(Benjamin, 1.2-1.5) David Reckhow CEE 680 #2 1 Elemental - PowerPoint PPT Presentation

Print version Updated: 22 January 2020 Lecture #2 Intro: Expressions of Concentrations and Natural Abundance (Stumm & Morgan, Chapt.1 & 3.4 ) (Pg. 4-11; 97-105) (Pankow, Chapt. 2.8) (Benjamin, 1.2-1.5) David Reckhow CEE 680 #2 1

Elemental abundance in fresh water From: Stumm & Morgan, 1996; Benjamin, fig 1.4; Langmuir figure 8.12 David Reckhow CEE 680 #2 2

Same, but for groundwater  From Langmuir, figure 8.13  Based on Rose et al., 1979, Geochemistry in mineral exploration David Reckhow CEE 680 #1 3

Connecticut River vs Vaal River  Connecticut River (CT CT Vaal Sea River) watershed Parameter River River Avg SW Precip Water  Ryder et al., 1981 Sodium 6.5 4.7 6.3 9.4 10,800  Vaal River, site C1H001 Potassium 1.3 0.9 2.3 395  Mohr, 2015 Calcium 13 7.1 15 0.8 408  Avg (surface water) SW Magnesium 3.2 5.5 4.1 1.2 1280  Turekian, 1977 & Langmuir, pg 294 Chloride 10 4.5 7.8 17 19,400  Precipitation & Sea Bicarbonate 22 50 58 2.0 72 Water Sulfate 27 7.4 3.7 7.6 2710  Benjamin, Table 1.1 TDS 113 79 120 38 35,000  All values in mg/L Na/(Na+Ca) 0.33 0.40 0.30 0.92 0.96 David Reckhow CEE 680 #1 4

Connecticut River  dfsr David Reckhow CEE 680 #1 5

 sda David Reckhow CEE 680 #1 6

Orange River  Vaal River sub basin David Reckhow CEE 680 #1 7

 ads David Reckhow CEE 680 #1 8

Connecticut River vs Vaal River  Connecticut River (CT CT Vaal Sea River) watershed Parameter River River Avg SW Precip Water  Ryder et al., 1981 Sodium 6.5 4.7 6.3 9.4 10,800  Vaal River, site C1H001 Potassium 1.3 0.9 2.3 395  Mohr, 2015 Calcium 13 7.1 15 0.8 408  Avg (surface water) SW Magnesium 3.2 5.5 4.1 1.2 1280  Turekian, 1977 & Langmuir, pg 294 Chloride 10 4.5 7.8 17 19,400  Precipitation & Sea Bicarbonate 22 50 58 2.0 72 Water Sulfate 27 7.4 3.7 7.6 2710  Benjamin, Table 1.1 TDS 113 79 120 38 35,000  All values in mg/L Na/(Na+Ca) 0.33 0.40 0.30 0.92 0.96 David Reckhow CEE 680 #1 9

Gaillardet Diagram  Median values for WQ monitoring stations along the Vaal River David Reckhow CEE 680 #1 10

Gibbs view of water composition  asd Gibbs, R. J. (1970). "MECHANISMS CONTROLLING WORLD WATER CHEMISTRY." Science 170(3962): 1088-1090 David Reckhow CEE 680 #1 11

Stoichiometry: Lake Example  Basic limnology tells us that phosphorus stimulates algal growth, they produce O 2 , and bacteria consume O 2 when the algae die  If we add 1 mg P to a small lake  How much algal biomass is produced?  How much O 2 is produced?  How much O 2 is later consumed?  Elemental analysis of algae: empirical formula  C 106 H 263 O 110 N 16 P David Reckhow CEE 680 #1 12

Solution  Balance equation - + HPO 4 -2 + 122 H 2 O + 18 H + =  106 CO 2 + 16 NO 3 C 106 H 263 O 110 N 16 P + 138 O 2  Use stoichiometric coefficients     − − −   1 mmole P 3551 mg a lg ae 1 mmole a lg ae  Biomass: = −     1 mg P       − − − 31 mg P 1 mmole a lg ae  1 mmole P      = − 115 mg a lg ae    −  − − 1 mmole P 32 mg O  138 mmole O   O 2 production:   = −     1 mg P 2 2     − − − 31 mg P 1 mmole O  1 mmole P      2 = − 142 mg O 2  O 2 consumption: David Reckhow CEE 680 #1 13

PFOA  Perfluorooctanoic acid ( PFOA ), also known as C8 and perfluorooctanoate , is a synthetic compound. One industrial application is as a surfactant in the production of fluoropolymers. It has been used in the manufacture of such prominent consumer goods as polytetrafluoroethylene (commercially known as Teflon). PFOA has been manufactured since the 1940s in industrial quantities.  PFOA persists indefinitely in the environment. It is an animan carcinogen. PFOA has been detected in the blood of more than 98% of the general US population in the low and sub-ppb range, and levels are higher in chemical plant employees and surrounding subpopulations. How general populations are exposed to PFOA is not completely understood. PFOA has been detected in industrial waste, stain resistant carpets, carpet cleaning liquids, home dust, microwave popcorn bags, water, food, some cookware and PTFE such as Teflon. Daily Hampshire Gazette, 27 January 2016 David Reckhow CEE 680 #1 14

