Environmental Chemistry ENVE 100
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Date Created: 10/29/15
ENVEESS 100 Environmental Chemistry Fall 2011 LN T bl 3 20 pts Pro em Set 1 DUE Thursday Sept 1 Reading Jensen Chapters 14 see Chapters 58 for examples of problem solving Pay attention to units and significant gures in your answers Show all of your work Practice Problems 1 Calculate the formula weight of the following minerals a calcite CaC03s b magnetite Fe304s c kaolinite a clay mineral Alei205OH4 a N wt Ca 1 4008 4008 C 1 12011 12011 0 3 159994 47997 total 100088 4 sig fig 1001 g mor1 b N wt Fe 3 55845 167535 0 4 159994 63976 total 231511 5 sig fig 23151 9 mor1 c N wt Al 2 26982 53964 Si 2 28086 56172 0 5 159994 79997 0 4 159994 639976 H 4 100794 403176 total 25816236 5 sig fig 25816 9 mor1 2 a The chemical formula for gypsum a common hydrated sulfate mineral is CaSO4 2HzO Gypsum dissolves easily in water How many moles of Ca2 are present in solution if 486 g of gypsum are dissolved in 1 liter of water ie what is the molar concentration of Ca2 in this solution Assume pure water formula weight gypsum moles of 9 9 mol gypsum CaSO4 2H20 486 17217 00282 1 mol CaSO42H20 dissociates to give one mole Ca2 one mole 804239 and two moles of H20 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 00282 mol x1 mol Ca2 Ca2 00282 M 282 x 10392 M molar molesliter 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 8 per mole of CaSO42H20 therefore 282 x10392 mol 8 L391x 32066 g s mol391 0905156 g L391 0905 g L391 3 sig fig 282 x10392 mol 8 L391 x 6022 x1023 atoms mol391 170 x1022 atoms L391 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 0250 M Na from NaCls b a 250 mL solution of 00400 M Fe2 solution from ferrous hydroxide FeOHzs MW NaCls 58443 g mon 0250 mol L391 146 g L391 x 1 L 146 g NaCls into 1 liter Mw FeOH2s 898597 g mon 00400 mol L391 3594 g L391x 0250 L 0899 g into 250 mL 4 Inorganic speciation At a former mining site the copper minerals malachite CuzOHzC03s and azurite Cu3OHzC03zs 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 ie 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 Cu2OH2C03s and azurite Cu3OH2C032s has the oxide components Cu2O H20 C02 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 CuOH2 CuOH339 CuOH4392 Cu2OH22 Cu3OH42 etc Whether or not these are found in a measureable concentration depends on the stability of the complex pH and Cu concentration C02 H2C03 or C02aq same species HCO3 C03239 H20 H OH39 The concentration of the complexes that might be present will depend on the total concentrations of copper and carbonate in the system the thermodynamic stability of the complexes ie their tendency to form assuming the system comes to equilibrium and pH Total concentrations will be determined by the mineral solubility and by any other input of copper or carbonate into the system eg equilibrium with air carbonate in water from dissolution other minerals It would be useful to have a measurement of pH total copper concentration and total carbonate in the system in order to predict which species dominate 5 Gas phase calculations Carbon dioxide gas COzg makes up about 0035 mole of the atmosphere The average molecular weight of gases in the atmosphere is 29 g mol391 and the total mass of gases the atmosphere is 52 x 102 g How many moles of COzg are in the atmosphere How many grams What is the concentration of COzg expressed as ppmm parts per million by mass 0035 mol 002 100 moles atm x 52 x1020 9 atm 29 9 atm mol atm 6276 x1016 mol co2 molecular wt of C02 12011 gmol 16 x 2 gmol 44 gmol C02 663 x1016 mol 002 x 44 gmol 002 2761x1018 9 C02 Expressed as ppm C02 276 x 1018g C02 531 x104g C02 XE 52 X 1020 g atm g atm 106 6 Equilibrium constant calculations Cadmium Cd is a very toxic contaminant at low concentrations that may be found in natural waters as a result of mining or industrial activities Its concentration in solution its solubility as the ion Cd2 could be controlled in some natural systems by equilibrium