Solution manual of environmental chemistry a global perspective 4th ch11

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 Sinc

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:

P3810 = 101 325 e-(0.029 x 9.81 x 3810) / (8.315 x 278)

P3810 = 63 394 Pa (1 atm = 101 325 Pa)

or Ph = 0.626 atm

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?

Ph  P eo M g h R Ta /

96

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 Latm 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

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 106), 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) ↔ H+ (aq) + HCO3- (aq) [HCO3-][H+]

Ka = - = 4.5 x 10-7 [CO2]

Assume [CO2] = 1.04 x 10-5 mol L-1, and that [HCO3-] = [H+]

Solve for [H+], [H+] = 2.16 x 10-6 pH = 5.67

Repeat the calculation for 406 ppmv: result is 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

98

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

2

where

2

O

P is the pressure of oxygen in the gas phase and

2

O

X is its mole fraction in solution Calculate the value of Henry’s law constant, KH , 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

or KH’ = 4290 MPa (the prefix M = 1 x 106, Appendix C.2)

Henry's Law (for O2) becomes: PO2 = 4290 MPa x XO2

7 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 SO2 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

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 H2S and HS- in water in an atmosphere containing that concentration of the gas? What is the solubility of H2S 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 = 1.0 x 10-7)

Ka = [HS-][H+]/[H2S] (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]

100

= 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 [HCO3-][H3O+]

Ka1 = - = 4.5 x 10-7 where [HCO3-] = [H3O+] [CO2]aq

Therefore [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 HCO3- 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 which gives [H+] = 1.26 x 10-7 and using

[HCO3-][H3O+]

Ka1 = - = 4.5 x 10-7 where [HCO3-] = 8.25 x10-3 [CO2]aq

[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

12 Methane and carbon dioxide are produced under anaerobic conditions in wetlands by fermentation of

organic matter approximated by the following equation

2 4

Calculate the total pressure (Pºplus 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 (CH4 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)

102

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 [HCO3-][H3O+]

Ka1 = - = 4.5 x 10-7 [HCO3-] = [H3O+] [CO2]

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

[CO32-][H3O+]

Ka2 = - = 4.7 x 10-11 assume [HCO3-] & [H3O+] don’t change

[HCO3-]

[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 HCO3- (see Chapter 1, Figure 1.2)

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

[CO32-][H+]

Using Ka2 = - = 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 [HCO3-][H3O+]

Ka1 = - = 4.5 x 10-7 [HCO3-] = [H3O+] [CO2]

[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

104

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,

÷ 64.0 g mol-1] = 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 m3 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 (for 1 m3) mixing ratio of SO2 is 3.125 x 10-6 mol ÷ 40.9 mol = 7.64 x 10-8

or 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