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Some Basic Terminology Solute: chemical species that is dissolved in water; element or compound, charged or neutral Total Dissolved Solids (TDS): total.

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Presentation on theme: "Some Basic Terminology Solute: chemical species that is dissolved in water; element or compound, charged or neutral Total Dissolved Solids (TDS): total."— Presentation transcript:

1 Some Basic Terminology Solute: chemical species that is dissolved in water; element or compound, charged or neutral Total Dissolved Solids (TDS): total amount of solids remaining when a water sample is evaporated to dryness (mg/L) – Related to salinity – Fresh water: < 1000 mg/L Drinking water standard = 500 mg/L – Brackish water: 1000 – 20,000 mg/L – Saline water: > 20,000 mg/L Seawater = 35,000 mg/L (on average) – Brines: > 100,000 mg/L 1

2 2 Ocean Salinity

3 Solute Concentration Terminology (mass) concentration: mass of solute / unit volume of solution (e.g. mg/L) Molarity (M): (mass of solute/formula weight) / liters of solution = moles/L Molality (m): moles of solute / kg of solution Equivalents/L: (moles of solute x valence of solute) / L of solution = eq/L – Used for ion balances, piper diagrams Normality (N): eq/L (primarily used for acids) Parts per million (ppm) by weight: 1 millionth g solute per g of solution = mass of solute / 10 6 g water – commonly used outside academia for most dilute water solutions, mg/L ≈ ppm and molarity ≈ molality 3

4 Rocks and Minerals 4

5 Mineralogy of Igneous Rocks: Bowen’s Reaction Series 5 Less Stable More Stable At/Near Earth’s Surface:

6 Mineralogy of Igneous Rocks Mafic – Olivines: fayalite (Fe 2 SiO 4 ) forsterite (Mg 2 SiO 4 ) – Pyroxenes: wollastonite (CaSiO 3 ) enstatite (MgSiO 3 ) augite ((Ca,Na)(Mg,Fe,Al) – Plagioclase feldspars: anorthite (CaAl 2 Si 2 O 8 ) – Amphiboles: hornblende ((Na,K) 0-1 Ca 2 (Mg,Fe,Al) 5 Si AlO(OH) 2 6

7 Mineralogy of Igneous Rocks Felsic – Feldspars: Albite (NaAlSi 3 O 8 ) orthoclase (KAlSi 3 O 8 ) anorthite (CaAl 2 Si 2 O 8 ) – Micas: muscovite (KAl 2 [Si 3 AlO 10 ](OH) 2 biotite (K(Mg,Fe) 3 AlSi 3 O 10 (OH) 2 ) – Quartz: SiO 2 7

8 Mineralogy of Metamorphic Rocks Mineral composition reflects parent rocks – e.g. marble from limestone (CaCO 3 ) 8

9 Mineralogy of Sedimentary Rocks Shale: low reactivity – Quartz, clay/mica, feldspars, calcite, organic matter Sandstone – Quartz, K/Na feldspars, micas, many other minerals Carbonates – Calcite: CaCO 3 – Dolomite: (Ca,Mg)(CO 3 ) 2 – Chert (SiO 2 ) – Clastic sediments Evaporites – Gypsum: CaSO 4 ·2H 2 O 9

10 Mineralogy of Secondary Minerals Form by chemical reactions (weathering/ diagenesis) between water and primary/ secondary minerals – Incongruent dissolution (solid products) Stable, may be in equilibrium with H 2 O Clays: – mineral form depends on reaction type and climate (precipitation, temperature) – garbage minerals: chemical composition highly variable, not well known 10

