Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 11

Gases in water

PROBLEMS/SOLUTIONS
1.

Lake Titicaca is situated at an altitude of 3810 m in the Bolivian Andes. Determine the atmospheric
pressure and calculate the solubility of oxygen in the lake at a temperature of 5ºC. (The Henry’s law
constant at 5ºC is 1.9 × 10–8 mol L–1 Pa–1)
Solution
Since we are given Henry's law constant, this suggests we need to calculate a partial pressure for oxygen,
which will then allow for the calculation of the concentration of oxygen in mol L -1. But first we need a
total pressure for the altitude given. Here we will assume the mixing ratio for oxygen remains constant (see
Chapter 2, Equation 2.3) with increasing altitude.

The total pressure at 3810 m (Ph) is:

or

Ph  P o e  Ma g h / R T

P3810 = 101 325 e-(0.029

x 9.81 x 3810) / (8.315 x 278)

P3810 = 63 394 Pa
Ph = 0.626 atm

(1 atm = 101 325 Pa)

Assume the mixing ratio of O2 is constant at 0.2095 (Chapter 2, Table 2.1a)
At 5ºC the vapour pressure of water is approximately 870 Pa (Chapter 9, Table 9.3).
63 394 Pa - 870 Pa = 62 524 Pa (dry atmosphere)
62 524 Pa x 0.2095 = 13 099 Pa (partial pressure of O2)
The partial pressure of oxygen is then 13 099 Pa at 3810 m.
To calculate the oxygen concentration use Henry’s Law;
[Gl] = KH x Pg
[O2] = 1.9 x 10-8 mol L-1 Pa-1 x 13 099 Pa
[O2] = 2.5 x 10-4 mol L-1

(or 8.0 mg L-1)

The solubility of oxygen at 3810 m and 5ºC is 8.0 mg L-1. This compares to a value of 12.8 mg L-1 at sea
level and 5ºC.

2.

The solubility of oxygen in water at 25ºC is approximately 8.5 mg L –1. At Pº and the same temperature,
what is the volume of oxygen gas occupied by 8.5 mg of oxygen?

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Gases in water

Solution

Use PV = nRT (at P, P = 1.00 atm, O2 mixing ratio is 20.95% giving an oxygen pressure of 0.2095 atm, T
= 298 K, R = 0.08206 L atm mol-1 K-1).
The mass of oxygen being considered is 8.5 mg, which is converted to moles.
8.5 mg  1000 mg g-1 = 0.0085 g
n = 0.0085 g  32.0 g mol-1 = 0.000265625 mol O2
The volume of O2 can be calculated using the ideal gas law:
V = nRT/P = 0.000265625 mol x 0.08206 L atm mol -1 K-1 x 298 K / 0.2095 atm
V = 0.006496 L atm / 0.2095 atm
V = 0.0310 L (multiply by 1000 mL L-1 to convert to mL)
The volume of gas from 8.5 mg of O2 (g) at P and 25C is 31 mL.
3.

At 30ºC, the solubility of oxygen in water is 7.5 mg L–1. Consider a water body at that temperature
containing 7.0 mg L–1 oxygen. By photosynthesis, 1.5 mg of carbon (as carbon dioxide) is converted to
organic biomass ({CH2O}) during a single hot day. Is the amount of oxygen produced at the same time
sufficient to exceed its aqueous solubility?

Solution
The process under consideration is
hν
CO2 (g) + H2O (l) → {CH2O} + O2 (aq)
1.5 mg of C (as CO2) gives:
0.0015 g ÷ 12.011 g mol-1 = 0.000125 mol C (or CO2, or O2)
0.000125 mol x 32 000 mg mol-1 = 4.0 mg (O2)
The amount of O2 produced from 1.5 mg of C is 4.0 mg. Starting from the original concentration of 7.0 mg
L-1 O2, the quantity of oxygen produced would only require 8 L of water to reach saturation. Any larger
volume would mean that the water would not be saturated. Therefore, the answer to the question requires
knowledge of the volume of water under consideration. The amount of oxygen produced at the same time
is not sufficient to exceed its aqueous solubility, assuming that the water body was larger than 8 L.
4.


Thermal power plants, whether powered by fossil fuel or nuclear energy, discharge large quantities of
cooling water to a lake or river. Discuss the meaning of thermal pollution in the context of this chapter.

