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Solutions: Environmental Chemistry - a Global Perspective 4Th Edition 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. o M g h / R T a Ph P e The total pressure at 3810 m (Ph) is: -(0.029 x 9.81 x 3810) / (8.315 x 278) P3810 = 101 325 e 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 -8 -1 -1 [O2] = 1.9 x 10 mol L Pa x 13 099 Pa -4 -1 -1 [O2] = 2.5 x 10 mol L (or 8.0 mg L ) 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? 95 Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 11 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 -1 n = 0.0085 g 32.0 g mol = 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 -1 V = 0.0310 L (multiply by 1000 mL L to convert to mL) The volume of gas from 8.5 mg of O2 (g) at P and 25C 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: -1 0.0015 g ÷ 12.011 g mol = 0.000125 mol C (or CO2, or O2) -1 0.000125 mol x 32 000 mg mol = 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 -1 L 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. 96 Solutions: Environmental Chemistry - a global perspective 4th Edition 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 106), and therefore, the partial pressure of carbon dioxide is given by: -6 310 ppmv CO2 = 310 x 10 x 101 325 Pa = 31.4 Pa -7 -1 -1 Henry’s Law constant for CO2 is 3.3 x 10 mol L Pa (Table 11.1) Using Henry’s Law : [G]l = KHPg -7 -1 -1 -5 -1 [CO2](aq) = 3.3 x 10 mol L Pa x 31.4 Pa = 1.04 x 10 mol L H2O (l) + CO2 (aq) → H2CO3 (aq) -7 Ka = 4.5 x 10 + - H2CO3 (aq) ↔ H (aq) + HCO3 (aq) - + [HCO3 ][H ] -7 Ka = ----------------- = 4.5 x 10 [CO2] -5 -1 - + Assume [CO2] = 1.04 x 10 mol L , 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. 97 Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 11 Gases in water 6. One of the alternative expressions for Henry’s law when applied to oxygen is P K X O2 H O2 where P is the pressure of oxygen in the gas phase and X is its mole fraction in solution. O2 O2 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. -4 the number of moles of O2 in 1 L = 2.66 x 10 , (total moles in 1L ~ 55.5 mol) -4 -6 XO2 = 2.66 x 10 ÷ 55.5 = 4.79 x 10 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 -6 9 KH’ = 20 557 Pa ÷ 4.79 x 10 = 4.29 x 10 Pa 6 or KH’ = 4290 MPa (the prefix M = 1 x 10 , 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 -8 -3 10.9 ppbv = 1.09 x 10 x 101325 Pa = 1.1 x 10 Pa SO2 + -5 - + given [H ] = 1.29 x 10 = [HSO3 ], assuming only source of H is from H2SO3 98 Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 11 Gases in water -2 -5 2 1.23 x 10 = (1.29 x 10 ) / [H2SO3] -8 -1 [H2SO3] = 1.35 x 10 mol L -8 -1 Assume [H2SO3] = [SO2] = 1.35 x 10 mol L [SO2] = KH x PSO2 -8 -1 -3 KH = 1.35 x 10 mol L / 1.1 x 10 Pa -5 -1 -1 = 1.23 x 10 mol L Pa -5 -1 -1 The Henry’s law constant for SO2 in water at this temperature was found to be 1.23 x 10 mol L Pa 8.
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