Natural Water Environment  Chemistry of:  Water column  Sediments  Soil  Groundwater  Atmosphere S&M: Fig. 1.1; Pg. 2 David Reckhow CEE 680 #2 15

Model Complexity: Phases  Single Phase to multi-phase Gas Gas Aqueous Aqueous Aqueous Aqueous Solution Solution Solution Solution Solid Solid Phase Phase Based on S&M: Fig. 1.2 Pg. 4 Gas Aqueous Aqueous Abundances? Solution Solution Solid Solid Solid Solid Solid Solid α β γ α β γ David Reckhow CEE 680 #2 16

 s David Reckhow CEE 680 #2 17

 O Elemental abundance in crust  Si  Al  Fe  Ca  Na  Mg  K  Ti  H  P  Mn  F David Reckhow CEE 680 #2 18

Elemental abundance in fresh water From: Stumm & Morgan, 1996; Benjamin, 2002; fig 1.1 David Reckhow CEE 680 #2 19

Rock-WQ Connection  Water Solutes reflect rock mineralogy; e.g.  Limestone  CaCO3, mostly  Dolomite  CaMg(CO3) 2 , mostly  Gypsum  CaSO4 “Stiff diagram” From Hounslow, 1995 Water Quality Data; Analysis and Interpretation David Reckhow CEE 680 #2 20

Review  Units  Chemical Stoichiometry  Mass based  mass balance  Molarity  balancing equations  Thermodynamics  Molality  Normality  law of mass action  Mole fraction  types of equilibria  Atmospheres David Reckhow CEE 680 #2 21

SI Unit prefixes Factor Prefix Symbol Factor Prefix Symbol 10 -1 deci d 10 1 deka da 10 -2 centi c 10 2 hecto d 10 -3 milli m 10 3 kilo k 10 -6 micro µ 10 6 mega M 10 -9 nano n 10 9 giga G 10 12 tera T 10 -12 pico p 10 15 peta P 10 -15 femto f 10 18 exa E 10 -18 atto a David Reckhow CEE 680 #2 22

Mass Based Concentration Units  Solid samples − 3 17 . 5 mg Pb 17 . 5 x 10 g Pb 17 . 5 g Pb = = 6 10 g soil 3 1 kg soil 1x10 g soil = 17 . 5 ppm Pb in soil m = 1 mg / kg 1 ppm m µ = 1 g / kg 1 ppb m David Reckhow CEE 680 #2 23

 Liquid samples Density of Water at 5ºC 0 . 35 mg Fe 1 L water x 3 1 L water 10 g water − 3 0 . 35 mg Fe 0 . 35 x 10 g Fe 0 . 35 g Fe = = = 6 10 g water 3 3 10 g water 10 g water = 0 . 35 ppm Fe in water m David Reckhow CEE 680 #2 24

Mass/Volume Units Mass/Mass Units Typical Applications g/L (grams/liter) (parts per thousand) Stock solutions mg/L (milligrams/liter) ppm (parts per million) Conventional pollutants 10-3g/L (DO, nitrate, chloride) µg/L (micrograms/liter) ppb (parts per billion) Trihalomethanes, Phenols. 10-6g/L ng/L (nanograms/liter) ppt (parts per trillion) PCBs, Dioxins 10-9g/L pg/L (picograms/liter) Pheromones 10-12g/L David Reckhow CEE 680 #2 25

Gas phase concentration  Gas samples (compressible) 0 . 056 mg Ozone 3 1 m air  Could be converted to a ppm m basis  But this would change as we compress the air sample  Could also be converted to a ppm v basis  Independent of degree of compression  But now we need to convert mass of ozone to volume of ozone David Reckhow CEE 680 #2 26

By definition: mass ( g ) = n GFW Ideal Gas Law  An ideal gas  Will occupy a certain fixed volume as determined by: = PV nRT RT mass ( g ) RT = = V n P GFW P regardless of the nature of the gas  Where:  P=pressure =22.4 L  V=volume at 1 atm, 273.15ºK  n=number of moles  T=temp  R=universal gas constant=0.08205 L-atm/mole-ºK  GFW=gram formula weight David Reckhow CEE 680 #2 27

Convert mass to moles  Now we know that ozone’s formula is O 3  Which means it contains 3 oxygen atoms  Therefore the GFW = 3x atomic weight of oxygen in grams or 48 g/mole 0 . 056 mg Ozone 3 1 m air  n=mass(g)/GFW  n=0.056x10 -3 g/(48 g/mole) − 3 0 . 00117 x 10 moles Ozone  n=0.00117x10 -3 moles 3 1 m air David Reckhow CEE 680 #2 28

Now determine ppm v volume = ppm ozone v volume air − 3 0 . 00117 x 10 moles x 22 . 4 L / mole = 3 1 m air − 3 0 . 026 x 10 L Ozone = 3 1x10 L air = 0 . 026 ppm O in air v 3 = 26 ppb O in air v 3 David Reckhow CEE 680 #2 29

RT = V n i i P Mole & volume fractions RT = V n total total P  Based on the ideal gas law:  The volume fraction (ratio of a component gas volume to the total volume) is the same as the mole fraction of that component V n = i i  Therefore: V n Defined as: total total mole fraction Defined as: Volume fraction  And since the fraction of the total is one-millionth of the number of ppm: V n − 6 ≡ = i i 10 ppm v V n total total David Reckhow CEE 680 #2 30