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Chapter 11

Gases in water

Solution
Increasing the temperature of the water (thermal pollution) by discharge of heat through cooling water will
cause gases to be less soluble. We saw for oxygen that the range of temperatures from 5ºC to 30ºC caused
its solubility to range from 12.4 mg L-1 to 7.5 mg L-1 (Section 11.1). At the higher temperature the
decreased amount of oxygen could have an impact on the water quality in terms of supporting fish life. The
heat added through the cooling water is thermal pollution that is causing the death of fish, the same way a
chemical toxin added to water could also cause an immediate response by killing fish.
5.

The estimated atmospheric carbon dioxide concentration in the Northern Hemisphere in 1950 was 310
ppmv. It may be predicted with some certainty that the concentration in 2020 will be 406 ppmv.
Calculate the pH of pure rain that would be in equilibrium with the carbon dioxide in each of the 2
years cited, and comment on the contribution that carbon dioxide makes towards precipitation acidity.

Solution
Assume total pressure is Pº = 101 325 Pa and temperature is 298 K.
The mixing ratios given (in units of ppmv) are also a mole fraction (x 10 6), and therefore, the partial
pressure of carbon dioxide is given by:

310 ppmv CO2 = 310 x 10-6 x 101 325 Pa = 31.4 Pa
Henry’s Law constant for CO2 is 3.3 x 10-7 mol L-1 Pa-1 (Table 11.1)
Using Henry’s Law :
[G]l = KHPg
[CO2](aq) = 3.3 x 10-7 mol L-1 Pa-1 x 31.4 Pa = 1.04 x 10-5 mol L-1
H2O (l) + CO2 (aq) → H2CO3 (aq)
H2CO3 (aq) ↔

Ka

H+ (aq) + HCO3 (aq)

[HCO3-][H+]
= ----------------- = 4.5 x 10-7
[CO2]

Assume [CO2] = 1.04 x 10-5 mol L-1, and that
Solve for [H+],

-

Ka = 4.5 x 10-7

[H+] = 2.16 x 10-6

Repeat the calculation for 406 ppmv: result is

[HCO3-] = [H+]
pH = 5.67
pH = 5.61


The increase in CO2 concentration in the atmosphere from 310 ppmv to 406 ppmv over the seventy year
period will cause a slight decrease (0.06 pH units) in the pH of pure rainfall. This decrease in pH (increase
in acidity) is negligible compared to the pH changes experienced from nitrogen and sulphur oxides emitted
from anthropogenic sources over this same time period, but the trend is concerning.

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Chapter 11

6.

Gases in water

One of the alternative expressions for Henry’s law when applied to oxygen is

PO2  K H X O2
where

PO2 is the pressure of oxygen in the gas phase and X O2 is its mole fraction in solution.

Calculate the value of Henry’s law constant,

K H , at 25ºC in units of MPa.

Solution
Assume both the air and water temperature are 25ºC and the total pressure is equal to 101 325 Pa.
PO2 = KH’ x XO2

At 25ºC there is 8.5 mg L-1 dissolved oxygen in water.
the number of moles of O2 in 1 L = 2.66 x 10-4, (total moles in 1L ~ 55.5 mol)
XO2 = 2.66 x 10-4 ÷ 55.5 = 4.79 x 10-6
The partial pressure of O2 is:
101 325 Pa - 3200 Pa (correction for moisture, using Figure 8.2 or Table 9.3)
= 98 125 Pa
PO2 = 98 125 x 0.2095 = 20 557 Pa
KH’ = 20 557 Pa ÷ 4.79 x 10-6 = 4.29 x 109 Pa
KH’ = 4290 MPa

or

Henry's Law (for O2) becomes:

7.