with its carbonate mineral form That is it dissolves according to the reaction CdC03s Cdz 003239 K1 1039 1 at 2500 1 atm where CdC03s is the solid mineral form and K1 is the equilibrium constant for the dissolution reaction a Write an expression in logarithmic form stating the Law of Mass Action for this reaction I09 K1 I09 3mm I09 aco32 IogaCdC03s full expression may make the assumption that the activity of a pure solid CdC03s 1 b Calculate the standard free energy of the reaction AGor at 25 C 1 atm note units Keq 1039121 7943 x103913 AG RT In K9 24789 kJmol In 7943 x 1039 691 kJmol c An alternative way of writing the reaction for the dissolution of CdC03s is to form the species HCOg39 rather than Cng39 Write a balanced chemical reaction for CdC03s dissolution with HC0339 rather than Cng39 as a product species Calculate an equilibrium constant and stande free energy of reaction for your new reaction see Chapter 4 New balanced chemical reaction CdC03s H Cd2 HCO339 which is the sum of the following reactions CdC03s Cd2 003239 K1 1039 1 C03239 H Hcos39 K2 1K 11039103 10103 adding reactions multiply K1 K2 also equivalent to log K1 log K2 103912391 10103 K 1039180 158 x10392 or log K 180 AG RT In Ks 24789 kJmol In 1048 10274 kJmol 103 kJmol sig fig d How would you decide which reaction to use to describe the equilibrium of CdC03s with water From the standpoint of chemical equilibrium does it matter Explain Both reactions are correct descriptions of the equilibrium of CdC03s with water The choice of particular species to use in reactions depends on which species are expected to be most abundant availability of thermodynamic data ie stability constants and how simple or complicated the user wants or needs the problem to be From an equilibrium standpoint all of the major species Cd2 H OH39 H2C03 HCO3 C032 would be present in solution and all reactions are assumed to come to equilibrium By convention the form of the first reaction given above is often used for dissolution of carbonate minerals However the second reaction is an equally valid representation as long as the correct equilibrium constant is used Solving the problem of the equilibrium distribution of species requires solving ALL reactions simultaneously equivalent to finding the global free energy minimum 7 Jensen Problem 68 Enumerating species in a system In preparing a cake recipe a baker adds some baking soda NaHC03 to water Generate a species list diagram such as Figure 62 for this system You will need to use at least some of the equilibria in Example 62 Purpose of problem Enumerating species Relevant sections of text Section 63 Solution The starting materials are H20 and NaHC03 Sodium bicarbonate NaHC03 is a sodium salt and likely dissociates to Na and HC0339 Water of course dissociates to Hand OH39 The pertinent equilibria from Example 62 or Appendix D are HCO339 003239 H H2CO3 HCO3 H H2CO3 H20 In an open system the species list diagram is Starting material NaH003s assume dissolution H20 Species list Na H OH H00339 003239 H2003 002g In a closed system ignore 002g S Jensen Problem 69 System description and interpretation Write a complete mathematical model for the system in Problem 68 Assume the system is closed is this reasonable and that NaHC03 dissolves completely Again you will need to use at least some of the equilibria in Example 62 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 sections of text Sections 63 and 64 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 NaH003s assume dissolution H20 Species list Na H OH H003 003239 H2003 Eguilibrium H20 H 0H39K KW H2CO3 HCO3 H K1 HC0339 003239 H K2 Mass balances Nam Na Ctot H2C03 Hcos39 C032 Charge balance H Na OH H00339 2003239 Other constraints activity of H20 1 NaH003s dissociates completely all concentrations 2 0 Thus the compete mathematical model is Unknowns NaH003s Na HC0339 C032 H2C03 H20 H and 0H39 Constraints assuming activities and concentrations are interchangeable KwH OH39 H20 K1 H HC0339 H2C03 K2 HC03239 HC0339 Nam Na