7 Questions with Solutions in Environmental Chemistry - Homework 1 | ENVE 100, Study notes of Engineering

Download 7 Questions with Solutions in Environmental Chemistry - Homework 1 | ENVE 100 and more Study notes Engineering in PDF only on Docsity! 1 ENVE/ESS 100: Environmental Chemistry, Fall 2011 SOLUTIONS (20 pts) Problem Set 1. DUE: Thursday, Sept. 1 Reading: Jensen: Chapters 1-4; see Chapters 5-8 for examples of problem solving. Pay attention to units and significant figures in your answers! Show all of your work. Practice Problems 1. Calculate the formula weight of the following minerals: a.) calcite, CaCO3(s) b.) magnetite, Fe3O4(s) c.) kaolinite (a clay mineral), Al2Si2O5(OH)4 a.) N wt. Ca 1 40.08 40.08 C 1 12.011 12.011 O 3 15.9994 47.997 total 100.088 4 sig. fig. = 100.1 g mol-1 b.) N wt. Fe 3 55.845 167.535 O 4 15.9994 63.976 total 231.511 5 sig. fig. = 231.51 g mol-1 c.) N wt. Al 2 26.982 53.964 Si 2 28.086 56.172 O 5 15.9994 79.997 O 4 15.9994 63.9976 H 4 1.00794 4.03176 total 258.16236 5 sig. fig. = 258.16 g mol-1 2. a) The chemical formula for gypsum, a common hydrated sulfate mineral, is CaSO4•2H2O. Gypsum dissolves easily in water. How many moles of Ca2+ are present in solution if 4.86 g of gypsum are dissolved in 1 liter of water (i.e., what is the molar concentration of Ca2+ in this solution)? Assume pure water. g formula weight gypsum (g mol-1) moles of gypsum 2 CaSO4•2H2O 4.86 172.17 0.0282 1 mol CaSO4•2H2O dissociates to give one mole Ca2+, one mole SO42-, and two moles of H2O. Since we are dissolving into 1 liter of pure water, the moles of water is insignificant when it dissolves, but they count for the formula weight. Therefore: 0.0282 mol x 1 mol Ca2+: Ca2+ = 0.0282 M = 2.82 x 10-2 M (molar = moles/liter) 3 significant figures b) How many grams of sulfur atoms are present in 1 liter of this solution? How many atoms of sulfur are present? There is one mole of S per mole of CaSO4•2H2O, therefore 2.82 x 10-2 mol S L-1 x 32.066 g S mol-1 = 0.905156 g L-1 = 0.905 g L-1 (3 sig. fig.) 2.82 x 10-2 mol S L-1 x (6.022 x 1023 atoms mol-1) = 1.70 x 1022 atoms L-1 3. Calculate the number of grams of solid needed to dissolve in the amount of water indicated in order to make solutions of the following concentrations: a.) a one liter solution of 0.250 M Na+ from NaCl(s) b.) a 250 mL solution of 0.0400 M Fe2+ solution from ferrous hydroxide, Fe(OH)2(s) MW NaCl(s) = 58.443 g mol-1 x 0.250 mol L-1 = 14.6 g L-1 x 1 L = 14.6 g NaCl(s) into 1 liter. MW Fe(OH)2(s) =89.8597 g mol-1 x 0.0400 mol L-1 = 3.594 g L-1 x 0.250 L = 0.899 g into 250 mL 4. Inorganic speciation At a former mining site, the copper minerals malachite (Cu2(OH)2CO3(s)) and azurite (Cu3(OH)2(CO3)2(s)) are dissolving into stream water, which is of concern because copper is harmful to fish at low concentrations. If the system consisted only of these two minerals dissolved in water, write down all of the possible aqueous species that might be present (i.e., what species or complexes might you analyze for, or consider in a speciation calculation?). Assume no change in oxidation state of the elements. Do you think it is likely that you would have all of these possible species present in solution at the same time? Why or why not? What properties of the solution would be useful to measure in order to guess which species might be present? (See the example for zinc at the end of Chapter 6). A solution made from malachite (Cu2(OH)2CO3(s)) and azurite (Cu3(OH)2(CO3)2(s)) has the oxide components: Cu2+O, H2O, CO2. If we ignore reduction and oxidation, the species list would be: Cu: Cu2+-- most important Cu species in solution if hydrolysis is considered, may form: CuOH+, Cu(OH)2, Cu(OH)3-, Cu(OH)4-2, Cu2(OH)22+, Cu3(OH)42+,… etc. Whether or not these are found in a measureable concentration depends on the stability of the complex, pH, and Cu concentration. 5 H2CO3 = HCO3– + H+ H2CO3 = CO2(g) + H2O In an open system, the species list diagram is: Starting material: NaHCO3(s) (assume dissolution), H2O Species list: Na+, H+, OH–, HCO3–, CO32–, H2CO3 , CO2(g) In a closed system, ignore CO2(g) 8. Jensen Problem 6.9: System description and interpretation Write a complete mathematical model for the system in Problem 6.8. Assume the system is closed (is this reasonable?) and that NaHCO3 dissolves completely. Again, you will need to use at least some of the equilibria in Example 6.2. This means that you need to determine: species list, equilibrium reactions, mass balances, charge balance, and any other constraints on the system. Purpose of problem; System description and interpretation Relevant section(s) of text; Sections 6.3 and 6.4 Solution The assumption of a closed system is not very reasonable, especially if the water solution is stirred and brought into contact with the air. For a closed system: Starting material: NaHCO3(s) (assume dissolution), H2O Species list: Na+, H+ , OH-, HCO3-, CO32-, H2CO3 Equilibrium: H2O = H+ + OH- ; K = Kw H2CO3 = HCO3- + H+; K1 HCO3-= CO32- + H+ ; K2 Mass balances: Natot = [Na+] Ctot = [H2CO3] + [HCO3-] + [CO32-] Charge balance: [H+] + [Na+] = [OH-] + [HCO3-] + 2[CO32-] Other constraints: activity of {H2O} = 1, NaHCO3(s) dissociates completely, all concentrations ≥ 0 Thus, the compete mathematical model is: Unknowns: [NaHCO3(s)], [Na+], [HCO3-], [CO32-], [H2CO3], [H2O], [H+], and [OH-] Constraints (assuming activities and concentrations are interchangeable): Kw =[H+][ OH-]/[ H2O] K1 = [H+][ HCO3-]/[ H2CO3] K2= [H+][CO32-]/[ HCO3-] Natot = [Na+] 6 Ctot = [H2CO3] + [HCO3-] + [CO32-] Charge balance: [H+] + [Na+] = [OH-] + [HCO3-]+ 2[CO32-] {H2O} = 1, [NaHCO3(s)] = 0, all concentrations ≥ 0