(the prefix M = 1 x 10 6, Appendix C.2)

PO2 = 4290 MPa x XO2

If sulfur dioxide (SO2) is bubbled through water, the following reactions take place

If the SO2 is in a gas stream at a concentration of 10.9 ppbv, while the total gas pressure is 1 atm and
the pH of the resulting solution is 4.89, what must the Henry’s law constant be for SO 2 in water at this
temperature?
Solution
10.9 ppbv = 1.09 x 10-8 x 101325 Pa = 1.1 x 10-3 Pa SO2
given [H+] = 1.29 x 10-5 = [HSO3-], assuming only source of H+ is from H2SO3

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1.23 x 10-2 = (1.29 x 10-5)2 / [H2SO3]
[H2SO3] = 1.35 x 10-8 mol L-1
Assume [H2SO3] = [SO2] = 1.35 x 10-8 mol L-1
[SO2] = KH x PSO2
KH = 1.35 x 10-8 mol L-1 / 1.1 x 10-3 Pa
= 1.23 x 10-5 mol L-1 Pa-1
The Henry’s law constant for SO2 in water at this temperature was found to be 1.23 x 10-5 mol L-1 Pa-1
8.

Hydrogen sulfide gas has toxicity comparable to hydrogen cyanide, but because of its intense odour its
presence is observed at concentrations well below toxic levels. A mixing ratio of 100 ppmv can be
lethal to humans over an extended time. What are the equilibrium concentrations of H 2S and HS- in
water in an atmosphere containing that concentration of the gas? What is the solubility of H 2S when
the gas is bubbled through water? Assume 25oC in both calculations.

Solution
Part 1)

100 ppmv = 1.0 x 10-4 x 101325 = 10.1 Pa H2S
[H2S] = 9.9 x 10-7 mol L-1 Pa-1 x 10.1 Pa = 1.0 x 10-5 mol L-1
H2S ↔ HS- + H+
Ka = [HS-][H+]/[H2S]


(Ka = 1.0 x 10-7)
(where [HS-] = [H+])

[HS-] = 1.0 x 10-6 mol L-1
The equilibrium concentrations of [H2S] and [HS-] are 1.0 x 10-5 mol L-1 & 1.0 x 10-6 mol L-1 respectively.
Part 2) When H2S gas is bubbled through the solution, it is assumed the partial pressure of the gas is equal
to the atmospheric pressure.
[H2S] = 9.9 x 10-7 mol L-1 Pa-1 x 101325 Pa = 0.10 mol L-1
The solubility of H2S when it is bubbled through water is significantly greater at 0.10 mol L-1.

9.

The solubility of oxygen in sea water at 15ºC and 35‰ is 7.9 mg kg–1. Calculate the Henry’s law
constant, KH, under these conditions.

Solution
[O2] = KH x PO2

where PO2 = 2.04 x104 Pa (see Example 11.1)

[O2] = 7.9 mg kg-1 x 1.025 kg L-1 = 8.098 mg L-1

[ / 1000 / 32 g mol-1]

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Chapter 11


Gases in water
= 2.53 x 10-4 mol L-1
KH = 2.53 x 10-4 mol L-1 / 2.04 x104 Pa
= 1.24 x 10-8 mol L-1 Pa-1

10. A septic tank leaks into the soil. The organic matter decomposes to produce carbon dioxide. Gas flow
is restricted in the soil so that there is no air exchange with the atmosphere above. The carbon dioxide
develops a pressure of 350 Pa. Calculate the pH of the soil water, assuming no buffering ability.
Solution
Use Henry’s Law to calculate the concentration of carbon dioxide in the pore water.
[CO2]aq = 350 Pa x 3.3 x 10-7 mol L-1 Pa-1 = 1.16 x 10-4 mol L-1
Ka1
Therefore

[HCO3-][H3O+]
= -------------------- = 4.5 x 10-7
[CO2]aq

where [HCO3-] = [H3O+]

[H3O+] = (1.16 x 10-4 mol L-1 x 4.5 x 10-7)½ = 7.22 x 10-6 mol L-1
pH = 5.14

The carbon dioxide pressure of 350 Pa causes the soil water pH (at 5.14) to be significantly more acidic
than the natural level of acidity in water in contact with the atmosphere. Clean rain water typically has a pH
of around 5.6 when in equilibrium with the current level of 41 Pa (402 ppmv) CO2 in the atmosphere.
11. A groundwater sample obtained near a septic tank has a pH of 6.90 and HCO 3- concentration of 8.25
mmol L-1. Calculate the partial pressure of carbon dioxide in the associated soil atmosphere.
Solution
Use Henry’s Law to calculate the pressure of carbon dioxide in the soil atmosphere, after first determining

the concentration of the aqueous carbon dioxide.
pH = 6.90

Ka1

which gives

[H+] = 1.26 x 10-7 and using

[HCO3-][H3O+]
= -------------------- = 4.5 x 10-7
[CO2]aq

where [HCO3-] = 8.25 x10-3

[CO2]aq = 2.31 x 10-3 mol L-1
PCO2 = 2.31 x 10-3 mol L-1 ÷ 3.3 x 10-7 mol L-1 Pa-1 = 7000 Pa
The partial pressure of carbon dioxide in the associated soil atmosphere is 7000 Pa

