –57 030x – 76.68
1/283 = 0.003534 K-1 = x
Solve for Kw

lnKw = –33.46
Kw = 0.2936 x 10-14

The calculated value of Kw at 283 K is 0.29 x 10-14.
A tabulated value for Kw at 10ºC is 0.29 x 10-14 (Skoog, West, Holler, 7thed.)
Calculation of the pH of pure water at 0ºC :
Kw = 0.29 x 10-14 = [H+][OH-]
[H+] = 5.42 x 10-8

pH = 7.27

where [H+] = [OH-]

A tabulated pH value for water at 10ºC is 7.3.

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 9


The hydrosphere

The Kw of pure water at 10ºC is 0.29 x 10-14 and its pH is calculated to be 7.3.
7.

In a particular fresh water sample, the concentrations of cations and anions are (in μmol L –1):
Na+
K

Cl–

33

+

120

NO 3–

4

Mg2+

31

HCO3–

Ca2+


160

CO32 –

13
270
0.67

SO24 –

11

Compare the concentration of total positive and negative charge in the solution. Assume that the
difference is due to hydronium or hydroxyl ion, and calculate the pH.
Solution
Assume the use of a 1.00 L volume so the values above can be considered to be µmol quantities of each
ion. (eq = equivalence)
total positive charge

= 1 eq x Na+ + 1 eq x K+ + 2 eq x Mg2+ + 2 eq x Ca2+
= 33 µmol + 4 µmol + 62 µmol + 320 µmol = 419 µmol

total negative charge

= 1 eq x Cl- + 1 eq x NO3- + 1 eq x HCO3- + 2 eq x CO32+ 2 eq x SO42= 120 µmol + 13 µmol + 270 µmol + 1.34 µmol + 22 µmol
= 426 µmol

The net charge for the solution using the above data is
426 µmol (– ve) – 419 µmol (+ ve) = 7 µmol (excess negative charge)
Therefore, in order for this solution to be neutral, there must also be an amount of H+ equal to 7 x 10-6 mol

L-1.
pH = –log[H+] = –log (7 x 10-6) = 5.15
The pH of the solution is approximately 5.2.
8.

The major ions and their concentrations (mmol L–1) in sea water are:
Na+
2+

470

Mg

53

Cl–

547

Br–

1

K+
Ca

10
2+

10


SO24 –

28

HCO3–  CO32 –

X

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 9

The hydrosphere

Assume that charges of these species balance and calculate the total concentration of negative charge
associated with the two carbonate species. With a pH of 8.2, calculate the concentrations of the two
individual carbonate species.
Solution
Assume 1.00 L of solution and the charge balance relationships described in the text (Section 11.2)
(eq = equivalence)
total positive charge

= 1 eq x Na+ + 1 eq x K+ + 2 eq x Mg2+ + 2 eq x Ca2+
= 470 mmol + 10 mmol + 106 mmol + 20 mmol
= 606 mmol

total negative charge


= 1 eq x Cl- + 1 eq x Br- + 2 eq x SO42- + Carbonate species
= 547 mmol + 1 mmol + 56 mmol + Carbonate species
= 604 mmol + Carbonate species
= 606 mmol (must equal total positive charge)

(Note: The contribution to the charge balance by OH- is negligible and ignored)
Therefore the combined carbonate species contribute a negative charge of 2 mmol L-1.
at pH = 8.2
carbonic acid :

[H+] = 6.31 x 10-9

Ka1 = 4.5 x 10-7

Ka2 = 4.7 x 10-11

Examination of Figure 1.2 (Chapter 1) shows that at pH 8.2 only a small amount of carbonic acid will
exist as the fully deprotonated species or as the fully protonated species. Most of the carbonate will exist as
HCO3-.
1)

Ka1

[H+][HCO3-]
= ----------------[CO2]

2)

Ka2 =


[H+][CO32-]
---------------[HCO3-]

the total charge of all carbonate species concentration is
3)

[HCO3- (aq)] + 2 [CO32- (aq)] = 2 x 10-3 mol L-1

There are 3 equations with three unknown values. Rearrange by substitution into equation 3 to give:
-

[HCO3 ]

2 Ka2 [HCO3-]
+ --------------= 2 x 10-3
+
[H ]

Plug in known values of [H+], Ka1 and Ka2 and solve for HCO3-.
[HCO3-]

= 1.97 x 10-3 mol L-1

As expected it accounts for most of the carbonate species present.

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Solutions: Environmental Chemistry - a global perspective 4th Edition

Chapter 9

The hydrosphere

Now, using equation 2 solve for [CO32-].

Ka2 =

[H+][CO32-]
---------------- , or
[HCO3-]

[CO32-] = 1.47 x 10-5 mol L-1

The concentration of the two carbonate species at pH 8.2 are approximately 2 mmol and 0.01 mmol for
HCO3- and CO32- respectively, as would be expected based on Figure 1.2.
9.

In the open oceans the concentration of iron is approximately 1 × 10 –4 ppm in the surface water and 4
× 10–4 ppm in the deep ocean. Corresponding values for concentration of aluminium are 9.7 × 10 –4
ppm and 5.2 × 10–4 ppm. Why are the concentrations so low? Why is the ratio surface concentration:
deep concentration < 1 for iron and > 1 for aluminium?