11 Mineralogy of Clays Mafics (high Mg, Ca, Fe) – Chlorite (Mg,Fe) 3 (Si,Al) 4 O 10 (OH) 2 ·(Mg,Fe) 3 (OH) 6  vermiculite  smectites (Na,Ca 0.5 ) 0.7 (Mg,Fe,Al) 4 (Al,Si) 8 O 20 (OH) 4  kaolinite Al 2 Si 2 O 5 (OH) 4  Fe/Al oxides [Al(OH) 3, Fe(OH) 3 ]  decreasing cation content Felsic (high Na, K, Si) – Illite K 15 (Mg 0.5 Al 3.5 )(AlSi 7 )O 20 (OH) 4 (close to muscovite)  smectites  kaolinte  Fe/Al oxides  degree of weathering – 80% muscovite, 20% smectite main clay in shales These are general trends observed. Often, several clay minerals exist, but dominant type can be predicted based on rock type and amount of precipitation Illite, kaolinite, chlorite, vermiculite, and montmorillonite common in Illinois 11

12 Silica Minerals Most Si released by weathering of primary silicates (feldspars, etc.) Quartz (SiO 2 ): not dominant secondary mineral, very resistant to weathering Secondary Si minerals (SiO 2 ) – Amorphous silica: non-crystalline – Chalcedony (agate): cryptocrystalline – Chert: nodules/beds – Opal: gem quality – Secondary SiO 2 is found in all rock types 12

13 Oxyhydroxides of Fe/Al Also found in all rock types Al: very low solubility (more soluble at high and low pH) – Amorphous Al(OH) 3 – Gibbsite: Al(OH) 3 Fe: Ferric iron (Fe 3+ ), low solubility – Amorphous Fe(OH) 3 – Goethite: FeOOH – Hematite: Fe 2 O 3 – Very common in Illinois sediments 13

14 Carbonates and Sulfates Carbonates – Found in almost all geologic environments – Formed by 2 most common dissolved ions Calcite: CaCO 3 Also dolomite [(Ca,Mg)(CO 3 ) 2 ], siderite [FeCO 3 ] Sulfates – Gypsum: CaSO 4 ·2H 2 O; layers or disseminated – Anhydrite: CaSO 4 14

15 Other sources of material to the subsurface Dust Organic matter 15

16 Dissolved Molecules Dissolved is (somewhat) arbitrarily defined as anything that passes through a 0.45 μm filter – > 0.45 μm defined as “suspended” material – Some colloids and all nano-particles are < 0.45 μm Aqueous Species refers to any molecule dissolved in water Ions are charged molecules only – HCO 3 - is an ion, H 2 CO 3 is not (uncharged) 16

17 Dissolved Molecules Major ions defined as being typically > 5 mg/L – Cations: Ca 2+, Mg 2+, Na + – Anions: HCO 3 -, SO 4 2-, Cl - – H 4 SiO 4 : almost all dissolved Si in this form In Groundwater, these typically account for > 90% of TDS regardless of TDS value 17

18 Dissolved Molecules Minor constituents typically between 0.01 and 10 mg/L, higher in unusual situations – K +, Fe 2+, Mn 2+, NO 3 -, F -, B(OH) 3, CO 3 2- Trace constituents usually < 0.1 mg/L – Most metals – Rare earth elements – Arsenic, bromide, iodide, phosphate, radium, uranium 18

19 Balancing Chemical Reactions Reactants Products 19

20 Balancing Reactions 1.Determine the species involved (reactants and products) and the state (solid, liquid, gas) that each species is in 2.Write an unbalanced equation that summarizes information from step 1 3.Balance: molar amounts on each side of reaction must be equal – Stoichiometry: refers to the mathematical balancing of reactants and products (moles) 20

21 Balancing Reactions Balancing (determining stoichiometric coefficients) – Start with most complicated species – Cations first, then Si, then C, then O, then H – Check charge balance – Assume that H 2 O, H +, and OH - are readily available since it’s an aqueous reaction In most weathering reactions, acid H + attacks the mineral Examples… 21

22 Reactions and Equilibrium 22

23 Chemical Reactions Chemical reactions usually need water When 2 ions or molecules approach each other closely and establish a bond, a chemical reaction has taken place – e.g., precipitating a solid Breaking bonds is also a chemical reaction – e.g., dissolving a solid 23