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Gases in water

12. Methane and carbon dioxide are produced under anaerobic conditions in wetlands by fermentation of
organic matter approximated by the following equation


2{CH2O}  CH 4  CO2
Calculate the total pressure (Pº plus that due to water) at a depth of 5 m. Gas bubbles are evolved at that
pressure and remain in contact with water at the sediment surface long enough so that equilibrium is
attained. Calculate the methane concentration (mol L–1) in the interstitial water at 25ºC.
Solution
Calculate the total pressure at 5 m depth. Use a 1 m2 surface area.
volume = 1 m2 x 5 m = 5 m3 = 5000 L
The mass of water is approximately 5000 kg. Force due to gravity is:
9.81 m s-2 x 5000 kg = 49 050 N
Including the 5 m depth with the 1 m2 area the pressure (P5m) becomes:
49 050 N ÷ 1 m-2 = 49 050 Nm-2 = 49 050 Pa
Total pressure at the 5 m depth is the sum of Pº and P5m:
P5m + Pº = 49 050 + (101 325 –3200) = 147 175 Pa
The gas pressure is a result of the production of two gases (CH 4 and CO2). These are formed in equal
amounts on a molar basis (see reaction above) and the mole fraction of the gases will be equal. Thus the
partial pressure of each gas will be half that of the total pressure calculated.
Therefore

PCH4 = 73 588 Pa

Use Henry’s Law with the value of the Henry’s Law constant at 25ºC taken from Table 11.1.
[CH4 (aq)] = 1.4 x 10-8 mol L-1 Pa-1 x 73 588 Pa = 1.0 x 10-3 mol L-1
The concentration of methane for the conditions described is 1.0 x 10-3 mol L-1.
13. Calculate the alkalinity of water that is in equilibrium with atmospheric carbon dioxide and contains no
additional species except those resulting from that equilibrium.
Solution
Use equation 11.21
Alkalinity = [OH-] + [HCO3-] + 2[CO32-] - [H3O+]
Mixing ratio of CO2 = 402 ppmv
Assume the total pressure is 1 atm (101 325 Pa).


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Chapter 11

Gases in water
101 325 Pa x 402 ppmv x 1 x 10-6 = 40.7 Pa (PCO2)

Using Henry’s Law, the concentration of carbon dioxide in the pore water is:
[CO2]aq = 40.7 Pa x 3.3 x 10-7 mol L-1 Pa-1 = 1.34 x 10-5 mol L-1
Ka1

[HCO3-][H3O+]
= -------------------- = 4.5 x 10-7
[CO2]

[HCO3-] = [H3O+]

Therefore [H3O+] = (1.34 x 10-5 mol L-1 x 4.5 x 10-7)½ = 2.46 x 10-6 mol L-1
pH = -log 2.46 x 10-6 mol L-1 = 5.6

Ka2

[CO32-][H3O+]
= -------------------- = 4.7 x 10-11
[HCO3-]

assume [HCO3-] & [H3O+] don’t change


[CO32-] = 4.7 x 10-11
The pH of this solution is 5.6 and the CO32- concentration will be negligible.
Alkalinity = 4.19 x 10-9 + 2.46 x 10-6 + 2 (4.7 x 10-11) - 2.46 x 10-6
= 4.2 x 10-9 mol L-1
= 0.004 μmol L-1
Simply put, there is essentially no alkalinity arising from atmospheric carbon dioxide dissolving into water
at equilibrium.
14. A sample of water from Lake Huron (one of the Great Lakes in North America) has a pH of 7.34 and
an alkalinity of 1.21 mmol L–1 as calcium carbonate. Assume equilibrium with the calcium carbonate
in the sediment and calculate the concentration of calcium ion (in mg L–1) in the water (25ºC). What
assumptions are made in carrying out the calculation?
Solution
Lake Huron
pH = 7.34, [H+] = 4.57 x 10-8, [OH-] = 2.19 x 10-7,
alkalinity = 1.21 mmol L-1 (as CaCO3, M.W. = 100 g mol-1)
1.21 x 10-3 mol L-1 CaCO3 means 1.21 x 10-3 mol L-1 of CO32- or an equivalent amount of proton
neutralizing capacity of 2.42 x 10-3 mol L-1.
Alkalinity = [HCO3-] + 2[CO32-] + [OH-] – [H+]
At pH 7.34 the contribution to the calculated alkalinity by OH- and H+ ions is negligible. Thus the
alkalinity could be attributed to the two carbonate anions.
2.42 x 10-3 mol L-1 = [HCO3-] + 2[CO32-]
However, at pH 7.34, the only species of any significance is HCO 3-. (see Chapter 1, Figure 1.2)