Solution
Why are the concentrations of iron and aluminium in the oceans so low?
Why are the ratios (surface to deep) opposite?
The concentrations of iron and aluminium are low because both of these metals have hydrous oxides with
very low Ksp values, resulting in extremely low solubility at the ocean water pH. Complexation by
inorganic ions like chloride and sulfate will increase the solubility somewhat. Greater increases could
occur where there is a large concentration of appropriate organic ligands.

The relatively high surface-water concentrations of aluminium are due in part to the input from
atmospheric sources, and to organic matter in the near surface water. Lower concentrations at depth result
from the removal of aluminium due to processes such as adsorption onto 'sinking' siliceous shells and
increasing distance from a source.
Subsurface waters are under-saturated with respect to oxygen. Under these conditions metals are reduced
so iron would tend to exist in the more soluble Fe2+ form, rather than the more insoluble Fe3+ that would
exist near the surface. Atmospheric sources would again be the primary route for iron to be introduced to
the open ocean. As deposited material settles the insoluble form of iron will have a chance to dissolve.
Aluminium has no similar oxidation change that affects its solubility.
As a result the ratios of ‘surface concentrations’ to ‘depth concentrations’ show opposite trends. For
aluminium the ratio is > 1 and for iron the ratio is < 1.

10. Manganese may precipitate as MnCO3 from aqueous solution according to the reaction

Mn2 (aq)  CO2 (aq)  3H2O  MnCO3 (s)  2H3O (aq)
The solubility product, Ksp, for manganese (II) carbonate is 5.0 × 10 –10. Use this value and the
equilibrium constant values for the carbonate system to determine the minimum pH required to

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 9

The hydrosphere

precipitate manganese (II) carbonate from a solution which contains 1.0 × 10 –3 mol L–1 manganese (II)
ion. Assume that aqueous carbon dioxide is in equilibrium with atmospheric carbon dioxide.
Solution
Mn2+ (aq) + CO2 (aq) + 3H2O (l) → MnCO3 (s) + 2H3O+ (aq)

Ksp = [Mn2+][CO32-] = 5x10-10
CO2 (aq) + H2O (l) ↔ H2CO3 (aq)

Ka1

H2CO3 (aq)

↔

H+ + HCO3-

HCO3- (aq)

↔

H+ + CO32-

[H+][HCO3-]
= ----------------- = 4.5 x 10-7
[H2CO3]

Ka2 =

[H+][CO32-]
---------------- = 4.7 x 10-11
[HCO3-]

If Mn2+ = 1 x 10-3 mol L-1 then the minimum amount of CO32- required to precipitate the Mn2+ is
5.0 x 10-10 = 1 x 10-3 x [CO32-]
[CO32-] = 5 x 10-7 mol L-1

Assume the atmosphere carbon dioxide concentration is 402 ppmv.
Using Henry’s Law and CO2 constant (KH = 3.3 x 10-7 mol L-1 Pa-1)
Pg x KH = [CO2 (aq)] (assume total pressure is 1 atm = 101 325 Pa)
402 ppmv = 0.0402 % = 0.000402 atm = 40.7 Pa
[CO2 (aq)] = 40.7 Pa x 3.3 x 10-7 mol L-1 Pa-1 = 1.34 x 10-5 mol L-1
The Ka1 value takes into account the conversion to H2CO3 (aq) (from CO2 (aq)) and the first deprotonation
step.
Assume

[CO2 (aq)] = [H2CO3 (aq)] = 1.34 x 10-5 mol L-1

Summing the first and second acid dissociation expressions, leads to:
Ka1 Ka2[H2CO3 (aq)]
---------------------------- = [H+]2
[CO32-]
Where

[H+] = 2.38 x 10-8

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 9

And

The hydrosphere

pH = 7.6


The minimum pH required to precipitate the manganese carbonate is approximately pH 7.6 under the
conditions described.
11.

The following are controlling factors for the ‘availability’ of different elements:


oxygen availability for iron;



sulfide concentration for zinc;



solution pH for chromium and silicon;



carbonate concentration for calcium;



sorption factors for copper.

Explain the chemical and environmental significance of these factors.
Solution
In an oxygen enriched environment iron will primarily exist in the oxidized form of Fe 3+. In low oxygen
environments, iron can exist as the more soluble iron form of Fe2+.

Of all the environmental ligands, sulfide forms particularly strong and insoluble complexes with zinc.
Therefore, in a sulfide-containing environment, zinc will be essentially insoluble and unavailable due to
this ligand metal interaction.
Chromium(III) is relatively immobile and unavailable due to its strong association with various solid
materials. At higher pHs it oxidizes more readily to chromate, which is more soluble and available (and
also toxic). SiO2 is only slightly soluble at low pH values, but when the pH exceeds 9, the solubility begins
to increase significantly as deprotonated species of silicic acid are formed.
Carbonate species are common in aqueous environments, being derived in part from carbon dioxide in the
air. Calcium carbonate is insoluble. The presence of carbonate species particularly when the pH is high
enough that a significant fraction is in the form of CO 32-, will cause calcium carbonate (calcite, limestone)
to precipitate from calcium-containing solution.
In the terrestrial and aquatic environments, copper is often found in association with organic matter.
Therefore, in the presence of insoluble humic material, much of the copper may be complexed via the
functional groups of the organic matter. This sorption removes the copper from solution.

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