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18 Chemistry of the Environment Solutions to Exercises (b) 125 1 kg mg soln Mg²⁺ X 1000 1gMg²⁺ mg Mg²⁺ 1.00 1.00 kg L soln soln 24.305 1 mol g Mg²⁺ Mg 2+ = 5.143 X 10⁻³ = 5.14 10⁻³ 4.0 1.0 g Lsoln 1 mol = = 0.038 ≈ = 2.6 10⁻³ M (This represents 6.9% hydrolysis, but the result will not be significantly different using the quadratic formula.) for Mg(OH)₂ = 1.6 10⁻¹²; > so Mg(OH)₂ will precipitate. 18.78 Because NO has an odd electron, like Cl(g), it could act as a catalyst for decomposition of ozone in the stratosphere. The increased destruction of ozone by NO would result in less absorption of short wavelength UV radiation now being screened out primarily by the ozone. Radiation in this wavelength range is known to be harmful to humans; it causes skin cancer. There is evidence that many plants don't tolerate it very well either, though more research is needed to test this idea. In Chapter 22 the oxidation of NO to NO₂ by oxygen is described. On dissolving in water, NO₂ disproportionates into (aq) and NO(g). Thus, over time the NO in the troposphere will be converted into which is in turn incorporated into soils. 18.79 Plan. Calculate the volume of air above Los Angeles and the volume of pure O₃ that would be present at the 84 ppb level. For gases at the same temperature and pressure, volume fractions equal mole fractions. Solve. = 4000 X mi² 1 10 m 1 10⁻³ 1L = 1.036 X 10¹⁴ = 1.0 10¹⁴ L air V (pure O₃) = 8.4 (1.036 10¹⁴ L air) = 8.702 10⁶ = 8.7 10⁶ LO₃ Values for P and T are required to calculate mol O₃ from volume O₃, using the ideal- gas law. Since these are not specified in the exercise, we will make a reasonable assumption for a sunny April day in Los Angeles. The city is near sea level and temperatures are moderate throughout the year, so P = 1 atm and T = 25°C (78°F) are reasonable values. n = 1.000 atm 8.702 L mol K = 3.558 10⁵ = 3.6 10⁵ mol O₃ 298 K 0.08206 L atm Check. Using known conditions to make reasonable estimates and assumptions is a valuable skill for problem solving. Knowing when assumptions are required is an important step in the learning process. 567