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Chapter 11


Gases in water

Assume that 2.42 x 10-3 mol L-1 = [HCO3-]
Using

Ka2

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

[CO32-] = 2.49 x 10-6 mol L-1
Ksp = [Ca2+][CO32-] = 5 x 10-9,

[Ca2+] = 2.01 x 10-3 mol L-1 (x 40.0 g mol-1)

Which yield the value 80.4 mg L-1 for the calcium ion (Ca2+) concentration.

15. In a soil atmosphere, the mixing ratio of carbon dioxide is often much higher than that in the normal
atmosphere due to respiration (release of carbon dioxide) by microorganisms. If the mixing ratio of
carbon dioxide in the air-occupying soil pores were 5000 ppmv, calculate the pH of the associated soil
solution (25ºC), assuming no other sources of proton donors or acceptors.
Solution
The mixing ratio of CO2 = 5000 ppmv. What is the resulting pH of the solution in contact with this air?
Assume the total pressure of air in the soil pores is the same as the atmosphere which we will assume is 1.0
atm (101 325 Pa).
101 325 Pa x 5000 ppmv x 1 x 10 -6 = 506.6 Pa (PCO2)
Using Henry’s Law, the concentration of carbon dioxide in the pore water is:
[CO2]aq = 506.6 Pa x 3.3 x 10-7 mol L-1 Pa-1 = 1.67 x 10-4 mol L-1
Ka1


[HCO3-][H3O+]
= -------------------- = 4.5 x 10-7
[CO2]

[HCO3-] = [H3O+]

[H3O+] = (1.67 x 10-4 mol L-1 x 4.5 x 10-7)½ = 8.67 x 10-6 mol L-1
pH = 5.06
This compares to the pH of approximately 5.6 for water in equilibrium with the natural level of 402 ppmv
CO2 current in the atmosphere.
16. At a depth of 200 m in Lake Nyos, what will be the pressure of the gas, carbon dioxide, as it is released
into the water? Use Henry’s law to calculate the solubility of carbon dioxide under these conditions?
Solution
Calculate the total pressure at 200 m depth. Use a 1 m2 surface area.
volume = 1 m2 x 200 m = 200 m3 = 200 000 L

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The mass of water is approximately 200 000 kg. Force due to gravity is:
9.81 m s-2 x 200 000 kg = 1 962 000 N
Including the 200 m depth with the 1 m2 area the pressure (P200m) becomes
1 962 000 N ÷ 1 m-2 = 1 962 000 Nm-2 = 1 962 000 Pa
Total pressure at the 200 m depth is the sum of Pº and P5m

P200m + Pº = 1 962 000 + (101 325 –3200) = 2 060 125 Pa
[CO2]aq = 2 060 125 Pa x 3.3 x 10-7 mol L-1 Pa-1 = 0.68 mol L-1
= 680 mmol L-1
The solubility of carbon dioxide under these conditions is 0.68 mol L-1.
17. In Mexico City, the concentration of sulfur dioxide sometimes reaches 200 μg m–3. Calculate the pH of
an aqueous aerosol in equilibrium with this gas at 25ºC. As a second case, assume that all of the sulfur
dioxide is oxidized in the aerosol by ozone and hydrogen peroxide to produce sulfuric acid. Calculate
the pH after such reactions have occurred.
Solution
Part1)
Mexico City