24 Reactions and Equilibria Let’s add salt (NaCl) to water at constant T, P; add more than can be dissolved – Salt crystals will dissolve, Na-Cl bonds are broken, ions (Na +, Cl - ) in solution – What if we measured the Na + (or Cl - ) concentration in solution over time? 24

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30 Reactions and Equilibria (NaCl) At equilibrium, Na + and Cl - concentrations in solution become invariant with time This does not mean that all reactions have stopped; both NaCl dissolution and precipitation are still occurring At equilibrium, the rate of NaCl dissolution = rate of NaCl precipitation – NaCl (s)  Na + + Cl - and Na + + Cl -  NaCl (s) have the same rates – NaCl (s) Na + + Cl - 30

31 Equilibrium The system maintains a constant state, and there are no observable changes in the constituents in the solution In Nature, reactions tend to progress to equilibrium – That is the preferred state – Disequilibrium is inherently unstable – But, there are plenty of reactions that are in disequilibrium near the Earth’s surface 31

32 La Châtelier’s principle When a reaction at equilibrium is disturbed, i.e., if conditions change, a new state of equilibrium will be established to counteract the disturbance – e.g., heat up NaCl solution, more NaCl will dissolve – Or, we add more water more NaCl will dissolve 32

33 33 Saturated solution: can’t dissolve any more

34 Solubility The amount of a compound dissolved to form a saturated solution – In our example, the amount of NaCl that dissolved – Measured in weight per volume (e.g., g/L) – Need to state T, P (usually 25°C, 1 atm, which is standard state) 34

35 Solubility Need equilibrium to define solubility e.g., CaCO 3 + 2HCl  Ca Cl - + 2H 2 O + CO 2 – In a closed system, reaction will (eventually) reach equilibrium – In an open system, this reaction cannot reach equilibrium because CO 2 does not accumulate Reaction will proceed until CaCO 3 or HCl is used up, i.e., reaction runs to completion Many reactions at/near Earth’s surface do not achieve equilibrium because products escape or are transported away Also many reactions occur slowly 35

36 Law of Mass Action Mathematical model that explains and predicts behaviors of solutions in equilibrium – Valid only for reversible reactions Consider the reaction: A + B  C + D – rate of forward reaction (  f ) = k f (A)(B) k f = proportionality constant for forward reaction (A) and (B) are in molar amounts Rate a function of reactant concentrations – rate of backward reaction (  b ) = k b (C)(D) 36

37 Law of Mass Action A + B  C + D – At equilibrium,  f =  b The ratio of products to reactants is fixed So k f (A)(B) = k b (C)(D) – = – = K eq = the equilibrium constant – K eq values have been determined in the lab for many reactions k f (C)(D) k b (A)(B) 37 kfkbkfkb

38 Law of Mass Action For reaction: aA + bB  cC + dD – Where a, b, c, and d are the stoichiometric coefficients for the reactants and products K eq = [(C) c (D) d ] [(A) a (B) b ] (A), (B), (C), (D) are concentrations at equilibrium [(C) c (D) d ] is the reaction quotient (Q) [(A) a (B) b ] Q changes until equilibrium is achieved, then it is constant – However, strictly speaking this is only valid for ideal gases 38

39 Solubility Index Calculations We can use the mass action equation to estimate the equilibrium status of a particular reaction. – Is the solution undersaturated, oversaturated, or at equilibrium with respect to a mineral? – i.e., do we expect it to dissolve or be precipitated? 39

40 Solubility Index Calculations SI = log IAP – log K eq – SI = saturation index – IAP = ion activity product = measured concentration (= Q) – K eq = equilibrium constant (IAP at equilibrium) If log IAP = log K eq, then SI = 0 (equilibrium): no net dissolution or precipitation of mineral If log IAP < log K eq, SI < 0, solution is undersaturated with respect to that mineral: expect active dissolution If log IAP > log K eq, SI > 0, solution is oversaturated with respect to that mineral: expect active precipitation In practice, if SI = 0 ± 0.5, then water at or close to equilibrium with mineral phase, due to uncertainty in the thermodynamic variables 40


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