200 µg m-3 SO2, Assume a volume of 1 m3, which gives 200 µg of SO2

Convert 200 µg SO2 to grams and then to moles of SO2.
÷ 106 μg g-1]
= 2.0 x 10-4 g SO2,
-1
÷ 64.0 g mol ] = 3.125 x 10-6 mol SO2
Assume 25ºC and 1 atm for polluted urban atmosphere. Calculate the total number of mol of gas in 1 m 3
volume using PV = nRT.
solve: 101 325 Pa x 1 m3 = n x 8.315 J mol-1 K-1 x 298 K
n = 40.9 mol
mixing ratio of SO2 is
or

(for 1 m3)

3.125 x 10-6 mol ÷ 40.9 mol = 7.64 x 10-8


7.64 x 10-8 x 109 ppbv = 76.4 ppbv SO2

The partial pressure of sulfur dioxide is:
101 325 Pa x 76.4 ppbv x 1 x 10 -9 = 0.007743 Pa (PSO2)
Using Henry’s Law, the concentration of sulfur dioxide in the aerosol is
[SO2]aq = 1.2 x 10-5 mol L-1 Pa-1 x 0.007743 Pa = 9.3 x 10-8 mol L-1

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All of the sulfur dioxide reacts to form sulfurous acid and the pH is then calculated using the first
dissociation constant:
SO2 + H2O → H2SO3
H2SO3 + H2O ↔ HSO3- + H3O+

Ka1 = 1.23 x 10-2

[H3O+] = (9.3 x 10-8 x 1.23 x 10-2)½ = 3.38 x 10-5

pH = 4.5

The pH of an aqueous aerosol in equilibrium with this gas at 25ºC would be 4.5.
Part 2)
If all SO2 becomes H2SO4 : In 1 m3 we started with 3.125 x 10-6 mol of SO2, so we would have 3.125 x 10-6
mol H2SO4 in 1 m3 of air. This would generate a mixing ratio of 76.4 ppbv for H 2SO4. Assume the amount

of water in 1 m3 is 0.023 L, which is based on 100 relative humidity at 25°C.
Assume complete uptake of H2SO4 in the water in the 1 m3. This yields a concentration of
3.125 x 10-6 mol H2SO4 / 0.023 L = 1.36 x 10-4 mol L-1
The total amount of acid from this would be twice the amount, giving
[H3O+] = 2.72x10-4 and pH = 3.6
The pH after such reactions have occurred would be 3.6.
18. Surface water in an open area of the Indian Ocean adjacent to Kenya has an alkalinity of 2320 μmol L–1
and a total carbonate concentration of 2.03 × 10 –3 mol L–1. Calculate the concentrations of

HCO3 and

CO 32 , and also the pH of this sea water.
Solution
Alkalinity = [HCO3-] + 2[CO32-] + [OH-] – [H+] = 2320 mol L-1
And

[CO2] + [HCO3-] + [CO32-] = 2.03 x 10-3 mol L-1

Assume the water is in equilibrium with the atmosphere where P CO2 = 40.7 Pa
[CO2]aq = 40.7.2 Pa x 3.3 x 10-7 mol L-1 Pa-1 = 1.34 x 10-5 mol L-1
Assume the hydroxide and hydronium ion concentrations are negligible in the total alkalinity (likely valid
since this is sea water and we expect the pH to be near 8.3). Therefore,
1]

Alkalinity = [HCO3-] + 2[CO32-] = 2.32 x 10-3 mol L-1 and

2]

[HCO3-] + [CO32-] = 2.03 x 10-3 - 1.34 x 10-5 = 2.02 x 10-3 mol L-1


Subtract equation 2 from equation 1 to get [CO32-] = 3.0 x 10-4 mol L-1

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Therefore

Ka1

Gases in water

[HCO3-] = 1.72 x 10-3 mol L-1

[HCO3-][H3O+]
= -------------------- = 4.45 x 10-7
[CO2]

[H3O+] = 4.45 x 10-7 x 1.34 x 10-5 mol L-1 / 1.72 x 10-3 mol L-1
[H3O+] = 3.46 x 10-9 mol L-1

and

pH = 8.5

In summary, the pH is 8.5, [HCO3-] = 1.72 x 10-3 mol L-1, and [CO32-] = 3.0 x 10-4 mol L-1.
19. Figure 1.2 was generated from solving for the fraction (alpha) of each of the carbonate species at the
defined variable of pH. A similar calculation was done in Chapter 10 for phosphate and is described in
detail in the text. Redo Figure 1.2 using the seawater values of 5.94 (pKa1) and 9.13 (pKa2) for

conditions of 15°C and salinity of 35.0‰.
Solution

There is a noticeable shift in the diagram to the left compared to Figure 1.2.

20. Consider the following fracking situation. Methane is extracted from deep shale using 12 000
3
m water pumped into deep shale at a pressure of 100 000 kPa. Assume one fifth of the water
returns to the surface as flow-back. How much methane would be released from the water
when it reaches the surface? For this calculation, assume Henry’s Law equilibrium applies
o
both in the extraction and at the surface, and assume a temperature of 25 C throughout.
Solution
12000 m3 water at 100 000 kPa (approx. 14 500 psi)
The pressure in Pa = 100 000 000 Pa

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20% of water returned: 12 000 m3 x 0.20 = 2400 m3
Constants: M.M. methane = 16.043 g mol-1, & KH = 1.4 x 10-8 mol L-1 Pa-1 (Table 11.1)
Use Equation 11.2,
[CH4] = KH x PCH4
= 1.4 x 10-8 mol L-1 Pa-1 x 100 000 000 Pa
= 1.4 mol L-1

Atmospheric methane:
1.834 ppmv ÷ 1 000 000 x 101 325 Pa = 0.186 Pa
[CH4] = KH x PCH4
= 1.4 x 10-8 mol L-1 Pa-1 x 0.186 Pa
= 2.6 x 10-9 mol L-1
Change in [CH4] = 1.4 mol L-1 - 2.6 x 10-9 mol L-1 = 1.4 mol L-1
Volume of water: 2400 m3 = 2 400 000 L
1.4 mol L-1 x 2 400 000 L = 3.36 x 106 mol CH4
3. 36 x 106 mol CH4 x 16.043 g mol-1 = 5.39 x 107 g CH4
= 53 900 kg (or 0.0539 Gg)
= 54 t
2545 Gg CH4 total per year (US EPA estimate, Sept 2013)
0.0539 / 2545 x 100% = 0.0021%
Approximately 54 t of methane will be released or about 0.0021% of the yearly total.
21. Old problem 19 from 3rd Edition (no longer relevant for 4th edition). In line 5 of the code supplied
for the PHREEQC calculation, replace calcite with aragonite. How does the graph differ using this
polymorph of CaCO3? Note: the database has Ksp of calcite = 3.31x10-9, and that of aragonite =
4.61x10-9.
Solution
The Ksp is slightly larger for aragonite and we should expect a slightly higher solubility for calcium. This is
shown to be the case with the data and graphs below.

Code for PHREEQC
1
2
3

# Returns the data and graph of calcite solubility vs. partial pressure of carbon dioxide
#
SOLUTION 1


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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 11
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Gases in water

EQUILIBRIUM_PHASES 1
Aragonite
REACTION

CO2 1
0.01 in 101 steps
USER_GRAPH
-headings CO2 Ca
-chart_title
-axis_scale x_axis 0 1500
-axis_scale y_axis 0 2
-initial_solutions
-axis_titles P_CO2/Pa Calcium/mM
-start
10 graph_x 10^si("CO2(g)")*101325
20 graph_y tot("Ca")*1e3
-end
END

Data Output for Aragonite
CO2 / Pa

Ca2+ / mmol L-1

0

0.17

2

0.24

6


0.33

13

0.42

25

0.51

40

0.60

61

0.69

87

0.78

118

0.87

155

0.95


198

1.04

246

1.12

299

1.20

358

1.28

422

1.36

492

1.43

566

1.51

645


1.58

729

1.65

817

1.72

Calcite Data (as in text)
CO2 / Pa
1
3
8
18
33
55
82
116
157
205
260
320
388
461
541
626
716
812

912
1018

Ca2+ / mmol L-1
0.16
0.23
0.32
0.41
0.50
0.59
0.68
0.77
0.85
0.93
1.01
1.09
1.17
1.24
1.31
1.38
1.45
1.52
1.58
1.65

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 11


Gases in water

910

1.79

1006

1.85

1107

1.92

1212

1.98

Graph for Calcite CO2 / Pa vs. Ca2+ / mmol L-1

Graph for Aragonite CO2 / Pa vs. Ca2+ / mmol L-1

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 11

Gases in water


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